Inter 2nd Year

Students get through AP Inter 2nd Year Chemistry Important Questions 11th Lesson Haloalkanes And Haloarenes which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 11th Lesson Haloalkanes And Haloarenes

Very Short Answer Questions

Question 1.
Write the structures of the following compounds. [IPE – 2015 (AP), 2016 (TS)]
i) 2-chloro-3-methylpentane,
ii) 1 -B romo4-sec-butyl-2-methylbenzene.
Answer:
i) 2-chloro-3-methyl pentane
Structure:

ii) 1-Bromo-4-sec-butyl 2-methyl benzene
Structure:

Question 2.
Which compound in each of the following pairs will react faster in S_N² reaction with -OH?
i) CH₃Br or CH₃I
ii) (CH₃)₃CCl or CH₃Cl.
Answer:
i) Among CH₃Br and CH₃I, CH₃ – I reacts faster in S_N2 reaction with OH^⊖ because bond dissociation energy of C -1 is less than the bond dissociation energy of C – Br.

ii) Among CH₃Cl and (CH₃)₃CCl, CH₃ – Cl reacts faster in S_N2 reaction with OH^⊖ because (CH₃)₃CCl has high steric hindrance than CH₃Cl.

Question 3.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed aqueous KOH ?
Answer:

• Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅ the 2nd one i.e., C₆H₅CHClC₆H₅ gets hydrolysed more easily than C₆H₅CH₂Cl.
• This can be explained by considering S_N1 reaction mechanism. In case of S_N1 reactions reactivity depends upon the stability of carbo cations.
• C₆H₅CHClC₆H₅ forms more stable carbo cation than C₆H₅CH₂Cl.
  

Question 4.
Treatment of alkyl halides with aq.KOH leads to the formation of alcohols, while in presence of alc.KOH what products are formed ?
Answer:
Treatment of alkyl halides with aq. KOH leads to the formation of alcohols. Here Nucleophillic substitution reaction takes place.
Eg. : C₂H₅Cl + aq.KOH → C₂H₅OH + KCl
Treatment of alkyl halides with alc.KOH leads to the formation of alkenes. Here elimination reaction takes place.
Eg. : C₂H₅Cl + alc. KOH → C₂H₄ + KCl + H₂O

Question 5.
What is Grignard’s reagents. How it is prepared.
Answer:
Alkyl magnesium halide is called Grignard reagent. It is prepared by the action of Mg on alkyl halide in ether solvent.
R – X + Mg → RMgX

Question 6.
What is the stereochemical result of S_N1 and S_N2 reactions? [T.S. Mar. 17 IPE 2015 (AP)]
Answer:

• The stereochemical result of S_N1 reaction is racemisation product.
• The stereochemical result of S_N2 reaction is inversion product.

Short Answer Questions

Question 1.
Define the following [A.P. Mar. 17] [IPE – 2014, 2016 (AP)]
i) Racemic mixture
ii) Retention of configuration
iii) Enantiomers.
Answer:
i) Racemic mixture: Equal portions of Enantiomers combined to form an optically inactive mixture. This mixture is called racemic mixture.

• Here rotation due to one isomer will be exactly cancelled by the rotation of due to other isomer.
• The process of conversion of enantiomer into a racemic mixture is called as racemisation.
  

ii) Retention of configuration: The preservation of integrity of the spatial arrangement of . bonds to an asymmetric centre during a chemical reaction (or) transformation is called Retention of configuration.
General Eg : Conversion of XC_abc chemical species into YC_abc.

Eg : (-) 2 – Methyl 1 – butanol conversion into (+) 1 – chloro 2. Methyl butane

Enantiomers : The stereo isomers related to each other as non-superimposable mirror images are called enantiomers. [A.P. Mar. 16]
These have identical physical properties like melting point, boiling points refractive index etc.
They differ in rotation of plane polarised light.

Question 2.
Explain the mechanism of Nucleophilic bimolecular substitution (S_N2) reaction with one example. [A.P. Mar. 18. 16] [Mar. 14]
Answer:
Nucleophilic Bimolecular substitution Reaction S_N2 :

1. The nucleophilic substitution reaction in which rate depends upon concentration of both reactants is called S_N2 reaction.
2.  It follows 2nd order kinetics. So it is called bimolecular reaction.
   Eg.: Methyl chloride reacts with hydroxide ion and forms methanol and chloride ion.
3. Here the rate of reaction depends upon the concentration of two reactants.
   
4. In the above mechanism the configuration of carbon atom under attack inverts in much the same way as an umbrella is turned inside out when caught in a strong wind. This process is called inversion of configuration.
5. In transition state the carbon atom is simultaneously bonded to the incoming nucleophile and out going group. It is very unstable.
6. The order of reactivity for SN2 reactions follows : 1°-alkyl halides > 2°-alkyl halides > 3°-alkyl halides.

Question 3.
Explain why allylic and benzylic halides are more reactive towards S_N1 substitution while 1-halo and 2-halobutanes preferentially undergoes S_N2 substitution.
Answer:

1. Allylic and benzylic halides show high reactivity towards the S_N1 reaction.
   Reason : The carbocation thus formed gets stabilised through resonance phenomenon as shown below.
   
2. 1-halo and 2-halo butanes preferentially undergoes S_N2 substitute.
   Reason : S_N2 reactions involve transition state formation. Higher the steric hindrance lesser the stability of transition state. The given 1 -halo and 2-halo butanes have less steric hindrance. So these are preferentially undergo S_N2 reaction.

Question 4.
Write the preparations of Alkyl halides.
Answer:

Preparation ofAlkyl Halides :
i) From Alcohols : Alkyl halides are prepared by the action of HX, PX₃, PX₃, PX₅, X₂ & red phosphurs or SOCl on alcohols.

Question 5.
Explain SN¹ and SN² reactions. [T.S. Mar. 18] [IPE – 2016, 2015, 2014 (TS)]
Answer:
i) SN¹: Substitution nucleophilic unimolecular reaction: In this reaction the first step is the formation of stable carbonium ion. This step is slow and is the rate determining step. The alkyl halides which form stable carbonium ion follow SN¹ reaction.

The order of reactivity of alkyl halides towards SN¹ reaction.
Tertiary halide > Secondary halide > Primary halide > CH₃ – X.
Benzyl halide and allyl halides are primary halides but participate in SN¹ mechanism due to the formation of stable benzyl and allyl carbonium ions (Benzyl and allyl carbonium ions are stabilised due to resonance).

ii) SN²: Substitution nucleophilic bimolecular reaction: In this reaction the rate of reaction depends on the concentration of alkyl halide and also on the concentration of nucleophile hence it is a bimolecular reaction.

The order of reactivity of alkyl halides towards SN2 reaction.
CH₃ – X > Primary halide> Secondary halide > Tertiary halide
For a given alkyl group, the reaction of the alkyl halide, R – X follows the same order in both the mechanisms R – I > R – Br> R – Cl > R – F

Question 6.
What is Wurtz reaction ? Give equation.
Answer:
Wurtz Reaction : Alkyl halides react with sodium in dry ether solvent give alkanes.

Question 7.
Explain different chemical properties of Alkyl halide. [IPE 2016 (TS)]
Answer:

Question 8.
Explain Wurtz – Fittig and Fittig reactions. [A.P. Mar. 17]
Answer:
i) Wurtz – Fittig reaction : The reaction of aryl halide with alkyl halide in the presence of sodium in ether to give alkyl benzene is called Wurtz Fittig reaction.

ii) Fittig reaction : Aryl halides react vith sodium in dry ether solvent to give diphenyl is called fitting reaction.

Question 9.
Write any one method for the preparation of chloro benzene.
Answer:
Chlorination of benzene in the pressence of Lewis acid like AlCl₃, FeCl₃ gives chioro benzene.

Question 10.
Explain electrophilic substitution reactions of chioro benzene.
Answer:
Electrophilic Substitution Reaction : Halogen atom is electron releasing group. it activates the benzene ring hence electrophilic substitution takes place at ortho and para positions.

Question 11.
Write the structures of the following organic halides. [IPE 2016 (T.S)]
i) 1 -Bromo-4-sec-butyl-2-methylbenzene,
ii) 2-Chioro- 1 -phenylbutane
iii) p-bromochlorobenzene,
iv) 4-t-butyl-3-iodoheptane.
Answer:
i) 1 -Bromo-4-sec-butyl-2-methylbenzene,
Structure :

ii) 2-Chioro- 1 -phenylbutane
Structure :

iii) p-bromochlorobenzene,
Structure :

iv) 4-t-butyl-3-iodoheptane.
Structure :

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(a) Group-15 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(a) Group-15 Elements

Very Short Answer Questions

Question 1.
Nitrogen molecule is highly stable – Why? (IPE 2014)
Answer:
Nitrogen Molecule is more stable because in between two nitrogen atoms of N₂, a triple bond is present. To break this triple bond high energy is required (941.4KJ/mole).

Question 2.
Why are the compounds of bismuth more stable in +3 oxidation state ?
Answer:
Bismuth compounds are more stable in +3 oxidation state because ‘Bi’ exhibits +3 stable oxidation state instead of +5 due to inert pair effect.

Question 3.
What is allotropy ? Explain the different allotropic forms of phosphorus. (IPE 2016(TS))
Answer:
Allotropy: The existence of an element in different physical forms having similar chemical properties is called allotropy.
→ Allotropes of ‘P’: → White ‘P’ (or) Yellow P.

• Red P
• Scarlet ‘P’
• Violet P
• α – black P’
• β – black ‘P’.

White phosphorus :

• It is poisonous and insoluble in water and soluble in carbon disulphide and glows in dark.
• It is a translucent white waxy solid.
• It dissolves in boiling NaOH solution and gives PH₃.
  
• It is more reactive than other solid phases.
• Bond angle is 60° and it readily catches fire,

Red phosphorus :

• Red P possesses iron grey lustre.
• In it odour less, non poisonous and insoluble in water as well as CS₂.
• Red P is much less reactive than white P.

Black P’:

• α – Black P : It is formed when red P is heated in a sealed tube 803K.
• β – Black P : It is prepared by heating white P at 473 K under high pressure.

Question 4.
Explain the difference in the structures of white and red phosphorus.
Answer:
White ‘P’ molecule has tetrahedral structure (discrete molecule). Discrete ‘P’ molecules are held by vander waal’s forces.

Red ‘P’ is polymeric consisting of chains of P₄ tetrahedron linked together through covalent bonds.

Question 5.
What is inert pair effect ?
Answer:
Inert pair effect: The reluctance of ns pair of electrons to take part in bond formation is called inert pair effect.

• Bi exhibits +3 oxidation state instead of +5 due to inert pair effect.

Question 6.
How do calcium phosphide and heavy water react ?
Answer:
Calcium phosphide reacts with heavy water to form Deutero phosphine.
Ca₃P₂ + 6D₂O → 3 Ca (OD)₂ + 2PD₃

Question 7.
Ammonia is a good complexing agent – Explain with an example.
Answer:
NH₃ is a lewis base and it donates electron pair to form dative bond with metal ions. This results in the formation of complex compound.
Eg :

Question 8.
NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states – Why ?
Answer:
In gaseous state NO₂ exists as a Monomer and contains one unpaired electron but in solid state it dimerises to N₂O₄ so it doesnot contain unpaired electron.
Hence NO₂ is para magnetic in geseous state but diamagnetic in solid state.

Question 9.
Iron becomes passive in cone. HNO₃ – Why ?
Answer:
Iron becomes passive in cone. HNO₃ due to formation of a passive film of oxide on the surface of iron.

Question 10.
Give the neutral oxides of nitrogen.
Answer:
Nitrous oxide (N₂O) and Nitric oxide (NO) are neutral oxides of nitrogen.

Question 11.
Give the paramagnetic oxides of nitrogen.
Answer:
Nitric oxide (NO) and nitrogen dioxide (NO₂) are the paramagnetic oxides of nitrogen due to the presence of odd number of electrons.

Question 12.
Why is white phosphorus is more reactive than red phosphorus ?
Answer:
In white phosphorus the P – P – P bond angle is 60° hence the bonds are in strain. As a result the bonds are broken easily. In red phosphorus the bond angle is 120°, hence it is stable and less reactive.

Question 13.
What happens when white phosphorus is heated with conc. NaOH solution in an inert atmosphere of CO₂ ?
Answer:
When white phosphorus heated with con. NaOH solution in an inert atmosphere of CO₂ forms PH₃.
P₄ + 3NaoH + 3H₂O → PH₃ + 3NaH₂PO₂.

Question 14.
Give the uses of
a) nitric acid and
b) ammonia.
Answer:
Nitric acid : It is used in the manufacture of fertilizers, explosives, nitro glycerine, nitro toluence etc.,
Ammonia : It is used to produce fertilizers like urea liquid NH₃ is used as a refrigerant.

Question 15.
Give the disproportionation reaction of H₃PO₃.
Answer:
Orthophosphoric acid (H₃PO₃) on heating disproportionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃.

Question 16.
Nitrogen exists as diatomic molecule and phosphorus as P₄ – Why ?
Answer:
Nitrogen exists as diatomic molecule :

• Nitrogen has small size and high electronegativity and nitrogen atom forms Pπ – Pπ multiple bonds with itself (triplebond). So it exists as a discrete diatomic molecule in elementary state.

Phosphorus exists as tetra atomic molecule :

• Phosphorus has large size and less electronegative and it forms P-P single bonds. So it exists as tetra atomic i.e., P₄.

Question 17.
Arrange the hydrides of group – 15 elements in the increasing order of basic strength and decreasing order of reducing character.
Answer:

• Increasing order of basic strength of Group – 15 elements hydrides is
  BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃.
• Decreasing order of reducing character of Group – 15 elements hydrides is
  BiH₃ > SbH₃ > AsH₃ > PH₃ > NH₃.

Question 18.
PH₃ is a weaker base than NH₃ – Explain.
Answer:
PH₃ is a weaker base than NH₃.

• In NH₃ nitrogen atom undergoes sp³ hybridisation and due to small size it has high electron density than in’P of PH₃.
• Due to large size of ‘P’ atom and availability of large surface area, lone pair of electron spread in PH₃. Hence PH₃ is weaker base than NH₃.

Question 19.
A mixture of Ca₃P₂ and CaC₂ is used in making Holme’s signal – Explain.
Answer:
A mixture of Ca₃P₂ and CaC₂ is used in Holme’s signal. This Mixture containing containers are pierced and thrown in the sea, when the gas is evolved bum and serve as a signal.
The spontaneous combustion of PH₃ is the technical use of Holme’s signal.

Question 20.
Which chemical compound is formed in the brown ring test of nitrate ions ?
Answer:
In the brown ring test of nitrate salts a brown ring is formed. It’s formula is [Fe(H₂O)₅NO]⁺².

Question 21.
Why does NH₃ act as a Lewis base ?
Answer:
Nitrogen atom in NH₃ has one lone pair of electrons with is available for donation. Therefore, it acts as a Lewis base.

Question 22.
Why does NO₂ dimerise ?
Answer:
NO₂ contains of odd number of electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N₂O₄ molecule with even number of electrons.

Question 23.
Why does PCl₃ fume in moisture ?
Answer:
PCl₃ hydrolyses in the presence of moisture giving fumes of HCl.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 24.
Are all the five bonds in PCl₅ molecule equivalent ? Justify your answer.
Answer:
PCl₅ has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent. While the two axial bonds are different and longer than equatorial bonds.

Question 25.
How is nitric oxide (NO) prepared ?
Answer:
Nitric oxide (NO) is prepared by the action of dilute nitric acid on copper.

Question 26.
Explain the following .
a) reaction of alkali with red phosphorus.
b) reaction between PCl₃ and H₂O.
Answer:
a) Red phosphorus reacts with alkalies slowly and forms phosphine and hypophosphite.
P₄ + 3NaOH + 3 H₂O → PH₃ + 3 NaH₂ PO₂
b) PCl₃ is hydrolysed by water and gives H₃PO₃

Question 27.
Write the oxidation states of phosphorus in solid PCl₅.
Answer:
In solid state PCl₅ exists as an ionic solid [PCl₄)⁺ [PCl₆]^–

• ‘P’ exhibits +5 oxidation state in [PCl₄)⁺ and [PCl₆]

Question 28.
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:

Question 29.
H₃PO₂ is a good reducing agent – Explain with an example.
Answer:
In H₃PO₂, two H-atoms are bonded directly to P-atom which imparts reducing character to the acid.

Question 30.
NH₃ forms hydrogen bonds but PH₃ does not – Why ?
Answer:
NH₃ forms hydrogen bonds but PH₃ does not.
Reason : Ammonia forms hydrogen bonds because it is a polar molecule and N-H bond is highly polar. Nitrogen has more electronegativity than phosphorus. In case of PH₃, P-H bond polarity decreases.

Question 31.
PH₃ has lower boiling point than NH₃. Why ?
Answer:
Unlike NH₃, PH₃ molecules are not associated through inter molecular hydrogen bonding. That is why the boiling point of PH₃ is lower than NH₃.

Long Answer Questions

Question 1.
How is ammonia manufactured by Haber’s process ? Explain the reactions of ammonia with
a) ZnSO_4(aq)
b) CuSO_4(aq)
c) AgCl_(s) (TS Mar. ’17 IPE 2015(AP))
Answer:
In Haber process ammonia is directly synthesised from elements (nitrogen and hydrogen). The principle involved in this is

This is a reversible exothermic reaction.
According to Le Chatelier’s principle favourable conditions for the better yield of ammonia are low temperature and high pressure. But the optimum conditions are :
Temperature : 720k
Pressure : 200 atmospheres
Catalyst: Finely divided iron in the presence of molybdenum (Promoter).
Procedure : A mixture of nitrogen and hydrogen in the volume ratio 1 : 3 is heated to 725 – 775K at a pressure of 200 atmospheres is passed over hot finely divided iron mixed with small amount of molybdenum as promotor. The gases coming out of the catalyst chamber consists of 10 – 20% ammonia. Gases are cooled and compressed, so that ammonia gas is liquified and the uncondensed gases are sent for recirculation.

a) Aq. ZnSO₄ reacts with ammonia aqueous solution to form white ppt of Zinc hydroxide.

Question 2.
How is nitric acid manufactured by Ostwald’s process ? How does it react with the following ? (A.P. Mar. ’17)
a) Copper
b) Zn
c) S₈
d) P₄
Answer:
Ostwald’s process : Ammonia, mixed with air in 1 : 7 or 1 : 8, when passed over a hot platinum gauze catalyst, is oxidised to NO mostly (about 95%).
The reaction is

The liberated heat keeps the catalyst hot. The ’NO’ is cooled and is mixed with oxygen to give the dioxide, in large empty towers (oxidation chamber). The product is then passed into warm water, under pressure in the presence of excess of air, to give HNO₃.
4NO₂ + O₂ + 2H₂O → 4HNO₃.
The acid formed is about 61% concentrated.

a) Copper reacts with dil. HNO₃ and cone. HNO₃ and liberates Nitric Oxide and Nitrogen dioxide respectively
3 Cu + 5 HNO₃ (dil) → 3 Cu (NO₃)₂ + 2 NO + 4 H₂O
Cu + 4 HNO₃ (conc.) → Cu (NO₃)₂ + 2 NO₂ + 2 H₂O

b) Zn reacts with dil. HNO₃ and Cone. HNO₃ and liberates Nitrous oxide and Nitrogen
dioxide respectively.
4 Zn + 10 HNO₃ (dil) → 4 Zn (NO₃)₂ + 5 H₂O + N₂O.
Zn + 4 HNO₃ (conc.) → Zn (NO₃)₂ + 2 H₂O + 2 NO₂

c) S₈ reacts with cone, nitric acid to form Sulphuric acid, NO₂ gas.
S₈ + 48 HNO₃ → 8 H₂SO₄ + 48 NO₂ + 16 H₂O
d) P₄ reacts with cone, nitric acid to form phosphoric acid and NO₂ gas.
P₄ + 20 HNO₃ → 4 H₃PO₄ + 20 NO₂ + 4 H₂O

Students get through AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits

Very Short Answer Questions

Question 1.
What is an n-type semiconductor? What are the majority and minority charge carriers in it?
Answer:

• If a pentavalent impurity is added to a pure tetravalent semiconductor, it is called an n-type semiconductor.
• In n-type semiconductor majority, of charge carriers are electrons and the minority charge carriers are holes.

Question 2.
What are intrinsic and extrinsic semiconductors? [A.P. Mar. 15]
Answer:

• Pure form of semiconductors are called intrinsic semiconductors.
• When impure atoms are added to increase their conductivity, they are called extrinsic semiconductors.

Question 3.
What is a p-type semiconductor ? What are the majority and minority charge carriers in it ? [A.P. & T.S. Mar. 17]
Answer:
If a trivalent impurity is added to a tetravalent semiconductor, it is called p-type semi-conductor.

In p-type semiconductor majority charge carriers are holes and minority charge carriers are electrons.

Question 4.
What is a p-n junction diode ? Define depletion’ layer.
Answer:
When an intrinsic semiconductor crystal is grown with one side doped with trivalent element and on the other side doped with pentavalent element, a junction is formed in the crystal. It is called p-n junction diode.

A thin narrow region is formed on either side of the p-n junction, which is free from charge carriers is called depletion layer.

Question 5.
How is a battery connected to a junction diode in i) forward and ii) reverse bias ?
Answer:
i) In p-n junction diode, if p-side is connected to positive terminal of a cell and n-side to negative terminal, it is called forward bias.

ii) In a p-n junction diode, p-side is connected to negative terminal of a cell and n-side to positive terminal, it is called reverse bias.

Question 6.
What is the maximum percentage of rectification in half wave and full wave rectifiers ?
Answer:

1. The percentage of rectification in half-wave rectifier is 40.6%.
2. The percentage of rectification in full-wave rectifiers is 81.2%.

Question 7.
What is Zener voltage (V_z) and how will a Zener diode be connected in circuits generally ?
Answer:

1. When Zener diode is in reverse biased, at a particular voltage current increases suddenly is called Zener (or) break down voltage.
2. Zener diode always connected in reverse bias.

Question 8.
Write the expressions for the efficiency of a full wave rectifier and a half wave rectifier.
Answer:

1. Efficiency of half wave rectifier (η) = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
2. Efficiency of full wave rectifier (η) = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)

Question 9.
What happens to the width of the depletion layer in a p-n junction diode when it is
i) forward-biased and ii) reverse biased ?
Answer:
When a p-n junction diode is forward bias, thickness of depletion layer decreases and in reverse bias, thickness of depletion layer increases.

Question 10.
Draw the circuit symbols for p-n-p and n-p-n transistors. [Mar. 14]
Answer:

Question 11.
Define amplifier and amplification factor.
Answer:

1. Rising the strength of a weak signal is known as amplification and the device is called amplifier.
2. Amplification factor is the ratio between output voltage to the input voltage.
   A = \(\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}}\)

Question 12.
In which bias can a Zener diode be used as voltage regulator ?
Answer:
In reverse bias Zener diode can be used as voltage regulator.

Question 13.
Which gates are called universal gates ? [T.S. Mar. 15]
Answer:
NAND gate and NOR gates are called universal gates.

Question 14.
Write the truth table of NAND gate. How does it differ from AND gate ?
Answer:

Short Answer Questions

Question 1.
What is a rectifier ? Explain the working of half wave and full Wave rectifiers with diagrams. [A.P. Mar. 17]
Answer:
Rectifier: It is a circuit which converts ac into d.c. A p-n junction diode is used as a rectifier.
Half-wave rectifier:

1. A half wave rectifier can be constructed with a single diode. The ac input signal is fed to the primary coil of a transformer. The output signal is taken across the load resistance R_L.
2. During positive half cycle, the diode is forward biased and current flows through the diode.
3. During negative half cycle, diode is reverse biased and no current flows through the load resistance.
4. This means current flows through the diode only during positive half cycles and negative half cycles are blocked. Hence in the output we get only positive half cycles.
5. Rectifier efficiency is defined as the ratio of output dc power to the input ac power.
   η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
   Where r_f = Forward resistance of a diode; R_L = Load resistance
   The maximum efficiency of half wave rectifier is 40.6%.

Full wave rectifier : The process of converting an alternating current into a direct current is called rectification.
The device used for this purpose is called rectifier.

1. A full wave rectifier can be constructed with the help of two diodes D₁ and D₂.
2. The secondary transformer is centre tapped at C and its ends are connected to the p regions of two diodes D₁ & D₂.
3. The output voltage is measured across the load resistance R_L.
4. During positive half cycles of ac, the diode D₁ is forward biased and current flows through the load resistance R_L. At this time D₂ will be reverse biased and will be in switch off position.
5. During negative half cycles of ac, the diode D₂ is forward biased and the current flows through R_L. At this time D₁ will be reverse biased and will be in switch off position.
6. Hence positive output is obtained for all the input ac signals.
7. The efficiency of a rectifier is defined as the ratio between the output dc power to the input ac power.
   η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
   The maximum efficiency of a full wave rectifier is 81.2%.

Question 2.
What is a junction diode ? Explain the formation of depletion region at the junction. Explain the variation of depletion region in forward and reverse-biased condition.
Answer:
p-n junction diode: When a p-type semiconductor is suitablyjoined to n-type semiconductor, a p-n junction diode is formed. ’
The circuit symbol of p-n junction diode is shown in figure.

Formation of depletion layer at the junction: When p-n junction is formed, the free electrons on n-side diffuse over to p-side, they combine with holes and become neutral. Similarly holes on p-side diffuse over to n-side and combihe with electrons become neutral.

This results in a narrow region formed on either side of the junction. This region is called depletion layer. Depletion layer is free from charge carriers.

The n-type material near the junction becomes positively charged due to immobile donor ions and p-type material becomes negatively charged due to immobile acceptor ions. This creates an electric field near the junction directed from n-region to p-region and cause a potential barrier.

The potential barrier stops further diffusion of holes and electrons across the* junction. The value of the potential barrier depends upon the nature of the crystal, its temperature and the amount of doping.

Forward bias :
“When a positive terminal of a battery is connected to p-side and negative terminal is connected to n-side; then p-n junction diode is said to be forward bias”.

The holes in the p-region are repelled by the positive polarity and move towards the junction. Similarly electrons in the n-region are repelled by the negative polarity and move towards the junction.

As a result, the width of the depletion layer decreases. The charge carriers cross the junction apd electric current flows in the circuit.
Hence in forward bias resistance of diode is low. This position is called switch on position.
Reverse bias:

“When the negative terminal of the battery is connected to p-side and positive terminal of the battery is connected to n-side of the p-n junction, then the diode is said to be reverse bias”.

The holes in the p-region are attracted towards negative polarity and move away from the junction. Similarly the electrons in the n-region are attracted towards positive polarity and move away from the junction.

So, width of the depletion layer and potential barrier increases. Hence resistance of p-n junction diode increases. Thus the reverse biased diode is called switch off position.

Question 3.
What is a Zener diode ? Explain how it is used as a voltage regulator.
Answer:
Zener diode : Zener diode is a heavily doped germanium (or) silicon p-n junction diode. It works on reverse bias break down region.
The circuit symbol of zener diode is shown in figure.

Zener diode can be used as a voltage regulator. In- general zener diode is connected in reverse bias in the circuits.

1. The zener diode is connected to a battery, through a resistance R. The battery reverse biases the zener diode.
2. The load resistance R_L is connected across the terminals of the zener diode.
3. The value of R is selected in such away that In the absence of load R_L maximum safe current flows in the diode.
4. Now consider that load is connected across the diode. The load draws a current.
5. The current through the diode falls by the same amount but the voltage drops across the load remains constant.
6. The series resistance ‘R’ absorbs the output voltage fluctuations, so as to maintain constant voltage across the load.
7. The voltage across the zener diode remains constant even if the load R_L varies. Thus, zener diode works as voltage regulator.
8. If I is the input current, I_Z and I_L are zener and load currents.
   I = I_Z + I_L; V_in = IR + V_Z
   But V_0ut = V_Z
   ∴ V_out = V_in – IR

Question 4.
Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
Answer:
Light emitting diode (LED) : It is a photoelectronic device which converts electrical energy into light energy.

It is a heavily doped p-n junction diode which under forward bias emits spontaneous radiation. The diode is covered with a transparent cover so that the emitted light may not come out. Working : When p-n junction diode is forward biased, the movement of majority charge carriers takes place across the junction. The electrons move from n-side to p-side through the junction and holes move from p-side to n-side.

As a result of it, concentration of majority carriers increases rapidly at the junction.

When there is no bias across the junction, therefore there are excess minority carriers on either side of the junction, which recombine with majority carriers near the junction.

On recombination of electrons and hole, the energy is given out in the form of heat and light.
Advantages of LED’s over incandescent lamp :

1. LED is cheap and easy to handle.
2. LED has less power and low operational voltage.
3. LED has fast action and requires no warm up time.
4. LED can be used in burglar alarm system.

Question 5.
Define NAND and NOR gates. Give their truth tables. [T.S. Mar. 17]
Answer:
NAND gate : NAND gate is a combination of AND gate and NOT gate.

NAND gate can be obtained by connecting a NOT gate in the output of an AND gate. NAND gates are called universal gates.

1. If both inputs are low, output is high.
   A = 0, B = 0, X = 1
2. If any input is low, output is high.
   A = 0, B = 1, X = 1
   A = 1, B = 0, X =1
3. If both inputs are high, output is low.
   A = 1, B = 1, X = 0
   

NOR gate : NOR gate is a combination of OR gate and NOT gate when the output of OR gate is connected to NOT gate. It has two (or) more inputs and one output.

1. If both inputs are low, output is high.
   A = 0, B = 0, X = 1
2. If any input is high, the output is low.
   A = 0, B = 1, X = 0
   A = 1, B = 0, X = 0
3. If both inputs are high, the output is low.
   A = 1, B = 1, X = 0
   NOR gate is also a universal gate.
   

Question 6.
Explain the working of a solar cell and draw its I-V characteristics.
Answer:
Solar cell is a p-n junction device which converts solar energy into electric energy.
It consists of a silicon (or) gallium – arsenic p-n junction diode packed in a can with glass window on top. The upper layer is of p- type semiconductor. It is very thin so that the incident light photons may easily reach the p-n junction.

Working : When light (E = hv) falls at the junction, electron – hole pairs are generated near the junction. The electrons and holes produced move in opposite directions due to junction field. They will be collected at the two sides of the junction giving rise to a photo voltage between top and bottom metal electrodes. Top metal acts as positive electrode and bottom metal acts as a negative electrode. When an external load is connected across metal electrodes a photo current flows.

I-V characteristics : I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes. Because it does not draw current.

Uses : They are used in calculators, wrist watches, artificial satellites etc.

Question 7.
Explain the operation of a NOT gate and give its truth table. [IPE 15, T.S.]
Answer:
NOT gate: NOT gate is the basic gate. It has one input and one output. The NOT gate is also called an inverter. The circuit symbol of NOT gate is shown in figure.

1. If input is low, output is high.
   A = 0, X = \(\overline{0}\) = 1
2. If input is high, output is low.
   A = 1, X = \(\overline{1}\) = 0
   

Question 8.
Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
Answer:
OR Gate : It has two input terminals and one output terminal. If both inputs are low the output is low. If one of the inputs is high, or both the inputs are high then the output of the gate is high. The truth tables of OR gate.

In truth table logic function is written as A or B ‘OR’ logic function is represented by the symbol ‘plus’.
Q = A + B
Logic gate ‘OR’ is shown given below.

Implementation of OR gate using diodes :
Let D₁ and D₂ be two diodes.

A potential of 5V represents the logical value 1.
A potential of OV represents the logical value 0.

When A = 0, B = 0 both the diodes are reverse biased and there is no current through the resistance. So, the potential at Q is zero i.e., Q = 0. When A = 0 or B = 0 and the other equal to a potential of 5 V the diode whose anode is at a potential of 5 V is forward – biased and that diode behaves like a closed switch. The output potential then becomes 5 V i.e., Q = 1. When both A and B are 1, both the diodes are forward-biased and the potential at Q is same as that at A and B which is 5 V i.e., Q = 1. The output is same as that of the OR gate.

Question 9.
Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors ?
Answer:
AND gate : It has two input terminals and one output terminal.

• If both the inputs are low (or) one of the inputs is low.
  • The output is low in an AND gate.
• If both the inputs .are high
  • The output of the gate is high.
• Note : If A and B are the inputs of the gate and the output is ‘Q’ then ‘Q’ is a logical function of A and B.
  AND gate Truth Tables
  
  The logical function AND is represented by the symbol dot so that the output, Q = A.B and the circuit symbol used for the logic gate AND is shown in Fig.
  
  The logical function AND is similar to the multiplication.

Implementation of AND gate using diodes : Let D₁ and D₂ represents two diodes. A potential of 5 V represents the logical value 1 and a potential of 0 V represents the logical value zero (0). When A = 0, B = ,0 both the diodes D₁ and D₂ are forward-biased and they behave like closed switches. Hence, the output Q is same as that A or B (equal to zero.) When A = 0 or B = 0, D₁ or D₂ is forward – biased and Q is zero. When A = 1 and B = T both the diodes are reverse – biased and they behave like open switches. There is no current through the resistance R making the potential at Q equal to 5V i.e., Q = 1. The output is same as that of an AND gate.

Doping increases the conductivity in Semiconductors: If a pentavalent impurity (Arsenic) is added to a pure tetravalent semiconductor it is called n-type semiconductor. Arsenic has 5 valence electrons, but only 4 electrons are needed to form covalent bonds with the adjacent Germanium atoms. The fifth electron is very loosely bound and become a free electron. Therefore excess electrons are available for conduction and conductivity of semiconductor increases.

Similarly when a trivalent impurity Indium is added to pure Germanium it is called p-type semi-conductor. In this excess holes in addition to those formed due to thermal energy are available for conduction in the valence band and the conductivity of semiconductor increases.

Question 10.
Explain how transistor can be used as a switch ?
Answer:
To understand the operation of transistor as a switch.

1. As long as V_i is low and unable to forward bias the transistor, V₀ is high (at V_cc).
2. If V_i is high enough to drive the transistor into saturation then V₀ is low, very near to zero.
   
3. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.
4. We can say that a low input to the transistor gives a high output and high input gives a low output.
5. When the transistor is used in the cutoff (or) saturation state it acts as a switch.
   

Problems

Question 1.
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier.
Solution:
Internal resistance (r_f) = 20Ω
R_L = 2kΩ = 2000 Ω
η = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.406 \times 2000}{20+2000} \times 100=\frac{812 \times 100}{2020}\)
η = 40.2%.

Question 2.
A full wave p-n junction diode rectifier uses a load resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Solution:
Given that R_L = 1300 Ω
r_f = 9Ω
η = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.812 \times 1300}{9+1300} \times 100 ; \eta=\frac{8120 \times 13}{1309}\)
η = 80.64%.

Question 3.
Calculate the current amplification factor β(beta) when change in collector current is 1mA and change in base current is 20μA.
Solution:
Change in collector current (∆I_C) = 1mA = 10^-3 A
Change in base current (∆I_B) = 20 μA = 20 × 10^-6 A
β = \(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{10^{-3}}{20 \times 10^{-6}}\); β = \(\frac{1000}{20}\)
β = 50

Question 4.
For a transistor amplifier, the collector load resistance R_L = 2k ohm and the input resistance R_i = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Solution:
R_L = 2kΩ = 2 × 10³ Ω
R_i = 1kΩ = 1 × 10³ Ω
β = 50.
Voltage gain (A_V) = β × \(\frac{\mathrm{R}_{\mathrm{L}}}{\mathrm{R}_{\mathrm{i}}}=\frac{50 \times 2 \times 10^3}{1 \times 10^3}\)

A_V = 100.

Textual Examples

Question 1.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors ?
Solution:
The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out and electron from these atoms(i.e., ionisation energy E_g) will be least for Ge, following by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.

Question 2.
Suppose a pure Si crystal has 5 × 10²⁸ atoms m^-3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n. = 1.5 × 10¹⁶ m^-3.
Solution:
Note that thermally generated electrons (n_i ~ 10¹⁶ m^-3) are negligibly small as compared to those produced by doping.
Therefore, n_i ≈ N_D.
Since n_en_h = n_i², The number of holes, n_h = (1.5 × 10¹⁶)² / 5 × 10²⁸ × 16^-6
n_h = (2.25 × 10³²)/(5 × 10²²) ~ 4.5 × 10⁹m^-3

Question 3.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction ?
Solution:
No ! Any slab, howsoever flat, will have roughness much large than the inter-atomic crystal spacing(~2 to 3 A°) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 4.
The V-I characteristics of a silicon diode are shown in the Fig. Calculate the resistance of the diode at (a) I_D = 15 mA and (b) V_D = -10 V.

Solution:
Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
r_fb = ∆V/∆I = 0.1V/10mA = 10Ω
b) From the curve at V = -10 V, I = -1 µA.
Therefore r_rb = 10V/1µA = 1.0 × 10⁷ Ω

Question 5.
In a Zener regulated power supply a Zener diode’with V_z = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10,0 V. What should be the value of series resistor R_s?
Solution:
The value of Rs should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., I_z = 20 mA. The total current through R_S is , therefore, 24 mA. The voltage drop across R_S is 10.0 – 6.0 = 4.0 V This gives R_S = 4.0V/(24 × 10^-3) A = 167Ω. The nearest value of carbon resistor is 150 Ω. So, a series resistor of 150 Ω is appropriate. Note that slight variation in the value of the resistor does not matter, what is important is that the current I_Z should be sufficiently larger than I_L.

Question 6.
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason then to operate the photodiodes in reverse bias ?
Solution:
Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n> > p). On illumination, let the excess eletrons and holes generated be ∆n and ∆p, respectively,
n’ = n + ∆n
p’ = p + ∆p
Here n’ and p’ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember ∆n = ∆p and n > > p. Hence, the fractional change in the majority carriers (i.e., ∆n/n) would be much less than that in the minority carriers (i.e,, ∆p/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes-are preferably used in the reverse bias condition for measuring light intensity.

Question 7.
Why are Si and GaAs are preferred materials for solar cells ?
Solution:
The solar radiation spectrum received by us is shown in figure.

The maxima is near 1.5 eV. For photo-excitation, hv > E_g. Hence, semiconductor with band gap ~1.5 eV or lower is likely to give better solar conversion efficiency. Silicon has E_g ~ 1.1 eV while for GaAS it is ~ 1.53 eV. In fact, GaAs is better On spite of its higher band gap) than Si because of its relatively higher absorption coefficient. If we choose materials like CdS or CdSe(E_g ~ 2.4 eV), we can use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use.

The question arises: why we do not use material like pbS(E_g ~ 0.4 eV) which satisfy the condition hv > E_g for v maxima corresponding to the solar radiation spectra ? if we do so, most of the solar radiation will be absorbed on the top-layer of solar cell mid will not reach in or near the depletion region. For effective electron-hole separation, due to the junction field, we want the photo-generation to occur in the junction region only.

Question 8.
From the output charactristics shown in fig, calculate the values of β_ac and β_dc of the transistor when V_CE is 10 V and I_C = 4.0 mA.

Solution:
β_ac = \(\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}\) ; β_dc = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}\)
For determining β_ac and β_dc at the stated values of V_CE and I_C one can proceed as follows. Consider any two characteristics for two values of I_B which lie above and below the given value of I_C. Here I_C = 4.0 mA. (Choose characteristics for I_B = 30 and 20μA.) At V _CE = 10V
we read the two values of Ic from the graph. Then
∆I_B = (30 – 20)μA = 10μA. ∆I_C
= (4.5 – 3.0) mA = 1.5 mA
Therefore, β_ac =1.5 mA/ 10μA = 150
For determining β_dc either estimate the value of I_B corresponding to I_C = 4.0 mA at V_CE = 10V or calculate the two values of β_dc for the two characteristics chosen and find their mean.
Therefore, for I_C = 4.5 mA and I_B = 30 μA
β_dc = 4.5 mA/30 μA = 150 and for I_C = 3.0 mA/ and I_B = 20 μA
β_dc = 3.0 mA / 20 μA= 150
Hence, β_dc = (150 + 150)/ 2 = 150

Question 9.
In Fig. the V_BB supply can be varied from OV to 5.0 V. The Si transistor has β_dc = 250 and R_B = 100 kΩ, R_C = 1 KΩ, V_CC = 5.0V. Assume that when the transistor is saturated, V_CE = 0V and V_BE = 0.8V. Calculate
(a) the minimum base current, for which the transistor will reach saturation. Hence,
(b) determine V₁ for when the transistor is ‘switched on’,
(c) find the ranges of V₁ for which the transistor is ‘switched of and ‘switched on’.

Solution:
Given at stauration
V_CE = OV, V_BE = 0.8V
V_CE = V_CC – I_CR_C
I_C = V_CC / R_C = 5.0V/1/0 kΩ = 5.0mA
Therefore, I_B = I_C/β
= 5.0 mA/250 = 20μA
The input voltage at which the transistor will go into saturation is given by
V_IH = V_BB = I_BR_B + V_BE
= 20μA × 100 kΩ + 0.8V = 2.8V
The value of input voltage below which the transistor remains cutoff is given by
V_IL = 0.6V, V_IH = 2.8V
Between 0.0V and 0.6V the transistor will be in the ‘switched off-state. Between 2.8V and 5.0V, it will be in ‘switched on’ state.
Note that the transistor is in active state when I_B varies from 0.0mA to 20mA. In this range, I_C = βI_B is valid. In the saturation range I_C ≤ βI_B.

Question 10.
For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kΩ is 2.0 V. Suppose the current amplification factor of the transistor is 100, what should be the value of R_B in series with V_BB supply of 2.0 V if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance. (Refer to Fig)

Solution:
The output ac voltage is 2.0 V. So, the ac collector current i_C= 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by i_B = i_C/β = 1.0 mA/100 = 0.010 mA. The dc base current has to be 10 × 0.010 = 0.10 mA
From V_BB = V_BE+ I_B R_B R_B = (V_BB – V_BE)/I_B. Assuming V_BE = 0.6V
R_B = (2.0 – 0.6)/0.10 = 14kΩ
The dc collector current I_C = 100 × 0.10 = 10 mA.

Question 11.
Justify the output waveform (Y) of the OR gate for the following inputs A and B given in fig.

Solution:
Note the following :

• At t ≤ t₁; A = 0, B = 0; Hence Y = 0
• For t₁ to t₂; A = 1, B = 0; Hence Y = 1
• For t₂ to t₃; A = 1, B = 1; Hence Y = 1
• For t₃ to t₄; A = 0, B = 1; Hence Y = 1
• For t₄ to t₅; A = 0, B = 0; Hence Y = 0
• For t₅ to t₆; A = 1, B = 0; Hence Y = 1
• For t > t₆; A = 0,B = 1; Hence Y = 1

Therefore the waveform Y will be as shown in the Fig.

Question 12.
Take A and B input waveforms similar to that in Example 11. Sketch the output waveform obtained from AND gate.
Solution:

• For t ≤ t₁; A = 0, B = 0; Hence Y = 0
• For t₁ to t₂; A = 1, B = 0; Hence Y = 0
• For t₂ to t₃; A = 1, B = 1; Hence Y = 1
• For t₃ to t₄; A = 0, B = 1; Hence Y = 0
• For t₄ to t₅; A = 0, B = 0; Hence Y = 0
• For t₅ to t₆; A = 1, B = 0; Hence Y = 0
• For t > t₆; A = 0,B = 1; Hence Y = 0

Based on the above, the output waveform for AND gate can be drawn as given below.

Question 13.
Sketch the output Y from a NAND gate having inputs A and B given below :
Solution:

• For t ≤ t₁; A = 1, B = 1; Hence Y = 0
• For t₁ to t₂; A = 0, B = 0; Hence Y = 1
• For t₂ to t₃; A = 0, B = 1; Hence Y = 1
• For t₃ to t₄; A = 1, B = 0; Hence Y = 1
• For t₄ to t₅; A = 1, B = 1; Hence Y = 0
• For t₅ to t₆; A = 0, B = 0; Hence Y = 1
• For t > t₆; A = 0,B = 1; Hence Y = 1
  

Students get through AP Inter 2nd Year Physics Important Questions 5th Lesson Electrostatic Potential and Capacitance which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 5th Lesson Electrostatic Potential and Capacitance

Short Answer Questions

Question 1.
Derive an expression for the electric potential due to a point charge. [T.S. Mar. 16]
Answer:
Expression for the electric potential due to a point charge:

1. Electric potential at a point is defined as the amount of workdone in moving a unit + ve charge from infinity to that point.
   
2. Consider a point P at a distance r from the point charge having charge + q. The electric field at P = E = \(\frac{q}{4 \pi \varepsilon_0 x^2}\)
3. Workdone in taking a unit +ve charge from B to A = dV = -E.dx (-ve sign shows that the workdone is +ve in the direction B to A, whereas the potential difference is +ve in the direction A to B.
4. Therefore, potential at P = The amount of workdone in taking a unit +ve charge from infinity to P
   

Question 2.
Derive an expression for the electrostatic potential energy of a system of two point charges and find its relation with electric potential of a charge.
Answer:
Expression for the electrostatic potential energy of a system of two point charges :

1. Let two point charges q₁ and q₂ are separated by distance ‘r’ in space.
2. An electric field will develop around the charge q₁.
3. To bring a charge q₂ from infinity to the point B some work must be done.
   
   Workdone = q₂ v_B
   But v_B = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r}\)
   W = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
4. This amount of workdone is stored as electrostatic potential energy (U) of a system of two charged particles. Its unit is joule.
   ∴ U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
5. If the two charges are similar then ‘U’ is positive. This is in accordance with the fact that two similar charges repel one another and positive work has to be done on the system to bring the charges nearer.
6. Conversely if the two charges are of opposite sign, they attract one another and potential energy is negative.

Question 3.
Derive an expression for the potential energy of an electric dipole placed in a uniform electric field.
Answer:
Expression for potential energy of an electric dipole placed in a uniform electric field :

1. Consider a electric dipole of length 2a having charges + q and -q.
2. The electric dipole is placed in uniform electric field E and it’s axis makes an angle θ with E.
3. Force on charges are equal but opposite sign. They constitute torque on the dipole.
   
   Torque τ = one of its force (F) × ⊥^r distance (BC)
   F = qE and sinθ = \(\frac{\mathrm{BC}}{2 \mathrm{a}} \) ⇒ BC = 2a sinθ
   ∴ Torque τ = qE × 2a sinθ = PE sin θ [∴ p = 2aq]
4. Suppose the dipole is rotated through an angle dθ, the workdone dw is given by
   dw = τdθ = PE sinθ dθ
5. For rotating the dipole from angle θ₁ to θ₂
   workdone W = \(\int_{\theta_1}^{\theta_2}\) PE sinθdθ = PE(cosθ₁ – cosθ₂)
6. This workdone (W) is then stored as potential energy(U) in the dipole.
   ∴ U = PE(cosθ₁ – cosθ₂)
7. If θ₁ = 90° and θ₂ = 0°, U = – PE cos₁.
   In vector form U = –\(\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the capacitance of a parallel plate capacitor. [A.P. Mar. 16; T.S. Mar. 14]
Answer:
Expression for the capacitance of a parallel plate capacitor:

1. P and Q are two parallel plates of a capacitor separated by a distance of d.
2. The area of each plate is A. The plate P is charged and Q is earth connected.
3. The charge on P is + q and surface charge density of charge = σ
   ∴ q = Aσ
   
4. The electric intensity at point x , E = \(\frac{|\sigma|}{\varepsilon_0}\)
5. Potential difference between the plates P and Q,
   \(\int {\mathrm{dV}}=\int_{\mathrm{d}}^0-\mathrm{Edx}=\int_{\mathrm{d}}^0 \frac{-\sigma}{\varepsilon_0} \mathrm{dx}=\frac{\sigma \mathrm{d}}{\varepsilon_0}\)
6. Capacitance of the capacitor C = \(\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{A} \sigma}{\frac{\sigma \mathrm{d}}{\varepsilon_0}}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\) Farads (In air)
   Note : Capacity of a capacitor with dielectric medium is C = \(\frac{\varepsilon_0 \mathrm{~A}}{\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}\right]}\) Farads.

Question 5.
Explain the behaviour of dielectrics in an external field.
Answer:

1. When an external field is applied across dielectrics, the centre of positive charge distribution shifts in the direction of electric field and that of the negative charge distribution shifts
   
   opposite to the electric field and induce a net electric field within the medium opposite to the external field. In such situation the molecules are said to be polarised.
2. Now consider a capacitor with a dielectric between the plates. The net field in the dielectric becomes less.
3. If E₀ is the external field strength and E_i is the electric field strength induced, then the net field, \(\overrightarrow{\mathrm{E}}_{\text {net }}=\overrightarrow{\mathrm{E}}_0+\overrightarrow{\mathrm{E}}_{\mathrm{i}}\)
   (E_net) = E₀ – E_i = \(\frac{E}{K}\) where K is the dielectric constant of the medium.

Question 6.
Define electric potential. Derive an expression for the electric potential due to an electric dipole and hence the electric potential at a point (a) the axial line of electric dipole (b) on the equatorial line of electric dipole.
Answer:
Electric potential (V) : The workdone by a unit positive charge from infinite to a point in an electric field is called electric potential.

Expression for the potential at a point due to a dipole:

1. Consider A and B having -q and + q charges separated by a distance 2a.
2. The electric dipole moment P = q × 2a along AB
3. The electric potential at the point’P’is to be calculated.
4. P is at a distance ‘r’ from the point ‘O’. θ is the angle between the line OP and AB.
5. BN and AM are perpendicular to OP.
6. Potential at ‘P’ due to charge +q at B,
   V₁ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{BP}}\right]\)
   ∴ V₁ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{NP}}\right]\) [∵ BP = NP]
7. Potential at P due to charge -q at A, V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{-q}}{\mathrm{AP}}\right]\)
   ∴ V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{MP}}\right]\) [∵ AP = MP]
8. Therefore, Resultant potential at P is V = V₁ + V₂
   V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{NP}}-\frac{\mathrm{q}}{\mathrm{MP}}\right]\) ………… (1)
9. In △le ONB, ON = OB cos0 = a cosθ; ∴NP = OP – ON = r – a cosθ ………………… (2)
10. In △le AMO, OM = AO cos0 = a cosθ; ∴ MP = MO + OP = r + a cosθ ………………. (3)
11. Substituting (2) and (3) in (1), we get
    
12. As r > >a, a² cos²θ can be neglected with comparision of r².
    ∴ V = \(\frac{\mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
13. (a) Electric potential on the axial line of dipole:
    
    • When θ = 0°, point p lies on the side of + q. ____
      ∴ V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 0° = 1]
    • When θ = 180°, point p lies on the side of — q.
      ∴ V = \(\frac{\mathrm{-P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 180° = -1]

Qeustion 7.
What is series combination of capacitors. Derive the formula for equivalent capacitance in series combination. [A.P.& T.S. Mar.15]
Answer:
Series combination : If a number of condensers are connected end to end between the fixed points then such combination is called series.
In this combination

1. Charge on each capacitor is equal.
2. P.D’s across the capacitors is not equal.

Consider three capacitors of capacitances C₁, C₂ and C₃ are connected in series across a battery of P.D V as shown in figure.

Let ‘Q’ be the charge on each capacitor.
Let V₁, V₂ and V₃ be the P.D’s of three
V = V₁ + V₂ + V₃ ……………… (1)
RD across I^st condenser V₁ = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\)
RD across III^nd condenser V₂ = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)
RD across IIII^rd condenser V₃ = \(\frac{\mathrm{Q}}{\mathrm{C}_3}\)
∴ From the equation (1),V = V₁ + V₂ + V₃
= \(\frac{\mathrm{Q}}{\mathrm{C}_1}+\frac{\mathrm{Q}}{\mathrm{C}_2}+\frac{\mathrm{Q}}{\mathrm{C}_3}=\mathrm{Q}\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right]\)
\(\frac{\mathrm{V}}{\mathrm{Q}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\)
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\) [∵ \(\frac{1}{C}=\frac{V}{Q}\)
For ‘n’ number of capacitors, the effective capacitance
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}+\ldots+\frac{1}{\mathrm{C}_{\mathrm{n}}}\)

Question 8.
What is parallel combination of capacitors. Derive the formula for equivalent capacitance in parallel combination. [T.S. Mar. 17; A.P.& T.S. Mar. 15]
Answer:
Parallel Combination: The first plates of different capacitors are connected at one terminal and all the second plates of the capacitors are connected at another terminal then the two terminals are connected to the two terminals of battery is called parallel combination.

In this combination,
1. The P.D’s between each capacitor is equal (or) same.
2. Charge on each capacitor is not equal.
Consider three capacitors of capacitance C₁, C₂ and C₃ are connected in parallel across a P.D V as shown in fig.
The charge on Ist capacitor Q₁ = C₁ V
The charge on IInd capacitor Q₂ = C₂ V
The charge on IIIrd capacitor Q₃ = C₃ V
∴ The total charge Q = Q₁ + Q₂ + Q₃
= C₁ V + C₂ V + C₃ V
Q = V(C₁ + C₂ + C₃) ⇒ \(\frac{\mathrm{Q}}{\mathrm{V}}\) = C₁ + C₂ + C₃
C = C₁ + C₂ + C₃ [∵ C = \(\frac{\mathrm{Q}}{\mathrm{V}}\)]
for ‘n’ number of capacitors connected in parallel, the equivalent capacitance can be written as
C = C₁ + C₂ + C₃ + ……………. + C_n

Question 9.
Derive an expression for the energy stored in a capacitor.
Answer:
Expression for the energy stored in a capacitor : Consider an uncharged capacitor of capacitance ‘c’ and its initial will be zero. Now it is connected across a battery for charging then the final potential difference across the capacitor be V and final charge on the capacitor be ‘Q’
∴ Average potential difference V_A = \(\frac{\mathrm{O}+\mathrm{V}}{2}=\frac{\mathrm{V}}{2}\)
Hence workdone to move the charge Q = W = V_A × Q = \(\frac{\mathrm{VQ}}{2}\)
This is stored as electrostatic potential energy ‘U’
∴ U = \(\frac{\mathrm{QV}}{2}\)
We know Q = CV then ‘U’ can be written as given below.
U = \(\frac{\mathrm{QV}}{2}=\frac{1}{2}\) CV² = \(\frac{Q^2}{2 C}\)
∴ Energy stored in a capacitor
U = \(\frac{\mathrm{QV}}{2}=\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

Qeustion 10.
What is the energy stored when the space between the plates is filled with dielectric.
a) With charging battery disconnected ?
b) With charging battery connected in the circuit ?
Answer:
Effect of Dielectric on energy stored :
Case (a) : When the charging battery is disconnected from the circuit:
Let the capacitor is charged by a battery and disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant ‘K’ then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge remains constant.
Capacity increases by’K’ times.
New capacity C’ = \(\frac{Q}{V}\) = \(\frac{\frac{Q}{V}}{K}\) = K\(\frac{Q}{V}\) = KC [V’ = \(\frac{V}{K}\); C = \(\frac{Q}{V}\)]
∴ C’ = KC
Energy stored U’ = \(\frac{1}{2}\) QV = \(\frac{1}{2}\) Q \(\frac{V}{K}\) = \(\frac{U}{K}\)
U’ = \(\frac{U}{K}\)
∴ Energy stored decreases by \(\frac{1}{\mathrm{~K}}\) times.

Case (b) : When the charging battery is connected in the circuit:
Let the charging battery is continue the supply of charge. When the dielectric is introduced then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge on the plates increases until the potential difference attains the original value = V
New charge on the plates Q’= KQ
Hence new capacity C’ = \(\frac{Q^{\prime}}{V}=\frac{K Q}{V}\) = KC
Energy stored in the capacitor U’ = \(\frac{1}{2}\) QV; = \(\frac{1}{2}\) (KQ) V = KU
U = KU
∴ Energy stored in the capacitor increases by ‘K times.

Problems

Question 1.
(a) Calculate the potential at a point P due to a charge of 4 × 10^-7 C located 9 cm away,
(b) Hence obtain the work done in bringing a charge of 2 × 10^-9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought ?
Solution:
(a) V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
= 9 × 10⁹ Nm² C^-2 \(\frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 × 10⁴ V

(b) W = qV = 2 × 10^-9 C × 4 × 10⁴V
= 8 × 10^-5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements. One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Qeustion 2.
Two charges 3 × 10^-8 C and -2 × 10^-8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Thke the potential at infinity to be zero.
Solution:
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be x-axis; the negative charge is taken to be on the right side of the origin fig.

Let P be the required point on the x-axis where the potential is zero. If x is the x – coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_0}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}}\right]\) = 0
where x is in cm. That is \(\frac{3}{x}-\frac{2}{15-x}\) = 0
which gives x = 9 cm
If x lies on the extended line OA, the required condition is \(\frac{3}{x}-\frac{2}{x-15}\) = 0
Which gives x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Question 3.
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when die slab is inserted between the plates ?
Solution:
Let E₀ = V_g/d be the electric field between the plates when there is no dielectric and the potential difference is V₀. If the dielectric is now inserted, the electric field in the dielectric will be E = E₀/K. The potential difference will then be
V = \(E_0\left(\frac{1}{4} d\right)+\frac{E_0}{K}\left(\frac{3}{4} d\right)\)
= \(\mathrm{E}_0 \mathrm{~d}\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)=\mathrm{V}_0 \frac{\mathrm{K}+3}{4 \mathrm{~K}}\)
The potential difference decreases by the factor (K + 3)/K while the free charge Q₀ on the plates remains unchanged. The capacitance thus increases,
C = \(\frac{\mathrm{Q}_0}{\mathrm{~V}}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \frac{\mathrm{Q}_0}{\mathrm{~V}_0}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \mathrm{C}_0\)

Question 4.
Four charges are arranged at the corners of a square ABCD of side d. as shown in fig. Find the work required to put together this arrangement, (b) A charge q₀ is brought to the centre of the square, the four charges being held fixed at its comers. How much extra work is needed to do this ?

Solution:
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A and then the charges -q, +q, and -q are brought to B, C and D, respectively. The total work needed can be calculated in steps.

1. Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.
2. Work needed to bring -q to B when + q is at A. This is given by (charge at B) × (electrostatic potential at B due to change +q at A) = -q × \(\left(\frac{q}{4 \pi \varepsilon_0 d}\right)\)
3. Work needed to bring charge +q to C when +q is at A and -q is at B. This is given by – (charge at Q × (potential at C due to charges at A and B) .
   
4. Work needed to bring -q to D when +q at A, -q at B, and +q at C.
   This is given by (charge at D) × (potential at D due to charges at A, B and C)
   
   Add the work done is steps (i), (ii), (iii) and (iv). The total work required is
   
   The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
   (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)

b) The extra work necessary to bring a charge q₀ to the point E when the four charges are at A, B, C and D is q₀ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D.Hence no work is required to bring any charge to point E.

Question 5.
a) Determine the electrostatic potential energy of a system consisting of two charges 7μC and -2μC (and with no external field) placed at (-9 cm, 0,0) and (9cm, 0, 0) respectively.
b) How much work is required to separate the two charges infinitely away from each ‘ other?
Solution:
(a) U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
= 9 × 10⁹ × \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\) = -0.7 J.

(b)W = U₂ – U₁ = 0 – U = 0 – (-0.7)
= 0.7J.

Question 6.
There is a uniform electric field in the XOY plane represented by (\(40 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}\)) Vm^-1. If the electric potential at the origin is 200 V, the electric potential at the point with coordinates (2m, 1m) is
Answer:
Given, uniform Electric field intensity,
\(\overrightarrow{\mathrm{E}}\) = (\(40 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}\)) Vm^-1
Electric potential at the origin = 200V
Position vector \(\mathrm{d} \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}+1 \hat{\mathrm{j}})\) m
We know that,

dV = \(-\vec{E} \cdot d \vec{r}=-(40 \hat{i}+30 \hat{j}) \cdot(2 \hat{i}+\hat{j})\)
V_p – V₀ = -(80 + 30) = -110Volt. ’
V_p = V₀ – 110 = (200 – 110) Volt = 90 Volt
∴ Potential at point P, V_p = 90Volt.

Question 7.
An equilateral triangle has a side length L. A charge +q is kept at the centroid of the triangle. P is a point on the perimeter of the triangle. The ratio of the minimum and maximum possible electric potentials for the point P is
Answer:
Charge at the centroid of an equilateral triangle = +q
The charge + q divides the line segment in ratio 2 : 1.
That means r_max = 2 and r_min = 1

Question 8.
A condenser of certain capacity is charged to a potential V and stores some energy. A second condenser of twice the capacity is to store half the energy of the first, find to what potential one must be charged ?
Solution:
(For first capacitor, C₁ = C; V₁ = V
And U₁ = \(\frac{1}{2}\) C₁V₁² = \(\frac{1}{2}\) CV² …………………. (1)
For second capacitor, C₂ = 2C₁ = 2C;
U₂ = \(\frac{\mathrm{U}_1}{2}=\frac{1}{4}\)CV²; Let potential difference across the capacitor = V₂

Then, U₂ = \(\frac{1}{2}\) C₂V₂²
⇒ \(\frac{1}{4}\) CV² = \(\frac{1}{2}\) × 2C × V₂²
⇒ V₂² = \(\frac{\mathrm{V}^2}{4}\)
∴ V₂ = \(\frac{V}{2}\)

Question 9.
Three Capacitors each of capaitance 9 pF are connected in series.
a) What is the total capacitance of the combination ?
b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Solution:
a) Resultant capacitance in series combination \(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \Rightarrow \frac{1}{C_S}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}\)
C_S = 3pF

b) Rd across each capacitor = \(\frac{\mathrm{V}}{3}=\frac{120}{3}\) = 40V

Question 10.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. [A.P. Mar. 17]
a) What is the total capacitance of the combination ?
b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Solution:
a) C_p = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF
b) For each capacitor, V is same = 100 Volt
q₁ = C₁V = 2 × 100 = 200pC
q₂ = C₂V = 3 × 100 = 300pC
q₂ = C₃V = 4 × 100 = 400pC

Textual Examples

Question 1.
(a) Calculate the potential at a point P due to a charge of 4 × 10^-7C located 9 cm away.
(b) Hence obtain the work done in bringing a charge of 2 × 10^-9C from infinity to the point P. Does the answer depend on the path along which the charge is brought ?
Solution:
(a) V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
= 9 × 10⁹ Nm² C^-2 \(\frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 × 10⁴ V

(b) W = qV = 2 × 10^-9 C × 4 × 10⁴V
= 8 × 10^-5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements. One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Question 2.
Two charges 3 × 10^-8 C and -2 × 10^-8C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.
Solution:
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be x-axis; the negative charge is taken to be on the right side of the origin fig.

Let P be the required point on the x-axis where the potential is zero. If x is the x – coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_0}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-\mathrm{x}) \times 10^{-2}}\right]\) = 0
where x is in cm. That is \(\frac{3}{x}-\frac{2}{15-x}\) = 0
which gives x = 9 cm.
If x lies on the extended line OA, the required condition is \(\frac{3}{x}-\frac{2}{x-15}\) = 0
Which gives x. = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the claculation required choosing potential to be zero at infinity.

Question 3.
Figure (a) and (b) shows the field lines of a positive and negative point charge respectively.

(a) Give the signs of the potential difference V_p – V_Q; V_B – V_A.
(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.
(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.
(d) Give the sign of the work done by the external agency in moving a small negative charge from B And A.
(e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A ?
Solution:
(a) As V ∝ \(\frac{1}{\mathrm{r}}\), V_p > V_Q. Thus, (V_p – V_Q) is positive. Also VB is less negative than V_A. Thus, V_B > V_A or (V_B – V_A) is positive.

(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly. (PE.)_A > (P.E.)_B and hence sign of potential energy differences is positive.

(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the eletric field. Therefore, work done by the field is negative.

(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.

(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going form B to A.

Question 4.
Four charges are arranged at the comets of a square ABCD of side d. as shown in fig. 5.15.(a) Find the work required to put together this arrangement, (b) A charge q0 is brought to the centre of the square, the four charges being held fixed at its comers. How much extra work is needed to do this ?

Solution:
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges -q, +q, and -q are brought to B, C and D, respectively The total work needed can be calculated in steps.

1. Work needed to bring charge +q to A when no charge is present elesewhere: this is zero.
2. Work needed to bring -q to B when +q is at A. This is given by (charge at B) x (electrostatic .potential at B due to charge +q at A) = -q × \(\left(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{~d}}\right)\)
3. Work needed to bring charge +q to C when +q is at A and -q is at B. This is given by – (charge at Q × (potential at C due to charges at A and B) .
   
4. Work needed to bring -q to D when +q at A, -q at B, and +q at C.
   This is given by (charge at D) × (potential at D due to charges at A, B and C)
   
   Add the work done is steps (i), (ii), (iii) and (iv). The total work required is
   
   The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
   (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)

b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A, B, C and D is q₀ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D.Hence no work is required to bring any charge to point E.

Question 5.
a) Determine the electrostatic potential energy of a system consisting of two charges 7µC and -2µC (and with no external field) placed at (-9 cm, 0,0) and (9cm, 0, 0) respectively.
b) How much work is required to separate the two charges infinitely away from each other ?
c) Suppose that the same system of charges is now placed in an external electric field E = A(1/r²); A = 9 × 10⁵ C m^-2. What would the electrostatic energy of the configuration be ?
Solution:
(a) U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
= 9 × 10⁹ × \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\) = -0.7 J.

(b)W = U₂ – U₁ = 0 – U = 0 – (-0.7)
= 0.7J.

c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,
q₁V(r₁) + q₂V(r₂) = A \(\frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}\) + A \(\frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
and the net electrostatic energy is
q₁V(r₁) + q₂V(r₂) + \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r_{12}}\)
= A \(\frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}\) + A \(\frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
= 70 – 20 – 0.7 = 49.3J

Question 6.
A molecule of a substance has a perma-nent electric dipole moment of magnitude 10^-29C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 10⁶ V m^-1. The direction of the field is suddenly changed by an angle of 60°. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.
Solution:
Here, dipole moment of each molecules = 10^-29C m
As 1 mole of the substance contains 6 × 10²³ molecules,
total dipole moment of all the molecules, p = 6 × 10²³ × 10^-29 Cm
= 6 × 10^-6 C m
Initial potential energy, U_t = -pE cos θ
=-6 × 10^-6 × 10⁶ cos 0° = -6 J
Final potential energy (when θ = 60°),
U_f = -6 × 10^-6 × 10⁶ × cos 60° = – 3J
Change in potential energy
= -3 J – (-6j) = 3 J
So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.

Question 7.
(a) A comb run through one’s dry hair attracts small bits of paper. Why ? What happens if the hair is wet or if it is a rainy day ? (Remember, a paper does not conduct electricity.)
(b) Ordinary rubber is an insulator. But special rubber types of aircraft are made slightly conducting. Why is this necessary ?
(c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?
(d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why ?
Solution:
(a) This is because the comb gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction, if the hair is wet, or if it is a rainy day. friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper.

(b) To enable them to conduct charge (produced by friction) to the ground; as too much of static electricity accumulated may result in spark and result in fire.

(c) Reason similar to (b).

(d) Current passes only when there is difference in potential.

Question 8.
A slab of material of dielectric constant K has the same area as the plates of a parallelplate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates ?
Solution:
Let E₀ = V_g/d be the electric field between the plates when there is no dielectric and the potential difference is V₀. If the dielectric is now inserted, the electric field in the dielectric will be E = E₀/K. The potential difference will then be
V = \(E_0\left(\frac{1}{4} d\right)+\frac{E_0}{K}\left(\frac{3}{4} d\right)\)
= \(\mathrm{E}_0 \mathrm{~d}\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)=\mathrm{V}_0 \frac{\mathrm{K}+3}{4 \mathrm{~K}}\)
The potential difference decreases by the factor (K + 3)/K while the free charge Q₀ on the plates remains unchanged. The capacitance thus increases,
C = \(\frac{\mathrm{Q}_0}{\mathrm{~V}}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \frac{\mathrm{Q}_0}{\mathrm{~V}_0}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \mathrm{C}_0\)

Question 9.
A network of four 10μF capacitors is connected to a 500V supply, as shown in Fig. Determine
(a) the equivalent capacitance of the network and
(b) the charge on each capacitor, (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)

Solution:
(a) In the given network, C₁, C₂ and C₃ are connected in series. The effective capacitance C’ of these three capacitors isgivgftby
\(\frac{1}{\mathrm{C}^{\prime}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\)
For C₁ = C₂ = C₃ = 10 μF. C = (10/3) μF. The network has C and C₄ connected in parallel. Thus, the equivalent capacitance C of the network is
C = C’ + C₄ = (\(\frac{10}{3}\) + 10) μF = 13.3 μF

(b) Clearly, from the figure, the charge on each of the capacitors, C₁, C₂ and C₃ is the same, say Q. Let the charge on C₄ be Q. Now, since the potential difference c across AB is Q/C₁ across BC is Q/C₂. across CD is Q/C₃, we have
\(\frac{\mathrm{Q}}{\mathrm{C}_1}+\frac{\mathrm{Q}}{\mathrm{C}_2}+\frac{\mathrm{Q}}{\mathrm{C}_3}\) = 500 V
Also Q’/C₄ = 500V.
This gives for the given value of the capacitances,
Q = 500 V × \(\frac{10}{3}\) μF = 1.7 × 10^-3C and
Q’ = 500 V × 10μF = 5.0 × 10^-3C

Question 10.
(a) A 900pF capacitor is charged by 100 V battery [Fig.a]. How much electrostatic energy is stored by the capacitor ?
(b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic y energy stored by the system ?

Solution:
The charge on the capacitor is
Q = CV = 900 × 10^-12F × 100 V
= 9 × 10^-8C
The energy stored by the capacitor is
= (1/2) CV² = (1/2) QV
= (1/2) × 9 × 10^-8C × 100 V
= 4.5 × 10^-6 J

(b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V. The charge on each capacitor is then Q’ = CV’. By charge conservation, Q’ = Q/2. This implies V’ = V/2. The total energy of the system is
= 2 × \(\frac{1}{2}\)Q’V’ = \(\frac{1}{4}\)QV = 2.25 × 10⁶J
Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone ?
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.

Students get through AP Inter 2nd Year Physics Important Questions 1st Lesson Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 1st Lesson Waves

Very Short Answer Questions

Question 1.
Write the formula for the speed of sound in solids and gases.
Answer:
Speed of sound in solids,
V_s = \(\sqrt{Y / \rho}\) [y = Young’s modulus of solid, ρ = density of solid]
Speed of sound in gases,
V_s = \(\sqrt{\gamma P / \rho}\) [γP = Adiabatic Bulk modulus of gas, ρ = density of gas] .

Question 2.
What does a wave represent?
Answer:
A wave represents the transport of energy through a medium from one point to another without translation of the medium.

Question 3.
Distinguish between transverse and longitudinal waves.
Answer:
Transverse waves

1. The particles of the medium vibrate perpendicular to the direction of wave propagation.
2. Crests and troughs are formed alternatively.

Longitudinal waves

1. The particles of the medium vibrate parallel to the direction of wave propagation.
2. Compressions and rare factions are formed alternatively.

Question 4.
What are the parameters used to describe a progressive harmonic wave ?
Answer:
Progressive wave equation is given y = a sin (ωt – kx)
Where ω = 2πv = \(\frac{2 \pi}{T}\); k = \(\frac{2 \pi}{\lambda}\)
Parameters:

1. a = Amplitude
2. λ = Wavelength
3. T = Time period
4. v = Frequency
5. k = Propagation constant
6. ω = Angular frequency.

Question 5.
What is the principle of superposition of waves ? .
Answer:
When two or more waves are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.
If y₁, y₂, ……………… y_n be the individual displacements of the particles,then resultant displacement
y = y₁ + y₂ + ……………… + y_n.

Question 6.
Under what conditions will a wave be reflected ?
Answer:

1. When the medium ends abruptly at any point.
2. If the density and rigidity modulus of the medium changes at any point.

Question 7.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary ?
Answer:
Phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary is radian or 180°.

Question 8.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transeverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 9.
What do you understand by the terms node’ and ‘antinode’ ?
Answer:
Node : The points at which the amplitude is zero, are called nodes.
Antinodes: The points at which the amplitude is maximum, are called antinodes.

Question 10.
What is the distance between a node and an antinode in a stationary wave ?
Answer:
The distance between node and antinode is \(\frac{\lambda}{4}\).

Question 11.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?
Answer:
When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 12.
What are harmonics ?
Answer:
The frequencies in which the standing waves can be formed are called harmonics..
(Or)
The integral multiple of fundamental frequencies are called harmonics.

Question 13.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?
Answer:
The possible frequencies of vibrations in a stretched string between two rigid supports is given by
v_n = (n + \(\frac{1}{2}\)) \(\frac{v}{2l}\) where n = 0, 1, 2, 3, ……….

Question 14.
If the air column in a long tube, closed at one end, is set in vibration, what harmonics are possible in the vibrating air column ?
Answer:
The possible harmonics in the vibrating air column of a long closed tube is given by
v_n = [2n +1]\(\frac{v}{4 l}\) where n = 0, 1, 2, 3,

Question 15.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible ?
Answer:
The possible harmonics in vibrating air column of a long open tube is given by no
v_n = \(\frac{\mathrm{nv}}{21}\)
where n = 1, 2, 3, ……………….

Question 16.
What are ‘beats’ ?
Answer:
Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Short Answer Questions

Question 1.
What are transverse waves ? Give illustrative examples of such waves.
Answer:
Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.
Illustration:

1. Waves produced in the stretched strings are transverse.
2. When a stretched string is plucked, the waves travel along the string.
3. But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
4. They can propagate only in solids and on the surface of the liquids.
5. Ex : Light waves, surface water waves.

Question 2.
What are longitudinal waves ? Give illustrative example of such waves.
Answer:
Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.
Illustration:

1. Longitudinal waves may be easily illustrated by releasing a compressed spring.
2. A series of compressions and rarefactions (expansions) propagate along the spring.
   
   C = Compression; R = Rarefaction.
3. They can travel in solids, liquids and gases.
4. Ex : Sound waves.

Question 3.
What are ‘beats’ ? When do they occur ? Explain their use, if any.
Answer:
Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.
It two vibrating bodies have slightly difference in frequencies, beats can occur.
No. of beats can be heard, ∆υ = υ₁ ~ υ₂
Importance:
1. It can be used to tune musical instruments.
2. Beats are used to detect dangerous gases.
Explanation for tuning musical instruments with beats:
Musicians use. the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 4.
What is ‘Doppler effect’ ? Give illustrative examples.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between .the observer and the source of sound is called doppler effect.
Examples:

1. The frequency of whistling, engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
2. Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Sample Problem on Doppler effect:
Two trucks heading in opposite direction with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330m/s). After the two trucks have passed each other, what frequency does the driver of the second truckhear?
Answer:
Speed of first truck = 60 kmph
= 60 × \(\frac{5}{18}\) = 16.66 m/s;
Speed of second truck = 70 kmph 5
= 70 × \(\frac{5}{18}\) = 19.44 m/s
Frequency of horn of first truck = 400 Hz;
Velocity of sound, (V) =330 m/s
Frequency of sound heard by the driver of the second truck when approaching each other,
v¹ = \(\left(\frac{V+V_0}{V-V_s}\right) v=\left(\frac{330+19.44}{330-16.66}\right)\) × 400 = 446 Hz
Frequency of sound heard by the driver of the second truck when approaching each other,
V¹¹ = \(\left(\frac{\mathrm{V}-\mathrm{V}_0}{\mathrm{~V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{v}=\left(\frac{330-19.44}{330+16.66}\right)\) × 400 = 358.5 Hz

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. [IPE]
Answer:
A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves. Let two transverse progressive waves of same amplitude a, wave length λ and frequency v, travelling in opposite direction be given by

y₁ = a sin (kx – ωt) and y₂ = a sin (kx + ωt)
where ω = 2πv and k = \(\frac{2 \pi}{\lambda}\)
The resultant wave is given by y = y₁ + y₂
y = a sin (kx – ωt) + a sin (kx + ωt)
y = (2a sin kx) cos ωt
2a sin kx = Amplitude of resultant wave.
It depends on ‘kx’. If x = 0, \(\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}\) ……………… etc, the amplitude = zero
These positions are known as “Nodes”.
If x = \(\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}\) …………… etc., the amplitude = maximum (2a)
These positions are called “Antinodes”.
If the string vibrates in ‘P’ segments and T is its length, then length of each segment = \(\frac{l}{\mathrm{P}}\)
Which is equal to \(\frac{\lambda}{2}\)
∴ \(\frac{l}{\mathrm{P}}=\frac{\lambda}{2} \Rightarrow \lambda=\frac{2 l}{\mathrm{P}}\)
Harmonic frequency v = \(\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}\)
v = \(\frac{v \mathrm{P}}{2 l}\) ………………. (1)
If’ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is
v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) …………… (2)
From the Eqs (1) and (2) :
Harmonic frequency v = \(\frac{\mathrm{P}}{2 l} \sqrt{\frac{\Gamma}{\mu}}\)
If P = 1 then it is called fundamental frequency (or) first harmonic frequency
∴ Fundamental Frequency v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) …………….. (3)

Laws of Transverse Waves Along Stretched String:
Fundamental frequency of the vibrating string v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)
First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.
∴ v ∝ \(\frac{1}{l}\) ⇒ vl = constant, when ‘T’ and ‘μ’ are constant. .

Second Law: When the length (l) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).
∴ v ∝ \(\sqrt{T}\) ⇒ \(\frac{v}{\sqrt{T}}\) = constant, when ‘l’ and ‘m’ are constant.
JT .
Third Law: When the length (l) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).
∴ v ∝ \(\frac{1}{\sqrt{\mu}}\) ⇒ \(v \sqrt{\mu}\) = constant, when ‘l’ and T are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. [A.P. 17; IPE 2015, 2016 (TS)]
Answer:
A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super-imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two anti nodes at two ends of the pipe with a node between them.

∴ The vibrating length (l) = half of the wavelength \(\left(\frac{\lambda_1}{2}\right)\)
l = \(\frac{\lambda_1}{2}\) ⇒ λ₁ = 2l
fundamental frequency v₁ = \(\frac{\mathrm{v}}{\lambda_1}\) where v is velocity,of sound in air, v₁ = \(\frac{v}{21}\) = v
For second harmonic (first overtone) will have one more node and antinode than the fundamental.
If λ₂ is wavelength of second harmonic l = \(\frac{2 \lambda_2}{2}\) ⇒ λ₂ = \(\frac{21}{2}\)
If ‘v₂’ is frequency of second harmonic then v₂ = \(\frac{v}{\lambda_2}=\frac{v \times 2}{2 l}\) = 2v
v₂ = 2v ……………… (2)
Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.
If λ₃ is wave length of third harmonic l = \(\frac{3 \lambda_3}{2}\)
λ₃ = \(\frac{2l}{3}\)
If ‘v₂‘ is frequency of third harmonic then
v₃ = \(\frac{v}{\lambda_3}=\frac{v \times 3}{2 l}\) = 3V
v₃ = 3v …………… (3)
Similarly we can find the remaining or higher harmonic frequencies i.e., v3, v4 etc., can be determined in the same way.
Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.
v : V₁ : v₂ = 1 : 2 : 3

Question 3.
How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. [IPE 2015, 2016(A.P.), (T.S) A.P. & T.S. Mar. 15]
Answer:
A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.

∴ l = \(\frac{\lambda_1}{4}\) ⇒ λ₁ = 4l
If ‘v₁‘ is fundamental frequency then
v₁ = \(\frac{v}{\lambda_1}\) where ‘υ’ is velocity of sound in air
v₁ = \(\frac{v}{4 l}\) = v …………….. (1)
To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.
Then length of the pipe is equal to \(\frac{3}{4}\) of the wavelength.
∴ l = \(\frac{3 \lambda_3}{4}\) where ‘λ₃‘ is wave length of third harmonic
λ₃ = \(\frac{4l}{3}\)
where ‘X3’ is wave length of third harmonic.
If ‘v₃‘ is third harmonic frequency (first overtone)
∴ v₃ = \(\frac{v}{\lambda_3}=\frac{3 v}{41}\)
v₃ = 3v …………….. (2)
Similarly the next overtone in the close pipe is only fifth harmonic, it will have three nodes and 3 antinodes between the closed end and open end.
Then length of the pipe is equal to \(\frac{5}{4}\) of wave length (λ₅)
∴ l = \(\frac{5 \lambda_5}{4}\) where ‘λ₅‘ is wave length of fifth harmonic.
λ₅ = \(\frac{4l}{5}\)
If ‘v₅‘ is frequency of fifth harmonic (second overtone)
v₅ = \(\frac{v}{\lambda_5}=\frac{5 v}{4 l}\)
v₅ = 5v …………….. (3)
∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.
Therefore the ratio of the frequencies of harmonics in closed pipe can be written as
v₁ : v₃ : v₅ = v : 3v : 5v
v₁ : v₃ : v₅ = 1 : 3 : 5

Question 4.
What is Doppler effect ? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest. [Mar. 17, BMP, 2016 (AP) Mar. 14, (TS)]
Answer:
Doppier effect: The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppier effect. When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:
Let S = Source of sound
O = Listener
Let ‘S’ be the source, moving with a velocity ‘υ_s‘ towards the stationary listener.
The distance travelled by the source in time period T = υ_s. T Therefore the successive compressions and rarefactions are drawn closer to listener.
∴ Apparent wavelength λ’ = λ – υ_sT.
λ’ = λ – \(\frac{v_s}{v}\) [∵ υ = \(\frac{1}{T}\)]
= \(\frac{\lambda v-v_s}{v}=\frac{v-v_s}{v}\) [∵ υ = vλ]
If “v'” is apparent frequency heard by the listener then v’ = \(\frac{v}{\lambda^{\prime}}\) where ‘υ’ is Velocity of sound in air
v’ = \(\frac{v . v}{v-v_S}\)
Therefore the apparent frequency is greater than the actual frequency.
Similarly, if the source is away from the stationary listener then apparent frequency v’ = \(\frac{v . v}{v+v_s}\), which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity.

Problems

Question 1.
A stretched wire of length 0.6 m is observed to vibrate with a frequency of 30 Hz in the fundamental mode. If the string has a linear mass of 0.05 kg / m find (a) the velocity of propagation of transverse waves in the string (b) the tension in the string. [IPE 2016 (T.S)
Answer:
v = 30 Hz; I = 0.6 m; μ = 0.05 kg m^-1
υ = ?; T = ?
a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s
b) T = υ²μ = 36 × 36 × 0.05 = 64.8 N .

Question 2.
A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?
Solution:
I = 0.4 m; M = 0.16g = 0.16 × 10^-3 kg;
μ = \(\frac{\mathrm{M}}{1}=\frac{0.16 \times 10^{-3}}{0.4}\) = 0.4 × 10^-3 kg/m;
T = 70 N; v_n = \(\frac{\mathrm{P}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}\)
v₁ = \(\frac{1}{21} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 \times 0.4} \sqrt{\frac{70}{0.4 \times 10^{-3}}}\) = 523 Hz
v₂ = 2v₁ = 2 × 523 = 1046 Hz
v₃ = 3v₁ = 3 × 523 = 1569 Hz

Question 3.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?
Solution:
l = 70 cm = 70 × 10^-2m; v = 331 m/s ;
v = ?
v = \(\frac{v}{4 l}=\frac{331}{4 \times 70 \times 10^{-2}}\) = 118.2 Hz.

Question 4.
A steel cable of diameter 3 cm is kept under a tension of lOkN. The density of steel is 7.8 g/ cm3. With what speed would transverse waves propagate along the cable ?
Solution:
T = 10 kN = 10⁴ N
D = 3 cm; r = \(\frac{D}{2}=\frac{3}{2}\) cm
= \(\frac{3}{2}\) × 10^-2m;

Question 5.
A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle. [T.S. Mar. 17]
Solution:
When a whistling train away from rest observer,
v’ = \(\left[\frac{v}{v-v_s}\right] v\) ……………… (1)
When a whistling train away from rest observer,
v” = \(\left[\frac{v}{v+v_s}\right] v\) ……………… (2)
Here v’ = 219 Hz; v” = 184 Hz;
v = 340 m/s

Question 6.
A rocket is moving at a speed of 200 m s-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 m s^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+v_{\mathrm{S}}}{v}\right]^{-1}\) and not the approximate
v = v₀ [1 – \(\frac{v_s}{v}\)]
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ [1 – \(\frac{v_s}{v}\)]^-1
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not v₀, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz

Textual Examples

Question 1.
Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both.
a) Motion of kink in a longitudinal spring produced by displacing one end of the spring sideways.
b) Waves produced in a cylinder containing a liquid by moving its piston back and forth.
c) Waves produced by a motorboat sailing in water.
d) Ultrasonic waves in air produced by a vibrating quartz crystal.
Solution:
a) Transverse and longitudinal
b) Longitudinal
c) Transverse and longitudinal
d) Longitudinal.

Question 2.
A wave travelling along a string is des-cribed by, y(x, t) = 0.005 sin (80.0 x – 3.01), in which the numerical constants are in SI untis (0.005 m, 80.0 rad m^-1, and 3.0 rad s^-1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ?
Solution:
On comparing this displacement equation with Eq. y (x, t)n = a sin(kx – ωt + Φ)
y(x, t) = a sin (kx – ωt).
We find
a) the amplitude of the wave is
0. 005 m = 5 mm.

b) the angular wave number k and angular frequency ω are k = 80.0 m^-1 and ω = 3.0 s^-1
We then relate the wavelength λ to k through Eq.
λ = \(\frac{2 \pi}{K}\)
= \(\frac{2 \pi}{80.0 \mathrm{~m}^{-1}}\) = 7.85 cm

c) Now we relate T to ω by the relation
T = \(\frac{2 \pi}{\omega}\)
= \(\frac{2 \pi}{3.0 \mathrm{~s}^{-1}}\)
= 2.09 s
and frequency, v = \(\frac{1}{T}\) = 0.48 Hz
The displacement y at x = 30.0 cm and time t= 20s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)
= (0.005 m) sin (-36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (97°) ≃ 5 mm

Question 3.
A steel wire 0.72 m long has a mass of 5.0 × 10^-3 kg. If the .wire is under a tension of 60 N, what is the speed of transverse waves on the wire ? [A.P. Mar. 19]
Solution:
Mass per unit length of the wire,
μ = \(\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}}\) = 6.9 × 10^-3 kg m^-1
Tension, T = 60 N
The speed of wave on the wire is given by
υ = \(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}\) = 93 m s^-1

Question 4.
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 × 10^-3 kg.
Solution:
We know that 1 mole of any gas occupies 22.4 litres at STP Therefore, density of air at STP is:
ρ₀ = (mass of one mole of air) / (Volume of one mole of air at STP)
= \(\frac{29.0 \times 10^{-3} \mathrm{~kg}}{22.4 \times 10^{-3} \mathrm{~m}^3}\) = 1.29 kgm^-3
According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
υ = \(\left[\frac{1.01 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}}{1.29 \mathrm{~kg} \mathrm{~m}^{-3}}\right]^{1 / 2}\) = 280 m s^-1

Question 5.
A pipe, 30.0 cm long is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source ? Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as 330 m s^-1.
Solution:
The first harmonic frequency is given by
v¹ = \(\frac{v}{\lambda_1}=\frac{v}{2 L}\) (open pipe)
Where L is the length of the pipe. The frequency of its n^th; harmonic is
v_n = \(\frac{n v}{2 L}\) for n = 1, 2, 3, ……………… (open pipe)
First few modes of an open pipe are shown in Fig.
For L = 30.0 cm. υ = 330 m s^-1
v_n = \(\frac{\mathrm{n} \times 330\left(\mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6(\mathrm{~m})}\) = 550 s^-1

Clearly, for a source of frequency 1.1 kHz the air column will resonate at υ₂, i.e. the second harmonic.
Now if one end of the pipe is closed (Fig.), the fundamental frequency is

and only to odd numbered harmonics are present :
v₃ = \(\frac{3 v}{4 L}/latex], v₅ = [latex]\frac{5 v}{4 L}\) and s0 0n.
For L = 30 cm and υ = 330 m s^-1;, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

Question 6.
Two sitar strings A and B playing the note ‘Dha’ are slightly put of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz ?
Solution:
Increase in the tension of a string increases its frequency. It the original frequency of B (v_B) were greater than that of A(v_A) .further increase in v_B should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that v_B < v_A. Since v_A – v_B = 5 Hz, and v_a = 427 Hz, we get v_B = 422 Hz.

Question 7.
A rocket is moving at a speed of 200 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 ms^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+\mathrm{v}_{\mathrm{S}}}{\mathrm{v}}\right]^{-1}\) and not the approximate
v = v₀ \(\left[1-\frac{v_s}{v}\right]\)
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ \(\left(1-\frac{v_{\mathrm{S}}}{v}\right)^{-1}\)
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v₀, the frequency intercepted by the target and not v0, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination

Very Short Answer Questions

Question 1.
Name the blood vessels that enter and exit the kidney.
Answer:
Renal artery enters kidney and renal vein comes out of the kidney.

Question 2.
What are renal pyramids and renal papillae?
Answer:
The conical shaped medullary regions of kidney are the renal pyramids. Tips of the renal pyramids which open into pelvis are renal papillae.

Question 3.
What are the columns of Bertin?
Answer:
Columns of Bertin are the medullary extensions of the renal cortex in between the renal pyramids.

Question 4.
Name the structural and functional unit of kidney. What are the two main types of structural units in it?
Answer:
The structural and functional unit of kidney is ‘Nephrons’. The two main parts are
1) Malpighian body (renal corpuscle),
2) Convoluted tube.

Question 5.
Distinguish between cortical and juxta medullary nephrons.
Answer:

1. Cortical nephrons have renal corpuscle in the superficial renal cortex. They have short loop of Henle but without vasa recta.
2. Juxta medullary nephrons re located near the renal medulla. They have loop of Henle and vasa recta.

Question 6.
Define glomerular Alteration.
Answer:
The process of Alteration of blood, which occur between glomerulus and lumen of the Bowman’s capsule due to difference in net pressure is called glomerular Alteration. The filtered fluid which entered the Bowman’s capsule is primary urine or glomerular filtrate which is hypotonic.

Question 7.
Define glomerular Alteration rate (GFR)?
Answer:
The amount of filterate formed by both the kidneys, per minute is called glomerular Alteration rate (GFR). GFR in a healthy individual is approximately 125 ml 11 minute.

Question 8.
What is meant by mandatory reabsorption? In which parts of nephron does it occur?
Answer:
In a healthy individual the GFR is 125 ml/1 minute or 180 ltr per day. About 85% of the filterate formed is reabsorbed in a constant, unregulated fashion by the proximal convoluted tubule and descending limb of Henle’s loop, called mandatory reabsorption.

Question 9.
Distinguish between juxtaglomerular cells and macula densa.
Answer:

1. The cells of the distal convoluted tubule are crowded in the region where distal convoluted tubule makes contact with afferent arteriole. These cells are known as Macula densa.
2. Along side of macula densa, the wall of the afferent arteriole contains the modified smooth muscle fibers called juxtaglomerular cells.

Question 10.
Whait is Juxtaglomerular apparatus?
Answer:
Macula densa along with juxtaglomerular cells form juxtaglomerular apparatus which releases an enzyme Called renin.

Question 11.
Distinguish between the enzymes reniri and rennin.
Answer:
Renin :
Renin is an enzyme produced by the JG cells. This enzyme catalyses the conversion of angiotensinogen into angiotensin -I.

Rennin :
Rennin is also an enzyme found in the gastric juice of infants. It acts on the milk protein casein in the presence of calcium ions and convert it into calcium para caseinate and proteoses.

Question 12.
What is meant by the term osmoregulation?
Answer:
The process of maintaining the quantity of water and-dissolved solutes in balance is referred to as osmoregulation.

Question 13.
What is the role of atrial-natriuretic peptide in the regulation of urine formation?
Answer:
A large increase in blood volume promotes the release of atrial natriuretic peptide from the heart. Atrial natriuretic peptide decreases the absorption of water, Na⁺ from proximal convoluted tubule.

Short Answer Questions

Question 1.
Terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic. Why?
Answer:
Ammonia is highly toxic and readily soluble in water, hence it should be eliminated from the body quickly in a very dilute solution.

Aquatic animals are surrounded by water, so water conservation is not a problem for them. In this manner, they are continuously eliminating ammonia.

On the other hand, terrestrial animals have to conserve water. They cannot waste water. So ammonia in diluted form can’t be eliminated continuously. Since ammonia is highly toxic, it has to be converted to less toxic form, like urea or uric acid.

Urea is 1,00,000 times less toxic than ammonia and requires less water for their excretion. Uric acid is less toxic than urea and being insoluble in water can be excreted as semi solid waste or pellets with very little water. This is the great advantage for animals with little access to water.

Question 2.
Differentiate vertebrates on the basis of the nitrogenous waste products they excrete, giving example.
Answer:
Vertebrates are divided into three categories based on nitrogenous waste excretory products. They are :
1) Ammonotelic animals:
The animals which excrete ammonia as nitrogenous waste products are called ammonotelic animals. These are aquatic organisms.
Ex : Some bony fishes.

2) Ureotelic animals:
The animals which excrete urea as their chief nitrogenous waste are called ureotelic animals.
Ex : Earth worms, cartilaginous fishes, most of the amphibians and mammals.

3) Uricotelic animals:
These animals excrete their nitrogenous waste products in the form of uric acid.
Ex : Reptiles, birds.

Question 3.
Draw labelled diagram of the V.S of kidney.
Answer:

Question 4.
Describe the internal structure of kidney of man.
Answer:

1. Kidney is bean shaped structure, the outer surface of kidney is convex and inner surface is concave where it has a deep notch called hilum.
2. A longitudinal sections of the human kidney shows two distinct regions namely the outer cortex and the inner medulla.
3. Medulla is divided into multiple cone shaped masses of tissue called renal pyramids. The renal pyramids are separated by the projections of the cortex called columns of Berlin.
4. The tips of the pyramids are renal papilla.
5. Renal papilla projects into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter. Ureter carries urine into urinary bladder.
6. In cortex and medulla, nearly one million nephrons are present. They are structural and functional units of kidney. They are embedded in the loose connective tissue of cortex and medulla.
7. In addition, kidney contains a network of blood capillaries, lymph sinuses and intestitial fluid in intra cellular spaces.
8. The kidney gets blood supply through renal artery and blood from kidney is carried out by renal vein.

Question 5.
Explain micturition.
Answer:
The process of passing out of urine is called micro nutrition and the neural mechanism involved is called micturition reflex.

Urine is formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This sighal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smodth muscles of the bladder and simultaneous relaxation of the urethral sphincter, causing the release of urine.

Question 6.
What is the significance of juxta glomerular apparatus (JGA) in kidney function?
Answer:
Macula densa together with JG cells form juxtaglomerular apparatus (JAG). JAG plays a complex regulating role. A fall in glomerular blood flow or glomerular blood pressure or GFR can activate JG cells to release an enzyme called renin into the blood. This catalyses the conversion of angiotensinogen into angotensin-I which is further converted into angiotensin-II by angiotensin converting enzyme. Angiotensin-II, being a powerful vasoconstrictor increase the glomerular blood pressure and there by GFR.

Angiotensin-II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na⁺ and water from distal convoluted tubule and collecting duct. To reduce loss through urine, and also promote secretion of K⁺ ions into distal convoluted tubule and collecting duct. It leads to increase in the blood pressure and GFR. This complex mechanism is generally known as renin – angiotensin- aldosterone system (RAAS).

Question 7.
Give a brief account of the counter current mechanism.
Answer:
Mammals have the ability to produce concentrated urine. The Henle’s loop and vasa recta plays an important role in this. The flow of the renal filterate in the two limbs of Henle’s loop is in opposite directions and thus form counter current. The flow of blood through vasa recta is also in counter current pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter currents of renal fluid and blood in them help in maintaining an increasing osmolarity towards the inner medullary interstitium.

This gradient is mainly caused by NaCl and urea. NaCl passes out the ascending limb of Henle’s loop, and it enters the blood of the descending limb of vasa recta. NaCl is returned to the intestitium from the ascending portion of the vasa recta. Similarly small amounts of urea enter the thin segment of ascending limb of Henle’s loop which is transported back to the interstitium, from the collecting duct. Transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta.is called the counter current mechanism.

This mechanism helps to maintain a concentration gradient .in the medullary interstitium. Presehce of such interstitium gradient help easy passage of water from the collecting duct, there by concentrating the urine.

Question 8.
Explain the auto regulatory mechanism of GFR.
Answer:
Auto Regulation of GFR: *The kidneys have built in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus. Juxta glomerular apparatus is a special region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact.

A fall in GFR can activate the juxta glomerular cells to release an enzyme called renin, which catalyses the conversion of angiotensinogen into angiotensin-I and further converted to angiotensin-II by action of an enzyme angiotensin converting enzyme. Angiotensin-II’ stimulate the adrenal cortex to secrete aldosterone. Aldosterone causes reabsorption of Na+ and water from DCT and collecting duct to reduce loss through urine and also promotes the secretion of K+ ions into the DCT and CD (collecting duct). It leads.an increase in the blood pressure and GFR.

Question 9.
Describe the role of liver, lungs and skin in excretion.
Answer:
In addition to the kidneys, liver, lungs and skin also play an important role in the elimination of excretory wastes.

a) Liver :
Liver is the largest gland in our body, secretes bile, containing substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drags. Most of these substances ultimately pass out along with digestive wastes.

b) Lungs :
Lungs remove large amounts of C02 (18 litres 1 day), various, volatile materials and significant quantities of water.

c) Skin :
Human skin possesses two types of glands namely sweat and sebaceous glands for the elimination of certain substances through their secretions.

• Sweat produced by the sweat glands is a watery fluid containing NaCl, small amount of urea, lactic acid etc.,
• Sebaceous glhnds eliminate certain substances like sterols, hydrocarbons and waxes through sebum. This secretion provides a protective oily covering for the skin.

Question 10.
Name the following.
Answer:
a) A chordate animal having protonephridial type excretatory structures Cephalo chordate.
b) Cortical portions projecting between the medullary pyramids in the human kidney. Columns of Bertini
c) Capillary network paralleing the loop of Henle. Vasa recta.
d) A non chordate animals having green glands as excretory structures. Crustaceans.

Long Answer Questions

Question 1.
Describe the excretory system of man, giving the structure of a nephron.
Answer:
In humans, the excretory system consists of a pair of kidney, a pair of ureters, a urinary bladder and urethra.

Kidney :
Kidneys are reddish brown, bean shaped structures, situated on either side of the vertebral column between the levels of last thoracic and third lumbar vertebrae in a retroperitoneal position. The right kidney is slightly lower than the left one due to the presence of liver.

The outer surface of the kidney is convex and the inner surface is concave, where it has a deep notch called hilum, the point at which the renal artery and nerves enter and renal vein and ureter leave. Each kidney is surrounded by a tough, fibrous tissue, called renal capsule.

Ureter :
These are slender whitish tubes, which emerges from the pelvis of the kidney. The ureter rundown and open into the urinary bladder.

Urinary bladder :
Urinary bladder is a pear shaped like muscular organ. It tempirarily stores the urine, situated in the lower abdominal cavity. The neck of the bladder leads into the urethra. Urethra opens near the vaginal orifice in the female and through the penis in the male.

Structure of nephron:
Each kidney has nearly one million nephrons. These are structural and functional units of kidney, embedded in the loose connective tissue of cortex and medulla. Nephron consist of malpighian body and renal tubule.

I) Malphigian body :
It begins in the cortex of the kidney. It contains Bowman’s capsule and glomerulus.

a) Bowman’s capsule :
It is a thin walled, double layered cup. The inner wall of the Bowman’s capsule has certain unique cells called podocytes.

b) Glomerulus :
It is a dense network of capillaries in the cup of Bowman’s capsule. Afferent arteriole of renal artery enter the cavity of Bowman’s capsule and split into five branches.

They unite and come out of the Bowman’s capsule as an afferent arteriole.

The podocytes of inner wall of Bowman’s capsule wrap around each capillary. The podocytes are arranged in an intricate manner so as to leave some minute spaces called filteration slits. The endothelium cells of capillaries have numerous pores called fenestrations.

II) Renal tubule:
It is narrow, delicate tubule arises from the posterior part of Bowman’s capsule known as neck. It opens into along narrow convoluted tubule with three parts like proximal convoluted tubule, Loop of Henle and Distal convoluted tubule.

a) Proximal convoluted tubule :
It is a lined by simple cuboidal epithelium with brush border to increase area of absorption.

b) Loop of Henle :
It is a hairpin like tubule present in medulla region. It consist of a descending limb and an ascending limb. The proximal part of the ascending limb is thin and the distal part is thick. The thick ascending limb continuous into the distal convoluted tubule.

c) Distal convoluted tubule (DCT) :
It is present in cortex. It is lined by simple cuboidal epithelium. The DCT continuous as the initial collecting duct in the cortex.

Collecting system :
Some initial collecting ducts unite to form straight collecting duct, which passes through the medullary pyramid. In the medulla, the tubes of each pyramid join and form duct of Bellini, which finally opens into tip of the renal papilla.

Capillary network of nephron :
The efferent arteriole emerging from the glomerulus forms a fine capillary network called the peritubular capillaries, around the renal tubule. The portion of the peritubular capillaries that surrounds the loop of Henle is called the vasa recta. The vasa recta is absent or highly reduced in the cortical nephrons. The juxta medullary nephrons possess well developed yasa recta.

Question 2.
Explain the physiology of urine formation.
Answer:
The formation of urine involves three main processes namely

1. Glomerular Alteration
2. Selective reabsorption
3. Tubular secretion.

1) Glomerular Alteration :
It is Arst step in urine formation. The process of Alteration of blood, which occurs between glomerulus and lumen of the Bowman’s capsule due to difference in netpressure is called glomerular Alteration.

The hydrostatic pressure of blood while Aowing in the glomerulus is 60 mm Hg. It is opposed by glomerular colloidal osmotic pressure of 32 mm Hg and Bowman’s capsule hydrostatic pressure of 18 mm Hg.

The net filterate pressure is 10 mm Hg ( 60 – 32 + 18 = 10). This causes the Alteration of blood through the 3 layered filterate membrane formed by endothelium cells of glomerular capillary together with the basement membrane and podocytes of the Bowman’s cup. By the result of glomerular Alteration primary urine or renal Auid is collected in lumen of the Bowman’s capsule.

The primary urine contains almost all the constituents of plasma, except the proteins. The primary urine is hypotonic to the cortical Auid, it passes into the next part of renal tubule.

2) Selective reabsorption:
During the process of glomerular Alteration 125 ml/minute df primary urine is formed. Nearly 99% of which and essential substances are reabsorbed* by renal tubules called selective reabsorption. About 85% of filterate formed (primary urine) is reabsorbed in a constant unregulated manner called obligatory reabsorption.

3) Tubular Secretion :
During the formation of urine, the tubular cells secrete substances such as H⁺, K⁺ and NH₃⁺ into the filterate. Tubular secretion is also an important step in the formation of urine as it helps in maintenance of ionic and acid base balance of the body fluids.

Mechanism of selective reabsorption and secretion takes place is different parts of nephrons.

a) In the proximal convoluted tubule :
Nearly all the essential nutrients and 70-80% of electrolytes and water are reabsorbed by this segment. Na⁺, glucose, amino acids, Cl^– and other essential substances are reabsorbed into blood.

PCT also helps to maintain the pH and ionic balance of body fluids by selective secretion of H⁺ and NH₃ into the filterate and by the absorption of HCO₃^– from it.

b) In the Henle’s loop :
Reabsorption in this segment is minmium.

• The descending loop of Henle is permeable to water and almost impermeable to electrolytes results the filterate concentration gradually increases.
• The ascending limb has two specialized regions, a proximal thin segment in which NaCl diffuses out into interstitial fluid passively, and distal thick segment, in which NaCl is actively pumped out.

The ascending limb is impermeable to water. Thus the filterate becomes progressively more dilute as it moves up to the cortex i.e., towards the DCT.

In the Distal convoluted tubule (DCT) :
It is permeable to water and ions. The reabsorption of water is variable depending on several conditions and is regulated by ADH. DCT is also capable of reabsorption of HCO₃^– and selective secretion of H⁺ and K⁺ ions and NH₃⁺ into DCT from peritubular network, to maintain the pH and sodium – potassium balance in tHe blood.

In the collecting duct (CD) :
Considerable’amount of water could be reabsorbed from this region to produce concentrated urine. This segment allows passage of small amount of urea to medullary interstitium to keep up its osmolarity. It also plays a role in the maintenance of pH and ionic balance of blood by the selective secretion of H⁺ and K⁺ ions.

The renal fluid after the process of facultative reabsorption in the CD, influenced by ADH, constitute the urine, that is sent out. Urine in the CD is hypertonic to the plasma of blood.

Andhra Pradesh BIEAP AP Inter 2nd Year Commerce Study Material 5th Lesson Consumer Protections Textbook Questions and Answers.

AP Inter 2nd Year Commerce Study Material 5th Lesson Consumer Protections

Essay Answer Questions

Question 1.
Explain the composition and jurisdiction of the state commission.
Answer:
The state commission settles consumer disputes at the state level. The state commission is headed by the judge of a high court and comprises other members not less than two and not more than as prescribed, one of whom shall be a woman.

The state commission shall have jurisdiction to entertain consumer complaints where the value of goods and services for which the compensation claimed exceeds ₹ 20 lakhs and less than ₹ 1 crore. The state commission is empowered to call for the records and pass appropriate orders in respect of any consumer dispute within the state jurisdiction. The state commission is empowered to transfer any complaint pending before on the district forum to another district forum within the state. The state commission has circuit Benches.

In case the aggrieved party is not satisfied with the order of the state commission, he can appeal to the national commission within 30 days of passing the order.

Question 2.
Describe the rights of a consumer as per CPA 1986.
Answer:
Although a businessman is aware of his social responsibilities even then we come across many cases of consumer protection. Hence Government of India provided the following six rights to all the Consumers under Consumer Protection Act.
1) Right to safety:
According to this right, the consumers have right to be protected against the marketing of goods and services which are hazardous to life and property. The right is important for safe and secure life.

2) Right to information :
According to this right, the consumer has right to get information about the quality, quantity, purity standard and piece of goods or services. The producer must supply all the relevant information at a suitable place.

3) Right to choice :
According to this right, every consumer has a right to choose the goods or services of his or her likings. The supplier should not force the consumer to buy a particular brand only. Consumer should be free to choose the most suitable product from his view point.

4) Right to consumer education:
According to this right, it is the right of the consumer to acquire knowledge and skill to be informed to customer. It is easier for the literate consumers to know their rights and take actions.

5) Right to seek redressal:
According to this right, the consumer has the right to get compensation or seek redressal against unfair trade practices or any other exploitation. The right assures justice to consumer against exploitation.

6) Right to heard / Right to represent:
According to this right, the consumer has the right to represents himself or to be heard or right to advocate his interest. In case a consumer has been exploited or has any complaint against the product or service then he has a right to be heard.

Question 3.
What are the responsibilities of a consumer?
Answer:
Various efforts have been made by government or non – government organisations to protect the interest of consumer, but exploitation of consumer will stop only when the consumer will come forward to safeguard his own interest. Consumer has bear the following responsibilities.

1) Be quality conscious :
To put to stop to adulteration and corrupt practices of the manufacturers and traders, it is the duty of every consumer to be conscious of the quality of the products they buy. They should look for the standard quality certification marks like ISI, Agmark, Wool mark, Ecomark, Hallmark, etc. While making the purchases.

2) Beware of misleading advertisements :
The advertisement often exaggerates the quality of the products. Hence the consumers should not rely on the advertisement and carefully check the product or ask the users before making a purchase.

3) Responsibility to inspect a variety of goods before making selection :
The consumer should inspect a variety of goods before buying the goods and services. For this purpose, he / she should compare their quality, price, durability, after sales service etc.

4) Collect proof of transaction :
The consumer should insist a valid documentary evidence (Cash memo / invoice) relating to purchase of goods or availing of any services and preserve it carefully. Such proof of purchase is required for filing a complaint. In case of durable goods the manufacturers generally provide the warrantee / guarantee card with the product. It is the duty of the consumers to obtain these documents and ensure that these are duly sighed, stamped and dated. The consumer must preserve them till the warrantee / guarantee period is over.

5) Consumers must aware of their rights:
The consumers must aware of their rights as stated above and exercise them while buying goods and services. For example, it is the responsibility of a consumer to insist on getting all information about the quality of the product and ensure himself / herself that it is free from any kind of defect.

6) Complaint for genuine grievances:
As a consumer, if you are dissatisfied with the product, you can ask for redressal of yoifr grievances. In this regard, you must file a proper claim with the company first. The manufacturer / company do not respond, then you can approach the forums. But your claim must state actual loss and the compensation claim must be reasonable. At no cost fictious complaints should be filed otherwise the forum may penalise you.

7) Proper use of product / service :
It is expected from consumers that they use and handle the product / service properly. It has been noticed that during guarantee period, people tend to reckless use of the product, thinking that it will be replaced during guarantee period. This practice should be avoided.

Question 4.
Explain the redressal mechanism available to consumers under the Consumers Proction Act, 1986.
Answer:
The judicial machinery set up under Consumer Protection Act (C.RA) 1986 consists of consumer courts (Forums) at the District, State and National levels. These are known as District Forum, State consumer disputes redressal commission (State Commission) and National consumer disputes redressal commission (National Commission).

1. District forum :
This is established by the state government in each of its districts.
a) Composition :
The district forum consists of a chairman and two other members one of whom shall be a woman. The district forms are headed by the person of the rank of a District Judge.

b) Jurisdiction:
A written complaint can be filed before the district forum where the value of goods or services and the compensation claimed does not exceed ₹ 20 lakhs.

c) Appeal:
If a consumer is not satisfied by the decision of the District Forum, he can challenge the same before state commission, within 30 days of the order.

2. State commission:
This is established by the state governments in their respective states.
a) Composition :
The state commission consists of a president and not less than two and not more such number of members as may be prescribed, one of whom shall be a woman. The commission is headed by a person of the rank of High Court Judge.

b) Jurisdiction :
A written complaint can be filed before the state commission where the value of goods or services and the compensation claimed exceeds ₹ 20 lakhs but does not exceed ₹ 1 Crore.

c) Appeal :
In case the aggrieved party is not satisfied with the order of the state commission he can a appeal to National Commission within 30 days of passing the order.

3. National Commission :
The national commission was constituted in 1988 by the central government; It is the apex body in the three tier judicial machinery set up by the government for redressal of consumer grievances. Its office is situated Janpath Bhawan in New Delhi.

a) Composition :
It consists of a president and not less than four and not more than such members as may be prescribed, one of whom shall be a woman. The National Commission is headed a sitting or retired judge of supreme court.

b) Jurisdiction:
All complaints pertaining to those goods or services and compensation whose value is more than ₹ 1 Crore can be filed directly before the National Commission.

c) Appeal:
An appeal can be filed against the order of the National Commission to the supreme court within 30 days from the date of order passed.

Question 5.
Who can file a complaint, what complaints can be filed, where to file the complaint, how to tile the complaints redressal of grievances under the Consumer Protection Act 1986?
Answer:
For redressal of consumer grievances a complaint must be filed with the appropriate form.

Who can complaint?
The following persons can file a complaint under Consumer Protection Act 1986.
a) a consumer.
b) Any recognised voluntary consumer association whether the consumer is a member of that association or not;
c) The central or any state government;
d) One more consumers where there are numerous consumers having same interest;
e) Legal heir or representative in case of death of consumer what complaints can be filed?

What complaints can be filed?
A consumer can complaint relating to any one or more of the following;
a) An unfair trade practice or a restrictive trade practice adopted by any trader or service provider;
b) Goods bought by him or agreed to bought by him suffer from one more defects ;
c) Services hired or availed of or agreed to be hired or availed of, suffer from deficiency in any respect;
d) Price charged in excess of the price
i) fixed by or under law for the time being in force
ii) displayed on the goods or the package
iii) displayed in the price list or
iv) agreed between the parties and
e) goods or services which are hazardous or likely to be hazardous to life and safety when used.

Where to file a complaint?
If the value of goods and services and the compensation claimed does not exceed ? 20 lakhs, the complaint can be filed in the district forum ; If it exceeds ₹ 20 lakhs but does not exceed ₹ 1 crore, the complaint can be filed before the State Commission ; and if it exceeds ₹ 1 Crore, the complaint can be filed before the National Commission.

How to file a complaint?
A complaint can be made in person or by any authorised agent or by post. The complaint can be written on a plain paper supported by documentary evidence in support ’of the allegation contained in the complaint. The complaint should clearly specify the relief sought. It should also contain the nature, description and address of the complaint as opposite party, and so also the facts relating to the complaint and when and where it arose.

Very Short Answer Questions

Question 1.
Give the meaning of consumer.
Answer:
Under the Consumer Protection Act 1986, The word consumer has been defined separately for the purpose of goods and services.

For the purpose of goods, a consumer is one who buys any goods for consideration and any user of such goods other than the person who actually buys it, provided such use is made with the approval of buyer.

For the purpose of services, a consumer is one who has any service or services for consideration ; and any benificiary of such services provided the service is availed with the approval of the person who had hired the service for a consideration.

Question 2.
What is consumerism?
Answer:
Consumerism is defined as a social force designed to protect consumer interest in the market place by organising consumer pressure on business. By consumerism we mean the process of realising the rights of the consumer as enrises in the Consumer Protection Act, 1986 and ensuring right standards for the goods and services for which one makes payment.

Question 3.
What is meant by consumer protection?
Answer:
Consumer protection means safe guarding the interest and rights of consumers. In other words, it refers to the measures adopted for the protection of consumers from redressal of their grievances. The most common business malpractices are sale of adulterated, spurious, substandard and duplicate goods, false and under weighting, hoarding and black marketing, charging more than MRP Price etc. .

Question 4.
District Forums.
Answer:
The state government in each district establishes District forum by notification. The district forum consists of a president nominated by the state government. The forum also comprises two other members who shall have atleast 10 years of experience in dealing problems of economics, law commerce and industry. Every member of the form shall have tenure of 5 years or 65 years whichever is earlier. The District collector acts as the chairman of the District Forum. The District forum shall have jurisdiction to entertain consumers complaints where the value of goods and services which the compensation claimed, should ₹ 20 lakhs.

Question 5.
State commission.
Answer:
The state commission settles the consumer dispute at state level. The state commission is headed by the judge of High Court and comprised of other members not less than two and not more than such members as prescribed. The state commission is empowered to call for the records and appropriate orders in respect of any consumer dispute within the state jurisdiction. The state commission shall have jurisdiction to entertain consumer complaints where the value of goods and services for which compensation claimed exceeds ₹ 20 lakhs.

Question 6.
National commission.
Answer:
National commission operates at National level. It settle the consumer disputes at in the country. The National Commission has a President, who should be a serving or retained Supreme Court Judge the commission also comprises other members of not less than four. The president and all the members of the commission are appointed by central government. The National Commission shall have jurisdiction to entertain consumers complaints where the value of goods and services and compensation exceeds ₹ 1 crore.

Question 7.
Who is consumer? In the opinion of Mahatma Gandhi.
Answer:
Mahatma Gandhi, the father of nation, attached great importance to what he described as the ‘poor consumer’, who according to him should be the principle benificiary of the consumer movement. He said “A consumer is the most important visitor on our premises. He is not dependent on us, we are on him. He is not an interruption to our work; he is the purpose of it. He is not an outsider to our business ; he is a part of it. We are not doing him a favour by serving him ; he is doing us a favour by giving an opportunity to do so.

Andhra Pradesh BIEAP AP Inter 2nd Year Commerce Study Material 3rd Lesson Business Services Textbook Questions and Answers.

AP Inter 2nd Year Commerce Study Material 3rd Lesson Business Services

Essay Answer Questions

Question 1.
Define banking. Explain the functions of banking.
Answer:
The word Bank is derived from the French word ‘Bancus’ means a bench. According to Banking Regulation Act of 1949, banking is defined as “Accepting for the purpose of lending or investment of deposits of money from the public, repayable on demand or otherwise and withdrawable by cheque, draft, order or otherwise”.

The basic functions of banks are classified as primary functions and secondary functions.

A) Primary functions:
i) Accepting Deposits :
Bank accept various types of deposits, such as

Fixed deposits :
Fixed deposits are also called as time deposits or term deposits. In this deposit amount cannot be withdrawn until the maturity. Interest rate is also high.

Current Deposits :
Current accounts bears no interest companies, institutions, government and business men hold the current account. The greatest advantage of having current account is that there is no restriction on the withdrawls.

Savings Deposits :
The aim of these accounts is to encourage small savings from the public. Certain restrictions are imposed on the depositors regarding the number of withdrawals and the amount to be withdrawn in a given period of time.

Recurring deposits :
The purpose of these accounts is to encourage regular savings particularly by the fixed income group. Generally money is deposited in these accounts on monthly instalments for a fixed period. It is repaid to the depositors along with interest on maturity.

ii) Advancing of Loans :
Lending is carried out purely on profit motive. Banks lend the amount which is mobilised through the deposits. The different forms of lending are :

Loans :
A specified amount sanctioned by bank is called a ‘Loan’. A loan is granted against the security of property or personal security. The loan may be repaid in lumpsum or in instalments. The loan may be classified into i) Demand loan ii) Term loan. Demand loan is repayable on demand. It is repayable at short notice. Medium and long term loans are called term loans. It is granted for more than year and repayment is done on longer period.

Cash credit:
Cash credit is an arrangement where by the bank agrees to lend money to the borrower upto a certain limit. The amount is credited to the borrowers account. The borrower draws the money as and when he needs. Interest will be charged only on the amount actually drawn. Banks may impose commitment charges on unutilised portion.

Overdraft:
Banks grants overdraft to current account holders by which he is allowed to draw an amount in excess of the balance held in his account. Interest is charged on the overdrawn amount.

Discounting of bills of exchange :
A holder of a bill of exchange may be in urgent need of cash before the due date. He may sell or discount the bill with the bank. He will receive lesser amount than the actual amount. On maturity, the bank gets it payment from the debtor.

B) Secondary Functions :
These services include agency services and general utility services.

1. Agency services :
Banks perform some agency services on behalf of their customers.

• Banks helps their customers in transferring funds from one place to another place through cheques, drafts etc.
• Banks collect and pay various credit instruments like cheques, bills of exchange, promissory notes etc.
• Banks undertake to purchase and sale of various securities like shares, bonds, debentures etc., on behalf of their customers.
• Banks preserve the wills of their customers and execute them after death.

2. General utility services :
These services are

• Letters of credit are issued by the banks to their customers certifying their credit worthiness.
• Banks issue travellers cheques to help to travel without fear of theft or loss of money.
• Banks provide safe deposit locker facilities to the public at selected branches.
• Accepting or collecting foreign bills of exchange.

Question 2.
Discuss the principles of insurance. [A.P. Mar. 17]
Answer:
Insurance means protection against risk of loss. It provides compensation against any loss or damage due to the happening of an event. It is a contract between the two parties by which one of them undertake to indemnity the other person against a loss which may arise due to some events.

Principles of insurance :
1. Insurable interest:
A person cannot enter into a contract of insurance unless he has insurable interest in the subject matter of insurance. It is essential feature of insurance. Without this insurable interest, the contract of insurance will be treated as a wager or gambling contract. A person has insurable interest on his own life or the life of his wife and a creditor has insurable interest in the debtor.

2. Utmost good faith :
Insurance is based on the principle of utmost good faith. It means both the parties of the contract must disclose all the facts relating to the subject matter of insurance. If the insured does not disclose all material facts, the contract between them is void.

A person who had suffered from T.B. in the past had not disclosed it in the proposal form. Later on the insurer comes to know of this fact. He may declare the contract as void.

3. Indemnity :
This is the chief principle of insurance. Indemnity means security against risk of loss. Under this principle, the insured gets only the loss suffered from the insurer but not profits out of the contract of insurance. The principle of indemnity applies to contracts of fire and marine insurances only, but not to life insurance contracts.

4. Contribution :
Sometimes, goods are insured with more than one company. It is double insurance. The insured can get compensation only for the total loss from all insurance companies put together, but not total loss from each company. The insurance companies will pay the compensation on prorata basis.

5. Subrogation :
According to this principle, the insurer after compensating the loss of insured, the right of ownership of the damaged goods is shifted from insured to insurance company. Ex : Mr. X owns a scooter worth ₹ 36,000 and it was insured with a insurance company for full value. Later it was met with an accident and damaged beyond repairs. The insurance company paid the full value as compensation. Then all the rights on the scooter will pass on to insurance company.

6. Causa proxima :
According to this principle, the loss is caused by nearest and direct factor, then only the insurer will have to bear the loss. Ex : Biscuits in a ship are insured and are destroyed because of the sea water entered through a hole made by the mouse in the bottom of the ship and water entered into the ship. The nearest and direct cause is sea water. Hence, the insurer will have to bear the loss.

7. Mitigation of loss :
It is duty of the insured to take necessary steps to minimise the loss happened due to some event. He should not act carelessly and negligently at the time of loss to the insured property.

Question 3.
Define Life Assurance Policy. What are the kinds of life assurance policies?
Answer:
A life insurance contract may be defined as “a contract whereby the insurer, in consideration of premium paid either in lumpsum or in periodical instalments, undertakes to pay an annuity or a certain sum of money, either on the death of the insured or the expiry of certain number of years”.

Kinds of life assurance policies:
The following are some of the popular life assurance policies.
1) Whole life policy :
Under this policy the premium is paid through out the life of the insured. The sum assured is payable only after the death of the assured. The premium payable is low and it is meant for the family.

2) Endowment policy:
This policy is taken up for a specific period called endowment period. The policy will mature at the end of the specified period or at the attainment of particular age or on the death of insured, whichever is earlier. This policy offers the advantage of both protection and investment.

3) With or without profits policies :
When the policy is insured with profits, the policy holders share the profits of the company. The profit is called Bonus. The amount of the policy and bonus is paid on the maturity of the policy. When the policy is insured without profits, the insured does not share any profits and the amount of policy only is paid on maturity.

4) Joint life policies:
A policy may be taken jointly on the lives of two or more persons. On the death of any one person, the policy is paid to other surviving policy holder as the case may be. This type of policy may be taken by husband and wife.

5) Convertible whole life policy :
The policy is insured for whole life policy with a provision to convert it into endowment policy after a specified period. The conversion is done at the request of the assured. The rate of premium is increased after conversion.

6) Janata policy:
Janata policy was introduced by life insurance corporation of India in May 1957. This policy was introduced for the benefit of lower income group people. It can be issued for a term of 5,10, 15, 20 and 25 years subject to the condition that it should mature at the age of 60 years. No loans are granted on such policies.

7) Annuity policy :
Under this policy, the insured would deposit a lumpsum amount with the insurance company. The amount of policy would be paid to the insured for a specified number of years or until the death of the insured.

8) Children endowment policy:
This policy is taken by a person for his / her children to meet the expenses of their education or marriage. The agreement states that a certain sum will be paid by the insurer when the children attain a particular age.

9) Group insurance policy :
Members of a family or the employees of a business concern can take this insurance policy.

Question 4.
What do you understand by the word Transport? Discuss the benefits and limitations of Transport.
Answer:
Transport is the physical means of moving goods and persons from one place to another. Transport creates place utility of goods by moving them from different centres of production to the places of consumption. Goods are now produced thousand miles away from places where the consumer resides. Transportation only help the business men to reach consumer. Not only does transport give place utility, but it also renders time utility in various ways. Transportation, in simple language can be defined as “a means through which goods are transferred from one place to another”.’

Benefits or Functions of transport:
1. Movement of goods:
The first and important function of transport is the movement of goods. The raw materials have to move from their sources to the factory. The manufactured goods have to move from the factory to the consuming areas.

2. Transport enhances mobility of labour and capital :
An efficient network of transport services encourages the movement of people from one place to another labour can migrate to the place where they can get better job opportunities which reduces exploitation of labour.

3. Creation of place utility:
It moves goods from places where they are abundant to the places where they are scarce and thus creates place utility.

4. Specialisation and division of labour :
Transportation facilitates optimum utilisation of natural resources of a country. For example, petroleum resources of Arab countries, watches of Switzerland etc.

5. Creation of time utility :
With the advancement of technology, transportation time is being shortened. So, it creates time utility.

6. Stability in prices :
Goods can be transported from the place where the goods are abundant, to the places where scarcity exists. In this way, prices are equalised throughout the country.

7. Contribution to national income :
The transportation also contributes national income of a nation. For example, our railways.

8. Economies of large scale production :
Transport has helped the development of large scale industries. Transport procure raw material, labour and sells the finished goods.

9. Improve standard of living :
Availability of wide variety of goods at reasonable prices improves standard of living.

10. National defence :
Transport strengthens the national defence transport system. During war period, all the personnel, material and equipment can be moved rapidly to the boarder areas.

Limitations of transport :
1. Cottage and small scale industries lost their glory :
With the development of transport, labour is showing interest to work in big factories. This has led to shortage of workers in tiny and small scale industries.

2. Accidents :
Improvement in transport facility has given rise to new problem viz. accidents.

3. High urbanisation:
Improved means of transport has helped in creating big cities, which have further resulted into concentration of population in these cities. This has given rise to many new problems such as housing, health and pollution.

Question 5.
Describe the road transportation. Explain the kinds of roads in India. [A.P. Mar. 17]
Answer:
Road transport is the oldest form of transport. The Indian road network is one of the largest in the world. Road transport plays an important role in trade and commerce. Road transport is very good for short distance. Door – to – door collection and delivery is possible in road transport. It is most suitable for perishable goods. In road transport both men and animals are used to carry goods and people.

Modes of road transport are bullock carts, Tonga, rickshaws and motor vehicles such as jeeps, buses, motor vans, trucks and other vehicles. Road transport is suitable for the goods such as paper goods, clothing, computers, livestock, cement etc.

Indian roads are classified into three types – National highway, State highway, District and Rural roads.

a) National highway:
These roads are meant for inter state transport and movement of defence men. These also connect the state capitals and major cities. The National Highway Authority of India (NHAI) has the responsibility of development, maintenance and operation of national highways. The national highways have a road length of about 65,000 kms or 2% of the length of the total road system but they carry nearly 40% of goods and passenger traffic.

b) State highways:
These are constructed and maintained by state government. They connect the state capital with district head quarters and other important towns. State highways constitute 4% of total road length in the country.

c) District roads :
These roads are the connecting link between district head quarters and the other important roads of the district. The account for 14 % of the total road length of the country.

d) Rural roads :
These roads provide link to the rural areas. There are about 80% of total length in India are categorised as rural roads.

e) Boarder roads :
These roads are in the northern and northern – eastern boundary of the country. The Boarder Road organisation constructs and maintains boarder roads. They construct roads in high altitude areas and undertakes snow clearance.

f) International highways :
There are meant td promote harmonious relationship with the neighbouring countries by providing effective links with India.

Question 6.
Explain the warehouse concept and its significance.
Answer:
The term warehousing is a combination of two terms, ‘ware’ and ’housing’. The word ware refers to goods. Therefore, warehousing can be defined as the ‘ place suitable for preserving the goods and warehousing is the activity involving storage of goods. In common parlance warehouse means a godown. Warehousing facilitates other marketing functions such as assembling, grading and transportation.

Warehousing performs two important functions with regarding to finished goods. They are movement function and storage function. Movement function refers to the receipt of the products from the manufacturing plant, their transfer into warehouse and transferring them to common carriers on their way to consumers. The storage function is performed by retaining storing products in the warehouse until they are sold because production and consumption cycles differ. Warehousing functions creates time utility at minimum cost.

Significance :

1. Some commodities are produced in a particular season only. To ensure their off – season availability, warehousing is needed.
2. Some products are produced throughout the year but their demand is seasonal. Warehousing is important in such cases.
3. For the companies which opt for large scale production and bulk supply, warehousing is unavoidable factor.
4. Warehousing help companies to ensure quick supply of goods in demand.
5. Production of goods and their movement of goods are important for the companies for continuous production of goods.
6. Warehousing is also important for price stabilisation. For necessary goods, the Government store them in the warehouse and control its supply in the market as per the price fluctuations.
7. Another important need of warehousing is for bulk breaking.

Short Answer Questions

Question 1.
Define services and goods.
Answer:
Services are those separately identifiable, essentially intangible, activities that provide satisfaction of wants and are not necessarily linked to the sale of a product or other services. Services are intangible as they are not seen or touched. Service is inconsistant, since there is no standard tangible product. Service is the simultaneous activities of production and consumption. Services cannot be stored for future. Service is the participation of the customer in the service delivery process.

Goods are physical objects and are homogeneous in nature. They are tangible. Ex: Medicine Different customers get standardised demands fulfilled. Ex: Mobile phones. There will be separation of production and consumption. Ex : Purchasing ice cream from a store. Goods can be kept in stock. Ex : Train journey ticket. Involvement at the time of delivery is not possible. Ex : Manufacturing a vehicle.

Question 2.
What are the advantages of E-Banking?
Answer:
E-Banking brings certain advantages.
1. It reduces costs :
The cost of banking transactions is considerably reduced. It increases the profitability of the banks.

2. Prompt in services :
There is high degree of personalisation and fast and flexible execution. Thus E-Banking prompt service and there is greater customer satisfaction.

3. Anywhere and any time banking :
It is 24 hours in a day and 7 days in a week banking service. Bank account can be accessed from anywhere. So the customer can obtain information his account and conduct transactions from his home or office.

4. Cashless banking :
Handling of cash is not necessary in E-Banking.

5. Global coverage :
It provides global network coverage of bank services. NRIs can monitor their bank account in Indian banks, from abroad.

6. Central data base :
The data base of each branch is centralised. Customer can deposit, withdraw or remit money from any branch of his bank.

7. Internet banking helps banks to reduce the workload of their branches, such as generation of statements, balance of enquiry etc.

Question 3.
What is Mobile Banking? What are the services can be obtained through mobile banking?
Answer:
The delivery of bank services to customers through mobile (cell) phone is called mobile banking. When compared to telephone banking the scope of mobile banking is more and effective also. Mobile banking can take the form of SMS banking, GSM SIM Toolkit and WAP.

a) SMS banking :
Short messages are sent to the customers mobile phones. SMS messages can be used for both passive and active banking operations. A client automatically receives information about his account balance after a certain operation is performed.

b) GSM SIM Tool kit :
The GSM SIM Toolkit service can be only be used from a mobile phone supporting this technology. GSM SIM Toolkit is a software that evolves arbitrary changes in the mobile phone menu. Mobile phones now on the market support GSM SIM Tool kit after buying a special sim card and activating the permanent bank branch. The client can use this service.

c) WAP (Wireless Application Protocol) :
WAP is often compared to web pages although it is simplified. Unlike pages appearing on computer monitor, WAP presents it output on a small mobile phone. WAP Banking is not very popular. Only few banks are providing this service. .

Question 4.
What are the facets of electronic banking?
Answer:
The following are the different facets of E-Banking.
1. ATM :
ATM is popularly known as Any Time Money Machine. The customer gets cash fast, withdrawal, transfer, payment of bills or cash deposit through ATM.

2. Tele banking (home banking) :
Customers can perform number of transactions from their telephone such as checking the balance in the accounts, transfer funds from one account to another, pay certain bills and order statements or cheque book etc.

3. E-mail banking :
Customers may communicate with bank by electronic mail or E-mail. The most frequently used service is sending account statement periodically to the clients mail box.

4. Network banking or online banking:
Internet or online banking is a facility provided by banks to enable the user to execute bank related transactions through internet. The people sitting at home can transact business and they need not visit bank.

5. Mobile banking:
The delivery of bank services to a customer through mobile (cell) phone is called mobile banking.

Question 5.
Explain the term Insurance? Explain the functions of Insurance.
Answer:
The method of sharing of risk through economic cooperation is called insurance. Insurance may described as a social device to reduce or eliminate risk of loss to life and property. Insurance renders valuable services to commerce as well as to the society. Insurance covers many risks and uncertainties in the world of business and act as a boon to business firms.

Functions of Insurance :
1. Providing certainty :
Insurance provides payment of the risk of loss. There are uncertainties of happening of time and amount of loss. Insurance removes these uncertainties and the assured receives payment of loss. The insurer charges premium for providing certainty.

2. Protection:
The second function of insurance is to provide protection from probable chances of loss. Insurance cannot stop the happening of a risk or event but can compensate the losses arising out of it.

3. Risk sharing:
On the happening of a risk event, the loss is shared by all the persons exposed to it. The share is obtained from every insured member by way of premiums.

4. Assist in capital formation:
The accumulated funds of the insurer received by way of premium payments made by the insured are invested in various income generating schemes.

Question 6.
Explain the costs and benefits of Insurance.
Answer:
The following are the benefits of Insurance :
i) Provides certainly :
Insurance helps the insured to convert his uncertainties into certainities by entering into contract of insurance. The payment of premium by insured enables to reduce the risk.

ii) Distribution of losses :
Insurance helps to distribute losses of any uncertain events among the large number of insurer’s. It enables to transfer the risks and spreads the financial loss of insured members over the whole insurers.

iii) Provides security:
It provides security to the insured against the risk of uncertain events. The insurance company guarantees the insured to compensate or indemnify the loss on the occurrence of an event in consideration for payment of premium.

iv) Generates capital :
Insurance reduces financial risks and losses by providing facilities of capital investment in various organisations.

v) Increases efficiency :
Insurance reduces the risk and increases the efficiency in business. It provides security for business community which in turn paves the way for growth and diversification of industry.

vi) Earns foreign exchange :
Insurance provides security to the international traders, shippers and banking institutions, thus paves the way for expansion of foreign trade. The increased foreign trade activities lead to securing foreign exchange which makes the country to become economically strong.

vii) Social security :
Insurance acts as an instrument to fight against evils of poverty, unemployment, disease, old age, sickness, disability, accidents, fire and similar other calamities of nature.

viii) Promotes thrift:
Insurance encourages the people to go for savings. It alter the people in their spending habits and makes them to save a certain sum of money regularly.

Costs of disadvantages of Insurance :
i) Sharing of loss:
The loss of one person should be shared by all other policy holders. But the sharing of loss is opposed by many people as their return on investment is reduced.

ii) Real value of money :
The maturity value of the policy after the specific period may be more but the real value of money is going to be less.

iii) Lack of confidence :
Many of the investors, who propose to save their money, prefer banks and other financial institutions. It is due to lack of confidence in the insurance companies and its policies.

Question 7.
What are the advantages of Life insurance policies?
Answer:
In life insurance contract the policy amount is definitely, it is a question of time. The policy may mature during the life time of the assured or it may be paid on his death.

Advantages of life assurance policies :
1. Encourages savings:
The insured has to pay premium to insurance company every year. Otherwise, the policy will be cancelled. So, the insurance is helpful in creating the habit of saving money.

2. Exemption from income tax :
The amount paid as premium on a life insurance policy is allowed as deduction from income for calculating income tax.

3. Protection:
Life insurance provides protection to the family members if the policy holder dies suddenly. Life insurance builds a fund for the benefit of the dependents.

4. Credit facilities :
Insured can get loans against their policies to meet emergency needs. Life Insurance Corporation itself gives a loan against the policy to the insured at a lower rate of interest.

5. Surrender:
The life insurance company can surrender the life policy if the insured is unable to continue it. The insurance company can return some premium known as’surrender value’.

6. Meets the future needs:
An insurance policy can be helpful in providing funds for marriage and educational needs of insured children.

Question 8.
Explain the characteristics of marine Insurance.
Answer:
Marine insurance is a contract whereby the insurer agrees to indemnify the insured against marine losses.

The following are the characteristics of Marine Insurance.
1. Fundamentals of general contract:
Marine insurance must have the fundamentals ieraj insurance i.e. insurable interest, utmost good faith, indemnity, subrogation, contribution, warranties, causa proxima etc.

2. Consideration :
Marine insurance is a contract between the insured and insurer. Hence, insured is under an obligation to pay certain amount periodically to the insurer in consideration for accepting risk.

3. Coverage for insurance:
In marine insurance, cargo ship and freight can be insured. It covers large number of risks such as sinking of the ship, burning of the ship, standing of the ship, collision of ships, sea decoits etc.

4. Mode of insurance :
In marine insurance, the insurance may be for a single journey or number of journeys or for specific period of time. Insurance must be renewed once the specific condition is lapsed.

5. Indemnify the losses :
In marine insurance, the insurers guarantees to indemnify the losses caused by sea perils only.

6. Condition for compensation :
In marine insurance the insured is compensated only when the loss is occurred to ship or cargo. It also includes third party insurance.

Question 9.
Define Fire Insurance. Explain characteristics.
Answer:
According to Asbury, Fire insurance is defined as “It is a contract of insurance by which the insurer agrees for consideration to indemnify the insured upto a certain extent and subject to certain terms and conditions against loss or damage by fire which may happen to the property of the insured during a specified period”.

Features of fire insurance :
1. Contract of indemnity:
The fire insurance contract is a contract of indemnity and insured cannot claim more than the value of goods lost or damaged by fire or the amount of policy whichever is less.

2. Lawful consideration :
There must be consideration in fire insurance contract. The consideration is paid by the insured, which is called premium. Thus the essential element of fire insurance contract is premium received from the insured.

3. Insurable interest:
The insured must have insurable interest in the property or goods insured against fire. He must have insurable interest at the time of taking the policy and also at the time when the loss occurs and claim is filed for compensation.

4. Claim over residue :
The scrap or damaged goods after the fire accident automatically pass on to the insurer after the payment of claim under fire insurance.

5. Cause of accident:
The loss must be the out come of fire or ignition. No other reason for loss of property is accepted for settlement of claim.

6. Utmost good faith:
In fire insurance contract, both insured and insurer must have utmost good faith on each other.

Question 10.
Briefly state the advantages and disadvantages of road transportation.
Answer:
Advantages of Road transport: The following are some of the merits of Road transport.
1. Low capital:
It requires lesser capital for constructing roads. Roads are maintained by government and local authorities.

2. Low maintenance :
The maintenance charges of the road carriers are much less than the cost of railways.

3. Flexible :
Road transport is flexible. The route and timing can be adjusted to the individual requirements.

4. Suitable for short distance :
It is more economical and quicker for carrying goods and people over short distance.

5. Door-to-door delivery:
Road transport provides door-to-door delivery service for industries. Goods can be loaded at sellers doors and unload at buyers door.

6. Service to rural areas :
Exchange of goods between villages and towns are made possible by road transport.

7. Feeder to other modes of transport:
All the movement of goods begin and ultimately end by making use of roads.

8. High speed :
Road transport reduces the effective duration of the transit.

Disadvantages :
1. Less reliable :
Road vehicles are less reliable for long distance to travel because of the breakdowns and road congetions.

2. Accidents and breakdown :
There are more chances of accidents and breakdown in case of motor transport.

3. Lesser speed :
The speed of motor transport is comparatively slow.

4. Limited carrying capacity :
Load carrying capacity of road transport is limited.

5. More expensive :
The road is more expensive than railway transport for long distance travel.

6. Instable rates:
The rates charged by the road carriers are not stable. They need to charge with market.

Question 11.
State various advantages and disadvantages of Railway transportation.
Answer:
Advantages of railway transport:

1. Railway transport facilitate long distance travel and transport bulky and heavy goods.
2. It is quick and more regular form of transport because it helps in the transportation of goods with speed and certainty.
3. It helps in the industrialisation process of a country by easy transportation of coal and raw materials at cheaper rates.
4. It helps in the quick movement of goods from one place to another at the time of emergencies like famines and scarcity.
5. It encourages mobility of labour and thereby provides a great scope for employment.
6. Railway is the safest form of transport. The chances of breakdown and accidents of railways are less as compared to other modes of transport.
7. The carrying capacity of the railways is extremely large. Moreover, its capacity is elastic which can be easily be increased by adding more wagons.

Disadvantages :

1. The railways require a large investment of capital. The cost of construction, maintenance* and overhead expenses are very high as compared to other modes of transport.
2. Railway transport is not flexible. The routes and timings cannot be adjusted to the individual requirements.
3. Rail transport cannot provide door-to-door service as it is tide to a particular track.
4. Railway transport is unsuitable and uneconomical for short distance and small tariff of goods.
5. It involves much time and labour for booking and taking delivery of goods through railways.

Very Short Answer Questions

Question 1.
ATM.
Answer:
ATM means Automatic Teller Machine. An ATM is an unmanned device located on or off the bank premises. The operation mechanism is that ATM is inserted into ATM, the terminal reads and transmits the tape data to a processor which activate the account. It works 24 hours a day, 7 days a week. ATM are being used to withdraw, deposit and transfer of funds.

Question 2.
Online Banking.
Answer:
It is also called as internet banking. It is a facility provided by banks that enable the user to execute bank related transactions through internet. People sitting at home can transact business.

Question 3.
Tele Banking.
Answer:
Customers can perform a number of transactions from their telephone such as they can check balances and statement information, transfer funds from one account to another, pay certain bills and order statements, cheque book etc.

Question 4.
Mobile Banking. [A.P. Mar 17]
Answer:
This type of service is provided free of cost to all the customers. Under this the customer can access his bank account on the mobile screen for the services such as checking the balance, ordering a demand draft, stop payment or viewing last five transactions. To avail this facility the customer requires to have a mobile phone with WAN facility.

Question 5.
Electronic Banking.
Answer:
The concept of E-banking will enable anyone to transact with bank from anywhere such as home or office at any time convenient to him, which can be beyond the banking hours. Electronic banking is banking with the use of electronic tools and facilities and through electronic delivery channels.

Question 6.
Differentiate Insurer and Insured.
Answer:
The party who agrees to pay money on the happening of an event is called insurer. The party who seeks protection against the risk by paying premium is called insured.

Question 7.
What is premium?
Answer:
It is the money which is paid periodically by the insured to the insurer in consideration for which the insurer gives protection to the insured.

Question 8.
Define Insurance.
Answer:
Insurance is the pooling of fortuitous losses by transfer of such risks to insurers, who agrees to indemnify insured for such losses, to provide other precuniary benefits on the occurrence or to render service connected with the risk”.

Question 9.
Re-Insurance.
Answer:
Insurance company undertakes the risk according to its capacity. If a company undertakes more risks than its capacity, then it tries to share the risks with some other insurance company. When insurance company insurer complete or part of the risk with other insurance company, then it is called re-insurance.

Question 10.
Double Insurance.
Answer:
When more than one insurance policy is taken on the same subject matter, it is called double insurance. In life insurance, any number of policies can be taken by the insured upon his life. He can collect full amount on all the policies. But this is not the case with fire and marine insurance. He is entitled to the compensation of the actual loss only.

Question 11.
What is subrogation?
Answer:
According to this principle, the insurer after compensating the loss of insured, the right of ownership on damaged goods is shifted from insured to insurer, i.e., insurance company.

Question 12.
What is proximate cause?
Answer:
According to this principle, the loss is caused by nearest and direct factor, then only the insurer will have to bear the loss. The principle is useful in deciding the actual cause of loss when number of causes have contributed for occurrence of loss.

Question 13.
What is insurable Interest?
Answer:
A person cannot enter into contract of insurance unless he has insurable interest. It is essential feature of insurance. Without the insurable interest the contract of insurance will be treated as a gambling contract. A person has insurable interest on own life or life of his wife.

Question 14.
Endowment polity.
Answer:
This policy is taken up for a specific period. The policy will mature at the expiry of specific period or attainment of particular age or on the death of the insured whichever is earlier.

Question 15.
Whole life policy.
Answer:
This policy runs throughout the life of the insured. The sum assured under this policy is payable only after the death of the insured. The premium is low and it is meant to protect the family. The insured will have to pay the premium throughout his life even at the old age when he is not earning.

Question 16.
Name the subject matters of marine insurance.
Answer:
Subject matter of marine insurance are

1. Cargo or the goods on transhipment
2. Hull
3. Freight

Question 17.
What is cargo insurance?
Answer:
The cargo while being transported by ship is subject to many risks. These may be at ports i.e., risk of theft, loss of goods on voyage etc. Thus an insurance policy can be issued to cover against such risk to cargo.

Question 18.
What is freight insurance?
Answer:
If the cargo does not reach the destination due to damage or loss in transit, the shipping company is not paid freight charges. Freight insurance is for reimbursing the loss of freight to the shipping company i.e., insured.

Question 19.
Essential of fire insurance.
Answer:
Essentials of fire insurance are :

1. It is a contract of insurance.
2. There must be consideration.
3. The object of the contract should be indemnify the assured for the loss caused by damage or destruction of property by fire.

Question 20.
National Highway.
Answer:
National highways are meant for internal transport. These roads also connect state capitals, major cities etc. The National Highway Authority has the responsibility of development, maintenance and operation of the national highways. These roads encompass a road length of about 65,000 kms.

Question 21.
Pipe Lines.
Answer:
Pipe line transport is used for the movement of liquid commodities. Crude oil, natural gas and other petroleum products are transported through pipe lines. Pipe lines offer continuous movement at a relatively low cost. They fuel are efficient, dependable and involves less losses and damage. It can be operated all around the clock (24 x 7).

Question 22.
Bonded warehouse.
Answer:
Bonded warehouses are owned and operated by port trust authorities. It is located near the port. It is a place where importers store goods till customs duties are paid or goods are re-shipped to other destination without being brought into the country.

Question 23.
Two significance of warehouse.
Answer:
Advantages of warehouses.

1. It serves the business men who have very limited space.
2. Some warehouses indirectly offer financial assistance.

Question 24.
Cash credit.
Answer:
A cash credit is an agreement where by a bank agrees to lend money to the borrower upto a certain limit. The amount is credited to the account of the borrower. The borrower draws money as and when he needs. Interest is charged on the amount actually drawn.

Question 25.
Bill discounting.
Answr:
The holder of a bill or drawer may be in urgent need of cash before the due date. In such circumstances, he can sell or discount the bill to the bank at lesser amount than the actual.

Question 26.
Recurring deposit.
Answer:
In Recurring deposit, the depositor is required to deposit a fixed amount of money every month for a specific period. After the completion of the specific period, the consumer gets back the deposited amount along with the cumulative interest accrued on the deposit.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 10th Lesson ఏకాంతర విద్యుత్ ప్రవాహం Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 10th Lesson ఏకాంతర విద్యుత్ ప్రవాహం

అతిస్వల్ప సమాధాన ప్రశ్నలు

ప్రశ్న 1.
10 ప్రాథమిక తీగచుట్లు ఉన్న ఒక పరివర్తకం (transformer) 200 Vac ని 2000 Vac కి మార్చగలిగితే, దాని గౌణ తీగచుట్లను లెక్కించండి. [TS. Mar. 16]
జవాబు:

ప్రశ్న 2.
6 Vబెడ్ లాంప్ ఎటువంటి పరివర్తకాన్ని ఉపయోగిస్తారు?
జవాబు:
6 V ల బెడ్ంప్ లో అవరోహణ పరివర్తకంను ఉపయోగిస్తారు.

ప్రశ్న 3.
పరివర్తకం పనిచేయడంలో ఏ దృగ్విషయం ఇమిడి ఉంది?
జవాబు:
అన్యోన్య ప్రేరణపై పరివర్తకం పనిచేస్తుంది.

ప్రశ్న 4.
పరివర్తక నిష్పత్తి అంటే ఏమిటి?
జవాబు:
గౌణ వి.చా.బ నికి, ప్రాథమిక వి.చా.బ గల నిష్పత్తిని (లేదా) గౌణ తీగ చుట్టలో సంఖ్యకు, ప్రాథమిక తీగచుట్టలో చుట్ల సంఖ్యకు గల నిష్పత్తిని పరివర్తకం నిష్పత్తి అంటారు.
పరివర్తకం నిష్పత్తి = \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\).

ప్రశ్న 5.
i) ప్రేరకం, ii) క్షమశీలి (కెపాసిటర్) ప్రతిరోధకానికి సమీకరణాలు వ్రాయండి.
జవాబు:
i) ప్రేరకం ప్రతిరోధకం (X_L) = ωL,
ii) క్షమశీలి (కెపాసిటర్) ప్రతిరోధకం (X_c) = \(\frac{1}{\omega \mathrm{C}}\)

ప్రశ్న 6.
ఏకాంతర విద్యుచ్ఛాలక బలం, విద్యుత్ ప్రవాహాల మధ్య దశాభేదం కింది వాటిలో ఏవిధంగా ఉంటుంది : శుద్ధ నిరోధం, శుద్ధ ప్రేరకం, శుద్ధ కెపాసిటర్.
జవాబు:

1. శుద్ధ నిరోధములో A.C. వి. బా.బ మరియు విద్యుత్ ప్రవాహం ఒకే దశలో ఉంటాయి.
2. శుద్ధ (ప్రేరకంలో వి.బా.బ కన్నా విద్యుత్ ప్రవాహం \(\frac{\pi}{2}\) (లేదా) 90°.
3. శుద్ధ కెపాసిటర్ వి. బా.బ కన్నా విద్యుత్ ప్రవాహం \(\frac{\pi}{2}\) (లేదా) 90° ముందు ఉంటుంది.

ప్రశ్న 7.
సామర్థ్య కారకాన్ని నిర్వచించండి. సామర్థ్య కారకం ఏ కారకాలపై ఆధారపడుతుంది? [TS. Mar. 15]
జవాబు:
నిజ సామర్థ్యానికి, దృశ్య సామర్థ్యానికి (మిథ్యా సామర్థ్యం) గల నిష్పత్తిని సామర్థ్య కారకం అంటారు.
సామర్థ్య కారకంcosΦ) = \(\frac{P}{P_{rms}}\) (∵ P_rms = V_rms . I_rms)

సామర్థ్య కారకం, r.m.s వోల్టేజి, rm.s విద్యుత్ ప్రవాహం మరియు సగటు సామర్థ్యంపై ఆధారపడుతుంది.

ప్రశ్న 8.
విద్యుత్ ప్రవాహం యొక్క వాట్లెస్ అంశ అంటే అర్థం ఏమిటి? [TS. Mar.’17]
జవాబు:
సగటు సామర్థ్యం (P) = V_rms (I_rms sin Φ) cos \(\frac{\pi}{2}\)

విద్యుత్ ప్రవాహం అంశము (I_rms sin Φ) వల్ల వలయంలో వినియోగించే సగటు సామర్థ్యం సున్నా. ఈ విద్యుత్ ప్రవాహ అంశాన్ని వాట్లెస్ విద్యుత్ ప్రవాహం అంటారు. ఇక్కడ (Irms sin థ) విద్యుత్ ప్రవాహం యొక్క వాట్లెస్ అంశం అంటారు.

ప్రశ్న 9.
LCR శ్రేణి వలయం కనిష్ఠ అవరోధం ఎప్పుడు కలిగి ఉంటుంది?
జవాబు:

ఈ పౌనఃపున్యమును అనునాద పౌనఃపున్యము అంటారు.

ప్రశ్న 10.
LCR శ్రేణి వలయం సామర్థ్య కారకం విలువ ఏకాంకం అయినప్పుడు వోల్టేజి, విద్యుత్ ప్రవాహాల మధ్య దశాభేదం ఎంత ఉంటుంది?
జవాబు:
LCR శ్రేణి వలయంలో సామర్థ్య కారకం (cos Φ) = 1
∴ వోల్టేజి మరియు విద్యుత్ ప్రవాహము మధ్య దశాభేదం సున్నా (Φ = 0).

స్వల్ప సమాధాన ప్రశ్నలు

ప్రశ్న 1.
ఏకాంతర విద్యుచ్ఛాలక బలం అనువర్తింపచేసిన ప్రేరకంలోని విద్యుత్ ప్రవాహానికి సమీకరణాన్ని పొందండి.
జవాబు:
వలయంలో శుద్ధ ప్రేరకం యొక్క ప్రేరకత L. దీనికి ac వి.చా.బ V = V_m sin ωt ని అన్వర్తించామనుకొనుము. i అనునది తాత్కాల విద్యుత్ ప్రవాహము.

ప్రశ్న 2.
ఏకాంతర విద్యుచ్ఛాలక బలం అనువర్తింపచేసిన కెపాసిటర్లోని విద్యుత్ ప్రవాహానికి సమీకరణాన్ని పొందండి.
జవాబు:

వలయంలో శుద్ధ కెపాసిటర్ యొక్క కెపాసిటీ C. దీనికి A.C వి. చా. బ V = V_m sin ωt ని అన్వర్తించామనుకొనుము. i మరియు q అనునది తాత్కాల విద్యుత్ ప్రవాహము మరియు ఆవేశాలు అనుకొనుము.
కెపాసిటర్ వద్ద పొటెన్షియల్ తేడా = –\(\frac{q}{C}\)
మొత్తం వి.చా. బ = V_m sin ωt – \(\frac{q}{C}\)
ఓమ్ నియమం ప్రకారం ఇది iR = 0 కు సమానం
V_m sin ωt – \(\frac{q}{C}\)= 0

i₀ అనునది శిఖర విద్యుత్ ప్రవాహము. ఇక్కడ విద్యుత్ ప్రవాహము వి.చా. బ కన్నా \(\frac{\pi}{2}\) (లేదా) 90° ముందుంటుంది.

ప్రశ్న 3.
పరివర్తకం (ట్రాన్స్ఫార్మర్) ఏ సూత్రంపై ఆధారపడి పనిచేస్తుందో తెలపండి. పరివర్తకం పనిచేసే విధానాన్ని తగిన సిద్ధాంతంతో వర్ణించండి.
జవాబు:
ఎక్కువ వోల్టేజి, తక్కువ విద్యుత్ ప్రవాహమున్న ఏకాంతర విద్యుత్ ప్రవాహంను తక్కువ వోల్టేజి, ఎక్కువ విద్యుత్ ప్రవాహంగా (లేదా) విపర్యంగా మార్చే పరికరాన్ని పరివర్తకం అంటారు.

నియమం :
పరివర్తకం రెండు తీగచుట్ల మధ్య అన్యోన్య ప్రేరణపై ఆధారపడి పనిచేస్తుంది.

పనిచేయు విధానం :
ఏకాంతర వి. చా. బను ప్రాథమిక తీగచుట్టకు అన్వర్తిస్తే నివేశ వోల్టేజి కాలంతోపాటు మారుతుంది. అందువలన కాలంతో పాటు అయస్కాంత అభివాహం కూడా మారుతుంది.

ఈ మారే అయస్కాంత అభివాహం గౌణ తీగచుట్టలో అనుసంధానం చెంది ఉంటుంది. కావున గౌణ తీగచుట్టలో వి. చా. బ ప్రేమవుతుంది.

సిద్ధాంతం :
N₁ మరియు N₂ అనునవి ప్రాధమిక మరియు గౌణ తీగచుట్టలలో చుట్ల సంఖ్య అనుకొనుము. V_P మరియు V_S అనునవి ప్రాథమిక మరియు గౌణ చుట్టలలో విద్యుత్చ్చాలక బలాలు అనుకొనుము.

దీర్ఘ సమాధాన ప్రశ్నలు

ప్రశ్న 1.
LCR శ్రేణి వలయంలో అవరోధానికి, విద్యుత్ ప్రవాహానికి సమీకరణాన్ని పొందండి. LCR శ్రేణి అనునాద వలయం పౌనఃపున్యానికి సమాసాన్ని రాబట్టండి.
జవాబు:
వలయంలో నిరోధకం యొక్క నిరోధం R, ప్రేరకం యొక్క ప్రేరకత L, కెపాసిటర్ యొక్క కెపాసిటి C లను శ్రేణిలో కలిపి, దానికి V = V_m sin ωt A.C వోల్టేజిని అన్వర్తించామనుకొనుము.

i మరియు q అనునవి తాత్కాల విద్యుత్ ప్రవాహము మరియు ఆవేశము అనుకొనుము.

అనునాద పౌనఃపున్యము (f₀) :
ఈ పౌనఃపున్యము వద్ద LCR శ్రేణీ వలయంలో అవరోధం కనిష్ఠంగా ఉంటుంది. ఇది R కు సమానం. ఈ పౌనః పున్యము వద్ద విద్యుత్ ప్రవాహం గరిష్ఠం.

అనునాద పౌనఃపున్యము (f₀) వద్ద విద్యుత్ ప్రవాహం గరిష్ఠం. ఈ శ్రేణీ అనునాద వలయాన్ని గ్రహీత వలయం

లెక్కలు Problems

ప్రశ్న 1.
20 mH ప్రేరకత్వం ఉన్న ఒక ఆదర్శ ప్రేరకాన్ని (తీగచుట్ట అంతర్నిరోధం శూన్యం) AC అమ్మీటర్కు శ్రేణిలో కలిపి, దీన్ని విద్యుచ్ఛాలకు బలం e = 20√2 sin(200t + π/3)V ఉన్న AC జనకానికి కలిపారు. ఇక్కడ t సెకనులలో గలదు. అమ్మీటర్ రీడింగును కనుక్కోండి.
సాధన:
L = 20 mH = 20 × 10^-3H

ప్రశ్న 2.
నిరోధకం, ప్రేరకం ఉన్న శ్రేణీవలయం చివరల తాక్షణిక విద్యుత్ ప్రవాహం, వోల్టేజి విలువలు 1 = √2 sin (100t – π/4)A, υ = 40 sin (100t) V అయితే నిరోధాన్ని లెక్కించండి.
సాధన:
i = √2 sin (100t – π/4)A. (∵ i = i₀ sin(ωt – Φ)
V = 40 sin(100t) V (∵ V = V₀sin(ωt))

ప్రశ్న 3.
ఒక AC వలయంలో ఒక కండెన్సర్, ఒక నిరోధకం, ఒక ప్రేరకంలు ఒక AC జనరేటర్, ఏకాంతరకానికి (alternator) అడ్డంగా శ్రేణిలో కలిపారు. వాటి చివరల వోల్టేజిలు వరసగా 20 V, 35 V, 20 V అయితే ఆల్టర్నేటర్ సరఫరా చేసిన వోల్టేజిని కనుక్కోండి
సాధన:
V_C = 20 V, V_R = 35 V, V_L = 20 V

ప్రశ్న 4.
ఒక ఏకాంతర విద్యుత్ వలయంలో నిరోధం R, ప్రేరకం L, కెపాసిటెన్స్ C లను శ్రేణిలో స్థిర వోల్టేజి, చర పౌనః పున్యం ఉన్న ఏకాంతరకం కొనల మధ్య కలిపారు. అనునాద పౌనఃపున్యం వద్ద ప్రేరకత్వ ప్రతిరోధం, క్షమత్వ ప్రతిరోధం, నిరోధం సమానం మరియు వలయంలోని విద్యుత్ ప్రవాహం i₀ అయితే, అనునాద పౌనఃపున్యానికి రెట్టింపు పౌనఃపున్యం వద్ద వలయంలోని విద్యుత్ ప్రవాహాన్ని కనుక్కోండి.
సాధన:

ప్రశ్న 5.
ఒక శ్రేణీ అనునాద వలయం L, R,, C లను కలిగి ఉంది. అనునాద పౌనఃపున్యం f. మరొక శ్రేణి అనునాద వలయం Lz, Rz, C, అను కలిగి ఉంది. దీని అనునాద పౌనఃపున్యం కూడా f. ఈ రెండు వలయాలను శ్రేణిలో కలిపితే అనునాద పౌనఃపున్యాన్ని లెక్కించండి.
సాధన:
అనునాద పౌనఃపున్యము (f)

ప్రశ్న 6.
ఒక LCR శ్రేణి వలయంలో నిరోధం R = 200 Q, ప్రధాన సరఫరా వోల్టేజి 200V, పౌనఃపున్యం 50 Hz. వలయం నుంచి కెపాసిటెన్స్ను బయటకు తీసినప్పుడు విద్యుత్ ప్రవాహం వోల్టేజి కంటే 45° వెనుకబడి ఉంది. ప్రేరకాన్ని వలయం నుంచి బయటకు తీసినప్పుడు విద్యుత్ ప్రవాహం వోల్టేజి కంటే 45° ముందు ఉంది. అయితే LCR వలయంలో దుర్వ్యయమైన సామర్థ్యాన్ని లెక్కించండి.
సాధన:

ప్రశ్న 7.
ప్రాథమిక, గౌణ చుట్ల నిష్పత్తి 1:2 ఉన్న ట్రాన్స్ఫార్మర్ ప్రాథమికాన్ని వోల్టేజి 200 ఉన్న ఏకాంతరానికి కలిపారు. ప్రాథమిక తీగచుట్ట ద్వారా 4A విద్యుత్తు ప్రవహిస్తున్నది. ట్రాన్స్ఫార్మర్ ఎటువంటి నష్టాలు కలిగి లేవనుకొన్నట్లయితే గౌణ వోల్టేజి, విద్యుత్ ప్రవాహాలను కనుక్కోండి.
సాధన:

అభ్యాసాలు Textual Exercises

ప్రశ్న 1.
100Ω నిరోధాన్ని 220V, 50 Hz ఉన్న ac సరఫరాకు కలిపారు.
(a) వలయంలో విద్యుత్ ప్రవాహ వర్గ మధ్యమ ‘మూలం (rms) విలువ ఎంత?
(b) ఒక పూర్తి చక్రంలో వినియోగమైన నికర సామర్థ్యం ఎంత?
సాధన:
నిరోధము R = 100Ω
V_rms = 220V
పౌనఃపున్యము (f) = 50Hz

a) వలయంలో విద్యుత్ ప్రవాహం.
(I_rms) = \(\frac{V_{rms}}{R}=\frac{220}{100}\) = 2.2A

b) వినియోగించిన మొత్తం సామర్థ్యం
(P) = V_rms × I_rms
= 220 × 2.2
= 484W

ప్రశ్న 2.
(a) ఒక ac సరఫరా శిఖర వోల్టేజి 300 V. వర్గ మధ్యమ మూలం వోల్టేజి ఎంత?
(b) ఒక ac వలయంలో విద్యుత్ ప్రవాహ rms విలువ 10 A. శిఖర విద్యుత్ ప్రవాహం ఎంత?
సాధన:
a) గరిష్ఠ వోల్టేజి విలువ (V₀) = 300V.

ప్రశ్న 3.
44 mH ప్రేరకాన్ని 220V, 50 Hz ac సరఫరాకి కలిపారు. వలయంలో విద్యుత్ ప్రవాహ rms విలువను నిర్ధారించండి.
సాధన:
ప్రేరకత (L) = 44 mH = 44 × 10^-3H
V_rms = 220 V

పౌనఃపున్యము (f) = 50 Hz
ప్రేరక ప్రతిరోధం (X_L) = 2πfL
= 2 × 3.14 × 50 × 44 × 10^-3
X_L = 13.83Ω
r.m.s విద్యుత్ ప్రవాహం

ప్రశ్న 4.
110V, 60Hz ఉన్నacసరఫరాకి 60 µF కెపాసిటర్ను కలిపారు. వలయంలో విద్యుత్ ప్రవాహ rms విలువను నిర్ధారించండి.
సాధన:
కెపాసిటర్ యొక్క కెపాసిటి
C = 60 µF = 60 × 10^-6F
V_rms = 110 V
పౌనఃపున్యం (f) = 60Hz
క్షమత్వ ప్రతిరోధం

ప్రశ్న 5.
3, 4 అభ్యాసాలలో ఒక పూర్తి చక్రంలో ప్రతీ వలయం శోషణం చేసుకొనే నికర సామర్థ్యం ఎంత? మీ సమాధానాన్ని వివరించండి.
సాధన:
i) సగటు సామర్థ్యము (P) = V_rms × I_rms × cos Φ
P = V_rms × I_rms × cos 90° = 0 (∴ Φ = 90°)
P = 0

ii) P = V_rms × I_rms × cos Φ
విద్యుత్ ప్రవాహము, వోల్టేజి మధ్య దశాభేదం క్షమత్వంలో 90°
P = V_rms × I_rms × cos 90° = 0

ప్రశ్న 6.
L = 2.0H, C = 32 µF, R = 10Ω లు ఉన్న శ్రేణి LCR వలయం అనువాద పౌనఃపున్యం అని పొందండి. ఈ వలయం Q-విలువ ఎంత?
సాధన:
L = 2H, C = 32µF, R = 10Ω
అనునాద కోణీయ పౌనఃపున్యము

ప్రశ్న 7.
30µF ఉన్న ఆవేశిత కెపాసిటర్ను 27 mH ఉన్న ప్రేరకానికి కలిపారు. వలయం స్వేచ్ఛా డోలనాల కోణీయ పౌనఃపున్యం ఎంత ?
సాధన:
C = 30µF = 30 × 10^-6F
ప్రేరకత (L) = 27mH = 27 × 10^-3H
అనువాద కోణీయ పౌనఃపున్యము

ప్రశ్న 8.
ఒకవేళ అభ్యాసం 7 లో కెపాసిటర్ మీద తొలి ఆవేశం 6 mC అనుకోండి.. తొలుత ఆ వలయంలో నిల్వ ఉండే మొత్తం శక్తి ఎంత ? ఆ తరవాత సమయంలో మొత్తం శక్తి ఎంత?
సాధన:
కెపాసిటర్పై ఆవేశం (Q) = 6mC = 6 × 10^-3C
C = 30µF = 30 × 10^-6F
వలయంలో నిల్వయున్న శక్తి

కొంత సేపటికిLమరియు Cలు శక్తిని పంచుకుంటాయి. కాని మొత్తం శక్తి స్థిరంగా ఉంటుంది. కావున శక్తి నష్టం ఉండదు.

ప్రశ్న 9.
R = 20Ω, L = 1.5 H, C = 35µF లు ఉన్న LCR శ్రేణి వలయాన్ని చర పౌనఃపున్యం ఉన్న 200 V ac సరఫరాకు కలిపారు. సరఫరా పౌనఃపున్యం వలయం సహజ పౌనఃపున్యానికి ఎప్పుడు సమానం అవుతుంది? ఒక పూర్తి చక్రంలో వలయానికి బదిలీ అయిన సగటు సామర్థ్యం ఎంత?
సాధన:
నిరోధం (R) = 20Ω,
ప్రేరకత (L) = 1.5H,
కెపాసిటెన్స్ (C) = 35 × 10^-6F
V_rms = 200V
అనునాదము వద్ద
Z = R = 20Ω
rms విద్యుత్ ప్రవాహము
I_rms = \(\frac{V_{rms}}{Z}=\frac{200}{20}\) = 10A
Φ = 0° (అనునాదం)
ఒక పూర్తి చక్రానికి వలయానికి బదిలీ అయిన సామర్థ్యం
P = I_rms × V_rms × cos Φ
= 10 × 200 × cos 0° = 2000 W
P = 2KW

ప్రశ్న 10.
ఒక రేడియో MW (Medium Wave – మధ్యంతర తరంగం) ప్రసార అవధిలోని భాగమైన 800 kHz నుంచి 1200 kHz ఉన్న పౌనఃపున్య అవధికి శృతి చేయగలదు. దాని వలయం 200 µH ప్రభావాత్మక ప్రేరకత్వాన్ని కలిగి ఉంటే, దానిలోని చర కెపాసిటర్ అవధి ఎంత ఉండాలి?
(Hint : శృతి చేయడానికి, సహజ పౌనఃపున్యం, అంటే LC వలయం స్వేచ్ఛా డోలనాల పౌనఃపున్యం, రేడియో తరంగ పౌనఃపున్యానికి సమానం కావాలి.)
సాధన:
కనిష్ఠ పౌనఃపున్యము
f₁ = 800KHz = 800 × 10³Hz
ప్రేరకత (L) = 200 µH = 200 × 10^-6H
గరిష్ట పౌనఃపున్యము
f₂ = 1200 KHz = 1200 × 10³Hz

కెపాసిటర్ల వ్యాప్తి 87.8pf నుండి 197.7pF వరకు ఉంటుంది.

ప్రశ్న 11.
I = 5.0 H, C = 80µE, R = 40Ω విలువలు ఉన్న శ్రేణి LCR వలయాన్ని చర పౌనఃపున్యం ఉన్న 230 v జనకానికి పటంలో చూపినట్లుగా కలిపారు.

(a) వలయంలో అనునాదం ఏర్పడాలంటే జనక పౌనఃపున్యం ఎంత ఉండాలి?
(b) అనునాద పౌనఃపున్యం వద్ద వలయం అవరోధం, విద్యుత్ ప్రవాహ కంపనపరిమితిని పొందండి.
(c) వలయంలోని మూడు ఘటకాల కొనల మధ్య ఉండే rms పొటెన్షియల్ను కనుక్కోండి. అనువాద పౌనః పున్యం వద్ద LC సంయోగం కొనల మధ్య ఉండే పొటెన్షియల్ శూన్యం అని చూపండి.
సాధన:
r.m.s వోల్టేజి (V_r.m.s) = 230V
ప్రేరకత (L) = 5H
కెపాసిటెన్స్ C = 80µF = 80 × 10^-6F
నిరోధం (R) = 40Ω
a) అనువాద కోణీయ పౌనఃపున్యం

అనునాద పౌనఃపున్యం

b) అనునాద పౌనఃపున్యము వద్ద X_L = X_C
వలయంలో అవరోధం Z = R = 40Ω
r.m.s విద్యుత్ ప్రవాహము

అదనపు అభ్యాసాలు Additional Exercises

ప్రశ్న 12.
ఒక LC వలయం 20 mH ఉన్న ఒక ప్రేరకం, 10 mC తొలి ఆవేశాన్ని కలిగి ఉన్న 50 µF ఉన్న ఒక కెపాసిటర్ను కలిగి ఉంది. వలయ నిరోధం ఉపేక్షించ దగింది. t = 0 సమయం వద్ద వలయం మూసి ఉందనుకోండి.
(a) తొలుత నిల్వ ఉన్న మొత్తం శక్తి ఎంత? LC డోలనాలు చేసేటప్పుడు ఇది నిత్యత్వమవుతుందా?
(b) వలయం సహజ పౌనఃపున్యం ఎంత?
(c) (i) ఏ సమయం వద్ద నిల్వ ఉన్న శక్తి పూర్తిగా విద్యుత్ శక్తిగా ఉంటుంది. (అంటే కెపాసిటర్లో నిల్వ ఉన్నది), (ii) ఏ సమయం వద్ద నిల్వ ఉన్న శక్తి పూర్తిగా అయస్కాంత శక్తి (అంటే ప్రేరకంలో నిల్వ ఉన్నది) గా ఉంటుంది?
(d) ఏ సమయం వద్ద మొత్తం శక్తి ప్రేరకం, కెపాసిటర్లలో సమానంగా పంచబడుతుంది?
(e) వలయంలో నిరోధాన్ని ఉంచినప్పుడు ఎంత శక్తి ఉష్ణ రూపంలో దుర్వ్యయమవుతుంది?
సాధన:
ప్రేరకత (L) : = 20mH = 20 × 10^-3H
కెపాసిటీ C = 50µf = 50 × 10^-6 F
కెపాసిటర్ మీద తొలి ఆవేశం
(Q₁) = 10mc = 10 × 10³C
a) కెపాసిటర్లో తొలిగా నిల్వయున్న శక్తి (E) = \(\frac{Q^2}{2C}\)

ఈ శక్తి నిత్యత్వముగా ఉండును.

b) అనునాద పౌనఃపున్యము

వలయంలో సహజ పౌనఃపున్యము
ω = 2πυ = 2л × 159.2
999.78 ≈ 1000 = 10³rad/s

c) i) కెపాసిటర్ మీద ఆవేశం Q = Q₀ cos ωt
Q = Q₀ cos \(\frac{2 \pi}{T}\) .t …………. (1)
Q = Q₀ అయితే,
cos \(\frac{2 \pi}{T}\)t = ±1 = cos nл (లేదా) t = \(\frac{nT}{2}\)
ఎక్కడ n = 0, 1, 2, 3…………
t = 0, T/2, T, 3T/2, ………….

ii) ఏదైనా కాలం వద్ద, నిల్వయున్న శక్తి పూర్తిగా అయస్కాంత శక్తి

e) వలయంలో నిరోధాన్ని చేర్చితే, మొత్తం శక్తి ఉష్ణ రూపంలో కోల్పోతుంది. డోలనాలు అవరుద్ధమైతే, కొంత సేపటికి కోల్పోయిన శక్తి ఉష్ణరూపంలోకి మారుతుంది.

ప్రశ్న 13.
0.50 H ప్రేరకత్వం ఉన్న ఒక తీగచుట్ట, 100Ω నిరోధాన్ని 50 Hz పౌనఃపున్యం ఉన్న 240 Vac సరఫరాకు కలిపారు.
a) తీగచుట్టలో గరిష్ఠ విద్యుత్ ప్రవాహం ఎంత?
b) గరిష్ట విద్యుత్ ప్రవాహం, గరిష్ఠ వోల్టేజిల మధ్య కాల విలంబనం (time lag) ఎంత?
సాధన:
ప్రేరకత (L) : = 0.50H
నిరోధము (R) = 100Ω
r.m.s వోల్టేజి V_rms
a) అవరోధము

ప్రశ్న 14.
అభ్యాసం 13లోని వలయాన్ని అధిక పౌనఃపున్యం ఉన్న సరఫరాకు (240 V, 10kHz) కలిపినప్పుడు (a), (b) లను కనుక్కోండి. అందువల్ల అధిక పౌనఃపున్యం వద్ద వలయంలో ప్రేరకం తెరచిన వలయంలాగా ఉంటుందనే ప్రవచనాన్ని వివరించండి. నిలకడ స్థితి తరవాతdcవలయంలో ప్రేరకం ఏవిధంగా ప్రవర్తిస్తుంది?
సాధన:
పౌనఃపున్యము (f) = 10kHz = 10⁴Hz
r.m.s వోల్టేజి (V_r.m.s) = 240V
నిరోధము R = 100Ω,
ప్రేరకత (L) = 0.5H

అల్ప పౌనఃపున్యం వద్ద I₀ = 1.82A, అధిక పౌనః పున్యం వద్ద I₀ = 0.0108 A కావున అధిక పౌనః పున్యము వద్ద ప్రేరకం అధిక నిరోధాన్ని కలిగించి వివృత వలయం అవుతుంది.

వలయంలో ω = 0 అందువలన X_L = ω_L = 0.

ప్రశ్న 15.
100µF కెపాసిటర్ను 40Ω నిరోధానికి శ్రేణిలో కలిపి 110V, 60 Hz సరఫరాకి కలిపారు.
(a) వలయంలో గరిష్ఠ విద్యుత్ ప్రవాహం ఎంత?
(b) గరిష్ఠ విద్యుత్ ప్రవాహం, గరిష్ఠ వోల్టేజిల మధ్య కాల విలంబనం ఎంత?
సాధన:
కెపాసిటర్ యొక్క కెపాసిటీ

C = 100µF = 100 × 10^-6F
నిరోధము (R) = 40Ω,
V_rms = 110 V
పౌనఃపున్యము (f) = 60 Hz
(a) అవరోధము (Z)

గరిష్ఠ వోల్టేజి మరియు గరిష్ఠ విద్యుత్ ప్రవాహం మధ్య కాల అవధి = 1.55 × 10^-3s.

ప్రశ్న 16.
సమస్య 15లోని వలయాన్ని 110 V, 12 kHz సరఫరాకు కలిపినప్పుడు (a), (b) లకు సమాధానాలను కనుక్కోండి. అందువల్ల అధిక పౌనఃపున్యాల వద్ద కెపాసిటర్ ఒక వాహకం లాగా ఉంటుందనే ప్రవచనాన్ని వివరించండి. ఈ ప్రవర్తనను నిలకడ స్థితి తరవాత dc వలయంలో కెపాసిటర్ ప్రవర్తనతో పోల్చండి.
సాధన:
r.m.s వోల్టేజి (V_rms) = 110V.
పౌనఃపున్యము (f) = 12kHz = 12000 Hz.
కెపాసిటీ (C) = 10^-4
నిరోధము (R) = 40Ω
క్షమత్వ ప్రతిరోధం

కావున ఇది వివృత వలయంలాగ పనిచేస్తుంది.

ప్రశ్న 17.
LCR శ్రేణి వలయంలోని అనునాద పౌనఃపున్యానికి సమాన పౌనఃపున్యాన్ని జనకానికి ఉంచి, L,C, R మూడు ఘటకాలను సమాంతరంగా అమర్చినప్పుడు ఈ పౌనఃపున్యం వద్ద LCR సమాంతర వలయంలో మొత్తం విద్యుత్ ప్రవాహం కనిష్ఠం అని చూపండి. ఈ పౌనఃపున్యానికి అభ్యాసం 11 వలయంలో నిర్దేశించిన జనకానికి ప్రతి శాఖలో (ఘటకాలకు) విద్యుత్ ప్రవాహ rms విలువలను పొందండి.
సాధన:
సమాంతర సంధానంలో ఉన్నాయి, కాబట్టి

దీనర్థం \(\frac{1}{Z}\) = కనిష్టం
Z = గరిష్ఠం అయితే, విద్యుత్ ప్రవాహం కనిష్టం.
ప్రేరకం ద్వారా విద్యుత్ ప్రవాహం

మొత్తం విద్యుత్ ప్రవాహం = I_rms
= నిరోధంలో విద్యుత్ ప్రవాహం = 5.75A

ప్రశ్న 18.
ఒక 800mH ప్రేరకం, 60 µF కెపాసిటర్లను శ్రేణిలో 230, 50 Hz సరఫరాకి కలిపారు. వలయం నిరోధం ఉపేక్షించదగినంతగా ఉంది.
(a) విద్యుత్ ప్రవాహ కంపన పరిమితి, rms విలువలను పొందండి.
(b) ప్రతీ మూలకం కొనల మధ్య ఉండే పొటెన్షియల్ భేదం rms విలువలను పొందండి.
(c) ప్రేరకానికి బదిలీ అయ్యే సగటు సామర్థ్యం ఎంత?
(d) కెపాసిటర్కు బదిలీ అయ్యే సగటు సామర్థ్యం ఎంత?
(e) వలయం శోషించుకొనే మొత్తం సగటు సామర్థ్యం ఎంత? (సగటు అంటే ఒక పూర్తి చక్రానికి తీసుకొన్న సగటు).
సాధన:
ప్రేరకత్వం L = 80mH = 80 × 10^-3H
కెపాసిటీ C = 60µF = 60 × 10^-6F
V_rms = 230V

పౌనఃపున్యము (f) = 50Hz, నిరోధము (R) = 0
a) వలయం యొక్క అవరోధము

c) ప్రేరకానికి బదిలీ అయిన సగటు సామర్థ్యం
(P) = I_rms · V_rms· cos Φ
దశాభేదం 90°, కాబట్టి P = 0

d) కెపాసిటర్ బదిలీ అయిన సగటు సామర్థ్యం
(P) = I_rms · V_rms . cos Φ
దశాభేదం 90°, P = 0

e) వలయంలోని నిరోధం లేదు కాబట్టి మొత్తం సామర్థ్యం, ప్రేరకం మరియు కెపాసిటర్ సగటు సామర్థ్యాల మొత్తానికి సమానం. కావున వినియోగించిన మొత్తం సామర్ధ్యం శూన్యం.

ప్రశ్న 19.
అభ్యాసం 18లోని వలయం 15Ω నిరోధం కలిగి ఉందనుకోండి. వలయంలో ప్రతి ఘటకానికి బదిలీ అయ్యే సగటు సామర్థ్యాన్ని శోషించుకొనే మొత్తం సామర్థ్యాన్ని పొందండి.
సాధన:
r.m.s వోల్టేజి (V_rms) = 230V
నిరోధము (R) = 15Ω
పౌనఃపున్యము (f) = 50Hz
విద్యుత్ ప్రవాహము వోల్టేజి మధ్య దశాభేదం 90°
మొత్తం సామర్థ్యం (P_av) = V_rms . I_rms

ప్రశ్న 20.
L = 0.12 H, C = 480 nE, R = 23Ω లతో ఉన్న LCR శ్రేణి వలయం 230 V చర పౌనఃపున్యం ఉన్న సరఫరాకి కలిపారు.
a) విద్యుత్ ప్రవాహ కంపనపరిమితి గరిష్ఠమవడానికి జనక పౌనఃపున్యం ఎంత ఉండాలి? ఈ గరిష్ఠ విలువను పొందండి.
b) వలయ శోషణం చేసుకొనే సగటు సామర్థ్యం గరిష్ఠం అవడానికి జనక పౌనఃపున్యం ఎంత ఉండాలి? ఈ గరిష్ఠ సామర్థ్యం విలువ పొందండి.
c) జనకం యొక్క ఏ పౌనఃపున్యాలకు, వలయానికి సరఫరా అయిన సామర్థ్యం అనునాదం వద్ద ఉండే సామర్థ్యంలో సగం ఉంటుంది. ఈ పౌనఃపున్యాల వద్ద విద్యుత్ ప్రవాహ కంపనపరిమితి ఎంత?
d) ఇచ్చిన వలయం Q-కారకం ఎంత?
సాధన:
ప్రేరకత్వం (L) = 0.12H,
కెపాసిటెన్స్ (C) = 480 = 480 × 10^-9 F
నిరోధము (R) = 23Ω
V_rms = 230V

a) అనునాదము వద్ద విద్యుత్ ప్రవాహం గరిష్ఠం
Z = R = 23Ω
గరిష్ఠ విద్యుత్ ప్రవాహం
(I₀) = √2 I_rms = 1.414 × 10 = 14.14A.
సహజ పౌనఃపున్యము వద్ద, విద్యుత్ ప్రవాహ కంపన పరిమితి గరిష్టం

b) అనునాదం వద్ద సగటు సామర్థ్యం గరిష్టం
P_av = I²_rms .R = 10 × 10 × 23 = 2300W.
R = 10 x 10 x 23 = 2300W.

c) బదిలీ అయిన సామర్థ్యం, అనునాదం వద్ద సామర్థ్యంలో సగం ఉంటుంది.

ప్రశ్న 21.
L = 3.0 H, C = 27µE, R = 10. 4Ω లు ఉన్న సాధన. Xc = 2xfc XL = 2xfL LCR శ్రేణి వలయం అనువాద పౌనఃపున్యం మరియు Q-కారకాలను పొందండి. వలయం యొక్క అర్ధ గరిష్ఠం వద్ద మొత్తం వెడల్పు (FWHM) ను తగ్గించడం ద్వారా వలయం అనునాద నైశిత్యాన్ని రెండు రెట్లు మెరుగుపరచాలని ఆశించడమైనది. తగిన విధానాన్ని సూచించండి.
సాదన:
ప్రేరకత్వం (L) = 3H
కెపాసిటీ (C) = 27 × 10^-6F
నిరోధము (R) = 7.4Ω
అనునాద పౌనఃపున్యము

ప్రశ్న 22.
క్రింది ప్రశ్నలకు సమాధానాలు ఇవ్వండి.
a) ఏదైనా acవలయంలో అనువర్తిత తాక్షణిక వోల్టేజి శ్రేణి వలయంలోని మూలకాల కొనల మధ్య ఉండే వోల్టేజిల బీజీయ మొత్తానికి సమానమవుతుందా? rms వోల్టేజి విషయంలో ఇది నిజమవుతుందా?
సాధన:
అవును. అనువర్తిత వోల్టేజి, వలయంలో శ్రేణిలో కలిపిన మూలకాల తాత్కాల వోల్టేజిల మొత్తానికి సమానం. లేదు r.m.s వోల్టేజికి సంబంధించి సరియైనది. కాదు. అందుకు కారణం వలయంలో మూలకాల వద్ద కొంత దశాభేదం ఉంటుంది.

b) ప్రేరక తీగచుట్ట ప్రాథమిక వలయంలో ఒక కెపాసిటర్ను ఉపయోగించారు.
సాధన:
ప్రేరక తీగచుట్టలో ప్రాథమిక వలయంలో కెపాసిటర్ను కలపాలి. కారణం వలయాన్ని ఛేదించినప్పుడు అధిక ప్రేరిత వోల్టేజి, కెపాసిటర్ను ఆవేశితం చేస్తారు. కాబట్టి ఎలాంటి నష్టం లేకుండా ఉంటుంది.

c) అనువర్తిత వోల్టేజి సంకేతం (సిగ్నల్) dc వోల్టేజి, అధిక పౌనఃపున్యంతో ఉన్న ac వోల్టేజిల అధ్యారోపణంగా ఉంది. వలయంలో ప్రేరకం, కెపాసిటర్లు శ్రేణిలో ఉన్నాయి. కెపాసిటర్ కొనల మధ్య de సంకేతం, ప్రేరకం కొనల మధ్య ac సంకేతం కనిపిస్తుందని చూపండి.
సాధన:

DC వద్ద కెపాసిటర్ ఆగిపోతుంది. C వద్ద AC ఉంటుంది.
అధిక AC పౌనఃపున్యం వద్ద, X_L బాగా అధికం, X_c = 0

d) ఒక బల్బుకు శ్రేణిలో ఉన్న చౌక్ కాయిల్ dc లైనుకు కలిపారు. అప్పుడు బల్బు చాలా ప్రకాశవంతంగా వెలిగినట్లు అనిపించింది. చౌక్ లో ఇనుప కోర్ను ఉంచినప్పుడు బల్బు ప్రకాశవంతతలో ఎటువంటి మార్పు రాలేదు. ఒకవేళ ac లైనుకు కలిపితే అనురూప పరిశీలనలను ఊహించండి.
సాధన:
చోక్ తీగచుట్టకు dc ని కలిపితే, వెలుగులో ఎలాంటి మార్పుండదు.
కారణం f = 0, X_L = 0.
Ac చోక్ అవరోధం (X) ను కలిగిస్తుంది. కాబట్టి డిమ్ గా వెలుగుతుంది.
ఇనుప కోర్ను చొప్పిస్తే అయస్కాంత క్షేత్రం పెరుగుతుంది. కాబట్టి ప్రేరకత పెరుగుతుంది.
BA = LI = Φ
L ∝ B
కాబట్టి X_L పెరిగి, బల్బు వెలుగు తీవ్రత తగ్గుతుంది.

e) ac మొయిన్స్కు కలిపిన ప్రతిదీప్తి బల్బులలో (fluore- scent bulbs) చౌక్ కాయిల్ బదులుగా సాధారణ నిరోధకాన్ని ఎందుకు ఉపయోగించరాదు?
సాధన:
నిరోధం బదులు చోక్ తీగచుట్టను ఉపయోగిస్తే నిరోధం వద్ద సామర్ధ్యం నష్టం గరిష్ఠం. చోక్ వద్ద సామర్థ్య నష్టం శూన్యం
నిరోధానికి Φ = 0
P = I_rms . V_rms . cos 0°
= I_rms . V_rms = గరిష్ఠం
ప్రేరకానికి Φ = 90°
P = I_rms . V_rms. cos 90° = 0

ప్రశ్న 23.
ఒక విద్యుత్ ప్రసార లైన్ 2300 V వద్ద నివేశ సామర్థ్యాన్ని 4000ప్రాథమిక తీగచుట్లు ఉన్న అవరోహణ పరివర్తకానికి అందిస్తుంది.230Vవద్ద నిర్గమ సామర్థ్యాన్ని పొందడానికి గౌణ తీగచుట్ల సంఖ్య ఎంత ఉండాలి?
సాధన:
ప్రాథమిక వోల్టేజి (V_P) = 2300V
N_P = 4000 చుట్లు
గౌణ వోల్టేజి (V_S) = 230v.

గౌణ చుట్ల సంఖ్య (N_S) = 400

ప్రశ్న 24.
జల విద్యుచ్ఛక్తి కేంద్రంలో నీటి పీడన స్తంభం 300 m ఎత్తులో ఉన్నది. అందుబాటులో ఉన్న నీటి ప్రవాహం 100 m³s^-1 గా ఉంది. టర్బైన్ జనరేటర్ దక్షత 60% అయితే కేంద్రం నుంచి అందుబాటులో ఉండే విద్యుత్ సామర్థ్యాన్ని అంచనా వేయండి. (g = 9.8 ms^-2).
సాధన:
నీటియొక్క ఎత్తు (h) = 300m
నీటి ప్రవాహ రేటు (V) = 100m³/s
దక్షత η = 60%
g = 9.8m/².

ప్రశ్న 25.
440 V వద్ద విద్యుచ్ఛక్తిని ఉత్పత్తి చేసే విద్యుదుత్పాదన కేంద్రం నుంచి 15 km దూరంలో ఉన్న చిన్న పట్టణంలో 220 V చొప్పున 800 kW విద్యుచ్ఛక్తి వినియోగ డిమాండ్ ఉన్నది. రెండు తీగలు ఉన్న ప్రసారిత వ్యవస్థ 1kmకి 0.5Ω చొప్పున నిరోధాన్ని కలిగి ఉంది. 4000-220 V అవరోహణ పరివర్తకాన్ని కలిగి ఉన్న ఉపకేంద్రం (సబ్-స్టేషన్) నుంచి తీగల ద్వారా పట్టణానికి విద్యుచ్ఛక్తి సరఫరా అవుతుంది.
(a) తీగలలో ఉష్ణరూపంలో నష్టపోయే శక్తిని అంచనా వేయండి.
(b) లీకేజి వల్ల నష్టం ఉపేక్షించదగినంతగా ఉన్నదనుకొని, విద్యుత్ కేంద్రం ఎంత శక్తిని సరఫరా చేయాల్సి ఉంటుంది?
(c) విద్యుత్ కేంద్రం వద్ద ఆరోహణ పరివర్తకం అభి లక్షణాలను తెలపండి.
సాధన:
విద్యుత్ ప్లాంట్లో జనించిన సామర్థ్యం = 800 kW
V = 220V
= 15km,
జనించిన వోల్టేజి = 440V
నిరోధము/పొడవు = 0.5Ω/km
ప్రాథమిక వోల్టేజి (V_p) = 4000V
గౌణ వోల్టేజి (V_s) = 220V

(a) సామర్థ్యం – I_p . V_p
800 × 1000 = I_p × 4000
I_p = 200A
ఉష్ణరూపంలో కోల్పోయిన సామర్థ్య నష్టం
= (I_p)² × నిరోధం × 2
= (200)² × 0.5 × 15 × 2
= 60 × 10⁴ W = 600KW

(b) లీకేజీ ద్వారా సామర్థ్య నష్టం లేకపోతే
= 800 + 600 = 1400KW

(c) లైన్ వద్ద వోల్టేజి = I_p. R (2 లైన్లు)
= 200 × 0.5 × 15 × 2
= 3000V

ప్రసారం నుండి వోల్టేజి = 3000 + 4000 = 7000
స్టెప్-అప్ ట్రాన్స్ ఫార్మర్ 440 7000 వరకు అవసరం.

ప్రశ్న 26.
పైన ఇచ్చిన అభ్యాసంలోని ట్రాన్స్ఫార్మర్ బదులు 40,000-220V అవరోహణ పరివర్తకాన్ని ఉపయోగించి అన్ని అభ్యాసాలను చేయండి. (ఇంతకు ముందు లాగా లీకేజి నష్టాలను ఉపేక్షించండి. లీకేజి నష్టాలను ఉపేక్షించడమనే ఊహన ఇక ఏ మాత్రం మంచిది కాదు. ఎందుకంటే ప్రసారం చాలా అధిక వోల్టేజితో కూడుకొని ఉంది.) అయితే, అధిక వోల్టేజి ప్రసారం ఎందుకు ప్రాధాన్యం గలదో వివరించండి.
సాధన:
ప్రాధమిక వోల్టేజి (V_P) = 40,000V
ప్రాథమిక విద్యుత్ ప్రవాహం = I_p
∴ V_P . I_P = P
800 × 1000 = 40000 × I_P
I_P = 20A

a) సామర్ధ్య నష్టం = (I_P)² × R (2 లైన్లు)
= (20)² × 2 × 0.5 × 15
= 6000W = 6KW

b) సామర్థ్యం = 800 + 6 =806KW
వోల్టేజి = I_P.R (2 లైన్లు)
= 20 × 2 × 0.5 × 15
= 300V

ప్రసార వోల్టేజి = 40000 + 300 = 40300V
ప్లాంట్ వద్ద స్టెప్ అప్ ట్రాన్స్ఫార్మర్
= 440 – 40300V అవసరం
= \(\frac{6}{100}\) × 100 = 0.74%

అల్ప వోల్టేజి వద్ద సామర్ధ్య నష్టం
\(\frac{600}{1400}\) × 100 = 42.8%

కాబట్టి అధిక వోల్టేజి వద్ద సామర్ధ్య నష్టం కనిష్టం. అందువలన ప్రసారం చేయుటకు అధిక వోల్టేజికి ప్రాధాన్యం ఇవ్వబడును.

సాధించిన సమస్యలు Textual Examples

ప్రశ్న 1.
220.V సరఫరాకు ఒక విద్యుత్ బల్బు రేటింగ్ 100W అయితే (a) బల్బు నిరోధం, (b) జనక శిఖర వోల్టేజి, (c) బల్బు ద్వారా ప్రవహించే rms విద్యుత్ ప్రవాహాలను చెక్కించండి. AP (Mar. ’15)
సాధన:
P = 100 W, V = 220 V అని ఇచ్చారు.

ప్రశ్న 2.
25.0mH ప్రేరకత్వం ఉన్న ఒక శుద్ధ ప్రేరకాన్ని 220V ఉన్న జనకానికి కలిపారు. జనకం పౌనఃపున్యం 50Hz అయితే వలయంలో ప్రేరకత్వ ప్రతిరోధం, rms విద్యుత్ ప్రవాహాన్ని కనుక్కోండి.
సాధన:
ప్రేరకత్వ ప్రతిరోధం X_L = 2πνL
= 2 × 3.14 × 50 × 25 × 10^-3 = 7.85Ω
వలయంలో rms విద్యుత్ ప్రవాహం

ప్రశ్న 3.
ఒక బల్బును కెపాసిటర్కు శ్రేణిలో కలిపారు. dc, ac సంధానాలకు ఏమిజరుగుతుందో ఆ పరిశీలనను తీసుకోండి. కెపాసిటర్ క్షమత్వాన్ని (కెపాసిటెన్స్) తగ్గించి నప్పుడు ప్రతీ సందర్భంలో ఏమి జరుగుతుంది?
సాధన:
కెపాసిటర్కు dc జనకాన్ని కలిపినప్పుడు కెసాసిటర్ ఆవేశితమవుతుంది. ఆవేశితమయిన తరవాత వలయంలో ఎటువంటి విద్యుత్ ప్రవాహం ఉండదు కాబట్టి బల్బు వెలగదు. క్షమత్వాన్ని తగ్గించినప్పటికీ ఎటువంటి మార్పు ఉండదు. ac జనకానికి కెపాసిటర్ని కలిపినప్పుడు కెపాసిటర్ క్షమత్వ ప్రతిరోధం (1/ωC)ని అందిస్తుంది. అందువల్ల వలయంలో విద్యుత్ ప్రవాహం ఉంటుంది. తద్వారా బల్బు వెలుగుతుంది. కెపాసిటన్స్ C విలువ తగ్గిస్తూ పోయేకొద్దీ ప్రతిరోధం పెరుగుతుంది. అందువల్ల బల్బు ఇంతకుముందు కంటే తక్కువ ప్రకాశవంతంగా వెలుగుతుంది.

ప్రశ్న 4.
220, 50 Hz ఉన్న ఏకాంతర జనకానికి 15.0 µF కెపాసిటర్ను కలిపారు. అయితే వలయంలో క్షమత్వ ప్రతిరోధం rms, శిఖర విద్యుత్ ప్రవాహాలను కనుక్కోండి. పౌనఃపున్యాన్ని రెట్టింపు చేసినట్లైతే క్షమత్వ ప్రతిరోధం, విద్యుత్ ప్రవాహాలు ఏమవుతాయి?
సాధన:

ఈ విద్యుత్ ప్రవాహం + 1.47A – 1.47 Aల మధ్య డోలనాలు చేస్తూ వోల్టేజి కంటే 7/2 రేడియన్లు ముందు ఉంటుంది.

ఒకవేళ పౌనఃపున్యం రెట్టింపయితే క్షమత్వ ప్రతిరోధం సగం అవుతుంది. తదనుగుణంగా విద్యుత్ ప్రవాహం రెట్టింపు అవుతుంది.

ప్రశ్న 5.
ఒక విద్యుత్ బల్బు, ఒక వివృత తీగచుట్ట ఉన్న ప్రేరకాన్ని కీ ద్వారా ఏకాంతర జనకానికి పటంలో చూపిన విధంగా కలిపారు.

కీ ని మూసి కొంత సమయం తరవాత ప్రేరకంలోపల ఒక ఇనుప కడ్డీని అమర్చారు. అప్పుడు విద్యుత్ బల్బు దీప్తి (a) పెరుగుతుంది; (b) తగ్గుతుంది; (c) మారదు. తగిన కారణాలతో మీ సమాధానాన్ని ఇవ్వండి.
సాధన:
తీగచుట్టలో ఇనుప కడ్డీని అమర్చగానే తీగచుట్టలోని అయస్కాంతక్షేత్రం ఇనుప కడ్డీని అయస్కాంతీకరిస్తుంది. తద్వారా లోపల అయస్కాంత క్షేత్రం పెరుగుతుంది. అందువల్ల తీగచుట్ట ప్రేరకత్వం పెరుగుతుంది. పర్యవసానంగా తీగచుట్ట ప్రేరకత్వ ప్రతిరోధం పెరుగుతుంది. ఫలితంగా అనువర్తిత ఏకాంతర వోల్టేజిలో అధికభాగం ప్రేరకం కొనల మధ్యే ఉండి వోల్టేజిలోని కొంత భాగం మాత్రమే బల్బుకు అందచేయబడుతుంది. అందువల్ల బల్బు దీప్తి తగ్గుతుంది.

ప్రశ్న 6.
220V, 50 Hz ఏకాంతర వోల్టేజి జనకానికి శ్రేణిలో 200 Ω నిరోధకం, 15.0 µF కెపాసిటర్ను కలిపారు. (a) వలయంలోని విద్యుత్ ప్రవాహాన్ని లెక్కించండి; (b) కెపాసిటర్, నిరోధకాల చివర వోల్టేజి (rms) ని లెక్కించండి. ఈ వోల్టేజిల జీయ మొత్తం జనక వోల్టేజి కంటే ఎక్కువా? అయితే విరోధాభాసాన్ని (paradox) విశ్లేషించండి.
సాధన:
R = 2002. C = 15.0pF = 15.0 × 10^-6F
V = 220V, ν = 50Hz అని ఇచ్చారు.

(a) విద్యుత్ ప్రవాహాన్ని లెక్కించడానికి వలయం అవరోధం తెలుసుకోవాలి.

(b) వలయం అంతటా విద్యుత్ ప్రవాహం సమానం కాబట్టి,
V_R = IR = (0.755 A) (200Ω) = 151V
V_C = IX_C = (0.755A) (212.3Ω) = 160.3V.

V_R, V_C ల బీజీయ మొత్తం 311.3 V. ఈ విలువ జనక వోల్టేజి 220 V కంటే ఎక్కువ. ఈ విరోధాభాసాన్ని ఏవిధంగా విశ్లేషణ చేయవచ్చు ? పాఠంలో ఇదివరకు చదివిన ప్రకారం రెండు వోల్టేజిలు 90° తో దశలో విభేధిస్తున్నాయి. రెండు వోల్టేజిలు ఒకే దశలో లేవు కాబట్టి వీటిని సాధారణ సంఖ్యల లాగా కలపలేం. కాబట్టి రెండు వోల్టేజీల మొత్తాన్ని పైథాగరియన్ సిద్ధాంతం ఉపయోగించి పొందవచ్చు.
V_{R + C} = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2}\) = 220V
కాబట్టి, రెండు వోల్టేజిల మధ్య దశాభేదాన్ని తగినట్లుగా లెక్కలోకి తీసుకొన్నట్లయితే నిరోధకం, కెపాసిటర్ చివరల ఏర్పడే వోల్టేజిల మొత్తం జనక వోల్టేజికి సమానం అవుతుంది.

ప్రశ్న 7.
a) విద్యుత్ శక్తిని రవాణా చేయడానికి ఉపయోగించే వలయాలకు తక్కువ సామర్థ్యకారకం ఉంటే ప్రసారంలో సామర్థ్య నష్టం అధికం, వివరించండి.
b) వలయంలో తగిన క్షమత్వం ఉన్న కెపాసిటర్ను ఉపయోగించితరచుగా సామర్థ్యకారకాన్ని మెరుగు చేస్తారు. వివరించండి.
సాధన:
a) P = IV cosΦ అని మనకు తెలుసు. ఇందులో cos ని సామర్థ్యకారకం అంటారు. ఇచ్చిన వోల్టేజి వద్ద సామర్ధ్యాన్ని సరఫరా చేయాలంటే, coso తక్కువగా ఉన్నట్లయితే, దానికి అనుగుణంగా విద్యుత్ ప్రవాహాన్ని పెంచాలి. కాని ఇలా చేయడం వల్ల సరఫరాలో ఎక్కువ సామర్థ్య నష్టం (I²R) జరుగుతుంది.

b) వలయంలో విద్యుత్ ప్రవాహం వోల్టేజి కంటే కోణంతో వెనుకబడి ఉన్నదనుకోండి. అప్పుడు సామర్థ్యకారకం cos Φ = R/Z.

Z ను Rకి సమీపింపచేసి సామర్థ్య కారకాన్ని (1కి సమీపించే విధంగా) మెరుగుపరచవచ్చు. దీన్ని ఏవిధంగా సాధించవచ్చో పటంలో చూపిన ఫేజర్ పటం ఆధారంగా అర్ధం చేసుకోవచ్చు. ఇప్పుడు I ని రెండు అంశాలుగా విభజిద్దాం. ఇందులో I_P అనేది అనువర్తిత వోల్టేజి V వెంబడి ఉండే అంశ, I_q అనేది అనువర్తిత వోల్టేజికి లంబదిశలో ఉండే అంశ. 10.7 సెక్షన్ లో చదివిన ప్రకారం I_q ని వాట్ లెస్ అంశ అంటారు. ఎందుకంటే ఈ అంశానికి అనురూపంగా విద్యుత్ ప్రవాహంలో ఎటువంటి సామర్థ్య నష్టం ఉండదు. అనువర్తిత వోల్టేజి I_P అంశదిశ ఉండటంవల్ల దీన్ని సామర్థ్య అంశం అంటారు. వలయంలోని సామర్థ్య నష్టాన్ని ఈ అంశ ఇస్తుంది.

ఈ విశ్లేషణ నుంచి స్పష్టమవుతున్నదేమంటే, సామర్థ్యకారకాన్ని మెరుగు పరచాలంటే వెనుకబడిన వాట్స్ విద్యుత్ ప్రవాహం I_q ని, దీనితో సమానంగా ఉండి, ముందుండే వాట్లెస్ విద్యుత్ ప్రవాహం I’_q తో తటస్థపరచాలి. దీనికోసం తగిన విలువ కలిగిన కెపాసిటర్ను సమాంతరంగా ఏర్పాటుచేయడం వల్లI_q I’_q లు ఒకదానికొకటి రద్దయి ప్రభావాత్మక సామర్థ్యం P విలువ I_p V అవుతుంది.

ప్రశ్న 8.
50 Hz పౌనఃపున్యం, జావక్రీయ శిఖర వోల్టేజి 283V ఉన్న జ్యావక్రీయ వోల్టేజిని R = 3Ω L = 25.48 mH. and C = 796µF విలువ ఉన్న ఒక LCR శ్రేణి వలయానికి అనువర్తించారు. (a) వలయం అవరోధం; (b) జనక వోల్టేజి, విద్యుత్ ప్రవాహాల మధ్య దశాభేదం; (9వలయంలో దుర్వ్యయమయిన సామర్థ్యం: (బి) సామర్ధ్య కారకాలను కనుక్కోండి.
సాధన:
a) వలయం అవరోధాన్ని కనుక్కోవడానికి ముందుగా X_L X_C లను లెక్కించాలి.
X_L = 2πνL
= 2 × 3.14 × 50 × 25.48 × 10^-3Ω = 8Ω

Φ రుణాత్మకం కాబట్టి జనక వోల్టేజి కంటే విద్యుత్ ప్రవాహం వెనుకబడి ఉంటుంది.
(c) వలయంలో దుర్వ్యయమయిన సామర్ధ్యం

d) సామర్థ్య కారకం = cos o = cos 53.1° = 0.6.

ప్రశ్న 9.
ఇంతకు ముందు సమస్యలో జనక పౌనఃపున్యాన్ని మార్చగలిగినట్లయితే (a) అనునాదం సంభవించాలంటే “జనక పౌనఃపున్యం ఎంత ఉండాలి? (b) అనునాద నిబంధన వద్ద అవరోధం, విద్యుత్ ప్రవాహం, దుర్వ్యయమైన సామర్థ్యాన్ని లెక్కించండి.
సాధన:
(a) అనునాదం జరిగే పౌనఃపున్యం

b) అనునాద నిబంధన వద్ద అవరోధం, నిరోధానికి సమానం. అంటే Z = R = 3Ω
అనునాదం వద్ద rms విద్యుత్ ప్రవాహం

అనునాదం వద్ద దుర్వ్యయమైన సామర్థ్యం
P = I² × R = (66.7)² × 3 = 13.35 kW
ఉదాహరణ 8లో కంటే ఈ సందర్భంలో ఎక్కువ సామర్థ్యం దుర్వ్యయం అవడాన్ని మీరు గమనించగలరు.

ప్రశ్న 10.
ఒక విమానాశ్రయంలో భద్రతా కారణాల దృష్ట్యా ఒక మనిషిని లోహ శోధక (metal detector) ద్వారంలో నుంచి నడిచేట్లు చేస్తారు. ఒకవేళ అతడు/ఆమె ఏదైనా లోహంతో తయారైన వస్తువులను కలిగి ఉన్నట్లయితే. లోహ శోధకం శబ్దాన్ని ఉద్గారం చేస్తుంది. లోహ శోధకం ఏ సూత్రంపై ఆధారపడి పనిచేస్తుంది?
సాధన:
ఏకాంతర వలయాల్లో అనునాదం అనే సూత్రంపై ఆధారపడి లోహ శోధకం పనిచేస్తుంది మీరు ఒక లోహ శోధకం ద్వారా నడుస్తున్నారంటే నిజానికి మీరు అనేక చుట్లు ఉన్న తీగచుట్టలో నుంచి నడుస్తున్నారని అర్థం. ఈ తీగచుట్ట ఒక కెపాసిటర్కు కలపబడి ఉంటుంది.

కాబట్టి వలయం అనునాదంలో ఉంటుంది. మీ జేబులో లోహాన్ని కలిగి లోహ శోధక ద్వారంలో నుంచి నచినట్లయితే వలయం అవరోధం మారడంవల్ల వలయంలోని విద్యుత్ ప్రవాహంలో చెప్పుకోదగినంత మార్పు కలుగుతుంది. విద్యుత్ ప్రవాహంలోని ఈ మార్పు గుర్తించబడి వలయం ఒక శబ్దాన్ని అలారం లాగా ఉద్గారం చేసేలా చేస్తుంది.

ప్రశ్న 11.
LC వలయంలోని స్వేచ్ఛా డోలనాలలో ఒక సమయం వద్ద కెపాసిటర్, ప్రేరకాలలో నిల్వ ఉండే శక్తుల మొత్తం స్థిరం అని చూపండి.
సాధన:
కెపాసిటర్పై తొలి ఆవేశం q₀ అనుకోండి. ఆవేశిత కెపాసిటర్ L ప్రేరకత్వం ఉన్న ప్రేరకానికి కలిపా మనుకోండి. సెక్షన్ 8లో చదివిన విధంగా ఈ LC వలయం ω(2πν = \(\frac{1}{\sqrt{LC}}\)) పౌనఃపున్యంతో కలపనాలు కొనసాగిస్తుంది.

ఏదైనా సమయం t వద్ద కెపాసిటర్పై ఉన్న ఆవేశం q విద్యుత్ ప్రవాహం లను
q(t) = q₀ cos ωt,
i(t) = -q₀ sin ωt గా రాయవచ్చు.
t సమయంవద్ద కెపాసిటర్లో నిల్వ ఉండే శక్తి,

q₀, Cలు కాలంపై ఆధారపడవు కాబట్టి మొత్తం శక్తి స్థిరం. ఈ విలువ కెపాసిటర్ తొలి శక్తికి సమానం అని గమనించగలరు. ఎందుకిలా అవుతుంది? ఆలోచించండి.

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Quadratic Expressions Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 1.
Form quadratic equation whose root 7 ± 2\(\sqrt{5}\) (Mar. ’11, ’05)
Solution:
α + β = 7 + 2\(\sqrt{5}\) + 7 – 2\(\sqrt{5}\) = 14
αβ = (7 + 2\(\sqrt{5}\)) (7 – 2\(\sqrt{5}\)) = 49 – 20 = 29
The required equation is
x² – (α + β)x + αβ = 0
x² – 14x + 29 = 0

Question 2.
Form quadratic equation whose root -3 ± 5i. (Mar. ’07)
Solution:
α + β = -3 + 5i – 3 – 51 = -6
αβ = (-3 + 5i)(-3 – 5i)
= 9 + 25 = 34
The required equation is
x² – (α + β)x + αβ = 0
x² + 6x + 34 = 0

Question 3.
For what values of x, 15 + 4x – 3x² expressions are negative? (AP Mar. ’15)
Solution:
The roots of 15 + 4x – 3x² = 0 are
\(\frac{-4 \pm \sqrt{16+180}}{-6}\) i.e., \(\frac{-5}{3}\), 3
∴ The expression 15 + 4x – 3x² is negative if
x < \(\frac{-5}{3}\) or x > 31 ∵ a = -3 < 0

Question 4.
If α, β are the roots of the equation ax² + bx + c = 0, find the value \(\frac{1}{\alpha^{2}}\) + \(\frac{1}{\beta^{2}}\) expressions in terms of a, b, c. (AP & TS Mar. ‘16, 08)
Solution:

Question 5.
Form quadratic equation whose root
\(\frac{p-q}{p+q}\), \(\frac{-p+q}{p-q}\), (p ≠ ±q) (Mar. ’06)
Solution:
α + β = \(\frac{p-q}{p+q}\) – \(\frac{p+q}{p-q}\)

Question 6.
Find the values of m for which the following equations have equal roots?
i) x² – 15 – m(2x – 8) = 0. (AP Mar. ’17) (TS Mar. ’15 13)
Solution:
Given equation is x² – 15 – m(2x – 8) = 0
x² – 2mx + 8m – 15 = 0
a = 1, b = -2m, c = 8m – 15
b² – 4ac = (-2m)² – 4(1) (8m – 15)
= 4m² – 32m + 60
= 4(m² – 8m + 15)
= 4(m – 3)(m – 5)
Hint: If the equation ax² + bx + c = 0 has equal roots then its discriminant is zero.
∵ The roots are equal b² – 4ac = 0
⇒ 4(m – 3) (m – 5) = 0
⇒ m – 3 = 0 or m – 5 = 0
∴ m = 3 or 5

Question 7.
(m + 1)x² + 2(m + 3)x + (m + 8) = 0.
Solution:
Given equation is
(m + 1)x² + 2(m + 3)x + (m + 8) = 0
a = m + 1, b = 2(m + 3), c = m + 8
b² – 4ac = (2(m + 3)]² – 4(m + 1) (m + 8)]
= 4(m² + 6m + 9) – 4(m² + 8m + m + 8)
= 4m² + 24m + 36 – 4m² – 36 m – 32
= -12m + 4
= -4(3m – 1)
∵ The roots are equal ⇒ b² – 4ac = 0
⇒ -4(3m – 1) = 0
⇒ 3m – 1 = 0
⇒ 3m = 1
∴ m = \(\frac{1}{3}\)

Question 8.
If x is real, prove that \(\frac{x}{x^{2}-5 x+9}\) lies between 1 and \(\frac{-1}{11}\). (Mar. ‘14, 13, ‘08, ‘02; May 11, ‘07)
Solution:
Let y = \(\frac{x}{x^{2}-5 x+9}\) ⇒ yx² – 5yx + 9y = x
⇒ yx² + (-5y – 1)x + 9y = 0
x ∈ R ⇒ (-5y – 1)² – 4y(9y) ≥ 0
⇒ 25y² + 1 + 10y – 36y² ≥ 0
⇒ -11y² + 10y + 1 ≥ 0 —— (1)
⇒ -11y² + 10y + 1 = 0 ⇒ -11y² + 11y – y + 1 = 0
⇒ 11y(-y + 1) + 1(-y + 1) = 0
⇒ (-y + 1)(11y + 1) = 0 ⇒ y = 1, \(\frac{-1}{11}\)
-11y² + 10y + 1 ≥ 0
∴ y² coeff is be, but the exp is ≥ 0 from (1)
⇒ \(\frac{-1}{11}\) ≤ y ≤ 1 ⇒ y lies between 1 and \(\frac{-1}{11}\)

Question 9.
Theorem : The roots of ax² + bx + c = 0 are

(Mar. ’02)
Proof:
Given quadratic equation is ax² + bx + c = 0
⇒ 4a(ax² + bx + c) = 0
⇒ 4a²x² + 4abx + 4ac = 0
⇒ (2ax)² + 2(2ax) (b) + b² – b² + 4ac = 0
⇒ (2ax + b)² = b² – 4ac

Question 10.
Find the maximum value of the function \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R.
Solution:
Let y = \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\)
⇒ yx² + 2yx + 3y = x² + 14x + 9
⇒ (y – 1)x² + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)]² – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y² – 14y + 49) [(3y² – 12y + 9)] ≥ 0
⇒ -2y² – 2y + 40 ≥ 0
⇒ y² + y – 20 ≤ 0
⇒ (y + 5) (y -4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
⇒ Maximum value of y = 4
∴ Maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R is 4.

Question 11.
If x² – 6x + 5 = 0 and x² – 12x + p = 0 have a common root, then find p. (TS Mar. ’17)
Solution:
Given x² – 6x + 5 = 0, x² – 12x + p = 0 have a common root.
If α is the common root then
α² -6α + 5 = 0, α² – 12α + p = 0
α² – 6α + 5 = 0 ⇒ (α – 1) (α – 5) = 0
⇒ α = 1 or 5
If α = 1 then α² – 12α + p = 0
⇒ 1 – 12 + p = 0 ⇒ p = 11
If α = 5 then α² – 12α + p = 0
⇒ 25 – 60 + p = 0 ⇒ p = 35
∴ p = 11 or 35

Question 12.
If x₁, x₂ are the roots of the quadratic equation ax₂ + bx + c = 0 and c ≠ 0, find the value of (ax₁ + b)^-2 + (ax₂ + b)_-2 in terms of a, b, c. (TS Mar. ’17)
Solution:
x₁, x₂ are the roots of the equation
ax² + bx + c = 0

Question 13.
Prove that \(\frac{1}{3 x+1}\) + \(\frac{1}{x+1}\) – \(\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4, if x is real. (AP & TS Mar. ’16, AP Mar. 15, ’11) (AP Mar. ‘17)
Solution:

⇒ 3yx² + 4yx + y = 4x + 1
⇒ 3yx² + (4y – 4) x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)² – 4(3y)(y – 1) ≥ 0
⇒ 16y² + 16 – 32y – 12y² + 12y ≥ 0
⇒ 4y² – 20y + 16 ≥ 0
4y² – 20y + 16 = 0
⇒ y² – 5y + 4 = 0
⇒ (y – 1)(y – 4) = 0
⇒ y = 1, 4
⇒ 4y² – 20y + 16 ≥ 0.
⇒ y ≤ 1 or y ≥ 4
⇒ y does not lie between 1 and 4
Since y² coeff is the and exp ≥ 0.

Question 14.
Solve the following equations: (T.S Mar. ‘15)
2x⁴ + x³ – 11x² + x + 2 = 0
Solution:
Dividing by x²

Substituting in (1)
2(a² – 2) + a – 11 = 0
⇒ 2a² – 4 + a – 11 = 0
⇒ 2a² + a – 15 = 0
⇒ (a + 3) (2a – 5) = 0
⇒ a + 3 = 0 or 2a – 5 = 0
a = -3 or 2a = 5 a = \(\frac{5}{2}\)

Case(i) : a = -3
x + \(\frac{1}{x}\) = -3
x² + 1 = -3
x² + 3x + 1 = 0
x = \(\frac{-3 \pm \sqrt{9-4}}{2}\) = \(\frac{-3 \pm \sqrt{5}}{2}\)

Case (ii) : a = \(\frac{5}{2}\)
x + \(\frac{1}{x}\) = \(\frac{5}{2}\)
⇒ \(\frac{x^{2}+1}{x}\) = \(\frac{5}{2}\)
⇒ 2x² + 2 = 5x
⇒ 2x² – 5x + 2 = 0
⇒ (2x – 1) (x – 2) = 0
⇒ 2x – 1 = 0 or x – 2 = 0
⇒ x = \(\frac{1}{2}\), 2
∴ The roots are \(\frac{1}{2}\), 2, \(\frac{-3 \pm \sqrt{5}}{2}\)

Question 15.
For what values of x, the following expressions are positive? (May ’11)
i) x² – 5x + 6
Solution:
x² – 5x + 6 = (x – 2) (x – 3)
Roots of x² – 5x + 6 = 0 are 2, 3 which are real.
The expression x² – 5x + 6 is positive if x < 2 or x > 3, ∴ a = 1 > 0.

ii) 3x² + 4x + 4
Solution:
Here a = 3, b = 4, c = 4,
Δ = b² – 4ac
= 16 – 48 = -32 < 0
∴ 3x² + 4x + 4 is positive ∀ x ∈ R,
∵ a = 3 >0 and Δ < 0
Hint: ax² + bx + c and ‘a’ have same sign ∀ x ∈ R, if Δ < 0

iii) 4x – 5x² + 2
Solution:
Roots of 4x – 5x² + 2 = 0 are

iv) x² – 5x + 14
Solution:
Here a = 1, b = -5, c = 14,
Δ = b² – 4ac
= 25 – 56 = -31 < 0
∴ Δ < 0 ∵ a = 1 > 0 and Δ < 0
⇒ x² – 5x + 14 is positive ∀ x ∈ R.

Question 16.
Determine the range of the \(\frac{x^{2}+x+1}{x^{2}-x+1}\) expressions. (Mar ’04)
Solution:
Let y = \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
⇒ x²y – xy + y = x² + x + 1
⇒ x²y – xy + y – x² – x – 1 = 0
⇒ x²(y – 1) – x(y + 1) + (y – 1) = 0
x is real ⇒ b² – 4ac ≥ 0
⇒ (y + 1)² – 4(y – 1)² ≥ 0
⇒ (y + 1)² – (2y – 2)² ≥ 0
⇒ (y + 1 + 2y – 2)(y + 1 – 2y + 2) ≥ 0
⇒ (3y – 1)(-y + 3) ≥ 0
⇒ -(3y – 1) (y – 3) ≥ 0
a = coeff. of y² = -3 < 0., But
The expression ≥ 0
⇒ y lies between \(\frac{1}{3}\) and 3
∴ The range of \(\frac{x^{2}+x+1}{x^{2}-x+1}\) is \(\left[\frac{1}{3}, 3\right]\)

Question 17.
Theorem : Let α, β be the real roots of ax² + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < β ⇒ ax² + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax² + bx + c and ‘a’ have the same sign. (Apr. ’96, ’93)
Proof:
α, β are the roots of ax² + bx + c = 0
⇒ ax² + bx + c = a(x – α) (x – β)
⇒ \(\frac{a x^{2}+b x+c}{a}\) = (x – α) (x – β)

i) Suppose x ∈ R, α < x < β
α < x < β ⇒ x – α > 0, x – β < 0
⇒ (x – α) (x – β) < 0
⇒ ax² + bx + c, a have opposite signs.

ii) Suppose x ∈ R, x < α
x < α < 13
⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign.
Suppose x ∈ R, x > β
x > β > α
⇒ x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign,
∴ x ∈ R, x < α or x > β
⇒ ax² + bx + c and a have the same sign.

Question 18.
Theorem: Let f(x) = ax² + bx + c be a quadratic function. (Apr. ‘01)
i) If a > 0 then f(x) has minimum value at x = \(\frac{-b}{2 a}\) and the minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
ii) If a < 0 then f(x) has maximum value at x = \(\frac{-b}{2 a}\) and the maximum value \(\frac{4 a c-b^{2}}{4 a}\)
Proof:
ax² + bx + c =

≤ \(\frac{4 a c-b^{2}}{4 a}\),
when a < 0
∴ If a < 0, then \(\frac{4 a c-b^{2}}{4 a}\) is the maximum fór f when x = \(\frac{-b}{2 a}\)
Second Proof : f(x) = ax² + bx + c
⇒ f'(x) = 2ax + b
⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = \(-\frac{b}{2 a}\)
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = \(-\frac{b}{2 a}\).
Minimum of

If a < 0,
then f”(x) < 0 and hence ‘f’ has maximum value at x = \(-\frac{b}{2 a}\) maximum of

Question 19.
Theorem : The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\).
Proof:
Given quadratic equation is ax² + bx + c = 0
⇒ 4a (ax² + bx + c) = 0
⇒ 4a²x² + 4abx + 4ac = 0
⇒ (2ax)² + 2(2ax) (b) + b² – b² + 4ac = 0
⇒ (2ax + b)² = b² – 4ac
⇒ 2ax + b = ±\(\sqrt{b^{2}-4 a c}\)
⇒ 2ax = -b ± \(\sqrt{b^{2}-4 a c}\)
⇒ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
The roots of ax² + bx + c = 0 are
\(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

Question 20.
Theorem : Let a, b, c ∈ R and a ≠ 0. Then the roots of ax² + bx + c = 0 are non-real complex numbers if and only if ax² + bx + c and a have the same sign for all x ∈ R.
Proof :
The condition for the equation
ax² + bx + c = o to have non-real complex roots is b² – 4ac < 0, i.e., 4ac – b² > 0.

Hence 4ac – b² > 0, so that b² – 4ac < 0. Thus b² – 4ac < 0 if and only if ax² + bx + c and a have the same sign for all real x.

Question 21.
Theorem: If the roots of ax² + bx + c = 0 are real and equal to α = \(\frac{-b}{2 a}\); then for α ≠ x ∈ R, ax² + bx + c and ‘a’ have the same sign.
Proof:
The roots are equal ⇒ b² – 4ac = 0
⇒ 4ac – b² = 0

∴ For α ≠ x ∈ R, ax² + bx + c and a have the same sign.

Question 22.
Theorem : Let α, β be the real roots of ax² + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < 13 ⇒ ax² + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax² + bx + c and ‘a’ have the same sign.
Proof.
α, β are the roots of ax² + bx + c = 0
⇒ ax² + bx + c = a(x – α)(x – β)
⇒ \(\frac{a x^{2}+b x+c}{a}\) = (x – α)(x – β)

i) Suppose x ∈ R, α < x < β
α 0, x – β < 0
⇒ (x – α)(x – β) < 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) < 0
⇒ ax² + bx + c, a have opposite signs

ii) Suppose x ∈ R, x < α
x < α < β ⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign.
Suppose x ∈ R, x > 3
x > β > α ⇒ x – α > 0, x – β > 0
⇒ (x – α) (x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign,
x ∈ R, x < α or x > β
⇒ ax² + bx + c and a have the same sign

Question 23.
Theorem : Let f(x) = ax² + bx+ c be a quadratic function.
i) If a > 0 then f(x) has minimum value at x = \(\frac{-\mathbf{b}}{2 a}\) and the minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
ii) If a < 0 then f(x) has maximum value at x = \(\frac{-\mathbf{b}}{2 a}\) and the maximum value = \(\frac{4 a c-b^{2}}{4 a}\) (Apr. ’01)
Proof.
ax² + bx+ c =

Second Proof: f(x) = ax² + bx + c
⇒ f'(x) = 2ax + b ⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = \(-\frac{b}{2 a}\)
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = \(-\frac{b}{2 a}\)
Minimum of

If a < 0 then f”(x) < 0 and hence ‘f’ has maximum value at x = –\(\frac{b}{2 a}\)
maximum of

Question 24.
Theorem : A necessary and sufficient condition for the quadratic equations
a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 to have a common root is (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁) (b₁c₂ – b₂c₁).
Proof:
Necessity
Let α be a common root of the given equations.
Then a₁α² + b₁α + c₁ = 0 ——(1)
a₂α² + b₂α + c₂ = 0 —— (2)
On multiplying euqation (1) by a₂, equation (2) by a₁ and then subtracting the latter from the former, we get

On multiplying euqation (1) by b₂, equation (2) by b₁ and then subtracting the latter from the former, we get
α² (a₁b₂ – a₂b₁) = b₁c₂ – b₂c₁ —— (4)
On squaring both sides of equation (3) and using (4) we obtain


Therefore the given equations have the same roots.

Case (ii): a₁b₂ – a₂b₁ ≠ 0

Similarly we can prove that a₂α² + b₂α + c₂ = 0
Thus α is a common root of the given equations.

Question 25.
Find the roots of the equation 3x² + 2x – 5 = 0.
Solution:
The roots of the quadratic equation
ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Here a = 3, b = 2 and c = -5.
Therefore the roots of the given equation are

Hence 1 and –[/latex]\frac{5}{3}[/latex] are the roots of the given equation.
Another method
We can also obtain these roots in thè following way.
3x² + 2x – 5 = 3x² + 5x – 3x – 5
= x(3x + 5) -1 (3x + 5)
= (x – 1) (3x + 5)
= 3(x – 1) \(\left(x+\frac{5}{3}\right)\)
Since 1 and –[/latex]\frac{5}{3}[/latex] are the zeros of 3x² + 2x – 5, they are the roots of 3x² + 2x – 5 = 0.

Question 26.
Find the roots of the equation 4x² – 4x + 17 = 3x² – 10x – 17.
Solution:
Given equation can be rewritten as x² + 6x + 34 = 0. The roots of the quadratic equation
ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Here a = 1, b = 6 and c = 34
Therefore the roots of the given equation are

Hence the roots of the given equation are -3 + 5i and -3 – 5i

Question 27.
Find the roots of the equation
\(\sqrt{3}\)x² + 10x – 8\(\sqrt{3}\) = 0.
Solution:
Here a = \(\sqrt{3}\), b = 10, c = -8\(\sqrt{3}\)

Question 28.
Find the nature of the roots of 4x² – 20x + 25 = 0
Solution:
Here a =4, b = -20, c = 25
Hence Δ = b² – 4ac
= (-20)² – 4(4) (25) = 0
Since Δ is zero and a, b, c are real, the roots of the given equation are real and equal.

Question 29.
Find the nature of the roots of 3x² + 7x + 2 = 0
Solution:
Here a = 3, b = 7, c = 2
Hence Δ = b² – 4ac
= 49 – 4(3) (2)
= 49 – 24
= 25 = (5)² > 0
Since a, b, c are rational numbers and Δ = 5² is a perfect square and positive, the roots of the given equation are rational and unequal.

Question 30.
For what values of m, the equation x² – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?
Solution:
The given equation will have equal roots iff its discriminant is 0.
Here Δ = [(-2(1 + 3m)]² – 4(1) [7 + (3 + 2m)]
= 4(1 + 9m² + 6m) – 28(3 + 28m)
= 9m² – 8m – 20
= (m – 2)(9m + 10)
Hence Δ = 0 ⇔ m = 2, \(\frac{-10}{9} .\)

Question 31.
If α and β are the roots of ax² + bx + c = 0, find the value of α² + β² and α³ + β³ in terms of a, b, c.
Solution:
α, β are the roots of ax² + bx + c = 0

Question 32.
Show that the roots of the equation x² – 2px + p² – q² + 2qr – r² = 0 are rational, given that p, q, r are rational.
Solution:
Here a = 1, b = -2p, c = p² – q² + 2qr – r²
∆ = b² – 4ac
= (-2p)² – 4(1) (p² – q² + 2qr – r²)
= 4p² – 4p² + 4q² + 8qr + 4r²
= 4(q – r)².
∵ p, q, r are rational, the coefficient of the given equation are rational and is a square of a rational number 2(q – r).
∴ The roots of the given equation are rational.

Question 33.
Form a quadratic equation whose roots are 2\(\sqrt{3}\) – 5 and -2\(\sqrt{3}\) – 5.
Solution:
Let α = 2\(\sqrt{3}\) – 5 and β = -2\(\sqrt{3}\) – 5
Then α + β = (2\(\sqrt{3}\) – 5) + (-2\(\sqrt{3}\) – 5) = -10
and αβ = (2\(\sqrt{3}\) – 5) + (-2\(\sqrt{5}\) – 5)
= (-5)² – (-2\(\sqrt{3}\))²
= 25 – 4 × 3
= 13
The required quadratic equation is
x² – (α + β)x + αβ = 0
⇒ x² – (-10)x + 13 = 0
⇒ x² + 10x + 13 = 0

Question 34.
Find the quadratic equation, the sum of whose roots is 1 and the sum of squares of the roots is 13.
Solution:
Let α, β be the roots of a required quadratic equation.
Then α + β = 1 and α² + β² = 13
Now αβ = \(\frac{1}{2}\)[(α + β)² – (α² + β²)]
= \(\frac{1}{2}\)[(1)² – (13)]
Therefore x² – (α + β)x + αβ = 0
⇒ x² – (1)x + (-6) = 0
⇒ x² – x – 6 = 0

Question 35.
Let α and β be the roots of the quadratic equation ax² + bx + c = 0, c ≠ 0, then form the quadratic equation whose roots are \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\).
Solution:
From the given equation

Question 36.
Solve x^2/3 + x^1/3 – 2 = 0
Solution:
(x^1/3)² + (x^1/3) – 2 = 0
Let x^1/3 = a, a² + a – 2 = 0
⇒ (a +2)(a – 1) = 0 ⇒ a = 1, -2
Now, x^1/3 = 1 ⇒ x = (1)³ = 1
x^1/3 = -2 ⇒ x = (-2)³ = -8
∴ roots are 1, -8.

Question 37.
Solve 7^{1 + x} + 7^{1 – x} = 50 for real x.
Solution:
The given equation can be written as,
7.7^x + \(\frac{7}{7^{x}}\) – 50 = 0
Let 7^x = a
7a + \(\frac{7}{a}\) – 50 = 0
⇒ 7a² + 7 – 50a = 0
⇒ 7a² – 49a – a + 7 = 0
⇒ 7a(a – 7) – 1(a – 7) = 0
⇒ (a – 7)(7a – 1) = 0
∴ a = 7, \(\frac{1}{7}\)
Now, if a = 7 then 7^x = 7¹ ⇒ x = 1
a = \(\frac{1}{7}\) then 7^x = \(\frac{1}{7}\) = 7^-1
x = -1
∴ x = -1, 1

Question 38.
Solve

Solution:
On taking \(\sqrt{\frac{x}{1-x}}\) = a
a + \(\frac{1}{a}\) = \(\frac{13}{6}\)
⇒ \(\frac{a^{2}+1}{a}\) = \(\frac{13}{6}\)
⇒ \(\frac{a^{2}+1}{a}\) = \(\frac{13}{6}\)
⇒ 6a² + 6 = 13a
⇒ 6a² – 13a + 6 = 0
⇒ 6a² – 9a – 4a + 6 = 0
⇒ 3a(2a – 3) – 2(2a – 3) = 0
⇒ (2a – 3)(3a – 2) = 0
a = \(\frac{3}{2}\), a = \(\frac{2}{3}\)
If a = \(\frac{3}{2}\) then \(\sqrt{\frac{x}{1-x}}\) = \(\frac{3}{2}\)
⇒ \(\frac{x}{1-x}\) = \(\frac{9}{4}\)
⇒ 4x = 9 – 9x
⇒ 13x = 9
⇒ x = \(\frac{4}{13}\)
∴ x = \(\frac{9}{13}\), \(\frac{4}{13}\)

Question 39.
Find all number which exceed their square root by 12.
Solution:
Let the required number be ‘x’
given, x = \(\sqrt{x}\) + 12
⇒ x – 12= \(\sqrt{x}\)
Squaring on both sides
(x – 12)² = (\(\sqrt{x}\))²
⇒ x² – 24x + 144 = x
⇒ x² – 25x + 144 = 0
⇒ (x – 9)(x – 16) = 0
⇒ x = 9, 16
But x = 9 does not satisfy the given condition
x = 16
∴ The required numbér = 16.

Question 40.
If x² + 4ax + 3 = 0 and 2x² + 3ax – 9 = 0 have a common root, then find the values
of a and the common root.
Solution:
The quadratic equations

Substitute in the above equation
[(3) (2) – (-9) (1)]² = [(1) (3a) – (2) (4a)] [(4a) (-9) – (3a) (3)]
(15)² = (-5a) (-45a)
⇒ 225 = 225a² ⇒ a² = 1 ⇒ a = ±1

Case (i) : If a = 1, the given equations become x² + 4x + 3 = 0 and 2x² + 3x – 9 = 0
⇒ (x + 1)(x + 3) = 0 and (2x + 3)(x + 3) = 0
⇒ x = 3, -1 and -3, \(\frac{3}{2}\)
In this case, the common root of the given equations is -3

Case (ii) : If a = -1, the given equations become x² – 4x + 3 = 0 and 2x² – 3x – 9 = 0
⇒ (x – 1) (x – 3) = 0 and (2x + 3) (x – 3) = 0
⇒ x = 1, 3 and x = 3, –\(\frac{3}{2}\)
In this case, the common root of the given equation is 3.

Question 41.
Prove that there is unique pair of consecutive positive odd integers such that the sum of their squares is. 290 and find it.
Solution:
The difference of two consecutive odd integers is 2.
So, we have to prove that there is a unique positive odd integer ‘x’ such that
x² + (x + 2)² = 290 —– (1)
x² + (x + 2)² = 290
⇒ x² + x² + 4x + 4 = 290
⇒ 2x² + 4x – 286 = 0
⇒ x² + 2x – 143 = 0
⇒ (x + 13) (x – 11) = 0
⇒ x = -13, 11
Hence 11 is the only positive odd integer satisfying equation (1).
∴ 11, 13 is the unique pair of integers which satisfies the given condition.

Question 42.
The cost of a piece of cable wire is Rs. 35/-, If the length of the piece of wire is 4 meters more and each meter costs, Rs. 1/— less, the cost would remain unchanged. What is the length of the wire?
Solution:
Let the length of the piece of the wire be ‘l’ meters and the cost of each meter be Rs. x.
given lx. = 35 —— (1)
and (l + 4) (x – 1) = 35
⇒ lx – l + 4x – 4 = 35
⇒ 35 – l + 4x – 4 = 35
⇒ 4x = l + 4
⇒ x = \(\frac{l+4}{4}\)
Substitute x in (1), we get l[latex]\frac{l+4}{4}[/latex] = 35
⇒ l² + 4l = 140
⇒ l² + 4l – 140 = 0
⇒ l² + 14l – 10l – 140 = 0
⇒ l(l + 14) – 10(l + 14) = 0
⇒ (l – 10)(l + 14) = 0
⇒ l = -14, 10
Since length cannot be negative, l = 10
∴ The length of the piece of wire is 10 meters.

Question 43.
One fourth of a herd of goats was seen in the forest. Twice the square root of the number in the herd had gone up the hill and the remaining 15 goats were on the bank of the river. Find the total number of goats.
Solution:
Let the number of goats in the herd be ‘x’.
The number of goats seen in the forest = \(\frac{x}{4}\)
The number of goats gone upto the hill = \(2 \sqrt{x}\)
The number of goats on the bank of a river = 15
∴ \(\frac{x}{4}\) + 2\(\sqrt{x}\) + 15 = x
⇒ x + 8\(\sqrt{x}\) + 60 = 4x
⇒ 3x – 8\(\sqrt{x}\) – 60 = 0
Put \(\sqrt{x}\) = y
⇒ 3y² – 8y – 60 = 0
⇒ 3y² – 18y + 10y – 60 = 0
⇒ 3y(y – 6) + 10(y – 6) = 0
⇒ (3y + 10)(y – 6) = 0
⇒ y = 6, –\(\frac{10}{3}\)
Since y cannot be negative, y = 6
⇒ \(\sqrt{x}\) = 36
∴ x = 36
∴ Total number of goats = 36

Question 44.
In a cricket match Anil took one wicket less than twice the number of wickets taken by Ravi. If the product of the
number of wickets taken by them is 15, find the number of wickets taken by each of them.
Solution:
Let the number of wickets taken by Ravi be ‘x’ and the number of wickets taken by Anil is 2x – 1.
Given x(2x – 1) = 15
⇒ 2x² – x – 15 = 0
⇒ 2x² – 6x + 5x – 15 = 0
⇒ 2x(x – 3) + 5(x – 15) = 0
⇒ (x – 3)(2x + 5) = 0
⇒ x = 3, –\(\frac{5}{2}\)
Since the number of wickets be integer,
x = 3 and 2x – 1 = 2(3) – 1 = 5
∴ The number of wickets taken by Anil and Ravi are 5 and 3 respectively.

Question 45.
Some points on a plane are marked and they are connected pairwise by line segments. If the total number of line
segments formed is 10, find the number of marked points on the plane.
Solution:
Let the number of points marked on the plane be ’x’.
The total number of line segments actually formed is

Given \(\frac{x(x-1)}{2}\) = 10
⇒ x² – x = 20
⇒ x² – x – 20 = 0
⇒ (x – 5)(x + 4) = 0
Since x cannot be negatives x = 5
∴ The number of points marked on the plane is 5.

Question 46.
Suppose that the quadratic equations ax² + bx + c = 0 and bx² + cx + a = 0 have a common root. Then show that a³ + b³ + c³ = 3abc.
Solution:
Let α be the common root of the equations

Question 47.
For what values of x1 the expression x² – 5x – 14 is positive?
Solution:
Since x² – 5x – 14 = (x + 2) (x – 7), the roots of the equation -2 and 7
Here the coefficient of x² is positive.
Hence x² – 5x – 14 is positive when x < -2 or x > 7.

Question 48.
For what values ofx. the expression -6x² + 2x³ is negative?
Solution:
-6x² + 2x – 3 = 0 can be written as 6x² – 2x + 3 = 0

∴ The roots of -6x² + 2x – 3 = 0 are non-real complex numbers.
Hence coefficient of x² is -6, which is negative.
∴ -6x² + 2x – 3 < 0 for all x ∈ R.

Question 49.
Find the value of x at which the following expressions have maximum or minimum.

i) x² + 5x + 6
Solution:
Here a = 1 > 0, the expression has minimum value at x = \(\frac{-b}{2 a}\)
= \(\frac{-5}{2(1)}\)
= \(\frac{-5}{2}\)

ii) 2x – x² + 7
Solution:
Here a = -1 < 0, the expression has maximum value at x
= \(\frac{-b}{2 a}\)
= \(\frac{-2}{2 x-1}\) = 1

Question 50.
Find the maximum or minimum value of the quadratic expression
(i) 2x – 7 – 5x²
(ii) 3x² + 2x + 17 (Mar. ’14)
Solution:
i) Compare the given equation with ax² + bx + c, we get
a = -5, b = 2, c = -7
Here a = -5 < 0, the given expression has maximum value at x = \(\frac{-b}{2 a}\) = \(\frac{-2}{2(-5)}\)
= \(\frac{1}{5}\)
and maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-5) \cdot(-7)-(?)^{2}}{4(-5)}\)
= \(\frac{140-4}{-20}\) = \(\frac{-34}{5}\)

(ii) Compare 3x² + 2x + 11 with
ax² + bx + c, we get a = 3, b = 2, c = 11
∵ a = 3 > 0, the given expression has minimum value at

Question 51.
Find the changes in the sign of 4x – 5x² + 2 for x ∈ R and find the extreme value.
Solution:
The roots of 4x – 5x² + 2 = 0
⇒ 5x² – 4x – 2 = 0
roots are = \(\frac{2 \pm \sqrt{14}}{5}\)
∴ \(\frac{2-\sqrt{14}}{5}\) < x < \(\frac{2+\sqrt{14}}{5}\) the sign of 4x – 5x² + 2 is positive.
x < \(\frac{2-\sqrt{14}}{5}\) or x > \(\frac{2+\sqrt{14}}{5}\), the sign of 4x – 5x² + 2 is negative.
Since a = -5 < 0, then maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-5)(2)-(4)^{2}}{4(-5)}\)
= \(\frac{-56}{-20}\) = \(\frac{14}{5}\)
Hence extreme value = \(\frac{14}{5}\)

Question 52.
Find the solution set of x² + x – 12 ≤ 0 by both algebraic and graphical methods.
Solution:
Algebraic Method : We have x² + x – 12 = (x + 4) (x – 3).
Hence -4 and 3 are the roots of the equation x² + x – 12 = 0.
Since the coefficient of x² in the quadratic expression x² + x – 12 = 0 is positive, x² + x – 12 is negative if -4 < x < 3 and positive if either x < -4 or x > 3.
Hence x² + x – 12 ≤ 0 ⇔ -4 ≤ x ≤ 3.
Therefore the solution set is
{x ∈ R : -4 ≤ x ≤ 3}.
Graphical Method: Let y = f(x) = x² + x – 12.
The values of y at some selected values of s are given in the following table:

The graph of the function y = f(x) is drawn using the above tabulated values. This is shown in Fig.

Therefore the graph of y = f(x) we observe that
y = x² – x – 12 < 0 if f -4 ≤ x ≤ 3.
Hence the solution set is {x ∈ R : -4 ≤ x ≤ 3}.

Question 53.
Find the set of values of x for which the inequalities x² – 3x – 10 < 0, 10x – x² – 16
> 0 hold simultaneously.
Solution:
∵ x² – 3x – 10 < 0
⇒ x² – 5x + 2x – 10 < 0
⇒ x(x – 5) + 2(x – 5) < 0
⇒ (x – 5)(x + 2) < 0
∴ x² coeff. is the and expression is <0
⇒ x ∈ (-2, 5) ——- (1)
Now 10x – x² – 16 > 0
⇒ x² – 10x + 16 < 0
⇒ (x – 2) (x – 8) < 0
∴ x² coeff is the ana expression is < 0
⇒ x ∈ (2, 8) ——- (2)
Hence x² – 3x – 10 < 0 and 10x – x² – 16 > 0
⇔ x ∈ (-2, 5) ∩ (2, 8)
∴ The solution set is {x/x ∈ R : 2 < x < 5}

Question 54.
Solve the inequation \(\sqrt{x+2}\) > \(\sqrt{8-x^{2}}\).
Solution:
When a, b ∈ R and a ≥ 0, b ≥ 0 then \(\sqrt{a}\) > \(\sqrt{b}\) ⇔ a > b ≥ 0
∴ \(\sqrt{x+2}\) = \(\sqrt{8-x^{2}}\)
⇔ x + 2 > 8 – x² ≥ 0 and x > -2, |x| < 2\(\sqrt{2}\)
We have (x + 2) > 8 – x²
⇔ x² + x – 6 > 0
⇔ (x + 3)(x – 2) > 0
⇔ x ∈ (-∞, -3)(2, ∞) .
We have 8 – x² ≥ 0
⇔ x² ≤ 8 ⇔ |x| < 2 \(\sqrt{2}\) ⇔ x ∈ [-2 \(\sqrt{2}\), 2 \(\sqrt{2}\)] Also x + 2 > 0 ⇔ x > -2
Hence x + 2 > 8 – x² ≥ 0
⇔ x ∈ ((-∞, -3) ∪ (2, ∞)) ∩ [-2 \(\sqrt{2}\), 2 \(\sqrt{2}\)] and x > -2
⇔ x ∈ (2, 2 \(\sqrt{2}\))
∴ The solution set is [x ∈ R : -2 < x ≤ 2 \(\sqrt{2}\)]

Question 55.
Solve the equation
\(\sqrt{(x-3)(2-x)}\) < \(\sqrt{4 x^{2}+12 x+11}\).
Solution:
The given inequation is equivalent to the following two inequalities.
(x – 3)(2 – x) ≥ 0 and
(x – 3)(2 – x) < 4x² + 12x + 11
(x – 3)(2 – x) ≥ 0 ⇔ (x – 2)(x – 3) ≤ 0
⇔ 2 ≤ x ≤ 3
-x² + 5x – 6 < 4x² + 12x + 11
⇔ 5x² + 7x + 17 > 0
The discriminant = b² – 4ac = 49 – 340 < 0 ⇒ x ∈ R : a = 5 > 0 and expressive is the
Hence the solution set of the given inequation is {x ∈ R : 2 ≤ x ≤ 3}

Question 56.
Solve the inequation

Solution:

⇔ either 6 + x – x² = 0, 2x + 5 ≠ 0 and x + 4 ≠ 0 or 6 + x – x² > 0
and \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
We have 6 + x – x² = -(x² – x – 6)
= -(x + 2) (x – 3)
Hence 6 + x + x² = 0 ⇔ x = -2 or x = 3
2x + 5 = 0 ⇔ x = –\(\frac{5}{2}\)
x + 4 = 0 ⇔ x = -4
∴ 6 + x – 2x² > 0 ⇔ -2 < x < 3 x ∈ (-2, 3), 2x + 5 > -4 + 5 = 1 > 0 and x + 4 > -2 + 4 = 2 > 0
x + 4 > -2 + 4 = 2 > 0
x ∈ (-2, 3), \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
⇔ 2x + 5 ≤ x + 4,
2x + 5 ≤ x + 4 ⇔ x ≤ 1
Hence 6 + x – x² > 0 and \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
⇔ -2 < x ≤ -1
∴ The solution set is {-2, 3} ∪ (-2, -1)
= [-2, -1] ∪ {3}

Question 57.
Find the maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R. (May ’05)
Solution:
Let y = \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\)
⇒ yx² + 2yx + 3y = x² + 14x + 9
⇒ (y – 1)x² + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)²] – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y² – 14y + 49) [(3y² – 12y + 9)] ≥ 0
⇒ -2y² – 2y + 40 ≥ 0
⇒ y² + y – 20 ≤ 0
⇒ (y + 5) (y – 4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
∴ Maximum value of y = 4
∴ Maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R is 4.

Question 58.
Show that none of the values of the function over \(\frac{x^{2}+34 x-71}{x^{2}+2 x-7}\) over R lies between 5 and 9. (Mar. 2005)
Solution:
Let y = \(\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\)
⇒ x² + 2x – 7 = x² + 34x – 71
⇒ (y – 1)x² + 2(y – 17)x + (-7y + 71) = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 17)]² – 4(y – 1)(-7y + 71)] ≥ 0
⇒ 4[(y² – 34y + 289) – (-7y² + 78y – 71)] ≥ 0
⇒ 8y² – 112y + 360 ≥ 0
⇒ y² – 14y + 45 ≥ 0
⇒ (y – 5) (y – 9) ≥ 0
⇒ y ≤ 5 or y ≥ 9
⇒ y does not lies between 5 and 9, since expression is ≥ 0 and y² coeff is the
∴ None of the values of the function
\(\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\) over R lies between 5 and 9.

Question 59.
Solve the inequation
\(\sqrt{x^{2}-3 x-10}\) > (8 – x)
Solution:
(i) \(\sqrt{x^{2}-3 x-10}\) > (8 – x)
⇒ x² – 3x – 10 ≥ 0 and (i) (8 – x < 0)
or (ii) x² – 3x – 10 > (8 – x)² and (8 – x ≥ 0)
Now x² – 3x – 10 = (x – 5) (x + 2)
Hence x² – 3x – 10 ≥ 0
⇒ x ∈ (-∞, -2] ∪ [5, ∞)
8 – x < 0 ⇒ x ∈ (8, ∞)
∴ x² – 3x – 10 ≥ 0 and 8 – x < 0
⇔ x ∈ (-∞, -2] ∪ [5, ∞) and x ∈ (8, ∞)
⇔ x ∈ (8, ∞)

(ii) x² – 3x – 10 > (8 – x)²
⇔ x² – 3x – 10 > 64 + x² – 16x
⇔ 13x > 74

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(a)

I.

Question 1.
If ⁿP₃ = 1320, find n.
Solution:
Hint: ⁿP_r = \(\frac{n !}{(n-r) !}\) = n(n – 1)(n – 2)…….(n – r + 1)
∵ ⁿP₃ = 1320
= 10 × 132
= 10 × 12 × 11
= 12 × 11 × 10
= ¹²P₃
∴ n = 12

Question 2.
If ⁿP₇ = 42 . ⁿP₅, find n.
Solution:
ⁿP₇ = 42 . ⁿP₅
⇒ n(n – 1) (n – 2) (n – 3) (n – 4) (n – 5) (n – 6) = 42 . n(n – 1) (n – 2) (n – 3) (n – 4)
⇒ (n – 5) (n – 6) = 42
⇒ (n – 5) (n – 6) = 7 × 6
⇒ n – 5 = 7 or n – 6 = 6
⇒ n = 12

Question 3.
If ⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7, find n.
Solution:
⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7
⇒ \(\frac{(n+1)_{P_5}}{n_{P_6}}=\frac{2}{7}\)
⇒ \(\frac{(n+1) n(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)(n-4)(n-5)}=\frac{2}{7}\)
⇒ 7(n + 1) = 2(n – 4) (n – 5)
⇒ 7n + 7 = 2n² – 18n + 40
⇒ 2n² – 25n + 33 = 0
⇒ 2n² – 22n – 3n + 33 = 0
⇒ 2n(n – 11) – 3(n – 11) = 0
⇒ (n – 11) (2n – 3) = 0
⇒ n = 11 or \(\frac{3}{2}\)
Since n is a positive integer
∴ n = 11

Question 4.
If ¹²P₅ + 5 . ¹²P₄ = ¹³P_r, find r.
Solution:
We have
^(n-1)P_r + r . ^(n-1)P_(r-1) = ⁿP_r and r ≤ n
¹²P₅ + 5 . ¹²P₄ = ¹³P₅ = ¹³P_r (given)
⇒ r = 5

Question 5.
If ¹⁸P_(r-1) : ¹⁷P_(r-1) = 9 : 7, find r.
Solution:
¹⁸P_(r-1) : ¹⁷P_(r-1) = 9 : 7

⇒ 18 × 7 = 9(19 – r)
⇒ 14 = 19 – r
⇒ r = 19 – 14 = 5

Question 6.
A man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons into the schools so that no two of them will be in the same school?
Solution:
The number of ways of admitting 4 sons into 5 schools if no two of them will be in the same school = ⁵P₄
= 5 × 4 × 3 × 2
= 120

II.

Question 1.
If there are 25 railway stations on a railway line, how many types of single second-class tickets must be printed, so as to enable a passenger to travel from one station to another?
Solution:
Number of stations on a railway line = 25
∴ The number of single second class tickets must be printed so as to enable a passenger to travel from one station to another = ²⁵P₂
= 25 × 24
= 600

Question 2.
In a class, there are 30 students. On New year’s day, every student posts a greeting card to all his/her classmates. Find the total number of greeting cards posted by them.
Solution:
The number of students in a class is 30.
∴ Total number of greeting cards posted by every student to all his/her classmates = ³⁰P₂
= 30 × 29
= 870

Question 3.
Find the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and consonants are not disturbed.
Solution:
Vowels – A, E, I, O, U
In a given, word, the number of vowels is 3
number of consonants is 5

Since the relative positions of the vowels and consonants are not disturbed,
the 3 vowels can be arranged in their relative positions in 3! ways and the 5 consonants can be arranged in their relative positions in 5! ways.
∴ The number of required arrangements = (3!) (5!)
= 6 × 120
= 720

Question 4.
Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8 without repetition.
Solution:
First Method:
The number of 4 digited numbers formed by using the digits 0, 2, 4, 7, 8 without repetition = ⁵P₄ – ⁴P₃
= 120 – 24
= 96
Out of these 96 numbers,
⁴P₃ – ³P₂ numbers contain 2 in the units place
⁴P₃ – ³P₂ numbers contain 2 in the tens place
⁴P₃ – ³P₂ numbers contain 2 in the hundreds place
⁴P₃ numbers contain 2 in the thousands place

∴ The value obtained by adding 2 in all the numbers = (⁴P₃ – ³P₂) 2 + (⁴P₃ – ³P₂) 20 + (⁴P₃ – ³P₂) 200 + ⁴P₃ × 2000
= ⁴P₃ (2 + 20 + 200 + 2000) – ³P₂ (2 + 20 + 200)
= 24 × (2222) – 6 × (222)
= 24 × 2 × 1111 – 6 × 2 × 111
Similarly, the value obtained by adding 4 is 24 × 4 × 1111 – 6 × 4 × 111
the value obtained by adding 7 is 24 × 7 × 1111 – 6 × 7 × 111
the value obtained by adding 8 is 24 × 8 × 1111 – 6 × 8 × 111
∴ The sum of all the numbers = (24 × 2 × 1111 – 6 × 2 × 111) + (24 × 4 × 1111 – 6 × 4 × 111) + (24 × 7 × 1111 – 6 × 7 × 111) + (24 × 8 × 1111 – 6 × 8 × 111)
= 24 × 1111 × (2 + 4 + 7 + 8) – 6 × 111 × (2 + 4 + 7 + 8)
= 26664 (21) – 666 (21)
= 21 (26664 – 666)
= 21 × 25998
= 545958

Second Method:
If Zero is one among the given n digits, then the sum of the r – digited numbers that can be formed using the given ‘n’ distinct digits (r ≤ n ≤ 9) is
^(n-1)P_(r-1) × sum of the digits × 111 …… 1 (r times) – ^(n-2)P_(r-2) × Sum of the digits × 111 ……… 1 [(r – 1) times]
Hence n = 5, r = 4, digits are {0, 2, 4, 7, 8}
Hence the sum of all 4 digited numbers that can be formed using the digits {0, 2, 4, 7, 8} without repetition is
^(5-1)P_(4-1) × (0 + 2 + 4 + 7 + 8) × (1111) – ^(5-2)P_(4-2) × (0 + 2 + 4 + 7 + 8) × (111)
= ⁴P₃ (21) × 1111 – ³P₂ (21) (111)
= 24 × 21 × 1111 – 6 (21) (111)
= 21 (26664) – 21 (666)
= 21 (26664 – 666)
= 21 (25998)
= 545958

Question 5.
Find the number of numbers that are greater than 4000 which can be formed using the digits 0, 2, 4, 6, 8 without repetition.
Solution:
While forming any digit number With the given digits, zero cannot be filled in the first place.
We can fill the first place with the remaining 4 digits.
The remaining places can be filled with the remaining 4 digits.
All the numbers of 5 digits are greater than 4000.
In the 4-digit numbers, the number starting with 4 or 6 or 8 are greater than 4000.
The number of 4-digit numbers that begin with 4 or 6 or 8 = 3 × ⁴P₃
= 3 × 24
= 72
The number of 5-digit numbers = 4 × 4!
= 4 × 24
= 96
∴ The number of numbers greater than 4000 is 72 + 96 = 168

Question 6.
Find the number of ways of arranging the letters of the word MONDAY so that no vowel occupies an even place.
Solution:
In the word MONDAY, there are two vowels, 4 consonants and three even places, three odd places.
Since no vowel occupies an even place, the two vowels can be filled in the three odd places in ³P₂ ways.
The 4 consonants can be filled in the remaining 4 places in 4! ways.
∴ The number of required arrangements = ³P₂ × 4!
= 6 × 24
= 144

Question 7.
Find the number of ways of arranging 5 different mathematics books, 4 different Physics books, and 3 different chemistry books such that the books of the same subject are together.
Solution:
The number of ways of arranging Mathematics, Physics, and Chemistry books are arranged 3! ways.
5 different Mathematics books are arranged themselves in 5! ways.
4 different Physics books are arranged themselves in 4! ways.
3 different Chemistry books are arranged themselves in 3! ways.
∴ The number of required arrangements = 3! × 5! × 4! × 3!
= 6 × 120 × 24 × 6
= 1,03,680

III.

Question 1.
Find the number of 5-letter words that can be formed using the letters of the word CONSIDER. How many of them begin with “C”, how many of them end with ‘R’ and how many of them begin with “C” and end with “R”?
Solution:

The number of 5 letter words that can be formed using the letters of the word CONSIDER = ⁸P₅
= 8 × 7 × 6 × 5 × 4
= 6720

The number of 5 letter words beginning with ‘C’ = 1 × ⁷P₄
= 7 × 6 × 5 × 4
= 840

The number of 5 letter words end with ‘R’ = 1 × ⁷P₄ = 840

The number of 5 letter words begins with ‘C’ and ends with ‘R’ = 1 × 1 × ⁶P₃
= 6 × 5 × 4
= 120

Question 2.
Find the number of ways of seating 10 students A₁, A₂, ………, A₁₀ in a row such that
(i) A₁, A₂, A₃ sit together
(ii) A₁, A₂, A₃ sit in a specified order
(iii) A₁, A₂, A₃ sit together in a specified order
Solution:
A₁, A₂, A3, ……….., A₁₀ are the ten students.
(i) Consider A₁, A₂, A₃ as one unit and A₄, A₅, A₆, A₇, A₈, A₉, A₁₀ as seven units.
These 8 units can be arranged in 8! ways.
The students A₁, A₂, A₃ in one unit can be arranged among themselves in 3! ways.
∴ The number of ways seating 10 students in which A₁, A₂, A₃ sit together = (8!) (3!)

(ii) To arrange A₁, A₂, A₃ to sit in a specified order,
A₁, A₂, A₃ can arrange in 10 positions in specific order in \(\frac{{ }^{10} P_3}{3 !}\) ways.
The remaining 7 people can be arranged in the remaining places in 7! ways.
∴ The number of ways of A₁, A₂, A₃ sit in a specific order = \(\frac{10 !}{7 ! \times 3 !} \times 7 !\)
= \(\frac{10 !}{3 !}\)
= ¹⁰P₇

(iii) To arrange A₁, A₂, A₃ sit together in a specified order.
Consider A₁, A₂, A₃ in that order as one unit.
Now there are 8 objects, they can be arranged in 8! ways.
∴ The number of ways of A₁, A₂, A₃ sit together in specified order = 8! ways

Question 3.
Find the number of ways in which 5 red balls, 4 black balls of different sizes can be arranged in a row so that
(i) no two balls of the same colour come together.
(ii) the balls of the same colour come together.
Solution:
No. of red balls = 5
No. of black balls = 4
(i) No. two balls of the same colour come together.
For the required arrangements first, we arrange 4 black balls it can be done in 4! ways.

There are 5 places to arrange 5 red balls it can be done in 5! ways.
∴ The number of required arrangements = 4! 5!

(ii) The balls of the same colour come together.
For the required arrangements 4 black balls of different sizes can be considered as one object and 5 red balls can be considered as one object. These can be arranged in 2! ways.
The 4 black balls can be permuted among themselves in 4! ways.
The 5 red balls can be permuted among themselves in 5! ways.
∴ The no. of required arrangements = 2! 4! 5!

Question 4.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 5, 6, 7. How many of them are divisible by
(i) 2
(ii) 3
(iii) 4
(iv) 5
(v) 25
Solution:
The number of 4 digited numbers that can be formed using the digits 1, 2, 5, 6, 7 is ⁵P₄ = 120.
(i) A number is divisible by 2 when its unit place must be filled with an even digit from among the given integers. This can be done in 2 ways.

Now, the remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ = 4 × 3 × 2 = 24 ways.
∴ The number of 4 digited numbers divisible by 2 = 2 × 24 = 48

(ii) A number is divisible by 3 only when the sum of the digits in that number is a multiple of 3.
Sum of the given 5 digits = 1 + 2 + 5 + 6 + 7 = 21
The 4 digits such that their sum is a multiple of 3 from the given digits are 1, 2, 5, 7 (sum is 15)
They can be arranged in 4! ways and all these 4 digited numbers are divisible by 3.
∴ The number of 4 digited numbers divisible by 3 = 4! = 24

(iii) A number is divisible by 4 only when the last two places (tens and units places) of it are a multiple of 4.

Hence the last two places should be filled with one of the following 12, 16, 52, 56, 72, 76.
Thus the last two places can be filled in 6 ways.
The remaining two places can be filled by the remaining 3 digits in ³P₂ = 3 × 2 = 6 ways.
∴ The number of 4 digited numbers divisible by 4 = 6 × 6 = 36.

(iv) A number is divisible by 5 when its units place must be filled by 5 from the given integers 1, 2, 5, 6, 7. This can be done in one way.

The remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ = 4 × 3 × 2 = 24 ways.
∴ The number of 4 digited numbers divisible by 5 = 1 × 24 = 24

(v) A number is divisible by 25 when its last two places are filled with either 25 or 75.

Thus the last two places can be filled in 2 ways.
The remaining 2 places from the remaining 3 digits can be filled in ³P₂ = 6 ways.
∴ The number of 4 digited numbers divisible by 25 = 2 × 6 = 12

Question 5.
If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the ranks of the words
(i) REMAST
(ii) MASTER
Solution:
(i) The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words beginning with A is 5! = 120
The number of words that begins with E is 5! = 120
The number of words begins with M is 5! = 120
The number of words beginning with RA is 4! = 24
The number of words beginning with REA is 3! = 6
The next word is REMAST
∴ Rank of the word REMAST = 3(120) + 24 + 6 + 1
= 360 + 31
= 391

(ii) The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words beginning with A is 5! = 120
The number of words that begins with E is 5! = 120
The number of words beginning with MAE is 3! = 6
The number of words that begins with MAR is 3! = 6
The number of words beginning with MASE is 2! = 2
The number of words begins with MASR is 2! = 2
The next word is MASTER.
∴ Rank of the word MASTER = 2(120) + 2(6) + 2(2) + 1
= 240 + 12 + 4 + 1
= 257

Question 6.
If the letters of the word BRING are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the 59th word.
Solution:
By using the letters of the word BRING in alphabetical order the 59th word must start with ‘I’.
Since the words start with B, G sum to 48.
i.e., Start with B = 4! = 24
Start with G = 4! = 24
Start with IB = 3! = 6
Start with IGB = 2! = 2
Start with IGN = 2! = 2
The next word is 59th = IGRBN

Question 7.
Find the sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.
Solution:
The number of 4 digited numbers formed by using the digits 1, 2, 4, 5, 6 without repetition = ⁵P₄ = 120
Out of these 120 numbers,
⁴P₃ numbers contain 2 in the units place
⁴P₃ numbers contain 2 in the tens place
⁴P₃ numbers contain 2 in the hundreds place
⁴P₃ numbers contain 2 in the thousands place

∴ The value obtained by adding 2 in all the numbers = ⁴P₃ × 2 + ⁴P₃ × 20 + ⁴P₃ × 200 + ⁴P₃ × 2000
= ⁴P₃ (2 + 20 + 200 + 2000)
= ⁴P₃ (2222)
= ⁴P₃ × 2 × 1111
Similarly, the value obtained by adding 1 is ⁴P₃ × 1 × 1111
the value obtained by adding 4 is ⁴P₃ × 4 × 1111
the value obtained by adding 5 is ⁴P₃ × 5 × 1111
the value obtained by adding 6 is ⁴P₃ × 6 × 1111
∴ The sum of all the numbers = ⁴P₃ × 1 × 1111 + ⁴P₃ × 2 × 1111 + ⁴P₃ × 4 × 1111 + ⁴P₃ × 5 × 1111 + ⁴P₃ × 6 × 1111
= ⁴P₃ (1111) (1 + 2 + 4 + 5 + 6)
= 24 (1111) (18)
= 4,79,952

Second Method:
The sum of the r-digited numbers that can be formed using the given ‘n’ distinct non-zero digits (r ≤ n ≤ 9) is ^(n-1)P_(r-1) × sum of all digits × 111 …… 1 (r times)
Hence n = 5, r = 4, digits = {1, 2, 4, 5, 6}
The sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition is ^(5-1)P_(4-1) × (1 + 2 + 4 + 5 + 6) × (1111)
= ⁴P₃ (18) (1111)
= 24 × 18 × 1111
= 4,79,952

Question 8.
There are 9 objects and 9 boxes. Out of 9 objects, 5 cannot fit in three small boxes. How many arrangements can be made such that each object can be put in one box only?
Solution:
No. of objects = 9
No. of boxes = 9
For the required arrangements, out of 9 objects, 5 cannot bit in three small boxes.
These five can be arranged in 6 boxes these can be done in 6P5 ways.
The remaining 4 objects can be arranged in the remaining 4 boxes it can be done in 4! ways.
∴ No. of required arrangements = ⁶P₅ × 4! ways

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive

Long Answer Questions

Question 1.
Explain the powers and functions of the President of India.
Answer:
Introduction:
The President of India is the constitutional head of the Indian Republic. He is the First citizen of India. He administers the affairs of the union Government either himself or through the officers subordinate to him. (Articles 52 and 53)

Qualifications :
A person to be eligible to contest the office of the President shall possess the following qualifications :

1. He should be a citizen of India.
2. He should have completed the age of 35 years.
3. He should be qualified for election as a member of the Lok Sabha.
4. He should not hold any office of profit under the union, state or local Governments (Article 59 (i))
5. Possess such other qualifications as prescribed by the Parliament.

Election Procedure :
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Oath of office :
The person who is elected as President assumes office only after he takes oath of office and secrecy by the Chief Justice of India.

Term of office :
The President continues in office for five years from the date of his assumption of office.

Salary and Allowances :
The President now gets a monthly salary of ₹ 1,50,000/-. His official residence is Rashtrapathi Bhavan at New Delhi. On retirement, he will get a monthly pension of ₹ 75,000/-.

Removal (or) Impeachment :
The President can be removed from the office by a process of impeachment for violation of the constitution. Impeachment is a quasi-judicial procedure adopted by the Parliament.

Powers and Functions :
The President shall exercise his powers with the help of the Council of Ministers headed by the Prime Minister. His powers may be analysed under the following heads :

1. Executive Powers :
An executive action of the Union Government shall be expressed in the name of the President. The President appoints the Prime Minister and other Ministers, Attorney General, Comptroller and Auditor General of India, State Governors, Judges of the Supreme Court and State High Courts, Finance Commission, Chairman and members of U.P.S.C., Election Commission, and Chief Commissioners of Unit Territories. He allocates portfolios to the Ministers.

2. Legislative Powers :
The President is an integral part of Parliament (Art. 79) and as such enjoys extensive legislative powers.
They are :

1. He summons from time to time each House of Parliament, adjours, and prorogues either or both the Houses.
2. He addresses either House separately or both the Houses Jointly.
3. He can dissolve the Lok Sabha on the advice of the Prime Minister.
4. He opens the first session of Parliament after the General Elections and at the commencement of the first session of each year.
5. He can send messages to the Parliament.
6. He arranges a joint session of both the Houses when there is a dead-lock over an ordinary bill.
7. All bills passed by Parliament require his assent for becoming in acts.
8. He nominates 12 members to Rajya Sabha and two Anglo Indian members to Lok Sabha.
9. He promulgates ordinances when the Parliament is not in session.
10. He sends the annual reports of Finance Commission, U.P.S.C etc., for the consideration and approval of Parliament.

3. Financial Powers :
The President also enjoys some financial powers. They are :

1. He recommends the financial bills to be introduced by the members in parliament. The Budget is caused to be laid down before the Parliament by the President.
2. He operates the Consolidated Fund of India.
3. He determines the shares of States in the proceeds of Income Tax.
4. No Money Bill can be introduced in the Parliament except on his recommendations.
5. He constitutes a Finance Commission for every five years etc.

4. Judicial Powers :

1. The President can grant pardons, reprieves, respites or remission of punishments.
2. He appoints the judges of the Supreme Court and State High Courts.
3. He can also remove them on an address by the Parliament.

5. Military Powers:
The President is the Supreme Commander of the Defence Forces of the Union. He appoints the Chiefs on the Staff for Army, Navy, and Air Force. He can declare war and conclude peace. But he has to take the approval of Parliament.

6. Diplomatic Powers :
The president appoints Ambassadors to foreign countries to represent India. He receives the credentials of the Ambassadors appointed in India. He represents our Nation in International affairs. He makes treaties and agreements with other countries subject to the ratification by the Parliament.

7. Emergency Powers :
In extraordinary conditions, the President can proelaim emergency to safeguard the security, integrity, and independence of our country. They are of three types :

• Emergency caused by war or external aggression or armed rebellion (Article 352).
• Emergency due to failure of Constitutional machinery in the States (Article 356)
• Emergency due to threat to the financial stability of India (Article 360).

Question 2.
Write briefly the Emergency powers of the President of India. [Mar. 16]
Answer:
Articles 352 to 360 of part XVIII of Indian constitution deals with three types of emergency powers of the Indian President. They are :

1. National Emergency,
2. Constitutional Emergency,
3. Financial Emergency

They may be explained as follows.
1) National Emergency : (Article 352)
The President exercises this power during the period of war, external aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are : 1. Chinese Aggression (1962), 2. Indo – Pak war (1965), 3. Indo – Pak war in the context of Bangladesh Liberation Movement (1971), 4. Opposition’s call for blocking Parliament (1975).

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise; is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule. So far this type of emergency was proclaimed for over 100 times.

3) Financial Emergency : (Article 360)
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatended, then he may proclaim financial emergency in the country. During the period of financial emergency, the President enjoys the following powers.

• The President may reserve all the money bills or other financial bills of the state after they are approved by the state legislature.
• He may reduce the salaries and allowances of all or any person serving in the states.
• The President can reduce the salaries allowances of the persons working at the union level including the judges of the Supreme Court and the State High Courts. But so far the financial emergency has not been yet imposed in the country.

Question 3.
Discuss the powers and functions of the Prime Minister of India. [Mar. 18, 16]
Answer:
The Prime Minister is the real executive head of the Union Government. He occupies an important position in the administration of our country. Since India has a Parliamentary form of Government the real power rests with him. He is the ‘uncrowned king’ and “the keystone of the Cabinet arch in the Union Government”.

Qualifications :

1. He should be citizen of India.
2. He should have completed the age of 25 ytears.
3. He should be qualified for election as a member of the Lok Sabha.
4. He should not hold any office of profit under the union or state or local governments.

Appointment:
The President appoints the Prime Minister. Generally the President has to summon the leader of the majority party in the Lok Sabha to form the Ministry. If no party gets an absolute majority, the President can use his discretion and summon the leader of the party, who in his opinion can manage to form a ministry. Afterwards the Prime Minister will be asked to prove his majority in the Lok Sabha.

Oath of Office :
The President of India will administers the oath of office of the Prime Minister.

Term of Office :
The Prime Minister shall remain in office during the pleasure of the President. But actually he assumes his powers as long as he retains the confidence of the majority members in the Lok Sabha. He resigns when the Lok Sabha accepts a no-confidence motion against his ministry.

Salary and Allowances:
The salary and allowances of Prime Minister are decided by the Parliament from time to time. He gets his salary and allowances that are payable to a member of Parliament. At present the Prime Minister gets a salary and allowances of ₹ 1,60,000/- per month.

Powers and Functions :
The Prime Minister is the head of the union government. He is the real executive. The Council of Ministers cannot exist without the Prime Minister. His powers are explained here under.

1) Leader of the Union Cabinet:
The Prime Minister is the leader of the Union Cabinet and Union Council of Ministers. He selects some eminent members of his party in parliament and sees that they are appointed as ministers by the President. He has a free choice of both allocating portfolios and reshuffling the ministry. All the ministers are personally and politically loyal to the Prime Minister. He decides the agenda, of the cabinet meetings. Further, he presides over the cabinet meetings.

2) Leader of the Union Government:
The Prime Minister acts as the leader of the union government. The union executive (Union Council of Ministers) initiates its business after the swearing in ceremony of the Prime Minister. All the ministers in the union ministry assume their office, owe their position and exercise their powers along with the Prime Minister. Infact, the Prime Minister influences the nature and working of the union government. He not only has a clear understanding but holds complete control over the affairs of the union government. All the high-level officers and the entire ministry in the union government behave and act according to the wishes of the Prime Minister.

3) Leader of the parliament:
The Prime Minister acts the leader of the Parliament in India. He is primarily a member of Parliament. He extends co-operation to the presiding officers in the smooth conduct of the two Houses. He wields complete control over his party members in the Parliament. He ensures that his party members maintain discipline during the sessions of the Parliament. He informs out the cabinet decisions to the Parliament. He communicates the major domestic and foreign policies of the union government to the members of Parliament. He maintains rapport with the opposition leaders and discusses the major issues confronted by the nation with them.

4) Link between the President and the Council of Ministers :
The Prime Minister acts as the main link between the President and the Union Council of Ministers. It is his duty to communicate to the President about the decisions of the Union Council of Ministers. He furnishes the every information required by the President concerning the affairs of union government. All the ministers shall formally meet the President only with the consent of the Prime Minister.

5) Leader of the Majority Party:
The Prime Minister acts as the leader of the majority party or group in the lower House of Parliament. He participates in the meetings of the party and acquaints his party members on various issues and steps taken by his ministry in implementing the party promises. He utilizes the services of the senior party leaders in running the government. He acts as the main link between the part and the government.

6) Leader of the Nation :
The Prime Minister is the leader of the nation. He takes initiative in finding solutions to several problems in the internal matters of the country. He plays an important role in the development of the nation.

7) Maker of Foreign Policy :
The Prime Minister plays a dominant role in shaping the foreign policy of the nation. He keeps in touch with the developments in all countries. He meets Heads of various countries and maintains friendly relations with them.

8) Chairman of NITI Aayog :
The Prime Minister heads the NITI Aayog (National Institution for Transforming India) NITI Aayog means policy commission. It is a policy think tank of government of India that replaces planning commission which aims to involve the states in economic policy making in India. It will provide strategic and technical advice to the central and state governments. It will have a governing council comprising Chief Ministers of all the states and it governors of Union Territories. Union government set up the NITI Aayog on January 1,-2015.

Question 4.
Explain the composition, powers and functions of the Union Council of Ministers. .
Answer:
Article 74 (1) of the Constitution provides for a Council of Ministers at the Centre. It’s main function is to aid and advice the President in the performance of his duties. It consists of Prime Minister and other Ministers. It is this body which runs the entire administration of our country. It is thefreal Executive authority of the country. It functions on the ‘Principle of Collective Responsibility1. It holds office till it continues to enjoy the confidence and support of the Lok Sabha.

Formation of Council of Ministers :
The formation of the Council of Ministers starts with the appointment of the Prime Minister. The President appoints the Prime Minister and on the advice of the Prime Minister, the other Ministers are appointed by the President.

Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers:

1. Cabinet Ministers
2. Ministers of State
3. Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State.” It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1. Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It coordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2. Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3. Financial Powers:
The Cabinet possess important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4. Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Collective Responsibility :
Article 75(3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of commissions and commissions. They act as a team under the leadership of the Prime Minister. They sail together, they swim together and they sink together.

Conclusion:
It is thus clear that the Council of Minister or Cabinet enjoys far reaching powers both with regard to the internal and external policies of the country. Internally it maintains law and order within the country and externally protects the country from foreign aggression. The progress of the country largely depends upon the ability of the Cabinet.

Short Answer Questions

Question 1.
How is the President of India elected?
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Each member of the Electoral College has one vote. But the value of the vote differes from State to State. The value of the vote of a Parliament member also differs from the value of the vote of a member of State Assembly.

The value of the vote of an M.L.A is worked out as follows. The total population of the State is divided by the total number of elected members of the Assembly. The quotient thus obtained is to be divided by 1,000. Fractions of half or more should be counted as one and added to the quotient. If it is less than half, it is ignored. This may be shown as follows.

1. Value of vote of an M.L.A = Population of State / number of elected members of the Assembly / 1,000.
2. Value of vote of M.P. = Total value of votes of all Assembly members / Total number of elected members of both Houses of Parliaments.

This method is followed to keep the election of the President above narrow political considerations.

Question 2.
Write briefly about the procedure of impeachment of President
Answer:
The President may be removed from the office for violation of the constitution by a process of impeachment. Impeachment is a quasi-judicial procedure adopted by the Parliament. Either House of Parliament shall prefer the charge for removal of the President. The other House shall investigate into the charges itself or cause the charge to be investigated.

There are four stages in the impeachment of the President.

Firstly the impeachment resolution has to be moved with a 14 days prior notice in writing signed by not less than 1/4^th of the total members of that House. Such a resolution has to be passed by a majority of not less than 2/3^rds of the total members of the House.

Secondly, the resolution approved by the first House will be sent to the second House for consideration and approval.

Thirdly, the second House investigates into the charges directly or constitutes a committee to enquire into the charges. The President has the right to present his views directly or through a deputy during such enquiry.

Fourthly, if the charges against the President are established and adopted by the second House with 2/3^rds majority of the total members, the President stands removed from the office. With regard to voting on the resolution for impeachment, only the elected members cast their vote. No president has so far been impeached in our country till today.

Question 3.
Mention any two Emergency powers of the Indian President.
Answer:
1) National Emergency : (Article 352)
The President exercises this power during the period of war, External aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are :

1. Chinese Aggression (1962) ‘
2. Indo – Pak war (1965) .
3. Indo – Pak war in the context of Bangaldesh Liberation Movement (1971)
4. Oppositions call for blocking Parliament (1975)

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution Empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise, is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency, It is also called as the President’s rule. So far this type of emergency was proclaimed for over 100 times.

Question 4.
Explain the role and position of the President in Union Government.
Answer:
The President of India is the head of the union executive. He is the first citizen of India. He could exercise many powers as enstined in the constitution in the following manner.

Position of the President:
There are different opinions on the actual position of the president of India in the administration of our country. The farmers of the constitution wanted him to be a nominal Head of the state.

Dr. Ambedkar compared his position to that of the British King Sri. M. C. Setalved, the farmer Attorney General of India mentioned that the position of the President of India is like the king in England and the Governor General in a Dominion. Sri Alladi Krishna Swamya Ayyar also said that it was perfectly clear that our presidents position was similar to that of the constitutional Monarch in England.

Jawaharlal Nehru, the first Prime Minister of India, said that “We have not given our President any real power but we have made his position one of great authority and dignity.” These opinions make it clear that our President is only a nominal figure head and he does not have any real powers. This is confirmed by the 42nd Amendment of the constitution of India.

However, the President exercises independent powers under some conditions. He utilises these powers in regard to the appointment of the Prime Minister, dissolving the Lok Sabha and ordering midterm poll to the Lok Sabha. Some presidents like Sanjiva Reddy, Zail Singh, R.Venkata Raman, Dr. S.D. Sharma etc., utilised their discretionary powers where there was political instability or Hung Parliament after the general elections in the country. Like the Monarch of England, he still enjoys three rights the right to be ! consulted the right to encourage and the right to warn.

Question 5.
Write about any two powers of the Vice – President of India.
Answer:
The Vice President of India occupies the second highest position in the union government. He is accorded a rank next to the President.

Qualifications :
A person to be eligible as vice-president should possess the following qualifications.

1. He should be a citizen of India.
2. He should have completed 35 years of age.
3. He should be qualified for election as a member of the eousil of states.
4. He should not hold any office of profit under the union, state or local Governments in India.

Election :
The election of the vice – president like that of the President shall be indirect and in accordance with the system of proportional representation by means of the single transferable vote system. He is elected by the members of an electoral college consisting of the members of both the houses of Parliament.

Term of Office :
The Vice-President holds office for a term of 5 years from the date on which he enters his office.

Removal:
The Vice President may be removed from his office by a resolution of the council of states passed by a majority of all the members of the council and agreed to by the house of the people.

Salary and Allowances :
The Vice President of India receives a monthly salary of ₹ 1,25,000/-in addition, he is entitled to daily allowance, free furnished residence, medical, travel and other facilities.

Powers and Functions :
The Vice – President is the ex-officio Chairman of the Rajya Sabha. As. such he enjoys the same powers like the Speaker of Lok Sabha, such as (1) presiding over the meetings of Rajya Sabha, (2) maintaining discipline, decency and decorum in the House, (3) exercising casting vote in case of a tie, (4) admitting visitors, (5) protecting the privileges and rights of the members. He has no right with regard to money bills.

Acting as President of India :
He discharges the functions of the President during the temporary absertce of the President. He may take over the office of the President under 4 situations like (1) Death of the President, (2) Resignation of the President, (3) Removal of the President, (4) Inability of the President due to absence, illness or any other cause.

Question 6.
How is the Prime Minister appointed?
Answer:
Article 75 (1) of the Indian constitution deals with the appointment of the Prime Minister of India.

Appointment:
The constitution simply lays down that the Prime Minister shall be appointed by the President. After the conduct of General Elections to the Lok Sabha, the President has to invite the majority party leader of the Ldk Sabha to form the Government.

When no single party is able to secure majority seats in Lok Sabha, the President invites the leader of a coalition to form the Government. The president uses his discretionary powers in this regard. The President appoints the leader of the coalition as the Prime Minister on the condition that he has to prove his majority in the Lok Sabha within a specified period. Being the leader of the majority in Lok Sabha to be the Prime Minister, the person has to be a member of Parliament. If he is not a member at the time of appointment, he has to acquire it within six months from the date of his appointment as Prime Minister.

The powers of the President in choosing, inviting and appointing the Prime Minister cannot be questioned in any court of Law.

Question 7.
Explain the role of the Prime Minister in Union Government.
Answer:
The prime Minister plays a predominant role in the affairs of the union government. He will have an indelible impression on every one in the administration of the union government. He is described as the Primus Interparus (first among equals). His role as the leader of the Union Council of Ministers, Union Cabinet, Party in power, Lok Sabha, Nation and as the link between the President and the Union Council of Ministers is unique. He wields tremendous political power and patronage. He enjoys enormous powers and fulfils innumerable tasks. It all depends on the image, influence, stature, and personality of the Prime Minister in the union government.

Jawaharlal Nehru, Dr. Ambedkar, and other eminent leaders of the National Movement and members of the Constituent Assembly described the Prime Minister as the linchpin of the union government. It is in this context that Sir william Harcourt remarked that every one expects from the Prime Minister dignity and authority, firmness to control, tact, practice and firmness, an impartial mind, a tolerant temper, a kind and prudent counsellorship, and accessibility to the people.

Question 8.
Describe the composition and powers of the Union Council of Ministers.
Answer:
Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers: 1) Cabinet Ministers 2) Ministers of State 3) Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They a&ist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet Or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State”. It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1) Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It co-ordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2) Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of. Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3) Financial Powers :
The Cabinet possesses important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4) Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Very Short answers

Question 1.
Composition of the Union Executive.
Answer:
The constitution of India provides for the Union Executive. Articles 52 to 78 in part V of the constitution deal with the union executive. The Union Executive consists.
i) The President
ii) The Vice-President
iii) The Prime Minister and
iv) The Union Council of Ministers

Question 2.
Qualifications required for contesting the Presidential elections.
Answer:
A person to be eligible to contest the office of the president shall possess the following qualifications.

1. He should be a citizen of India.
2. He should have completed the age of 35 years.
3. He should be qualified for election as a member of the Lok Sabha.
4. He should not hold any office of profit under the Union, State or Local Governments.
5. Possess such other qualifications as prescribed by the Parliament.

Question 3.
Election of President.
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both houses of Parliament, State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Question 4.
Important appointments of President.
Answer:
The president has the power to appoint the following high dignitaries :

• The Prime Minister of India.
• Members of the union council of Ministers.
• The Affomey General of India.
• The Comptroller and Auditor general of India.
• The judges of the Supreme Court and the High Court.
• State Governors etc.

Question 5.
Judicial Powers of the President.
Answer:

1. The President can grant pardons, reprieves, respites or remission of punishments.
2. He appoints the Judges of the Supreme Court and State High Courts.
3. He can also remove them on an address by the Parliament.

Question 6.
Article 352. [Mar. 18, 16]
Answer:
Article 352 of the Indian constitution empowers the President to impose National Emergency during the period of war, External Aggression, Armed Rebellion or internal disturbance. So far National Emergency was proclaimed on Four occasions. They are :

1. Chinese Aggression (1962)
2. Indo – Pak War (1965)
3. Indo – Pak war in the context of Bangladesh Liberation movement (1971) and
4. Opposition’s call for blocking Parliament (1975).

Question 7.
Article 356.
Answer:
Article 356 of Indian Constitution empowers the President to proclaim the constitutional emergency. If the President on receipt of a report from the governor or otherwise is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule.

Question 8.
Financial Emergency.
Answer:
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatened then he may proclaim financial emergency in the country as per Article 360 of the Indian Constitution.

Question 9.
National Emergency.
Answer:
Article 352 of the Indian constitution empowers the president to impose National Erpergency during the period of war, External Aggression, Armed Rebellion or Internal disturbances. When National Emergency is in force, the Federal provisions of our constitution cease to operate. So far, National Emergency was imposed four times in 1962, 1965, 1971 and 1975.

Question 10.
Qualifications required for contesting as Vice-President. [Mar. 16]
Answer:
A person to be eligible for election as Vice-president should possess the following qualifications : ,

1. He should be a citizen of India.
2. He should have completed 35 years of age.
3. He should be qualified for election as a member of the council of states.
4. He should not hold any office of profit under the Union, State or Local Governments in India.

Question 11.
Chairman of Rajya Sabha,
Answer:
The Vice President is the ex-officio chairman of the Rajya Sabha. As such he enjoys the same powers like the speaker of Lok Sabha such as

1. Presiding over the meetings of Rajya Sabha.
2. Maintaining discipline, decency and decorum in the House.
3. Exercising casting vote in case of a tie.
4. Protecting the privileges and rights of the members.

Question 12.
Appointment of Prime Minister.
Answer:
After the conduct of General elections of the Lok Sabha, the President has to invite the majority party leader of the Lok Sabha to form the Government. When no single party is able to secure majority seats in Lok Sabha, the President invites the Leader of a coalition to form the Government. The president uses his discretionary powers in this regard.

Question 13.
Categories of Union Council of Ministers.
Answer:
There are three kinds of ministers in the union council of ministers. They are

1. Cabinet Minister.
2. Ministers of State.
3. Deputy Ministers.

1) The Cabinet Ministers are entrusted with the maintenance of some important ministries like Finance, Home, Defence etc.

2) The Ministers of state act as the heads of some important sections in the Ministry. They are directly responsible to the Prime Minister for their activities.

3) The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers.

Question 14.
Any two functions of the Union Cabinet.
Answer:

1. The Union Cabinet formulates the policies of the union government. It finalizes the domestic as well as foreign policies of the nation after having serious deliberations.
2. It pilots several bills in the Parliament at various stages and strives to secure the approval of the latter.

Question 15.
Collective Responsibility. [Mar. 18]
Answer:
Article 75 (3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of omissions and commissions. They act as a team tinder the leadership of the Prime Minister. They sail together, they swim together and they sink together.