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AP State Board Syllabus AP SSC 10th Class Telugu Solutions 10th Class Telugu Grammar Samasalu సమాసాలు Notes, Questions and Answers.

AP State Syllabus SSC 10th Class Telugu Grammar Samasalu సమాసాలు

సమాసాలు

సమాసం :
వేరు వేరు అర్థాలు కల రెండు పదాలు కలసి, ఏకపదంగా ఏర్పడితే దాన్ని ‘సమాసం’ అంటారు.

గమనిక :
అర్థవంతమైన రెండు పదాలు కలిసి, కొత్త పదం ఏర్పడడాన్ని సమాసం అంటారు. సమాసంలో మొదటి పదాన్ని పూర్వ పదం అంటారు. రెండవ పదాన్ని ఉత్తర పదం అంటారు.
ఉదా :
‘రామ బాణము’ అనే సమాసంలో, ‘రామ’ అనేది పూర్వ పదం. ‘బాణము’ అనేది ఉత్తర పదం.

1. ద్వంద్వ సమాసం :
రెండు కాని, అంతకంటే ఎక్కువ కాని నామవాచకాల మధ్య ఏర్పడే సమాసాన్ని, “ద్వంద్వ సమాసం” అంటారు.

ఈ కింది వాక్యాల్లోని ద్వంద్వ సమాస పదాలను గుర్తించి రాయండి.

1) ఈ అన్నదమ్ములు ఎంతో మంచివాళ్ళు,
జవాబు:
అన్నదమ్ములు

2) నేను మార్కెట్ కు వెళ్ళి కూరగాయలు తెచ్చాను.
జవాబు:
కూరగాయలు

3) ప్రమాదంలో నా కాలుసేతులకు గాయాలయ్యాయి.
జవాబు:
కాలుసేతులు

అభ్యాసం:

I. ఈ కింది ద్వంద్వ సమాసాలకు విగ్రహవాక్యాలు రాయండి.

II. ఈ కింది విగ్రహవాక్యాలను సమాసపదాలుగా మార్చండి.

2. ద్విగు సమాసం :
సమాసంలో మొదటి (పూర్వ) పదంలో సంఖ్య గల సమాసాన్ని ద్విగు సమాసం అంటారు.

అభ్యాసం :
కింది సమాస పదాలను ఉదాహరణలో చూపిన విధంగా వివరించండి.
ఉదా : నవరసాలు – నవ (9) సంఖ్య గల రసాలు
1) రెండు జడలు – రెండు (2) సంఖ్య గల జడలు
2) దశావతారాలు – దశ (10) సంఖ్య గల అవతారాలు
3) ఏడు రోజులు – ఏడు (7) సంఖ్య గల రోజులు
4) నాలుగువేదాలు – నాలుగు (4) సంఖ్య గల వేదాలు

గమనిక :
పైన పేర్కొన్న సమాసాలలో సంఖ్యావాచకం పూర్వ పదంగా ఉండటాన్ని గమనించండి. ఇలా మొదటి పదంలో సంఖ్య గల సమాసాలు ద్విగు సమాసాలు.

3. తత్పురుష సమాసం :
విభక్తి ప్రత్యయాలు విగ్రహవాక్యంలో ఉపయోగించే సమాసాలు “తత్పురుష సమాసాలు.” అభ్యాసము : కింది పదాలను చదివి, విగ్రహవాక్యాలు రాయండి.

గమనిక :
‘రాజభటుడు’ అనే సమాసంలో ‘రాజు’ పూర్వ పదం. ‘భటుడు’ అనే పదం ఉత్తర పదం. ‘రాజభటుడు’ కు విగ్రహవాక్యం రాస్తే ‘రాజు యొక్క భటుడు’ అవుతుంది. దీంట్లో యొక్క అనేది షష్ఠీ విభక్తి ప్రత్యయం. భటుడు రాజుకు చెందినవాడు అని చెప్పడానికి షష్ఠీ విభక్తి ప్రత్యయాన్ని వాడారు. ఈ విధంగా ప్రత్యయాలు విగ్రహవాక్యంలో ఉపయోగించే సమాసాలు తత్పురుష సమాసాలు.

గమనిక :
పూర్వ పదం చివర ఉండే విభక్తిని బట్టి తత్పురుష సమాసాలు వస్తాయి.

అభ్యాసం :
కింది సమాసాలు చదివి, విగ్రహవాక్యాలు రాయండి. అవి ఏ తత్పురుష సమాసాలో తెలపండి.

తత్పురుష సమాసాలు :
విభక్తులు ఆధారంగా ఏర్పడే తత్పురుష సమాసాలను గూర్చి తెలిసికొన్నారు. కింది వాటిని కూడా పరిశీలించండి.
1) మధ్యాహ్నము – అహ్నము యొక్క మధ్యము (మధ్య భాగం)
2) పూర్వకాలము – కాలము యొక్క పూర్వము (పూర్వ భాగం)

గమనిక :
పై వాటిలో మొదటి పదాలైన మధ్య, పూర్వ అనే పదాలకు ‘ము’ అనే ప్రథమా విభక్తి ప్రత్యయం చేరడం వల్ల
‘మధ్యము’, ‘పూర్వము’గా మారతాయి. ఇలా పూర్వపదానికి ప్రథమా విభక్తి ప్రత్యయం రావడాన్ని ‘ప్రథమా తత్పురుష సమాసం’ అంటాము. కింది వాటిని పరిశీలించండి.
1) నఞ్ + సత్యం = అసత్యం – సత్యం కానిది
2) నఞ్ + భయం = అభయం – భయం కానిది
3) నఞ్ + అంతము = అనంతము – అంతము కానిది
4) నఞ్ + ఉచితం = అనుచితం – ఉచితము కానిది

గమనిక :
సంస్కృతంలో ‘నఞ్’ అనే అవ్యయం, వ్యతిరేకార్థక బోధకం. దీనికి బదులు తెలుగులో అ, అన్, అనే ప్రత్యయాలు వాడతారు. పై ఉదాహరణల్లో వాడిన ‘నం’ అనే అవ్యయాన్ని బట్టి, దీన్ని “నః తత్పురుష సమాసం” అంటారు.

అభ్యాసము :
కింది పదాలకు విగ్రహవాక్యాలు రాసి, సమాస నామము పేర్కొనండి.

4. కర్మధారయ సమాసం :
‘నల్లకలువ’ అనే సమాస పదంలో ‘నల్ల’, ‘కలువ’ అనే రెండు పదాలున్నాయి. మొదటి పదం ‘నల్ల’ అనేది, విశేషణం. రెండో పదం ‘కలువ’ అనేది నామవాచకం. ఇలా విశేషణానికీ, సామవాచకానికీ (విశేష్యానికీ) సమాసం జరిగితే, దాన్ని కర్మధారయ సమాసం అంటారు.

4. అ) విశేషణ పూర్వపద కర్మధారయ సమాసం :
విశేషణం పూర్వపదంగా (మొదటి పదంగా) ఉంటే, ఆ సమాసాన్ని “విశేషణ పూర్వపద కర్మధారయ సమాసం’ అంటారు.
ఉదా :
1) తెల్ల గుర్రం – తెల్లదైన గుర్రం.
తెలుపు (విశేషణం) (పూర్వపదం) – (మొదటి పదం) గుర్రం – (నామవాచకం) (ఉత్తరపదం)- (రెండవ పదం)

ఆ) విశేషణ ఉత్తరపద కర్మధారయ సమాసం :
మామిడి గున్న’ అనే సమాసంలో మామిడి, గున్న అనే రెండు పదాలున్నాయి. మొదటి పదం (పూర్వపదం) ‘మామిడి’ సోమవాచకం, రెండో పదం (ఉత్తరపదం) గున్న అనేది విశేషణం. ఇందులో విశేషణమైన ‘గున్న’ అనే పదం ఉత్తరపదంగా – అంటే రెండో పదంగా ఉండడం వల్ల, దీన్ని ‘విశేషణ ఉత్తరపద కర్మధారయ సమాసం’ అంటారు. అభ్యాసము : కింది పదాలను చదివి, విగ్రహవాక్యాలు రాసి, ఏ సమాసమో రాయండి.

ఇ) సంభావనా పూర్వపద కర్మధారయ సమాసం :
‘తమ్మివిరులు’ అనే సమాసంలో, మొదటి పదమైన ‘తమ్మి’, ఏ రకం విరులో తెలియజేస్తుంది. ఇలా పూర్వపదం, నదులు, వృక్షాలు, ప్రాంతాలు మొదలైన వాటి పేర్లను సూచిస్తే దాన్ని సంభావనా పూర్వపద కర్మధారయ సమాసం అంటారు.
ఉదా : మట్టి చెట్టు – మట్టి అనే పేరు గల చెట్టు – సంభావనా పూర్వపద కర్మధారయ సమాసం
గంగానది – గంగ యనే పేరు గల నది – సంభావనా పూర్వపద కర్మధారయ సమాసం
భారతదేశం – ‘భారతం’ అనే పేరు గల దేశం – సంభావనా పూర్వపద కర్మధారయ సమాసం

ఈ) ఉపమాన పూర్వపద కర్మధారయ సమాసం :
‘కలువ కనులు’ అనే సమాసంలో కలువ, కనులు అనే రెండు పదాలున్నాయి. దీనికి ‘కలువల వంటి కన్నులు’ అని అర్థం. అంటే కన్నులను కలువలతో పోల్చడం జరిగింది. సమాసంలోని మొదటి పదం (పూర్వపదం) ఇక్కడ ‘ఉపమానం’ కాబట్టి దీన్ని ఉపమాన పూర్వపద కర్మధారయ సమాసం అంటారు.

ఉ) ఉపమాన ఉత్తరపద కర్మధారయ సమాసం :
‘పదాబ్దము’ అనే సమాసంలో పద (పాదం) మరియు, అబ్జము (పద్మం) అనే రెండు పదాలున్నాయి. వీటి అర్థం పద్మము వంటి పాదము అని. ఇక్కడ పాదాన్ని పద్మం (తామరపూవు)తో పోల్చడం జరిగింది. కాబట్టి పాదం ఉపమేయం. పద్మం ఉపమానం. ఉపమానమైన అబ్జము అనే పదం, ఉత్తరపదంగా (రెండవపదం) ఉండడం వల్ల దీన్ని ఉపమాన ఉత్తరపద కర్మధారయ సమాసం అంటారు.

అభ్యాసం :
కింది సమాసాలకు విగ్రహవాక్యాలు రాసి, సమాస నామాలు పేర్కొనండి.

5. రూపక సమాసం :
‘విద్యాధనం’ – అనే సమాసంలో విద్య, ధనం అనే రెండు పదాలున్నాయి. పూర్వపదమైన విద్య, ధనంతో పోల్చబడింది. కాని ‘విద్య అనెడి ధనం’ అని దీని అర్థం కనుక, ఉపమాన, ఉపమేయాలకు భేదం లేనంత గొప్పగా చెప్పబడింది. ఈ విధంగా ఉపమాన, ఉపమేయాలకు భేదం లేనట్లు చెబితే అది ‘రూపక సమాసం’.

6. బహుప్రీహి సమాసం :
అన్య పదార్థ ప్రాధాన్యం కలది. కింది ఉదాహరణను గమనించండి. ”
చక్రపాణి – చక్రం పాణియందు (చేతిలో) కలవాడు. ‘విష్ణువు’ అని దీని అర్థం. దీంట్లో సమాసంలోని రెండు పదాలకు అనగా “చక్రానికి” కాని “పాణికి” కాని ప్రాధాన్యం లేకుండా, ఆ రెండూ మరో అర్థం ద్వారా “విష్ణువును” సూచిస్తున్నాయి. ఇలా సమాసంలో ఉన్న పదాల అర్థానికి ప్రాధాన్యం లేకుండా, అన్యపదముల అర్థాన్ని స్ఫురింపజేసే దాన్ని బహుప్రీహి సమాసం అంటారు. అన్య పదార్థ ప్రాధాన్యం కలది. ‘బహుబ్లి హి సమాసం’. అభ్యాసం : కింది పదాలకు విగ్రహవాక్యాలు రాసి, సమాసం పేరు రాయండి.

7. అవ్యయీభావ సమాసం :
అవ్యయం పూర్వపదముగా ఉన్న సమాసాలను, “అవ్యయీభావ సమాసాలు” అంటారు.

అవ్యయం :
అవ్యయాలు అనగా లింగ, వచన, విభక్తి లేని పదాలు. ఈ విధమైన భావంతో ఉన్న సమాసాలను అవ్యయీభావ సమాసాలు అంటారు. ఈ క్రింది సమాస పదాలను, వాటికి రాయబడిన విగ్రహవాక్యాలను పరిశీలించండి.

గమనిక : ‘మధ్యాహ్నము” అనే సమాస పదానికి, విగ్రహం ‘మధ్యము – అహ్నము’ అని చెప్పాలి. ఇది ‘ప్రథమా తత్పురుష సమాసం’ అవుతుంది. అవ్యయీభావ సమాసం కాదు.

పాఠ్యపుస్తకంలోని ముఖ్య సమాసాలు – విగ్రహవాక్యాలు :

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 11 Electric Current.

AP State Syllabus SSC 10th Class Physics Important Questions 11th Lesson Electric Current

10th Class Physics 11th Lesson Electric Current 1 Mark Important Questions and Answers

Question 1.

Find the quantity of current in the above circuit. (AP March 2017)
Answer:
R = 3 + 5 + 2 = 10 Ω
I = \(\frac{1.5}{10}\) = 0.15 A.

Question 2.

Three resistors A, B and C are connected as shown in the figure. Each of them dissipates energy to a maximum of 18 W. Find the maximum current that can flow through the three resistors. (TS March 2015)
Answer:

Question 3.
What happens if we use a fuse made up of same wire which is used to make the electric circuit? (TS March 2017)
Answer:
It doesn’t work as a fuse. If high voltage occurs fuse do not melt and circuit will not be opened / breaked. So home appliances will be damaged.

Question 4.
Write any two differences between ohmic and non- ohmic conductors. (TS June 2018)
Answer:

Question 5.
Draw the electric circuit with the help of a Battery, Voltmeter, Ammeter, Resistance and connecting wires. (TS March 2018)
Answer:

Question 6.
What happens, if the household electric appliances are connected in series? (TS March 2019)
Answer:
If all household appliances are connected in series, then if one appliances stop working due to failure then all the appliances stops working due incomplete circuit.

Question 7.
Define lightning.
Answer:
Lightning is an electric discharge between two clouds or between cloud and earth.

Question 8.
Define drift speed or drift velocity.
Answer:
The electrons in the conductor move with a constant average speed called drift speed or drift velocity.

Question 9.
Define conductors.
Answer:
The materials which can conduct electricity are called conductors.
Eg : Copper, Silver, Aluminium.

Question 10.
Define insulators.
Answer:
The materials which can’t conduct electricity are called insulators or non – conductors.
Eg: Wood, Rubber.

Question 11.
Define semi-conductors.
Answer:
The materials whose resistivity is 10⁵ to 10¹⁰ times more than that of metals and 10¹⁵ to 10¹⁶ times less than that of insulators.
Eg: Silicon, Germanium,

Question 12.
How does a battery work?
Answer:
In a circuit, the battery stores chemical energy and this energy converts into electric energy. Thus a battery works.

Question 13.
Define lattice.
Answer:
Conductors like metals contain a large number of free electrons while the positive ions are fixed in their locations. The arrangement of the positive ions is called lattice.

Question 14.
Define potential difference.
Answer:
Work done by the electric force on unit positive charge to move it through a distance is called potential difference.

Question 15.
Define electromotive force.
Answer:
Electromotive force (emf) is defined as the work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

Question 16.
Define resistance of a conductor.
Answer:
The obstruction to the motion of the electrons in a conductor is called resistance of a conductor.

Question 17.
Define resistivity (ρ).
Answer:
Resistivity is a constant.

Question 18.
Write Ohm’s formula.
Answer:
V = IR, where V is the potential difference (voltage), I is the electric current and R is the resistance.

Question 19.
Define resistor.
Answer:
The material which offers resistance to the motion of electrons is called a resistor.

Question 20.
What is a multimeter?
Answer:
It is an electronic measuring instrument that combines several measurement functions in one unit.

Question 21.
Define electric power.
Answer:
The product of voltage and electric current is called electric power.

Question 22.
Define electric energy.
Answer:
The product of electric power and time is called electric energy.

Question 23.
What is lattice?
Answer:
The arrangement of positive ions in a conductor is known as lattice.

Question 24.
Name two special characteristics of fuse wire.
Answer:
High resistivity and low melting point.

Question 25.
Name two special characteristics of heating coil.
Answer:
High resistivity and high melting point.

Question 26.
How does resistivity vary with material of conductor?
Answer:
The resistivity is less for a good conductor and is large for a bad conductor.

Question 27.
If length of a particular conductor increased by two times and its area of crosssection decreased by four times, then what happens to its resistivity?
Answer:
The resistivity of the conductor does not change because resistivity does not depend on dimensions of conductor.

Question 28.
What does the slope of V -1 graph for an Ohmic conductor represent?
Answer:
For an Ohmic conductor the slope of V -1 graph represents the resistance.

Question 29.
Express the units of ohm in terms of volt and ampere.
Answer:
1) The SI unit of resistance is ‘Ohm’.
2) Ohm = \(\frac{\text { Volt }}{\text { Ampere }}\)

Question 30.
What is the resultant resistance of this combination?

Answer:
R, R, R Ω resistances are in parallel.
⇒ Resultant resistance = \(\frac{R}{3}\)

Question 31.
What are the quantities are conserved in Kirchhoff’s and Is’ laws?
Answer:
According to Ist law, charge, and 2^nd law, energy are conserved.

Question 32.
If the length and radius of a conductor are both halved. What happend resistance of wire?
Answer:
Length and radius are halved.

Resistance is doubled.

Question 33.
If work done is W and the charge that flows through is Q, then what is the equation 1 of potential difference?
Answer:

Question 34.
How many electrons constitutes a current of one ampere?
Answer:
6.25 × 10¹⁸ electrons in one second.

Question 35.
What are the maximum and minimum resistances are prepared by 30Ω, 30Ω, 30Ω?
Answer:
Maximum resistance = 30 + 30 + 30 = 90Ω
Minimum resistance = \(\frac{30}{3}\) = 10 Ω

Question 36.
Electric current I = nqA,v_d. Write the representation of letters.
Answer:
n = Electron density
A = Area of cross – section
v_d = drift velocity
q = charge of electron .

Question 37.
What is the resistance of bulb marked 60W and 120V?
Answer:

Question 38.
From figure, if V_A = 10V, then V_B = ?

Answer:
V_A – IR – E = V_B
10 – 1 × 5 – 3 = V_B
10 – 5 – 3 = V_B
2 = V_B
∴ V_B = 2 volts

Question 39.
Write the examples of non-ohmic conductors.
Answer:
Non-ohmic conductors are electrolytes, semi conductors, vacuum tubes.

Question 40.
What is meant by electric shock?
Answer:
When the current flows through the body the functioning of organs inside the body gets disturbed. This disturbance inside the body is felt as electric shock.

Question 41.
There is no electric shock on bird, when stand on the Electric wires. Why?
Answer:
When the bird stands on a high voltage wire, there is no potential differences between the legs of the bird, so no current passes through the bird.

Question 42.
Which factors are influence the resistivity of wire?
Answer:
a) Temperature,
b) Nature of material.

Question 43.
The length of wire is doubled and area of cross-section also doubled. What is the change in resistivity.
Answer:
Resistivity is independent of length and Area of cross-section.
Resistivity is not change.

Question 44.
A battery of 6V is applied across a resistance of 15Ω. Find the current flowing through the circuit.
Answer:

Question 45.
The formula V = I R is applicable for what substances?
Answer:
V = IR is applicable for
a) Ohmic conductors,
b) Non-ohmic conductor.

Question 46.
Resistance of a conductor of length 75 cm is 3.250. Calculate the length of a similar conductor, whose resistance is 16.25Ω.
Answer:

Question 47.
If four wires of each resistance R are joined to form a square, then what is the resistance between its opposite vertices?
Answer:

Resistance of ABC = 2R
Resistance of ADC = 2R
2R & 2R are parallel ⇒ Resultant = ‘R’

Question 48.
How much power consumption in a day of 100W television utilised 10 hours?
Answer:
Power consumption = \(\frac{100 \mathrm{~W} \times 10 \mathrm{~h}}{1000}\)
1 KWH or 1 unit.

Question 49.
How should we connect a voltmeter to measure voltage?
Answer:
The voltmeter must be connected in parallel to the electric device to measure the potential difference across the ends of the electric device.

Question 50.
How are ammeter and voltmeter connected in a circuit?
Answer:
Ammeter is always connected in series and voltmeter is always connected in parallel in a circuit.

Question 51.
Is the voltmeter connected in series or parallel in circuit? Why?
Answer:
Voltmeter should be connected parallel in the circuit to measure the potential difference between two points of conductor.

Question 52.
State whether the home appliances like Fridge, TV, Computer are connected in series or parallel Why?
Answer:
They are connected in parallel because if any one device is damaged, the rest will work as usual because the circuit does not break.

Question 53.
Why are copper wires used as connecting wires?
Answer:
Copper is a good conductor of electricity so copper wires are used as connecting wires.

Question 54.
Why is the fuse wire fitted in a porcelain casing?
Answer:
Porcelain is an insulator of electricity. So fuse wire is fitted in a porcelain casing.

Question 55.
Draw the diagram of potential – current of a copper conductor.
Answer:

Temperature T = 27°C
It is a straight line passing through origin.

Question 56.
Draw the shape of V – I graph for a silicon.
Answer:

Silicon is a semi conductor.
It is not obey the Ohm’s law.

Question 57.
Is there any application of Ohm’s Law in daily life?
Answer:

• Electrical device like electric bulb, iron box and regulators are some applications of Ohm’s Law.
• Fuse in household circuits is also another application of Ohm’s Law.

Question 58.
Two wires of the same material and same length have radii r₁ and r₂ respectively. Compare (i) their resistance, (ii) their resistivities.
Answer:

Resistivity for a same material is same. So their ratio =1:1.

Question 59.
A wire of resistance is doubled on itself, then what is its new resistance?
Answer:
Suppose length is l, area of cross-section is A and resistance is R.

Question 60.
Calculate effective resistance between points A and B.

Answer:
R₁ = 1 Ω, R₂ = 2 Ω are in series. So R’ = 1 + 2 = 3 Ω.
Now R₃ and R’ are in parallel.

Question 61.
A fuse is rated 8A. Can it be used with an electrical appliance of rating 5 KW and 200 V?
Answer:
Given P = 5 KW = 5 × 1000 = 5000 W; V = 200 V.

So a fuse of rate 8 A is not suitable because it uses current of 25 A.

Question 62.
A current of 2A is passed through a coil of resistance 75 Ω for 2 minutes. How much heat energy is produced?
Answer:
Given i = 2A, R = 75 Ω and t = 2 min. = 2 × 60 sec. = 120 sec.
Heat energy produced H = i² Rt = 2² × 75 × 120 = 36000 J = 36 KJ.

Question 63.
What is ,the resistance under normal working conditions of a 240 V electric lamp rated at 60 W?
Answer:
P = 60 W; V = 240V.

Question 64.
State the use of Ammeter. How should the Ammeter be connected in electric circuit?
Answer:
Ammeter should be connected in series in a circuit.

Question 65.
What is current value of x?

Answer:
2 + 3 = 5 of ‘A’
5 – 1 = 4 of ‘B’
2x = 4
x = 2 A at ‘C’

Question 66.
What is the value of V_A, when V_B = 8V?

Answer:
V_A – 6 × 1 – 3 = V_B
V_A – 6 – 3 = 8
V_A = 8 + 9
V_A = 17 volts

Question 67.
In the figure, how much current passing through 6Ω resistor?

Answer:
Answer:
R₁ : R₂ = 6 : 4
R₁ : R₂ = 3 : 2
i₁ : i₂ = R₂ : R₁ = 2 : 3
∴ i₁= 2A
i₂ = 3A

10th Class Physics 11th Lesson Electric Current 2 Marks Important Questions and Answers

Question 1.
Observe the graph of potential difference (V) drawn between two ends of a conductor and current (I) passing through it. Answer the following questions :

a) Which law is used to explain the graph? State it.
b) What is the resistance of the conductor? (AP June 2017)
Answer:
a) Ohm’s law is used to explain the given graph.
Ohm’s Law :
The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.

Question 2.
Draw the experimental set-up to verify that V/I is constant for a conductor. (TS March 2016)
Answer:

Question 3.
A house has 3 tube-lights of 20 watts each. On the average, all the tube-lights are kept on for five hours. Find the energy consumed in 30 days. (AP SCERT : 2019-20)
Answer:
Number of tube lights = 3
Wattage = 20 watts each
Consumed hours = 5 hrs

Consumed energy for 30 days = 0.3 × 30 = 9 KWH (or) 9 units.

Question 4.
How do the resistors 6 Ω, 10 Ω to be connected in a circuit to get minimum resistance? Find the resultant resistant of the circuit. (TS June 2019)
Answer:

1. Resistors 6Ω, 10Ω should be connected in parallel connection in a circuit to get minimum resistance.
2. Resultant resistance
   

Question 5.
Write two examples for ohmiq and non-ohmic materials each. (TS June 2019)
Answer:
Ohmic materials : Copper, Aluminium
Non – ohmic materials : Germanium, Silicon

Question 6.
Give reasons for using lead in making fuses.
Answer:

• Lead is used in making fuses because it has low melting point EK resistivity.
• If the current in the lead wire exceeds certain value, the wire will heat up and melt, so the circuit in the households is opened and all the electric devices are saved.

Question 7.
How can we decide the direction of electric current in a conductor?
Answer:
We know I = nqAv_d. In this n and A are positive. Hence the direction of current is determined by the signs of the charge ‘q’ and drift speed v_d.

1. For electrons, q is negative and v_d is positive. Then the product of q and v_d is negative. So the direction of electric current is opposite to the flow of negative charge.
2. For positive charge, the product of q and v_d is positive. Hence, the direction of electric current can be taken as the direction of flow of positive charges.

Question 8.
What are the devices used in a circuit?
Answer:
1) Ammeter :
It is used to measure current.

2) Volt meter:
It is used to measure potential difference across the ends of conductor.

3) Rheostat:
It varies current in a circuit.

4) Switch :
It is useful to make a circuit or break a circuit.

5) Cell:
It is source of electric energy in the circuit.

6) Multimeter :
It is useful to measure current, voltage and resistance in the circuit.

Question 9.
A student says “Potential difference and Emf are same.” Justify your answer.
Answer:
Both are different because potential difference is the work done by the electric force on unit positive charge to move it through a distance between two points whereas emf is the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

Question 10.
Define Ohmic and non-ohmic conductors and give two examples each of them.
Answer:
Ohmic conductors :
The conductors which obey Ohm’s law are called ohmic conductors, e.g, : Copper, Iron.

Non-ohmic conductors :
The conductors which do not obey Ohm’s law are called non-ohmic conductors, e.g. : Semiconductors, Electrolytes.

Question 11.
Explain the Junction law of Kirchhoff.
Answer:

1. At any junction point the sum of the currents into the junction must be equal to the sum of currents leaving the junction.
2. There is no accumulation of electric charges at any junction in a circuit.
   I₁ + I₄ + I₆ = I₅ + I₂ + I₃

Question 12.
Write differences between overloading and short circuiting.
Answer:
Current chooses a path which has least resistance. So sometimes electrical appliances get damaged by passage of excess of current due to short circuit.

When so many electrical appliances are connected to the same electrical main point, maximum current can be drawn from the mains which causes overheating and may cause a fire which is called overloading.

Question 13.
The V-I graph for a series combination and for parallel combination of two resistors is shown in figure. Which of the two A or B, represent parallel combination? Give reason for your answer.

Answer:
i) For same change in I, change in V is less for the straight line A than for the straight line B (i.e., straight line A is less steeper than B).
ii) So the straight line A represents small resistance, while straight line B represents more resistance.
iii) In parallel combination, the resistance decreases, while in series combination, the resistance increases, so A represents a parallel combination.

Question 14.
Two resistors are joined with a battery such that
a) same current flows in each resistor.
b) potential difference is small across each resistor.
c) equivalent resistance is less than either of the two resistors.
d) equivalent resistance is more than either of the resistors.
State how are the resistors connected in each of the above case.
Answer:
a) Since same current is flowing in each resistor, they are connected in series.

b) Potential difference is same across same resistor that shows they are connected in parallel.

c) Equivalent resistance is less than the either of the two resistances which shows they are connected in parallel because in parallel connection resistance decreases.

d) Equivalent resistance is more than the individual resistances which indicate that they are connected in series. Since in series connection resistance increases.

Question 15.
Which of the cables, one rated 5A and other 15 A will be thicker wire? Give reason for your answer.
Answer:
i) To carry larger current, the resistance of wire should be low, so its area of cross-section should be large.

ii) Because the second cable carrying 15 A ampere current which indicates it has low resistance that is more surface area which implies it is thicker wire.

Question 16.
When a potential difference 30V is applied across a resistor, it draws a current of 3A. If 20V is applied across the same resistor, what will be the current?
Answer:
Situation : 1

Situation : 2

Question 17.
Is Ohm’s Law universally applicable for all conducting elements? If not, give example of elements which do not obey the Ohm’s Law.
Answer:

• Ohm’s Law is not applicable for all conducting elements.
• For example, some semi-conductors like silicon, germanium do not obey the Ohm’s Law.
• Those which do not obey Ohm’s Law are called non-ohmic materials.
• LED’s are non-ohmic materials.

Question 18.
Alloys are used in electrical heating devices rather than the pure metal. Why?
Answer:
Alloys are homogeneous mixtures of two or more metals. Generally alloys have more resistivity than the metals from which they have been prepared. As the resistivity increases heating effect of conductor also increases. So alloys are preferred in heating devices.

Question 19.
A switch should not be touched with wet hands. Why?
Answer:

• A switch should not be touched with wet hands.
• If water reaches the live wire, it forms a conducting layer between the hand and the live wire of the switch through which the current passes to the hand, and the person may get a total shock.

Question 20.
Which material is used for power transmission? Why?
Answer:

• The wires which are used for connections and for power transmission are made of material such as copper or aluminium.
• The reason is their resistivity is very small and they are made thick so that their resistance can be considered to be negligible.
• Further, the loss of electrical energy due to heating is also negligible in them.

Question 21.
Which material is preferred for heating element? Why?
Answer:

• The heating elements or resistance wires (or standard resistors) are made of material such as nichrome, manganin, constantan, etc. for which the resistivity is quite large and the effect of change in temperature on their resistance is negligible.
• So electrical energy is converted into heat energy when current passes through the wire.

Question 22.
Draw the symbols of the following.
i) Battery
ii) Resistance
iii) Ammeter
iv) Voltmeter
v) Key
vi) Rheostat
Answer:

Question 23.
Draw V-I graphs of Ohmic and non-ohmic conductors.
Answer:

Question 24.
Draw a circuit diagram with a cell, an electric bulb, an ammeter and plug key.
Answer:

Question 25.
Draw a circuit diagram to verify the Ohm’s Law.
Answer:

Question 26.
If 60 C of charge passes through a conductor for 1 minute, find the current through the conductor.
Answer:

Question 27.
The work done in moving 6 C of charge through a circuit is 12 J. Find the potential difference in the circuit.
Answer:

Question 28.
Resistance of two resistors are 6 Ω, 12 Ω respectively. Find the resultant resistance if the resistors are connected (1) in series (2) in parallel.
Answer:

Question 29.
10 equal resistors of resistance 20 are connected in a circuit. Find the resultant resistance if they are connected in series or in parallel ?
n = 10; R = 20 Ω
In series connection resultant resistance R’ = nR = 10 × 20 = 200 Ω

Question 30.
From the figure find the current through 6 Ω, 12 Ω resistors and find the resultant resistance in the circuit and also find current in the circuit.

Answer:
R₁ =6 Ω; R₂ = 12 Ω; V = 6 V

Question 31.
Find which has greater resistance. 1 KW heater or a 100 W tungsten bulb, both marked for 230 V.
Answer:

Question 32.
Two wires (one is copper and other is aluminium) have equal area of cross-section and have the same resistance. Find which one is longer.
Answer:
Suppose the resistance of copper and aluminium wires are R₁ and R₂. Suppose their area of cross-section is A.
The resistivity of copper (ρ₁) = 1.68 × 10^-8
The resistivity of aluminium (ρ₂) = 2.82 × 10^-8

Question 33.
Three equal resistances are connected In series, then in parallel. What will be the ratio of their resultant resistances?
Answer:
Suppose the resistance of equal resistors is ‘R’. Suppose they are connected in series. Then their equivalent resistance R’ = R + R + R = 3R
If they are connected in parallel their equivalent resistance R” = \(\frac{R}{3}\)
∴ Ratio of resultant resistances = R’: R” = 3R : \(\frac{R}{3}\) = 9 : 1

Question 34.
How is Ideal earthing helpful during short circuiting?
Answer:

• During short circuiting an excessive current flows through the live wires.
• It will pass to earth through the earth wire if there is local earthing otherwise it may cause a fire due to overheating of the live wires.

Question 35.
You have three resistor values 2Ω, 3Ω and 5Ω. How will you join them Sd that the total resistance Is less than 2Ω? Find resultant resistance.
Answer:
Given R₁ = 2Ω, R₂ = 3Ω, R₃ = 5Ω.
The resistors should be joined in parallel.

Question 36.
An electric kettle Is rated 3 KW, 250 V. Give reason whether this kettle can be used in a circuit which contains a 13 A fuse?
Answer:
V = 250 V, P = 3KW = 3000 W

The fuse is suitable because safe limit of current for kettle is 12 A.

Question 37.
Two resistors of 4 Ω and 6 Ω are connected parallel. The combination is connected across 6 V battery of negligible resistance. Find, i) the power supplied by the battery and ii) the power dissipated in each resistance.
Answer:
i) R₁ = 4 Ω, R₂ = 6 Ω, V = 6V.

ii) In parallel connection the resultant resistance

Question 38.
In the circuit shown below calculate the value of x if the equivalent resistance between A and B is 4 Ω.

Answer:
Given R₁ = 4 Ω, R₂ = 8Ω, R₃ = x Ω and R₄ = 5 Ω.
And resultant resistance R = 4 Ω.
R₁ and R₂ are in series.
Resultant resistance R’ = R₁ + R₂ = 4Ω + 8Ω = 12 Ω.
R₃ and R₄ are also in series.
Resultant resistance R” = R₃ + R₄ = x + 5
R’ and R” are in parallel.

Question 39.
A wire of 9 Ω resistance having 30 cm length is tripled on itself. What is its new resistance?
Answer:
Given R = 9 Ω, l = 30 cm and suppose area of cross-section A.

10th Class Physics 11th Lesson Electric Current 4 Marks Important Questions and Answers

Question 1.
A house has four tube-lights, three fans and a television. Each tube-light draws 40 W. The fan draws 80 W and the television draws 100 W. On an average, all the tube-lights are kept on for 5 hours, all fans for 12 hours and the television for 6 hours everyday. Find the cost of electric energy used in 30 days at the rate of Rs. 3.00 per KWH. (AP March 2015)
Answer:
The power used by
1) Four tube-lights of 40 W for 5 hours in 30 days
= 4 × 40 × 5 × 30 = 2400 WH

2) Three fans of 80 W for 12 hours in 30 days
= 3 × 80 × 12 × 30 = 86400 WH

3) One television of 100 W for 6 hours in 30 days
= 1 × 100 × 6 × 30 = 18000 WH

4) Total electric energy used
= 24000 + 86400 + 18000 = 128400 WH

W.H. converted into K.W.H. = 128.4 K.W.H. \(\left[\because \frac{128400}{1000}\right]\)
Cost of 1 unit = Rs. 3.00
Amount to be paid for 128.4 units = 128.4 × 3 = Rs. 385.20

Question 2.
Observe the given circuit. (AP June 2017)
R₁ and R₂ are two resistors and R₁ = R₂ = 4Ω. Emf of the battery E is 10 V.

Answer the following questions.
a) How are the resistors R₁ and R₂ connected in the circuit ?
b) What is the potential difference across R₁?
c) What is the effective resistance of the circuit?
d) What is the total current drawn from the battery?
Answer:
a) Resistors R₁ and R₂ are connected in parallel in the given circuit.
b) The potential difference across R₁ is ‘E’ volts = 10 V.

Question 3.
State Kirchhoff’s Loop law and explain. (AP June 2018)
Answer:
Loop law :
The algebraic sum of the increases and decreases in potential difference across various components of the circuit in a closed circuit loop must be zero.

Explanation :
Let us imagine in a circuit loop the potential difference between the two points at the beginning of the loop has a certain value. As we move around the circuit loop and measures the potential difference across each component in the loop, the potential difference may decrease or increase depending upon the nature of the element like a resistor or battery. But when we have completely traversed the circuit loop and arrive back at starting point. The net change in the potential difference must be zero. Thus the algebraic sum of changes in potential difference is to be zero.
(OR)
Let us apply loop law to a circuit as below.

for the loop ACDBA
– V₂ + I₂R2₂ – I₁R₁ + V₁ = 0

for the loop EFDCE
-(I₁ + I₂)R₃ – I₂R₂ + V₂ = 0

for the loop EFBAE
-(I₁ + I₂)R₃ – I₁R₁ + V₁ = 0

Question 4.

Observe the above diagram and answer the following. (AP March 2018)
a) Are all the resistors connected in parallel or series?
b) What is the equivalent resistance of the combination of three resistors?
c) In this system, which physical quantity is constant?
d) If R₁ = 2Ω, R₂ = 3Ω, R₃ = 4Ω. find equivalent resistance.
Answer:
a) Connected in series.
b) R_eq = R₁ + R₂ + R₃
c) Current (I)
d) R_eq = R₁ + R₂ + R₃
=2 + 3 + 4
= 9Ω

Question 5.
How do you verify that resistance of a conductor of uniform cross-section area is proportional to the length of the conductor at constant temperature? (TS March 2015)
Answer:
Aim :
To verify the relation between re.M+iance and lenath of the conductor.

1) Required material :
Wires or spokes of different lengths with same cross-section area of the same metal.
2) Battery
3) Ammeter
4) Key
5) Connecting wires.

Procedure :
1) Construct a circuit with Battery, Ammeter, Switch (key) and connecting wires, keeping some space at the both ends.
2) Connect the selected wires or spokes at the ends to complete the circuit.

3) Connect the wires or spokes individually and record the current using ammeter.

Conclusion :
If the current flowing in the circuit decreases with an increase in the length of the wire or spokes (Resistance increases), we can say that the resistance of the conductor is proportional to the length of the conductor.

Question 6.
What are the factors affecting the resistance of an electric conductor? Explain any two factors. (TS June 2015)
Answer:
The factors affecting the resistance of an electrical conductor are

1. Nature of material
2. Temperature
3. Length of the conductor
4. Area of cross-section of conductor

Explanation :

1. As the temperature increases the resistance increases and vice versa.
2. As the material changes resistance changes.
3. Resistance is directly proportional to length of the conductor (if T and A are kept constant). R ∝ l
4. Resistance is inversely proportional to area of cross-section (if 1 and T are kept constant). R ∝ \(\frac{1}{A}\)

Question 7.
What is the relationship between length of a conductor and its resistance? Write the experimental procedure to verify that relationship. (TS Junc 2017)
Answer:

• The resistance of a conductor is directly proportional to its lenght for a constant potential difference.
• Take iron spokes of different lengths with same cross-sectional arreas.
• Make a circuit by connecting an iron spoke with battery, ammeter and switch in series.
• Put the switch on and allow the current to pass in the circuit. Measure the ammeter reading.
• Repeat this for other lengths of the iron spokes. Note the corresponding values of currents.
• The resistance of each spoke increases with increase in the length of the spoke.

Question 8.
Find the resultant resistance for the following given arrangement. Find the current when this arrangement is connected with 9 V battery. (TS March 2017)

Answer:
The diagram is not clear so award 4 marks in the public examinations.

Question 9.
12 V battery is conncected in a circuit and to this 4Ω, 12Ω resistors are connected in parallel, 3Ω resistor is connected in series to this arrangement. Draw the electric circuit from this information and find the current in the circuit. (TS June 2018)
Answer:

Question 10.
In a circuit,60V battery, three resistances R₁ = 10 Ω, R₂ = 20 Ω and R₃ = x Ω are connected in series. If 1 ampere current flows in the circuit, find the resistance in R₃ by using Kirchhoffs loop law. (TS March 2018)
Answer:

Given
I = 1 amp
R₁ = 10Ω
R₂ = 20Ω
R₃ = xΩ
V = 60V
In ADCBA loop
– IR₃ – IR₂ – IR₁ + V = 0
– lx – 1 × 20- 1 × 10 + 60 = 0
– x – 30 + 60 = 0
– x + 30 = 0
x = 30Ω

Question 11.
List out the material required for the experiment “The effect of increasing of cross-section of a conductor upon its resistance” and write the experimental procedure. (TS June 2019)
Answer:
Aim :
To show that the effect of increasing of cross-section of a conductor upon its resistance.

Required Material :
Battery eliminator, Key, Ammeter, Manganin wires of equal lengths but different cross sectional areas, Copper connecting wires.

Procedure :

• Collect manganin wires of equal lengths but different cross sectional areas.
• Make a circuit as shown in the figure.

• Connect one of the wires between points ‘P’ and ‘Q’.
• Note the value of the current using the ammeter connected to the circuit and note it.
• Repeat this with other wires.
• Note the corresponding values of currents in each case and note them.

Conclusion :
We can notice that the current flowing through the wire increases with increasing their cross sectional area.

Question 12.
Derive an expression to find drift velocity of electrons.
Answer:

• Consider a conductor with cross sectional area A. Assume that the two ends of the conductor are connected to a battery to make the current flow through it.
• Let ‘v_d‘ be the drift speed of the charges and ‘n’ be the number of charges present in the conductor in a unit volume.
• The distance covered by each charge in one second is ‘v_d‘

• Then the volume of the conductor for this distance = Av_d
• ∴ The number of charges contained in that volume = n.Av_d
• Let q be the charge of each carrier.
• Then the total charge crossing the cross sectional area at position D in one second is ‘n q A_d‘.
  This is equal to electric current.
  ∴ Electric current I = n q Av_d.
  ∴ v_d = \(\frac{I}{nqA}\)

Question 13.
Derive an expression to measure emf of a battery.
Answer:
Electromotive force (emf) is defined as the work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

1. Let this chemical force be F_c.

2. This chemical force does some work to move a negative charge ‘q’ from positive terminal to negative terminal against the electric force Fe. Let this work be ‘W’.

3. ∴ The work done by the chemical force to move 1 coloumb of charge from the

This S.I unit of emf is ‘volt’ and is measured using voltmeter.

Question 14.
What are the factors on which the resistance of conductor depends? Give the corresponding relation.
Answer:

• The value of resistance of a conductor depends on temperature for constant potential difference.
• Resistance of a conductor depends on the material of the conductor.
• Resistance of a conductor is directly proportional to its length, i.e., R ∝ l.
• Resistance of a conductor is inversely proportional to the area of cross-section of the material, i.e., R ∝ \(\frac{1}{A}\)

Question 15.
What do you mean by (i) short circuit (ii) overloading? What are the safety precautions taken to avoid these problems in domestic electric circuits?
Answer:
Short circuit:
Sometimes current chooses a path which has least resistance which is called short circuit.

Overloading :
The over heating due to drawing excess of current from a single main is called overloading.
Precautions to avoid damage due to short circuit and overloading.

1. Using fuse
2. Connecting electrical appliances in various mains.
3. Earthing

Question 16.
A circuit is set up as shown in the figure. Calculate the current and potential difference across R₁, R₂ and R₃, when
a) keys K₁ and K₂ are both closed,
b) key K₁ is closed and K₂ is open,
c) key K₁ is open and K₂ is closed.

Answer:
a) When both the keys K₁ and K₂ are closed :
The resistors R₁ and R₃ are parallel.
So resultant resistance R_p = \(\frac{R_{1} R_{3}}{R_{1}+R_{3}}=\frac{6 \times 4}{6+4}\) = 2.4 Ω.
The resistor R₁ and R_p are in series as shown in figure.

Total resistance of circuit R_s= R₁ + R_p = 6 + 2.4 = 8.4 Ω
Current I = \(\frac{V}{R_{S}}=\frac{4.2}{8.4}\) = 0.5 =A.
Potential difference across R₁ is V₁ = IR₁ = 0.5 × 6 = 3V.
Potential difference across the combination of R₂ and R₃ is
V’ = V – V₁ = 4.2 – 3 = 1.2 V.
Now since R₂ and R₃ are in parallel,
Potential difference across R₂ = Potential difference atross R₃ = V’ = 1.2 V

b) When the key K₁ is closed and key K₂ is open : The resistor R₃ will not be in circuit.
The resistors R₁ and R₂ are in series.
Total, resistance R_s = R₁ + R₂ = 6 + 6=12 Ω.
Current I = \(\frac{V}{R_{s}}=\frac{4.2}{12}\) = 0.35 A
The same current will flow through each resistor R₁ and R₂.
Potential difference across R₁ is V₁ = IR₁ = 0.35 × 6 = 2.1 V.
Potential difference across R₂ is V₂ = IR₂ = 0.35 × 6 = 2.1 V.
The current and potential difference across R₃ will be zero.

c) When key K₁ is open and key K₂ is closed :
No current flows through R₁ R₂ and R₃ since the circuit is incomplete. Hence potential difference across R₁, R₂ and R₃ is zero.

Question 17.
For the combination of resistance shown in figure, find the equivalent resistance between (a) C and D, (b) A and B.

Answer:
a) Between C and D :
The resistors R₂, R₃ and R₄ are in series. They can be replaced by an equivalent resistance R_s where
R_s = R₂ + R₃ + R₄ = 3 + 3 + 3 = 9 Ω
The resistance R₅ and R_s are in parallel between the points C and D.
The equivalent resistance between C and D then

Thus the equivalent resistance between C and D is 2.25 Ω.

b) Between A and B :
Now the resistors R₁, R_p and R₆ are in series between the points a and B.

The equivalent resistance between A and B is
R_AB = R₁ + R_p + R₆ = 3 +2.25 + 3 = 8.25 Ω.

Question 18.
How does resistance and resistivity vary with temperature?
Answer:

• For a metallic conductor, the resistance increases w ith the increase in temperature. The resistance of filament of bulb is more w hen it is glowing than when it is not glowing. The specific resistance or resistivity also increases with increase in temperature.
• For alloys (such as constantan and manganin), the resistance and the resistivity remains practically unchanged with the increase in temperature.
• For semi conductors (such as silicon, germanium, etc.), the resistance decreases with the increase in temperature.
  eg.: the resistance of carbon also decreases with the increase in its temperature.

Question 19.
Why is a copper wire unsuitable for making fuse wire?
Answer:

• A fuse is a short piece of wire made up of a material of high resistivity and of low melting point so that it may easily melt due to overheating when current in excess to the prescribed limit passes through it.
• The thickness of wire is different in different fuses depending on the amount of current which is permitted to flow through them.
• Generally an alloy of lead and tin is used as the material of the fuse wire because it has a high resistivity and low’ melting point.
• A copper wire is unsuitable for using as fuse wire because copper has low resistivity and high melting point.
• Therefore the use of an ordinary’ wire as a fuse must be avoided.

Question 20.
Why is current rating of a fuse required?
Answer:
1) The electric wiring for light and fan circuit uses a thin fuse wire of low current carrying capacity because the line wire has a current carrying capacity of 5A.

2) Thicker fuse wires of higher current carry ing capacity (15 A) are used for large current consuming appliances such as air conditioner, geyser, washing machine, etc. because the line wire for such dev ices have current carrying capacity of 15A.
The current rating of a fuse in a circuit can be obtained by the following relation.

Question 21.
Describe the activity with the help of diagram to establish the relationship between Current (I) flowing in a conductor and potential difference (V) maintained across its ends.
Answer:
Aim :
To establish the relationship between Current (I) flowing in a conductor and potential difference (V) maintained across its ends.

Material required :
5 dry cells of 1.5 V each, conducting wires, an ammeter, a volt meter, thin iron spoke of length 10 cm, LED and key.
Diagram :

Procedure :

• Connect a circuit as shown in the figure.
• Solder the conducting wires to the ends of the iron spoke.
• Close the key.
• Note the readings- of current from ammeter and potential difference from volt meter in the given table,

• Now connect two cells (in series) instead of one cell in the circuit.
• Note the respective readings of the ammeter and voltmeter and record the values in the table.
• Repeat the same for three cells, four cells and five cells respectively.
• Record the values of V and I corresponding to each case in the table.
•  Find \(\frac{V}{I}\) for each set of values.
  Conclusion : The ratio of \(\frac{V}{I}\) is constant.
• From this activity we can conclude that the potential difference between the ends of the iron spoke (conductor) is directly proportional to the current passing through it. This means V ∝ I.

Question 22.
In an experiment to verify Ohm’s Law the following values are given below. Draw a graph of ‘I’ versus ‘V’. Show that the graph conforms Ohm’s Law and find the resistance of the resistor.

Answer:
1) Graph between ‘V’ and ‘I’.

2) From the above graph,
Straight line §hows that the relation between potential difference (V) and current (I) as \(\frac{V}{I}\) is constant.

3) V ∝ I.

4) The potential difference between the ends of a conductor is directly prbportional to the electric current passing through it at constant temperature.

5) This is the Ohm’s Law and the graph conforms it.
Resistance of the resistor

Question 23.
Identify the defects 1ft the circuit. Redraw it.

Answer:
Defects in the circuit:

1. Ammeter was connected in parallel in the circuit.
2. Volt meter was connected in series in the circuit.
3. Positions of resistor and battery were reversed.

Correct diagram :

Question 24.
What is the advantage of MCB over fuse?
Answer:

• These days instead of fuses, Miniature Circuit Breakers (MCB) are used for each lighting circuit.
• They switch off the circuit in a very short time duration in case of short-circuiting or some fault in the line.
• After repairing the fault in the circuit, the MCB is again switched on.
• Thus, the use of MCB is better than a fuse. It avoids the inconvenience of connecting a new fuse wire and it is much safer due to its quick response.

Question 25.
A circuit is shown in the picture.

The current passing through A is I.
a) What is the potential difference between A and B?
b) What is the equivalent resistance between A and B?
c) What amount of current is flowed through C and D?
Answer:
a) According to KirchhofPs loop law the algebraic sum of increase and decrease in potential difference across various components of the circuit in a closed circuit loop must be zero.
So the potential difference across CD is zero because it is a closed loop.

b) Here 20 Ω, 5 Ω are parallel to each other and resultants are in series to each other. Resultant resistance of 20 Ω and 5 Ω.

Question 26.

Observe the picture. The potential values at A, B, C are 70 V, 0 V, 10 V
a) What is the potential at D?
b) Find the ratio of the flow of current in AD, DB, DC.
Answer:

a) By following Ohm’s law potential difference is (V) = IR
In the given circuit we are applying junction laws.
‘D’ works as junction so, I = I₁ + I₂
Let p.d at D is V₀.

b)

Question 27.
Observe the circuit R₁ = R₂ = R₃ = 200 Ω. If reading of voltmeter is 100 V, resistance of voltmeter is 1000 Ω.
Find the Emf of the battery.

Answer:
Given values are R₁ = R₂ = R₃ = 200 Ω.
and Voltmeter reading = V = 100 V.
Resistance of Voltmeter = R_v = 1000 Ω.
In the given circuit R₁ and R₂ are in series R = R₁ + R₂ = 200 + 200 = 400 Ω
Resultant resistance (400 Ω) and voltmeter (1000 Ω) are always in parallel.

Question 28.
A circuit is made with a copper wire as shown in the diagram. We know that conductor’s resistance is directly proportional to its length. Calculate the equivalent resistance between points 1 and 2.

Answer:

Let the resistance of the wire be ‘R’ and length of the wire be ‘l’.
The shape of the circuit be square length of the side (l) = R
In a square, diagonal is \(\sqrt{2}\) times its length = \(\sqrt{2}\)l
Resistance towards diagonal is \(\sqrt{2}\)R
The circuit diagrams for the given arrangement are along PTR and QTS is ineffective as no current flows through it.
PQ and PS are in series so effective resistance is R₁ + R₂ = R + R = 2R.
QR and SR are in series so effective resistance is R₁ + R₂ = R + R = 2R.
Redrawn of the circuit again as resultant resistance between the points 1 and 2 is

Question 29.
From the adjacent figure,

i) Find the potential at D.
ii) Find the current that passes through AD, DB and DC.
Answer:
Suppose from A and C current is flowing through the circuit and at B current is flowing away from the circuit. Suppose potential at D is V.
∴ I₁ + I₂ = I₃ (where I₁ I₂ are current into the junction. I₃ is the current away from the junction)

Question 30.
Find the electric current drawn from the battery of emf 8V from the given circuit.
Answer:

According Kirchhoffs loop law
6I₁ + 3I₁ – 8 = 0
9I₁ = 8
I₁ \(\frac{8}{9}\) = 0.89 A
∴ Cuttent in 8 V is 0.89 A

Question 31.
A household uses the following electric appliances.
i) Refrigerator of rating 400 W each for ten hours each day.
ii) Two electric fans of rating 80 W each for 12 hours each day.
iii) Six electric tubes of rating 18 W each for 6 hours each day. Calculate the electric bill of the household in a month if the cost per unit electric energy is Rs. 3.00.
Answer:
i) Electrical energy consumed by Refrigerator in a month (in KWH)

ii) Electrical energy consumed by two electrical fans in a month (in KWH)

iii) Electrical energy consumed by electric tubes in a month (in KWH)

Total electrical energy consumed by all the electrical appliances
= 120 + 57.6 + 19.44 = 197.04 units
Cost of 1 unit = ₹ 3
Cost of 197.04 units = 197.04 × 3 = ₹ 591.12

Question 32.
What is the reason for connecting the fuse in the live wire?
Answer:

• The fuse is always connected in the live wire of the circuit.
• If the fuse is put in the neutral wire, due to excessive flow of current the fuse burns, current stops flowing in the circuit.
• But the appliance remains connected to the high potential point of the supply through the live wire.
• Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

Question 33.
In the diagram given below, two resistors R₁ and R₂ of 3Ω and 6Ω respectively are connected in parallel across a battery of potential difference 12 V. Calculate the electrical energy consumed in 1 minute in each resistance.

Answer:
Given R₁ = 3 Ω, R₂ = 6 Ω, V= 12 V, t = 1 min 60 sec.
The resistors R₁ and R₂ are connected in parallel, so the voltage V across each resistor is equal to 12 V.

Question 34.
Calculate the electrical energy consumed in a month, in a house using 2 bulbs of 100 W each and 2 fans of 60 W each, if the bulbs and fans are used for an average of 10 hours each day.
If the cost per unit is ₹ 3, calculate the amount of electrical bill to be paid per month.
Answer:
Given power of each bulb = 100 W
Power of each fan = 60 W
Time t =10 hours each day
Power of 2 bulbs = 2 × 100 = 200 W
Power of 2 fans = 2 × 60 = 120 W
Total power = 200 + 120 = 320 W = \(\frac{320}{1000}\) = 0.32 KW
Time duration of consumption 10 hours per day per month
= 10 hours × 30 = 300 hours
∴ Total energy consumed = Total power x Time duration
= 0.32 KW × 300 h = 96 KWh
∴ Total cost = 96 x 3 = ₹ 288.

Question 35.
What resistance must be connected to a 15 Ω resistance to provide an effective resistance of 6 Ω?
Answer:
Given R₁ = 15 Ω, R₂ = ?, Effective resistance R_p = 6 Ω.
Since the effective resistance has decreased, R2 must be connected in parallel with R₁

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 7 Human Eye and Colourful World.

AP State Syllabus SSC 10th Class Physics Important Questions 7th Lesson Human Eye and Colourful World

10th Class Physics 7th Lesson Human Eye and Colourful World 1 Mark Important Questions and Answers

Question 1.
Define the power of lens. (AP June 2016)
Answer:
The reciprocal of focal length is called “power of lens”.

Question 2.
What physical quantity can be found in an experiment done with prism? (AP June 2017)
Answer:

1. Angle of deviation.
2. Refractive index of a prism.

Question 3.
What is the relation between Power and Focal length of the lens? (AP June 2018)
Answer:
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 4.
Draw the diagram of a lens which will be recommended by an eye doctor to a long sighted patient. (TS June 2015)
Answer:

The lens is convex lens.

Question 5.
What is the cause of Presbyopia? (TS March 2015)
Answer:
Presbyopia is vision defect when the ability of accommodation of the eye decreases with ageing.

Question 6.
Draw a ray diagram to show the angle of deviation when a ray of light passes through a glass prism. (TS March 2015)
Answer:

d = angle of deviation

Question 7.
Suggest reasons for the phenomenon associated with the following. The sky appearing blue. (TS March 2015)
Answer:
The reason for blue sky is due to the scattering of light by the molecules of N₂ and O₂ whose size is comparable to the wavelength of blue light.

Question 8.
+50 cm focal length bi-convex lens is recommended to correct the defect of vision of a mart. Find the power of the lens. (TS June 2016)
Answer:
f = +50 cm
Power (P) = \(\frac{100}{\mathrm{f}}\) D (in cm); P = \(\frac{100}{50}\) = 2 D
Power of the bi-convex lens is 2D.

Question 9.
What happens if the eye lens of a person cannot accommodate its focal length more than 2.4 cm? (TS March 2017)
Answer:
The person can able to see certain distance only he cannot see distance objects. For correction, he should use concave lens.

Question 10.
Write the reason for Sun appears red during the Sun-rise and Sun-set. (TS June 2018)
Answer:
Due to the high velocity (wave length) of red right, it reaches our eye without under go scattering. So, sun appears red during sunrise and sunset.

Question 11.
A person is unable to see distant objects. Show the defect of vision of the person with the help of ray diagram. (TS March 2018)
Answer:
1) His vision defect is myopia.
2) Ray diagram

Question 12.
Which molecules of atmosphere act as scattering centres are responsible for the blue sky? (TS June 2019)
Answer:
Oxygen and Nitrogen molecules.

Question 13.
Write any one application of a prism. (AP SA-I; 2019-20)
Answer:

1. Prism is used in scopes like binoculars, telescopes and light houses.
2. Prism is used to create artificial rainbow.

Question 14.
Mention the function of retina in a human eye.
Answer:
It acts as a screen, (which the image is formed) for image formed.

Question 15.
State the role of ciliary muscles in accommodation.
Answer:
It can adjust the focal length of the eye lens.

Question 16.
Why is normal eye not able to see clearly the objects kept closer than 25 cm?
Answer:
The maximum accommodation of a normal human eye is reached when the object is at a distance of 25 cm from the eyes. The focal length of the eye lens cannot be decreased this minimum limit.

Question 17.
What is “Power of accommodation of the eye” (or) “Least distance of clear vision”?
Answer:
The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.

Question 18.
What is “Iris” and “Pupil”?
Iris :
The muscular diaphragm between the aqueous humour and the lens is called ‘iris’. Iris is the coloured part that we see in an eye.

Pupil:
The small hole in iris is called ‘pupil’.

Question 19.
What are three common defects of vision?
Answer:
The common defects of vision are i) Myopia ii) Hypermetropia iii) Presbyopia.

Question 20.
What is “Far point”?
Answer:
The point of maximum distance at which the eye lens can form an image on retina is called ‘far point’.

Question 21.
What is power of lens?
Answer:
The reciprocal of focal length is called power of lens. Power of lens focal length = \(\frac{1}{\text { focal length }}\)

Question 22.
What is the function of pupil in human eye?
Answer:
It allows the light falling on iris.

Question 23.
What is the purpose of human eye?
Answer:
The purpose of eye is to see and perceive the objects around us.

Question 24.
What is the principle of the working of human eye?
Answer:
It acts as camera having a lens system forming an invented real image on the light sensitive screen, retina.

Question 25.
What is the nature of the image formed on the retina?
Answer:
Real, inverted and same sized.

Question 26.
What is meant by dispersion of light?
The splitting of white light into its component colours when it passes through the prism is called dispersion of light.

Question 27.
On which factors does the colour of the scattered white light depend?
Answer:

1. Angle of scattering
2. Distance travelled by light
3. Size of the molecules.

Question 28.
Why is normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The focal length of eye lens cannot decrease below 25 cm.

Question 29.
A person is advised to wear spectacles with concave lenses. What type of defect of vision is he suffering from?
Answer:
Myopia (or) short sightedness.

Question 30.
A person suffering from an eye-defect uses lenses of power – 1D. Name the defect he is suffering from and the nature of lens is used.
Answer:
Defect:
Myopia, Nature of lens, Concave/divergence.

Question 31.
Name the two phenomena involved in the formation of rainbow.
Answer:
Dispersion of light and internal reflection.

Question 32.
Identify the following part of the human eye.
1. Where is image of an object formed?
2. Which controls size of pupil?
Answer:

1. Image of an object is formed on retina.
2. Iris controls the size of pupil.

Question 33.
What is the relation between power of lens and focal length (f)?
Answer:
Power of lens (concave/convex)
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 34.
Explain about “Bending of light”.
Answer:

1. When a light travels from rarer to denser medium, it bends towards the normal.
2. And from denser to rarer medium, it moves away from the normal.

Question 35.
What is sclerotic?
Answer:
It is the outermost covering of the eye. It protects the vital internal parts of the eye.

Question 36.
What is “Retina”?
Answer:
Retina is the internal part and the light sensitive surface of the eye. It is equivalent to the photographic film in a camera.

Question 37.
What is “Atmospheric refraction”?
Answer:
When the light rays pass through the atmosphere having layers of different densities and refractive indices, refraction of light takes place. This refraction of light by the earth’s atmosphere is called atmospheric refraction.

Question 38.
What is a “Telescope”?
Answer:
The instrument which is used to see the distant objects such as a star, planet (moon, sun) or distant tree is called telescope.

Question 39.
Have you seen a rainbow in the sky after rain? How is it formed?
Answer:

1. A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on raindrops after the rain.
2. It is formed due to reflection, refraction and dispersion of sunlight by tiny water droplets present in the atmosphere.

Question 40.
“To look at the twinkling of stars is a wonderful experience.” How does it happen?
Answer:
The continuous changing atmosphere (due to varying atmospheric temperature and density) refracts the light from the stars by varying amounts and in different directions from one moment to the next.

Question 41.
Some things appear blue on a misty day. Give two examples.
Answer:

1. The long distance hills covered with thick growth of trees appear blue.
2. The smoke coming from a cigarette or an incense stick (agarbatti) appears blue on a misty day.

Question 42.
Which coloured suits do rescue workers wear?
Answer:
Rescue workers wear orange coloured suits during any rescue operations.

Question 43.
Which colour is best for school buses?
Answer:
Orange or yellow colour is best for school busses.

Question 44.
What is persistence of vision?
Answer:
The time for which the sensation of vision (of an object) continues in the eye is called persistence of vision. It is about 1/16^th part of a second.

Question 45.
Why can’t some people identify some colours?
Answer:
Rods identify the colours in the retina. If some rods are absent, the distinction of colours is not possible. In such cases, persons can’t identify some colours.

Question 46.
Write the reasons for colour blindness.
Answer:

1. Absence of colour responding rod cells in the retina.
2. Due to genetic disorder.

Question 47.
Why does it take some time to see objects in a dim room when we enter the room from bright sunlight outside?
Answer:
In bright light, the size of the pupil is small to control the amount of light entering the eye. When we enter a dim room, it takes some time so that the pupil expands and allows more light to enter and helps to see things clearly.

Question 48.
What is tyndall effect?
Answer:
The phenomenon of scattering of white light by colloidal particles is known as ‘Tyndall effect”.

Question 49.
Give two examples illustrating “Tyndall effect”.
Answer:

• A fine beam of sunlight entering a smoke filled room through a hole. Smoke particles scatter the white light and hence the path of light beam becomes visible.
• Sunlight passing through the trees in forest.
• Tiny water droplets through the trees in forest.

Question 50.
An eye camp was organised by the doctors in a village. What were the benefits to organise such camps in rural areas?
Answer:

1. To make people aware of eye diseases
2. To take proper and balanced diet.

Question 51.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
In the eye, the image distance (distance between eye lens and retina) is fixed and it cannot be changed. So when we increase the distance of an object, there is no change in the image distance.

Question 52.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is at a distance of 25 cm from the eye.
The focal length of the eye lens cannot be decreased below this. Thus an object placed closer than 25 cm cannot be seen clearly by a normal eye.

Question 53.
Write the relation between intensity of scattered light (I) and wavelength (λ).
Answer:
Light of short wavelength is scattered more than the light of long wavelength.
i.e., Intensity of scattered light (I) ∝ \(\frac{1}{\text { wavelength }(\lambda)}\)

Question 54.
Why would the sky look dark if the earth had no atmosphere?
Answer:
If the earth has no atmosphere, no particles present either. Thus no scattering of light. Then, the sky appears dark.

Question 55.
Why do different coloured rays deviate differently in the prism?
Answer:
Because the angle of refraction of different colours is different while passing through the glass prism.

Question 56.
What prevents rainbow from being seen as complete circles?
Answer:
The earth comes in the way of the rainbow and prevents it to form a complete circle.

Question 57.
When a monochromatic light passes through a prism will it show dispersion?
Answer:
No, it will not show any dispersion but show deviation.

Question 58.
Will a star appear to twinkle if seen from free space (say moon)?
Answer:
No, because there is no atmosphere in free space for refraction to take place.

Question 59.
A short-sighted person may read a book without spectacles. Comment.
Answer:
The statement is true, because a short-sighted person has difficulty in observing far off objects.

Question 60.
What is angle of vision?
Answer:
The maximum angle, at which we can see the whole object is called angle of vision.

Question 61.
What is cornea?
Answer:
The front curved portion of eye, which is covered by a transparent protective membrane is called the cornea.

Question 62.
Which lens do you use to correct the eye defect, Myopia?
Answer:
Bi-concave lens are used to correct the eye defect, Myopia.

Question 63.
What happens to power of lens if (i) focal length is increased, (ii) focal length is decreased?
Answer:

1. If focal length is increased, then power of lens decreases.
2. If focal length is decreased, then power of lens increases.

Question 64.
What type of image is formed by magnifying glass?
Answer:
It forms virtual, erect and magnified image.

Question 65.
How is power of lens related to its focal length?
Answer:
Power of lens is inversely proportional to its focal length.

Question 66.
What is “Scattering of light”?
Answer:
The process of re-emission of light in all directions with different intensity is called scattering of light. The re-emitted light is called scattered light.

Question 67.
Ramu is unable to see letters on the blackboard sitting at the last bench in the class¬room. What is the defect from which Ramu is suffering?
Answer:
Ramu is suffering from Myopia.

Question 68.
Vinay is able to read letters in a book beyond certain distance from least distance of distinct vision. What is the eye defect of Vinay?
Answer:
Vinay is suffering from Hypermetropia.

Question 69.
How can an eye lens accommodate its focal length?
Answer:
To see an object comfortably and distinctly, one must hold at a distance about 25 cm from his/her eyes. This distance is called least distance of distinct vision.

Question 70.
Write the uses of “rods” and “cones”.
Answer:
Retina contains about 125 million receptors called rods and cones. Rods identify the colour and cones identify the intensity of light.

Question 71.
Frame some questions, that you are going to ask your friend who is suffering from eye sight.
Answer:

1. Are you not able to see near objects or far objects?
2. Are you not able to see both near and far objects?

Question 72.
Write some more examples to find dispersion of light as VIBGYOR.
Answer:

1. Formation of rainbow.
2. When white light passes through water drop.

Question 73.
Why do stars twinkle?
Answer:
Due to change in atmospheric conditions, density changes so position keeps on changing.

Question 74.
“Sky appears dark to passengers flying at very high altitudes.” Why?
Answer:
At very high altitude, there is no atmosphere. So there is no scattering of light at such heights. So sky appears dark to passengers.

Question 75.
Why are danger signals red? (OR) Why are danger signals shown in red colour?
Answer:
Among the colours of visible light, red has more wavelength and least scattered. Thus, red colour can easily go through fog or mist or smoke without getting scattered. It can be seen from long distance. So red colour is used in universal danger signal.

Question 76.
How can you identify regarding the type of defect a person is suffering from by physically touching his spectacles?
Answer:
By touching the spectacles we can find out whether the lens is concave or convex lens and hence the defect from which he suffers.

Question 77.
A person cannot see objects beyond 1-2 m distinctly. What should be the nature of the defect and what type of lens should be used to correct the defect?
Answer:
The person is suffering from Myopia. It can be corrected by using concave lens.

Question 78.
How do you appreciate the working of “Retina”?
Answer:

1. It is the innermost delicate membrane having a large number of receptors called ‘rods’ and ‘cones’.
2. The rods identify the colour and the cones identify the intensity of light.
3. The retina is a part on which the image of an object is formed.

Question 79.
How do you appreciate the working of “Optic nerve”?
Answer:

1. Optic nerve consists of a large number of fibres.
2. These optic nerve fibres are connected to the rods and cones.
3. Optic nerve fibers transmit the light signals to the brain.

Question 80.
We see advertisements for eye donation on television or in newspaper. Write the importance of such advertisements.
Answer:
Eye donation advertisements are important as :

1. The people aw;are about donation of organs after their death.
2. Sympathetic nature towards others.

Question 81.
An eye donation camp is being organised by social workers in your locality. How and why would you help in this cause?
Answer:

1. We can intimate other people to participate in the camp.
2. As a human being, we should also register our eyes for donation after death.

Question 82.
A short sighted person cannot see clearly beyond 2m. Calculate the power of lens required to correct his vision.
Answer:

Question 83.
How does the power of a lens change if its focal length is doubled?
Answer:
The power get halved.

Question 84.
A person cannot see distinctly objects kept beyond 2m. Which among these lens is useful to correct the defect
a) – 0.2D
b) – 0.5 D
c) + 0.2D
d) +0.5D?
Answer:
The person is suffering from Myopia.
The lens is concave and its focal length f = – 2m.
\(P=\frac{1}{5}=\frac{1}{-2}=-0.5 D\)
So to correct the defect concave lens of – 0.5 D power should be used (b).

Question 85.
Express the power of concave lens of focal length 20 cm with its sign.
Answer:

10th Class Physics 7th Lesson Human Eye and Colourful World 2 Marks Important Questions and Answers

Question 1.
How do you appreciate the working of iris in the eye? (AP June 2018)
Answer:
Iris helps in controlling the amount of light entering the eye through pupil.

Question 2.
What is the reason for the blue colour of the sky? (AP March 2018)
How do you appreciate the role of molecules in the atmosphere in this regard?
Answer:
Blue colour of sky :

1. Atmosphere contains more O₂ and N₂ molacules and they are caused to blue colour of the sky.
2. The size of molecules of O₂ and N₂ are comparable with the blue colour and scatter blue colour only.

Role of molecules in the atomosphere in scattering :

1. Light of certain frequency falls on that atom or molecule.
2. This molecule responds to the light whenever the size of the molecule is comparable to the wavelength of light and vibrates.
3. Due to these vibration, molecule reemits a certain fraction of absorbed energy in all directions. The emitted light is called scattered light.
4. The atoms or molecules are called scattering centres.
5. I appreciate the role of the molecules in scattering.

Question 3.
In which conditions does a rainbow form? Why? (TS June 2015)
Answer:
1) Rainbow forms and appears when,
i) tiny water droplets present in the atmosphere (after rain shower),
ii) sunlight falls on the droplets,
iii) observer watches the rainbow in a specific direction.
2) Rainbow forms due to dispersion of sunlight by tiny droplets, present in the atmosphere, which acts as small prisms.

Question 4.
Draw the ray diagram, showing the correction of defect of vision hyper metropia by using a convex lens. (TS June 2016)
Answer:

Question 5.
Least distance of distinct vision of a person is observed as 35 cm. What lens is useful for him to see his surroundings clearly? Why? (TS March 2016)
Answer:
The least distance of distinct vision is 35 cm. This is more than the least distance of distinct vision of an ordinary person. Hence the person is suffering from Hypermetropia. He has to use double convex lens to see his surroundings clearly.

Question 6.
What happens, if Ciliary muscles do not perform contraction and expansion? Guess and write. (TS June 2018)
Answer:

1. If Ciliary muscles do not perform contraction and expansion, foal length of eye lens do not change.
2. Human eye can see the objects at specific distance only, eye cannot see the object either nearer or far distance.

Question 7.
Write any two situations to observe dispersion of light in your daily life. (TS June 2018)
Answer:
We can observe the dispersion of light in the following situations in our daily life.

1. In the formation of rainbow.
2. When observing sun light through the triangular transparent material like prism, scale edge.
3. At the time of curing of walls of new houses with water.
4. Dispersion of light by inclained plane mirror which is in water.

Question 8.
Write the material that you use to find out the value of refractive index of a prism. What is the necessity of the graph in this experiment? (AP March 2019)
Answer:
The material used to find out the value of refractive index of a prism:
Prism, Piece of white chart, Pencil, Pins, Scale and Protractor.

Necessity of the graph :
To find the angle of minimum deviation graph is required.

Question 9.
Draw a ray diagram showing the correction of myopia eye defect. (TS March 2019)
Answer:
Diagram of Myopia correction :
Note : Draw the diagram using Bi Concave Lens and show the far point (M). Image should form on Retina.

Question 10.
What happen if dispersion and scattering of light do not occur? (TS March 2019)
Answer:
If dispersion does not occur in nature, then there is no rainbow formation and splitting of light into seven different colors. If there is no scattering then the oceans and sky appears to be black. The sun appears white all the time (Including Sunrise and Sunset).

Question 11.
When Mohan viewed white light through a transparent scale, he observed some colours. Predict and write the phenomenon involved in his observation. (AP SCERT: 2019-20)
Answer:

1. The phenomenon involved in his observation is dispersion of light.
2. Splitting of white light into different colours (VIBGYOR) is called dispersion.

Question 12.
A boy who is suffering from eye defect has been given a prescription as -2D. Based on the information given, answer the following questions.
a) Identify the eye defect he is suffering.
b) Write the nature and focal length of the lens. (AP SCERT: 2019-20)
Answer:
a) The boy is suffering from myopia.
b) Nature of the lens :
The lens is biconcave lens.
It is thin at the middle and thicker at the edges.
Focal length of the lens :
Given that power is – 2D

Question 13.
How does eye lens change its focal length? (AP SA-I:2019-20)
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 14.
Kishore wore spectacles. When you saw through his spects the size of his eyes seemed bigger than their original size.
a) Which lens did he use?
b) Explain that defect of vision.
Answer:
a) When we saw through Kishore’s spects the size of his eyes seemed bigger than their original size. This is possible with convex lens only because magnification of the lens is greater than T.

b) The defect he suffers is hypermetropia. This is also called farsightedness.

A person who suffers with this type of defect, he can’t see the objects clearly which are placed near distance because the image is formed beyond the retina. So by using convex lens the rays can be converged on retina.

Question 15.
“God has given the gift for us to see the sunrise and sunset.” Explain the feeling of it.
Answer:

• The sun is visible two minutes before the actual sunrise and remains visible two minutes after the actual sunset.
• The actual sunrise takes place when the sun is just above the horizon.
• The actual sunset takes place when the sun is just below the horizon.

Question 16.
“Smoke coming out of coal fired chimney appears blue on a misty day.” Why?
Answer:

• On a misty day, the air has large amount of tiny particles of water droplets, dust and smoke.
• These tiny particles present in the air scatter blue colour of the white light passing through it.
• When this scattered blue light reaches our eyes the smoke appears blue.

Question 17.
“Motorists use orange light on a foggy day rather than normal white light.” Why?
Answer:

• On a foggy day, the air has large amount of water droplets.
• If a motorist uses white light, the water droplets present in the air scatter large amount of the blue light.
• This on reaching our eyes decreases visibility and hence driving becomes extremely difficult.
• Whereas orange light has longer wavelength and hence it is least scattered.

Question 18.
A rainbow viewed from an airplane may form a complete circle. Where will the shadow of the airplane appear? Explain.
Answer:

• A rainbow viewed from an airplane form a complete circle because the earth does not come along the way of the airplane and rainbow.
• A rainbow is a three dimensional cone of dispersed light it appear as a complete circle.
• The shadow of the airplane appears within the circle of the rainbow.

Question 19.
How do we see colours?
Answer:

• The retina of human eye has a large number of receptors.
• These receptors are of two types i.e., rods and cones.
• The rod cells recognise the colour of light rays, while the cones identify the intensity of light.
• It is these cone cells, which make it possible for a men to see different colours and distinguish between them.

Question 20.
Why do we use lenses in spectacles to correct defects of vision?
Answer:
The process of adjusting focal length is called “accommodation”. This process has to be done by eye itself. Sometimes the eye may gradually lose its power of accommodation. In such condition, the person cannot see the object clearly and comfortably. In this situation, we have to use lenses in spectacles to correct defects of vision.

Question 21.
What is a) far point of the eye and b) near point of the eye?
Answer:
a) The farthest point up to which the eye can see objects clearly without strain (in the eye) is called the far point of the eye. For a normal eye, the far point is at infinity.

b) The minimum distance at which objects can be seen most clearly without strain (in the eye) is called the least distance of distinct vision or the near point of the eye.

Question22.
Write the difference between “Myopia” and “Hypermetropia”.
(OR)
Distinguish between Myopia and Hypermetopia.
Answer:
The eye defect in which people cannot see at long distances but can see nearby objects clearly is called Myopia. The eye defect in which people cannot see near distant objects but can see distant objects is called hypermetropia.

Question 23.
Define the following words.
a) Prism
b) Dispersion of light
c) Scattering of light
Answer:
a) Prism :
A prism is a transparent medium separated from the surrounding medium by at least two plane surfaces which are inclined at a certain angle.

b) Dispersion of light :
The splitting of white light into different colours is called dispersion of light.

c) Scattering of light:
The process of reemission of absorbed light in all directions with different intensities by atoms or molecules is called scattering of light.

Question 24.
Define the words associated with prism with the help of figure.

1) Angle of incidence :
The angle between incident ray and normal is called angle of incidence.

2) Angle of emergence :
The angle between normal and emergent ray is called angle of emergence.

3) Normal:
Perpendicular drawn to the surface of prism.

4) Angle of deviation:
The angle between extended incident ray and emergent ray is called angle of deviation.

Question 25.
State the cause of dispersion, when white light enters a glass prism. Explain with a diagram.
Answer:

• Light is made up of different colours. Each colour travels at its own speed inside a prism.
• Due to this different colours of light bends through different angles with respect to the incident ray, as it passes through a prism.
• The red light bends the least while the violet most.
• Thus, the rays of each colour emerge along different paths and become distinct.
• It is the bond of distinct colours that we see in a spectrum.

Question 26.
What happens to the lens and the ciliary muscles when you are looking at distant objects and near objects?
Answer:
a) The ciliary muscles become relaxed and the lens becomes thin, i.e. its radius of curvature increases. So focal length of eye lens increases for distant object.

b) The ciliary muscles contract and the lens becomes thick, i.e. its radius of curvature decreases. So focal length of eye lens decreases for near objects.

Question 27.
Why does it take some time to see objects in cinema hall when we just enter the hall from bright sunlight? Explain in brief.
Answer:

• The pupil regulates and controls the amount of light entering eye.
• In bright sunlight, the size of pupil is small and when we enter the cinema hall it takes some time for the pupil to expand in size due to dim light.

Question 28.
What are the factors which influence the total angle of deviation?
Answer:

• The angle of incidence at the first surface (i).
• The angle of prism (A).
• Refractive index of the material.

Question 29.
How does eye change its focal length take place in the eyeball?
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 30.
Stars twinkle while planets do not. Why?
Answer:
1) Continuously changing atmosphere refracts light from the stars by different amounts from one moment to the other, when atmosphere refracts more starlight towards us and the stars appear to be bright and when the atmosphere refracts less star-light then the stars appear to be dim.

2) However the planets are nearer to us than the stars, they appear to be comparatively bigger to us so they cannot be considered as a point source, hence no twinkling is seen.

Question 31.
How do earth and stars appear for a person who is on the moon?
Answer:

• For the person who is on the moon, the earth appears blue due to blue colour of sunlight scattered by the earth’s atmosphere reaching him.
• Stars and other heavenly bodies are seen as usual, but without twinkling.

Question 32.
Why does the sky appear dark and black to an astronaut instead of blue?
(OR)
Why does the sky appear dark to the passenger flying at high altitudes?
Answer:

• This is because there is no atmosphere containing air in the outer space to scatter light.
• Since there is no scattered light which can reach our eyes in outer space, the sky looks dark and black there.
• This is why the astronauts who go to outer space find the sky to be dark and black instead of blue.

Question 33.
A person is able to see objects clearly only when these are lying at distance between 60 cm and 250 cm from his eye. What kind of defect of vision is he suffering from?
Answer:
For a normal eye, the near point is at 25 cm and the far point is at infinity. The given person cannot see object clearly either close to the eye or far away from the eye. So he is suffering from presbyopia.

Question 34.
Write the material required in finding the refractive index of a prism.
Answer:
Materials required:
Prism, piece of white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Question 35.
Draw the graph between angle of incidence and angle of deviation.
Answer:

Question 35.
Draw the diagram of scattering of sunlight.
Answer:

Question 35.
How do you appreciate the “Eye Donor”?
Answer:
The human eye is one of the most important sense organs. Without eye we are unable to see the beautiful world. So we have to appreciate “eye donor” for his kindness to give sight to blind people.

Question 36.
How can we appreciate the working of “Iris”?
Answer:

• Iris consists of muscles which help in controlling the amount of light entering the eye through pupil.
• In case of low light the iris makes the pupil to expand and allow more light to enter the eye.
• In case of bright or excess light the iris makes the pupil to contract in order to decrease the amount of light entering the eye. The iris consists of muscles that expand and contract the pupil.

Question 37.
The power of lens is 2.0 D. Find its focal length and state what kind of lens it is.
Answer:
P = \(\frac{1}{f}\) (f in metres)
Given P = 2D
∴ f= \(\frac{1}{P}\) = \(\frac{1}{2}\) = 0. 5 m = 50 cm.
The focal length positive indicates it is a convex lens.

Question 38.
Two convex lenses of powers 1D and 2D are combined together to form a new lens. Then what is the resultant power and focal length of lens?
Answer:
P₁ = 1 D ; P₂ = 2D
Resultant power P = P₁ + P₂ = 1 + 2 = 3D
P = \(\frac{1}{f}\) (f in metres)
∴ f= \(\frac{1}{P}\) = \(\frac{1}{3}\) = 0.3333 m = 33.33 cm.

Question 40.
Figure shows the refraction of light through an equilateral prism. Incident at an angle of 30°. The ray suffers a deviation of 37°. What are the angles marked at A, e and f respectively?
Answer:

Since the prism is an equilateral prism,
A = 60°, also D = 37° and i = 30° (i.e., i₁), e = i₂ = ?
We know that i₁ + i₂ = A + D
30° + e = 60 + 37
e = 97 – 30 = 67°
Also A + f = 180°
f = 180°-60° = 120°

10th Class Physics 7th Lesson Human Eye and Colourful World 4 Marks Important Questions and Answers

Question 1.
Kavya can see distant objects clearly but cannot see objects at near distance. With what eye defect is she suffering? Draw the diagrams showing the defected eye and its correction. (AP June 2016)
Answer:
Kavya is suffering from hypermetropia.
The following diagram shows the defective eye and its correction.

Question 2.
Revathi is a front bench student. She is unable to draw the picture drawn on the blackboard. She got permission from the teacher and sat in the back row. What could be the defect that Revathi is suffering from? Draw the diagram, which shows the correction of the above defect? (AP Mareh 2017)
Answer:
Hypermetropia

Question 3.
An eye specialist suggested a + 2D lens to the person with defect in vision. Which kind of defect in vision does he have? Draw the diagrams to show the defect of vision and its correction with a suitable lens. (TS June 2017)
Answer:
Eye specialist suggested +2D lens, that is convex lens is used to correct Hypermetropia. So the person has Hypermetropia.
Deffect of Vision:

Correction :

Question 4.
How will you calculate the focal length of a biconvex lens that is used to correct the defect of Hypermetropia? Explain it mathematically. (TS March 2017)
Answer:
The person who has hypermetropia cannot see near objects. He can see the objects those are beyond near point (H). For correction of this eye defect the image of the object placed at “least distance of distinct vision (L)” should be at near point (H).
u = -25 cm; v = -d cm

here d > 25, so ‘f gets positive value.
Hence, convex lens should be used.

Question 5.
Mention the required material and chemicals for the experiment of “scattering of light.” Write the experiment procedure. (TS March 2018)
(OR)
Write the required apparatus and chemicals to show the scttering of light experimentally and write the experimental process.
(OR)
How can you demonstrate scatteing of light by an experiment?
Answer:
Aim :
To show the scattering of light.

Material required :
Beaker, sodium thiosulphate, sulphuric acid.

Procedure:

1. Take a solution of sodium-thio-sulphate (hypo) and sulphuric acid in a glass beaker.
2. Place the beaker in an open place where abundant sun light is available.
3. Watch the formation of grains of sulphur and observe the changes in the beaker.
4. We will notice that sulphur participates as the reaction is in progress. At the beginning, the grains of sulphur are smaller in size and as the reaction progress, their size increases due to precipitation.
5. Sulphur grains appear blue in colour at the beginning and slowly their colour becomes white as their size increases.
6. The reason for this is scattering of light.
7. At the beginning, the size of grains is small and almost comparable to the wavelength of blue light. Hence they appear blue in colour.
8. As the size of grains increases, their size becomes comparable to wavelength of other colours.
9. As a result, they acts as scattering centres of all colours.

Question 6.
A boy has been playing games in mobile phone and is suffering from eye defect. The doctor prescribed him to use spectacles of power – 5D. What eye defect is he suffering from? (AP SA-I:2018-19)
Draw a neat diagram which shows the correction of above eye defect.
Answer:
Doctor suggested the boy – 5D lens, that is concave lens. Concave lens is used to correct myopia. So the boy has myopia.
Defect of Vision :

Correction:

Question 7.

Answer the following questions from the above information. (TS June 2019)
i) What is the defect of vision in ‘D’ suffering from? Why does it happen?
ii) Whose defect of vision can be corrected by using Biconcave lens?
iii) Who is suffering with similar defect of vision as of ‘B’?
iv) Who among the above do not have any defect of vision?
Answer:
i) 1) The vision defect of person ‘D’ is presbyopia.
2) Presbyopia happens due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens. This effect can be seen in aged people.
ii) Person A’ defect of vision can be corrected by using Biconcave lens.
iii) Person ’C’ is suffering with similar defect of vision as of ’B’.
iv) Person ‘E’ has no defect of vision.

Question 8.
A prism causes dispersion of white light while a rectangular glass block does not. Explain.
Answer:

• In a prism the refraction of light occurs at two plane surfaces.
• The dispersion of white light occurs at the first surface of prism where its constituent colours are deviated through different angles.
• At the second surface, these split colours suffer only refraction and they get further separated.
• But in a rectangular glass block, the refraction of light takes place at the two parallel surfaces.
• At the first surface, although the white light splits into its constituent colours on refractions, but they split colours on suffering refraction at the second surface emerge out in the form of a parallel beam, which gives an impression of white light.

Question 9.
A convex lens of power 4D is placed at a distance of 40 cm from a wall. At what distance from the lens should a candle be placed so that its image is formed on the wall?
Answer:

So candle should be placed 66.66 cm from the lens.

Question 10.
Explain briefly the reason for the blue of the sky.
Answer:

• Our atmosphere contains different types of molecules and atoms.
• The reason for blue sky is the molecules N₂ and O₂.
• The sizes of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.

Question 11.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:

• For a normal eye, image distance in the eye is fixed.
• This is equal to the distance of retina from the eye lens.
• When we increase the distance of the object from the eye, focal length of eye lens is changed on account of power of accommodation of the eye so as to keep the image distance constant.

Question 12.
A person is able to see objects clearly only when they are lying at distance between 60 cm and 250 cm from his eye. What kind of lenses will be required to increase his range of vision from 25 cm to infinity? Explain.
Answer:
A bi-focal lens consists of concave and convex lens of suitable focal lengths will be required to correct the defect and to increase his range of vision from 25 cm to infinity. In bi-focal lens, the upper half of the lens is concave lens which corrects distant vision and the lower half is convex which corrects near vision.

Question 13.
Discuss why sun is visible before actual sunrise and after actual sunset?
Answer:

• The sun is visible to us about 2 minutes before the actual sunrise and 2 minutes after the actual sunset because of atmospheric refraction.
• By actual sunrise we mean the actual crossing of the horizon by the sun.

• Figure shows the actual and apparent positions of the sun with respect to the horizon.
• The time difference between actual sunset and the apparent sunset is about 2 minutes.
• The apparent flattering of the sun’s disc at sunrise and sunset is also due to the same phenomenon.

Question 14.
When does the colour of sky appear black for an observer?
Answer:

• In the absence of atmosphere, there will not be any scattering of light and so light will reach our eye, i.e. the sky will appear black instead of blue at night in the absence of light.
• On the moon, since there is no atmosphere, there is no scattering of sunlight reaching the moon surface. Hence to an observer on the surface of moon, no light reaches except the light directly from sun. Thus sky will have no colour and will appear black to an observer on the moon surface. This is applicable for any planet which does not have atmosphere.
• When an astronaut goes above the atmosphere of the earth in rocket he sees the sky black.

Question 15.
Why does the colour of clouds appear black?
Answer:

• The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of size bigger than the wavelength of visible light.
• Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light does not reach our eye, the clouds seem black.

Question 16.
Give daily life examples of scattering of light by earth’s atmosphere.
Answer:
Some daily life effects of scattering of sunlight by earth’s atmosphere are

1. Red colour of sun at sunrise and sunset.
2. White colour of sky at noon.
3. Blue colour of sky is due to molecules N₂ and O₂.
4. Black colour of sky is due to the absence of atmosphere.
5. Use of red light for the danger signal because it is least scattered by particles due to its greater wavelength.
6. White colour of clouds is due to rise in temperature.

Question 17.
What is the relation between scattering and wavelength of light? Explain.
Answer:

• Scattering is the process of absorption and then re-emission of light energy.
• The air molecules of size smaller than the wavelength of incident light absorb the energy of incident light and then re-emit it without change in its wavelength.
• The intensity of scattered light is found to be inversely proportional to fourth power of wavelength of light.
• The wavelength of violet is least and red light is most, therefore from the incident white light, violet light is scattered the most and red light is scattered the least.

Question 18.
How can we get this (in human eye) same image distance for various positions of objects?
Answer:

• The ciliary muscle attached with eye lens helps to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• F_(max) – 2.5 cm,u = oo, v = 2.5 cm. The parallel rays coming from a distant object are focussed on the retina with 2.5 cm image distance.
• When the eye is focussed on a nearer object the muscles are strained. So the focal length of the eye lens decreases its minimum value.
• F_(min)= 2.27 cm, u = 25 cm, v = 2.5 cm. The rays from an object (u = 25 cm) at L (point of least distance of distinct vision) are focussed on the retina with 2.5 cm image distance.

Question 19.
Does eye lens form a real image or virtual image? Explain it.
Answer:

• Eye lens forms a real image.
• The light that enters the eye forms an image on the retina.

• The image is obtained on a screen (retina) and it is an inverted image.
• So, we can say eye lens forms a real image.

Question 20.
How does the image formed on retina help us to perceive the object without change its shape, size and colour?
Answer:

• The eye-lens forms a real and inverted image of an object on the retina.
• The retina is a delicate membrane which contains about 125 million receptors called ‘rods’ and ‘cones’ which receive the light signals.
• Rods identify the colour.
• Cones identify the intensity of light.
• These signals are transmitted to the brain through about 1 million optic nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 21.
How do you prove that a prism does not produce colours itself?
Answer:

• A white light from a slit ‘S’ is made to pass through prism P which forms spectrum on a white screen AB.
• A narrow slit H is made on the screen AB, parallel to slit S to allow the light of particular colour to pass through it.
• The light of a particular colour is made to fall on the second prism Q placed with its base in opposite direction to that of the prism P.
• The light after passing through the second prism Q is received on another white screen M.
• It is observed that the colour of light obtained on the screen M is same as that of the light incident on the second prism Q through the slit H.

Question 22.
Draw the structure of human eye and explain its parts.
Answer:

Question 23.
What is the part indicated by an arrow mark? What is its working function?
Answer:

• The part indicated by the arrow mark is ciliary muscle.
• Ciliary muscles help to change the focal length of the eye lens.
• When it relaxes the focal length of the eye lens is maximum.
• When it strains, the focal length of the eye lens is minimum.
• In this way the ciliary muscles to which eye lens is attached help to give us clear vision.

Question 24.
Two observers standing apart from each other do not see the “same” rainbow. Explain.
Answer:

• All the rain drops that disburse the light to form rainbow lie within a cone of semi vertical angle 40° to 42°.
• If two observers are standing at a distance apart, they will observe rainbow at different parts on the surface of the cone.
• So the portion of the rainbow observed by an observer depends on the position of the observer.
• Two different observers will form two different cones with the observer standing at the vertex of the cone, therefore rainbow seen by them will be different.

Question 25.
A prism with an angle A = 60° produces an angle of minimum deviation of 30°. Find the refractive index of material of the prism.
Answer:
Given, A = 60° and D = 30°

Question 26.
A person cannot see the objects distinctly, when placed at a distance less than 50 cm. Calculate the power and nature of the lens he should be using to see clearly the object placed at a distance of 25 cm from his eyes.
Answer:

Question 27.
A person cannot see the objects distinctly, when placed beyond 2 m.
Calculate the power and nature of the lens he should be using to see the distant objects clearly.
Answer:
For myopia the focal length = – far point distance = – 2m
Power = \(\frac{1}{\mathrm{f}}=\frac{1}{-2}\) = 0.5 D.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 4 Acids, Bases and Salts.

AP State Syllabus SSC 10th Class Chemistry Important Questions 4th Acids, Bases and Salts

10th Class Chemistry 4th Lesson Acids, Bases and Salts 1 Mark Important Questions and Answers

Question 1.
Take some water in a test tube and add concentrated H₂SO₄ to it. Shake the test tube well. If you touch the bottom of the test tube, you feel it as hot. Now, instead of H₂SO₄, if you add NaOH pellets to water in another test tube and touch the bottom, what do you observe? (TS June 2015)
Answer:
The bottom of test tube is also hot because reactions of acids, bases with water are exothermic reactions.

Question 2.
What happens if the copper sulphate crystals taken into dry test tube are heated? (TS June 2016)
Answer:

• When copper sulphate crystals are heated, water present in crystals is evaporated and the salt turns white.
• Evaporated water appears as droplets on the walls of the test tube.
• Blue coloured copper sulphate (CuSO₄ 5H₂O) is turned into white colour because 5H20 molecules are evaporated from crystals.

Question 3.
Why does the soil of agricultural lands get tested for pH?
Answer:
Plants require a specific pH range for their healthy growth. So, finding pH of a soil suggested the farmers to treat the fields with acidic or basic substances to maintain the required pH range.

Question 4.
Write the molecular formulae of common salt and baking soda which are widely used at home. (TS June 2017)
Answer:
Common Salt: NaCl; Backing soda : NaHCO₃

Question 5.
Mention the precautions to take while conducting an experiment to prove acids produce ions only in aqueous solutions. (TS June 2018)
Answer:

• Testing of the evolved gas by using dry litmus paper first. Then with wet litmus paper.
• Use gaurd tube containing calcium chloride.

Question 6.
What are antacids?
Antacids are mild alkalies. These are used for getting relief from acidity and indigestion and sometimes even headache. When taken orally, it reacts with hydrochloric acid present in the stomach and reduces its strength by consuming some of it.
Ex: Milk of magnesia.

Question 7.
Tap water conducts electricity whereas distilled water does not. Why?
Answer:
Tap water contains some impurities in the form of salts. Due to presence of salts, it conducts electricity. Distilled water is free from all kinds of salts and hence does not conduct electricity.

Question 8.
What do you mean by dilution of an acid or base? Why is it done?
Answer:
Dilution of an acid or base means mixing an acid or base with water. This is done to decrease the concentration of ions per unit volume. In this way the acid or the base is said to be diluted.

Question 9.
What is a universal indicator?
Answer:
An indicator which passes through a series of colour changes over a wide range of H₃O⁺ ions concentration is called universal indicator.

Question 10
What is tooth decay?
Answer:
Tooth enamel is chemically calcium phosphate Ca₃(PO₄)₂. It starts corroding when pH falls below 5.5. Food particles left in the mouth degrade to produce acid which lower the pH of the mouth. This is called tooth decay.

Question 11.
Define Alkalis and give some examples.
Answer:
Alkalis : An alkali is a base that dissolves in water.

Examples :
i) Sodium Hydroxide (NaOH),
ii) Potassium Hydroxide (KOH),
iii) Magnesium Hydroxide (Mg(OH)₂)

Question 12.
Why should we not taste or touch alkalis?
Answer:
We should not taste or touch alkali. Because they are corrosive.

Question 13.
Salts conduct electricity. Why?
Answer:
Salts contains ions. So they conduct electricity.

Question 14.
Why are calcium sulphate hemihydrates called Plaster of Paris?
Answer:
Calcium sulphate hemihydrates are used as plaster for supporting fractured bones in the right position. So, it is called Plaster of Paris.

Question 15.
Why does an aqueous solution of acid conduct electricity?
Answer:
An aqueous solution of acid liberates H⁺ ions. This makes the aqueous solution of acid to conduct electricity.

Question 16.
How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?
Answer:
When a solution of an acid is diluted, concentration of H₃O⁺ ions decreases.

Question 17.
What is pH?
A. pH is a scale for measuring hydrogen ion concentration in a solution. It is the negative logarithm of H⁺ concentration.
pH = – log [H⁺].

Question 18.
How is pH of a solution related to the [H₃O⁺] of that solution?
Answer:
The presence of H₃O⁺ ions indicate us whether it is a strong acid or weak acid.

Question 19.
There are two solutions of pH values 6 and 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:

• The solution whose pH value 6 is acid and has more hydrogen ion concentration.
• The solution of pH value 8 is basic and has less hydrogen ion concentration.

Question 20.
Can you give example for use of olfactory indicators in daily life?
Answer:
Examples of olfactory indicators : Onion, vanilla extract.

Question 21.
How do acids neutralize bases?
(OR)
How do acids and bases react with each other?
Answer:
According to Arrhenius theory acids produce H⁺ ions and bases produce OH^– ions in aqueous media.
The combination of H⁺ and OH^– ions is called ‘neutralization’.
Thus acids neutralize bases.

Question 22.
How strong are acids and base solutions?
Answer:
The acids of pH value as much less as possible have more concentration [pH < 7], Basic nature increases as pH value increases.

Question 23.
What do you say about salts of both weak acid and weak base?
Answer:
The pH of aqueous solutions of salt obtained from both weak acid and weak base is nearly 7.

Question 24.
Which base is used for removing permanent hardness of water?
Answer:
Sodium carbonate is used for removing permanent hardness of water.

Question 25.
Name two antacids used to get rid of our indigestion problem.
Answer:
Magnesium hydroxide and a mild base (baking soda).

Question 26.
Under what soil conditions would a farmer would treat the soil of his fields with quicklime (calcium hydroxide) or calcium carbonate?
Answer:
When the field has acidic nature, the farmer uses quicklime or calcium carbonate to neutralize it.

Question 27.
Write the formulas of Gypsum and Plaster of Paris.
Answer:
The formulae of Gypsum is CaSO₄ . 2H₂O and Plaster of Paris is CaSO₄ . ½H₂O.

Question 28.
Write any two Acid Base indicators.
Answer:

1. Methyl orange
2. Phenolphthalein.

Question 29.
Which salt is used in the manufacture of borax?
Answer:
Washing soda (Na₂CO₃.10H₂O)

Question 30.
What is family of salts? Give examples.
Answer:
Salt having the same positive or negative radical is called family of salts.
Eg : Family of sodium salts : NaCl, Na₂SO₄
Family of chloride salts : NaCl, KCl.

Question 31.
‘A’ is a substance which is acidic and it is added in solution to preserve pickles. What is A and what is the name given to its dilute solution?
Answer:
‘A’ is acetic acid and its dilute solution is called vinegar.

Question 32.
What are the chemical names of the following?
1) Baking soda
2) Gypsum
Answer:

• The chemical name of baking soda is sodium hydrogen carbonate (NaHCO₃).
• The chemical name of gypsum is calcium sulphate dihydrate (CaSO₄ • 2 H₂O).

Question 33.
Write the water of crystallisation of following compound.
a) Hydrous copper sulphate
b) Washing soda
c) Gypsum
d) Plaster of Paris
Answer:

Question 34.
Why don’t we use a strong base like NaOH as antacid?
Answer:
Strong bases like potassium hydroxide (KOH), sodium hydroxide (NaOH) are corrosive in nature. So they can harm the internal organs. Therefore we should not use them as antacid.

Question 35.
What do you mean by HsO⁺ ion?
Answer:
Hydrogen ions cannot exist as base ions. They associate with water molecules and exist as hydrated ions with each H⁺ attached by 4 to 6 water molecules. For this we represent H⁺ as hydronium ion, H³O⁺.
H⁺ + H₂O → H₃O⁺

Question 36.
Which indicator is useful at all pH? Why?
Answer:
Universal indicator is useful to test solutions of all pH because it gives different colours at different pH range.

Question 37.
Which substance is useful in removing permanent hardness of water?
Answer:
The permanent hardness of water is due to chloride and sulphate salts of magnesium and calcium, which can be removed by adding washing soda.

Question 38.
Given two examples for strongest bases.
Answer:
Sodium hydroxide – NaOH
Potassium hydroxide – KOH

Question 39.
What is the confirmation test for hydrous and anhydrous salt?
Answer:

• On heating hydrous salt in a test tube it will form water droplets on the sides of test tube.
• On heating anhydrous salt in a test tube it will not form water droplets on the sides of test tube.

Question 40.
P.O.P, cement, calcium chloride should be stored in moisture proof containers. Why?
Answer:

• P.O.P, cement and calcium chloride react with moisture (H₂O) in the atmosphere and set into hard solid masses.
• To avoid availability of moisture they should be stored in moisture proof containers.

Question 41.
Give some examples for hydrous and anhydrous salts.
Eg : For hydrous salts :

1. CuSO₄ . 5H₂O
2. Na₂CO₃ . 10H₂O
3. CaSO₄ . 2H₂O

Eg : for anhydrous salts :

1. NaCl
2. MgCl₂
3. Na₂CO₃

Question 42.
How are bitter and sour taste substances tested without testing?
Answer:

• Sour taste substances turn blue litmus to red.
• Bitter taste substances turn red litmus to blue. By these tests we can test them as acids and bases.

Question 43.
Do the metallic oxides react with acids?
Answer:
Yes, metallic oxides are basic in nature. They react with acids and form salt and water.

Question 44.
Does non-metallic oxide react with base?
Answer:
Yes, non-metallic oxide is acidic in nature. It reacts with base and forms salt and water.

Question 45.
Why does dry HCl gas not change the colour of the dry litmus paper?
(OR)
Prove that dry HCl gas is not an acid but HCl aqueous solution is an acid using an activity.
Answer:

• Dry hydrogen chloride gas is not an acid. Hence it can’t turn blue litmus into red.
• Hydrochloric acid is an aqueous solution. Hence it can turn blue litmus into red.

Question 46.
Do you know that the atmosphere of Venus is made up of thick white and yellowish clouds of sulphuric acid? Do you think life can exist on this planet?
Answer:

1. No, it is not possible.
2. When pH value decreases, the survival of living organisms becomes difficult.
3. Hence there is not any possibility of life on Venus.

Question 47.
Why do acids not show acidic behaviour in the absence of water?
Answer:
Acids don’t show acidic behaviour in the absence of water as H⁺ ions are absent in them.

Question 48.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
When a base like NaOH (Sodium hydroxide) is dissolved in water, it liberates (OH^–) ions.
Equation :

OH^– ion concentration increases.

Question 49.
How does the nature of the solution change with change in concentration of H⁺_(aq) ions?
Answer:
The concentration of H⁺ ions is responsible for the acidic nature of a substance.
If [H⁺] > 1 × 10^-7 mol/lit the solution is acidic.
If [H⁺] < 1.0 × 10^-7 mol/lit the solution is basic.

Question 50.
Do basic solutions also have H⁺_(aq) ions? If yes, then why are these basic?
Answer:
Yes. Basic solutions have OH^–_(aq) ions and bases have less number of H₃O⁺ ions. H⁺_(aq) ions are less in base. In basic solution [OH^–] > [H⁺].

Question 51.
What do acids have in common?
Answer:
(Acids have sour taste and conduct electricity. They release H₂ (Hydrogen) gas on reacting with metals). All acids have H⁺_(aq) ions.

Question 52.
What do bases have in common?
Answer:
Bases are slippery to touch and bitter to taste. All bases have OH^–_(aq) ions.

Question 53.
Why does pure acetic acid not turn blue litmus to red?
Answer:
Pure acetic acid is a weak acid so it does not have sufficient H⁺_(aq) ion to change the colour of blue litmus to red.

Question 54.
What will happen if the pH value in your body increases?
Answer:
It affects our digestion system.

Question 55.
A student checked pH of a salt solution and found that its pH is more than 7. How is that type of salt formed?
Answer:

• When a strong base reacts with weak acid then the solution is basic in nature. So its pH is more than 7.
• For example when acetic acid reacts with sodium hydroxide the salt formed has basic nature.

Question 56.
Explain the procedure that you follow to reduce water from a given salt.
Answer:
Procedure to reduce water from a given salt :

1. Take a boiling test tube.
2. Drop given salt in the test tube.
3. Heat the test tube gently.
4. Water from salt evaporates.
5. In this way we can reduce the water from salt.

Question 57.
Write the observations, when the hydrated salt or unhydrated salt is heated.
Answer:

• When hydrated salt is heated water droplets form inside the walls of test tube and sometimes blue or green colour salt turns into white colour.
• When unhydrated salt is heated it does not form water droplets inside the test tube walls and colour also does not change.

Question 58.
On heating the hydrated salt it loses water molecules present in it. To show this what are the equipment required?
Answer:
1) On heating the hydrated salt it loses water molecules present in it.
2) To show this the given equipment are required

1. Boiling tube
2. Test tube holder
3. Burner

Question 59.
Try to collect the information to reasons for calling calcium sulphate hemihydrates as Plaster of Paris (POP).
Answer:

• Gypsum plaster (or) Plaster of Paris (POP) is produced by heating gypsum to about 300°F.
• A large gypsum deposit is found at Montmartre in Paris (France).
• This gave the name Plaster of Paris to calcium sulphate hemihydrates.
• The term plaster can refer to gypsum.

Question 60.
Is the substance present in antacid tablet acidic or basis?
Answer:
The substance present antacid is weakly basic.

Question 61.
Give pH of neutral, acid and base.
Answer:

Question 62.
Which nature of Plaster of Paris makes its importance? Appreciate it.
Answer:
Plaster of Paris is a white powder. It is very soft and can be used to make toys, materials for decoration and to make surfaces smooth.

But on mixing with water, it changes to a hard solid mass (Gypsum). This is the important character of Plaster of Paris (POP).

Question 63.
What is acid rain? How does it affect our aquatic life?
Answer:
When the pH of rain water is less than 5.6 it is called acid rain. When acid rain flows into the rivers, it lowers the pH of the river water. Since our body works within a narrow pH range close to 7 the survival of aquatic life in river water mixed with rain water becomes difficult.

Question 64.
Why are pickles and sour substances not kept in brass and copper vessels?
Answer:
Pickles and sour substances contain acidic nature which may react with brass and copper vessels to produce toxic substances.
So, we don’t keep them in brass arid copper vessels.

Question 65.
Can you suggest some examples of use of pH in everyday life?
Answer:
Uses of pH in everyday life :

1. pH value helps us to identify acids, bases and neutrals.
2. If pH value is less in our mouth, it leads to tooth decay. We can find it as the reason for our tooth decay.
3. pH value helps us to know about acid rain.

Question 66.
Write any two uses of Bleaching powder.
Answer:

• It is used for disinfecting drinking water to make it free of germs.
• It is used as a reagent in the preparation of chloroform.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 2 Marks Important Questions and Answers

Question 1.
What value of pH in the mouth leads to tooth decay? Why? (TS June 2015)
Answer:

• Tooth decay starts when the pH of the mouth is lower than 5.5.
• Tooth enamel, made of calcium phosphate is the hardest substance in the body.
• It does not dissolve in water, but is corroded when the pH in the mouth is below 5.5.
• Bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth.

Preventions :

1. Clean the mouth after eating food.
2. Using tooth pastes, which are generally basic neutralize the excess acid and pre¬vent tooth decay.

Question 2.
Equal lengths of Magnesium ribbons are taken in two test-tubes X and Y. Hydro¬chloric acid is added to test-tube X and Acetic acid is added to test-tube Y. In which test-tube, the reaction will be more vigorous? Why? (TS March 2015)
Answer:
The speed of the reactions is higher in X test tube than Y test tube.

Reason :
Due to strong acidic nature, Hydrochloric acid reacts very fast with magnesium ribbon.

Question 3.
Name the four chemicals that are obtained from common salt and write their molecular formulae. (TS March 2015)
Answer:
Chemicals that can be obtained from common salt are

1. Sodium Hydroxide – NaOH
2. Baking soda / Cooking soda / Caustic soda / Sodium bicarbonate / Sodium Hydrogen carbonate. – NaHCOv
3. Washing soda / Sodium carbonate – Na₂CO₃ 10H₂O
4. Bleaching powder / Calcium Oxy Chloride – CaOCl₂

Question 4.
Observe the information given in the table and answer the questions given below the table. (TS March 2017)

i) Which one of them may be the neutral salt among A, B, C?
ii) What may happen when some drops of phenolphthalein is added to the substance B?
Answer:
i) C
ii) Pink Colour

Question 5.
Why do we use antacids? Write it’s nature. (TS March 2018)
Answer:
Pain and irritation will be caused in stomach during the acidity problem/indigestion problem. Antacids used to neutralize the excess acid in the stomach and gives relief from acidity Antacids are basic in nature.

Question 6.
Which product will form when CaO is dissolved in water? How do you find the nature of product? (TS March 2018)
Answer:
CaO reacts with water and gives calcium hydroxide [Ca(OH₂)]. The nature of the calcium hydroxide will be tested with red litmus paper or pH paper.

Calcium Hydroxide turns red litmus into blue. Thus we can say that ca(OH)₂ is basic in nature.

Ca(OH)₂ shows pH value more than 7. Thus we can say that Ca(OH)₂ is basic in nature.

Question 7.
How do you test the nature of the solution formed by dissolving CaO in water? What is the nature of the solution? (TS June 2019)
Answer:

• The solution formed by dissolving CaO in water is tested with red litmus paper, it turns into blue colour, (or) It is tested with methyl orange, it turns into yellow in colour.
• The solution of CaO and water is basic in nature.

Question 8.
Write the experimental procedure to test carbon dioxide gas. (AP SCERT: 2019-20)
Answer:

1. Pass the CO₂ gas through lime water [Ca(OH)₂].
2. The lime water appears as milky white.
3. The reaction is Ca(OH)₂ + CO₂ → CaCO₃ ↓ + H₂O.
4. The milky white is caused by CaCO₃.

Question 9.
Write two reactions of acids with carbonates and metal hydrogen carbonates. (AP SA-I : 2019-20)
Answer:
Reactions :

Question 10.
What do acids have in common?
Answer:
Common characteristics of acids :

1. Similar chemical properties.
2. Acids generate hydrogen gas on reacting with metals.
3. Hydrogen is common to acids.
4. Acids are sour in taste and turn blue to red when react with bases form salt and water.

Question 11.
What do bases have in common?
Answer:
Common characteristics of bases :

1. Bitter in taste.
2. Soapy in nature.
3. Turn red to blue colour.
4. On heating decompose into metal oxides and water.
5. React with acids to form salt and water.
6. Produce OH^– ions in aqueous solution.

Question 12.
How is bleaching powder produced?
Production of bleaching powder :
Bleaching powder is produced by the action of chlorine on dry slaked lime (Ca(OH)₂).
Equation :

Question 13.
What can you conclude about the ideal soil pH for the growth of plants in your region?
Answer:
1) Soil is considered the ‘skin of the earth’. The soil pH plays a vital role in the growth of a plant and it influences plant nutrition.
2) Soil pH strongly affects the nutrients required for the plant growth.
3) The nutrients may be stored on soil colloids, and live or dead organic matter, but may not be accessible to plants due to extremes of pH.

Conclusion :
For optimum plant growth, the generalized content of soil components by volume should be roughly 50% solids (45% mineral & 5% organic matter), and 50% voids of which half is occupied by water and half by gas.

Question 14.
How can you prepare turmeric indicator? What is the use of it?
Answer:
i) Turmeric indicator is prepared from turmeric.
ii) It has red colour in basic solution.

Question 15.
Name two salts and write their formulae which possess water of crystallization.
Answer:

1. Hydrous copper sulphate. Its formula is CuSO₄ . 5H₂O
2. Gypsum. Its formula is CaSO₄ . 2H₂O

Question 16.
What is neutralization reaction? Give two examples.
Answer:
When an acid reacts with base it forms salt and water. This reaction is called neutralisation reaction.
e.g.: NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
H₂SO_4(aq) + 2Na0H_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

Question 17.
All alkalis are bases but all bases are not alkalis. Do you agree with the statement? If yes, why?
Answer:
Yes, I agree with the statement. Because alkalis are those bases which are soluble in water. So all alkalis are bases but all bases are not alkalis.

Question 18.
Why are solutions of acids, bases and salts good conductors of electricity?
Answer:
For passage of electricity through a material or substance charged particles are required. In metals charged particles are electrons whereas in solutions ions are charged particles which carry electrical energy. Solutions of acids, bases and salts undergo ionisation and produce ions. So they are good conductors of electricity.

Question 19.
What is strength of acid ? What are the factors that influence strength of acid?
Answer:
The extent which an acid undergoes ionisation is called strength of acid.
Factors influence strength of acid :

1. Degree of ionisation.
2. Concentration of hydronium ions produced by acid.

Question 20.
Why are organic acids weak acids when compared with mineral acids?
Answer:

• Strength of acid depends on extent of ionisation.
• Organic acids do not undergo 100% ionisation. Their ionisation is less than 30%. There is equilibrium between ionised and unionised molecules whereas mineral acids undergo complete ionisation.
• So mineral acids behave like strong acids when compared with organic acids.

Question 21.
“Acids do not contain OH^– ions”. Do you agree with this statement? If not, why?
Answer:
No. I do not agree with the statement because all acids also contain OH^– ions but in acid solutions, H⁺ ions are more than OH^– whereas in bases OH^– ions are more than H⁺.

Question 22.

If B is calcium chloride, what are A, C, D and E?
Answer:
A is calcium carbonate or calcium hydrogen carbonate.
C is water and D is carbon dioxide.
E is calcium carbonate and F is water.

Question 23.
Some salts are given below. Classify them into hydrous and anhydrous salts. Sodium carbonate, Sodium chloride, Sodium hydrogen carbonate, Copper sulphate, Hypo, Magnesium Sulphate (epsum salt)
Answer:
Hydrous salts :

1. Hypo (Na₂S₂O₃ • 2H₂O)
2. Epsum (MgSO₄ • 7H₂O)
3. Copper sulphate (CuSO₄ • 5H₂O)

Anhydrous salts :

1. Sodium chloride (NaCl)
2. Sodium carbonate (Na₂CO₃)
3. Sodium hydrogen Carbonate (NaHCO₃)

Question 24.
If someone in the family is suffering from a problem of acidity, which of the following would you suggest as a remedy : lemon juice, vinegar or baking soda solution? Which property do you think of while suggesting the remedy?
Answer:

• I suggest baking soda solution. As acidity can be neutralized by baking soda solution, we can use it.
• Neutralizing property of baking soda solution.

Question 25.
Why are curd and sour substances not kept in copper vessels?
Answer:
Curd and sour substances contain acids which react with copper vessels and form poisonous substances. So curd and sour substances should not be kept in copper vessels.

Question 26.
Which gas is liberated when acids react with metals? Give one example.
Answer:
When acids react with active metals they release hydrogen gas.
Zn_(s) + 2HCl_(aq) → Zncl_2(aq) + H₂ ↑_(g)

Question 27.
Why should pickles not be stored in metallic containers?
Answer:
Pickles contain acids which react with metallic containers and form poisonous substance. So they are kept in plastic containers.

Question 28.
Solution x turned blue litmus red and Solution y turned red litmus blue.
a) What products could be formed when x and y are mixed?
b) Which gas is released when we put magnesium pieces in solution x?
c) Will any chemical reaction take place when zinc pieces are put in solution y?
d) Which of the above solutions contain more hydrogen ions?
Answer:
Given solution Y turned blue litmus into red so, Y is an acid.
Given solution ‘y’ turned red litmus into blue so, ‘y’ is a base.
a) The reaction of an acid (x) with a base to give a salt and water.
b) When we put magnesium pieces in solution releases hydrogen gas.
c) When zinc pieces are put in solution y, a chemical reaction will take place there.
d) Acids contain more H+ ions in the given solutions, Y has more H+ ions because it is an acid.

Question 29.
Acid should be added to water but not water to the acid. Why?
Answer:

• The dissolving of an acid or base in water is an highly exothermic process. Care must be taken while mixing concentrated HNO₃ or concentrated H₂SO₄ with water.
• The acid must always be added slowly to water with constant stirring.
• If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause bums.
• The glass container may also break due to excessive local heating.

Question 30.
Explain the procedure to confirm the given salt is a hydrous or anhydrous.
Answer:

1. Take given salt in a test tube
2. Observe the colour of salt
3. Heat the test tube gently
4. Observe the colour of salt and also moisture (droplets) inside of the test tube walls.
5. If its colour changes or forms water droplets, it is hydrous salt.
6. Otherwise, it is anhydrous salt.

Question 31.
Categorize the following as acids, bases, and salts :
Lemon juice, salt water, soap water, tamarind juice, surf water, lime water.
Answer:
Acids :

1. Lemon juice
2. Tamarind juice

Bases :

1. Soap water
2. Surf water
3. Lime water

Salts :

1. Salt water

Question 32.
Classify the following salts as family of salts having same cation or anion and prepare a table.
Potassium sulphate, Sodium sulphate, Calcium sulphate, Magnesium sulphate, Copper sulphate, Sodium chloride, Sodium nitrate, Sodium carbonate and Ammonium chloride.
Answer:

Question 33.
Observe the table and answer the following questions.

1) Which of the solutions among these is a strongest base?
2) Which body fluid has slightly basic nature?
3) What is the nature of pure water?
4) Which body fluid is strongest acid?
Answer:

1. Sodium hydroxide because its pH is 13.
2. Blood because its pH is 7.3.
3. Pure water is neutral in nature because its pH is 7.
4. Gastric juice because its pH is 1.2.

Question 34.
The diagram given below shows the removal of water crystallisation. Find error in the diagram.

Answer:
Error in the diagram

1. Test tube is placed towards observer. It causes bums on his hands.
2. So, test tube should be placed away from the observer.

Question 35.
What are the uses of Plaster of Paris?
Answer:

• The substance which doctors use as plaster for supporting fractured bones in the right position.
• Plaster of Paris is used for making toys, materials for decoration and for making surfaces smooth.

Question 36.
What are the applications of pH in daily life?
Answer:
1) In medical science :
The pH values of urine and blood are taken for diagnosis of various diseases.

2) In dairies :
Milk has pH of 6.6. A change in the pH of milk indicates that milk has turned sour.

3) In agriculture :
For better growth of crops the pH of the soil is regularly tested.
For examples :

1. Citrus fruits require slightly alkaline soil.
2. Rice requires acidic medium.
3. Sugarcane requires neutral soil.

4) In technology:
Organic and biochemical reactions are carried out under control pH.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 4 Marks Important Questions and Answers

Question 1.
Draw a neat diagram showing a base solution in water conducts electricity. Why the solution of sugar/glucose in water do not conduct electricity? (AP March 2017)
Answer:

The solution of sugar/glucose in water do not conduct electricity because there is no H⁺ ions in the solution.

Question 2.
Explain an activity to show the water of crystallisation in CuSO₄ • 5H₂O. (AP June 2018)
Answer:

• Take a few crystals of copper sulphate in a dry test tube and heat the test tube.
• We observe water droplets on the walls of the test tube and salt turns white.
• Add 2 – 3 drops of water on the sample of copper sulphate obtained after heating.
• We observe the blue colour of copper sulphate crystals is restored.

Question 3.
Read the information given in the table and answer the following questions. (TS March 2016)

a) List out the acids in the above table.
Answer:
The acids are HCl and lemon juice.

b) What is the nature of the solution which gives pink colour with Phenolphthalene solution?
Answer:
The nature of the solution which turns pink colour with phenolphthalene solution is basic.

c) List out the neutral solutions in the above table.
Answer:
The neutral solutions are distilled water and NaCl.

d) Name the strongest acid and the strongest base among the given solutions.
Answer:
The strongest acid is HCl and the strongest base is NaOH.

Question 4.
Observe the following table and answer the questions given below. (TS June 2o17)
The table contains the aqueous solutions of different substances with the same concentrations and their respective pH values.

i) Which one of the above acid solutions is the weakest acid? Give a reason.
Answer:
Weakest acid is ‘C’. Because its pH value is less than 7 and it is nearer to 7.

ii) Which one of the above solutions is the strongest base? Give a reason.
Answer:
Strongest base is ‘D’. Because it’s pH value is near to 14.

iii) Which of the above two produce maximum heat when they react ? What does that heat energy called?
Answer:
B, D produce maximum heat when they react. This heat energy is known as neutralization energy.

iv) Which one of the above solutions has the pH equal to that of the distilled water? What is the name given to solutions of that pH value?
Answer:
‘G’ has the pH equal to that of the distilled water. These type of solutions are known as neutral solution.

Question 5.
List out the materials required to test whether the solutions of given acids and bases contain ions or not. Explain the procedure of the experiment. (TS March 2017)
Answer:
Required Materials :
Beaker, Bulb, Graphite rods, connecting wires, 230 V AC current, water, different acids, bases.

Experimental Procedure :

1. Connect the two connecting wires to the graphite rods.
2. Keep the graphite rods into the beaker, take care that two graphite rods do not touch each other.
3. Arrange a bulb in the circuit.
4. Pour dilute acid into the beaker.
5. Connect the ends of the connectors to 230 V AC.
6. In this way, change the acid / base and do the experiment.

The bulb glows in the experiment when the beaker contains acid or base. Hence, when the bulb glows we can say that acid or base contain ions.

Question 6.
List out the material for the experiment “when Hydrochloric acid reacts with NaHCO₃ and evolves CO₂“. Write the experimental procedure. (TS March 2018)
Answer:
Required material : Stand, test tubes, delivery tube, thistle funnel, two hole rubber corks, Ca(OH)₂, NaHCO₃, HCl.

Experimental procedure :

1. Take NaHCO₃ in a test tube and fix two hole cork to the test tube.
2. Fix thistle funnel in one hole of cork and insert delivery tube in the second hole of the cork. Insert the second end of the delivery tube in the other test tube which is containing Ca(OH)₂/lime water.
3. Pour dil. HCl into the test tube using thistle funnel.
4. Due to chemical reaction, gas is evolved and pass into the Ca(OH)₂ through delivery tube. It turns in to milky. We can conclude that it is CO₂ gas.

Question 7.
Prepare a table based on the colour responses of acid, base and salt with indicators such as indicators. (AP SA-I:2018-19)
Answer:

Question 8.
Draw universal pH value indicator and identify different substances. (AP SA-I : 2019-20)
(OR)
Draw a neat diagram showing variation of pH with the chage in concentration of H⁺_(aq) ion and OH^–_(aq) ions.
Answer:

Variation of pH with the change in concentration of H⁺_(aq) ions and OH^–_(aq) ions.

Question 9.

Answer the following questions by using above information. (TS June 2019)
1) Which of the above is neutral solution?
2) Which of the above is used to neutralize the acidity in stomach?
3) Which is the strong acid among the above solutions?
4) What is the colour of Phenolphthalein indicator in NaOH solution?
Answer:

1. Distilled water
2. Milk of Magnesia
3. Gastric juice
4. Pink

Question 10.
If the pH values of solutions X, Y and Z are 13, 6 and 2 respectively, then
a) Which solution is a strong acid? Why?
b) Which solution contains ions along with molecules of solution?
c) Which solution is a strong base? Why?
d) Does the pH value of a solution increase or decrease when a base is added to it? Why?
Answer:
The strength of an acid (or) an alkali can be tested by using pH value of a solution. If the value of a pH of a solution is less, then that solution exhibits acidic nature.

If the value of a pH of a solution is more, then that solution exhibits basic nature.
pH value of a solution “X” is 13
pH value of a solution “Y” is 6
pH value of a solution “Z” is 2

a) Solution ‘Z’ is strong acid because its pH is 2.
b) Among given solutions, solution X is weakest acid. Weak solution contains ions along with molecules of solution. So X exhibits like this character.
c) Solution X is strong base. Because its pH is 13.
d) If base is added to solution ‘Z’, then its pH will increase.

Question 11.
Distinguish between acids and bases.
Answer:

Question 12.
Explain chlor-alkali process.
Answer:
When electricity is passed through an aqueous solution of sodium chloride, it decomposes to form sodium hydroxide. The process is called chloralkali process because of the products formed chlor for chlorine and alkali for sodium hydroxide.
2 NaCl (aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)

Chlorine gas is given off at the anode and hydrogen gas at the cathode and sodium hydroxide is formed near the cathode.

Question 13.
Define the following. Give one example for each.
a) Strong acid
b) Strong base
c) Weak acid
d) Weak base.
Answer:
a) Strong acid :
The acid which undergoes 100% ionisation is called strong acid.
e.g.: HCl, H₂SO₄

b) Strong base :
The base which undergoes 100% ionisation is called strong base.
e.g.: NaOH, KOH

c) Weak acid:
The acid which undergoes less than 100% ionisation is called weak acid.
e.g.: CH₃COOH, H₂CO₃

d) Weak base:
The base which undergoes less than 100% ionisation is called weak base.
e.g.: NH₄OH, Mg(OH)₂

Question 14.
Write any four chemical properties of acids.
Answer:
Chemical properties of acids :
1) Active metals react with acids and liberate hydrogen gas.
Zn + HCl → ZnCl₂ + H₂ ↑

2) Acids react with bases to form salt and water.
HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

3) Acids react with metallic oxides to form salt and water.
MgO_(s)+ 2HCl_(aq) → MgCl_2(aq) + H₂O_(l)

4) Acids react with carbonates and hydrogen carbonates and release carbon dioxide gas.
CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq)+ H₂O_(l) + CO_2(g) ↑
Ca(HCO₃)_2(l) + 2HCl_(aq) → CaCl_2(aq) + 2H20((| + 2CO_2(g) ↑

Question 15.
Write the formulae of the following salts.
a) Sodium sulphate
b) Ammonium chloride
Identify the acids and bases for which the above salts are obtained. Also write chemical equations for the reactions between such acids and bases. Which type of chemical reactions are they?
Answer:
a) Formula of sodium sulphate is Na2S04. When sulphuric acid reacts with sodium hydroxide it forms sodium sulphate.
H₂SO_4(aq)+ 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

b) Formula of Ammonium chloride is NH4C/. When Ammonium hydroxide reacts with hydrochloric acid it forms Ammonium chloride.
NH₄OH_(aq) + HCl_(aq) → NH₄Cl_(aq) + H₂O_(l)

Question 16.
Write balanced equations to satisfy each statement,
a) Acid + Active metal → Salt + Hydrogen
Answer:
Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

b) Acid + Base → Salt + Water
Answer:
NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)

c) Acid + Carbonate / Hydrogen carbonate → Salt + Water + Carbon dioxide
Answer:
CaCO_3(s) + 2 HCl_(aq) → CaCl_2(aq) + H₂O_(l) + CO_2(g) ↑
NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO_2(g) ↑

d) Metal oxide + Acid → Salt + Water
Answer:
CaO_(s) + 2 HCl_(aq) → CaCl_2(aq) + H₍₂₎O_(l)

e) Non metal oxide + base → Salt + Water
Answer:
CO_2(g) + 2 NaOH_(aq) → Na₂CO_3(aq) + H₂O_(l)

Question 17.
Give important products obtained from chloralkali process.

Answer:
Result:
Five water molecules are present in one formula unit of copper sulphate. Water of crystallization proves that the crystals contain a fixed quantity of water in them.

Question 18.
Give the equations for the preparation of each of the following.
i) Copper sulphate from copper (II) oxide.
Answer:
CuO + H₂SO₄ → CuSO₄ + H₂O

ii) Potassium sulphate from potassium hydroxide solution.
Answer:
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

iii) Lead chloride from lead carbonate.
Answer:
PbCO₃ + 2 HCl → PbCl₂ + H₂O + CO₂

Question 19.
How are the following salts prepared?
1) Calcium sulphate from calcium carbonate
Answer:
When calcium carbonate is treated with sulphuric acid it forms calcium sulphate.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

2) Lead carbonate from lead nitrate
Answer:
When lead nitrate is treated with carbonic acid we will get lead carbonate.
Pb(NO₃)₂ + H2CO₃ → PbCO₃ + 2 HNO₃

3) Sodium nitrate from sodium hydroxide
Answer:
When sodium hydroxide is reacted with nitric acid it will form sodium nitrate.
NaOH + HNO₃ → NaNO₃ + H₂O

4) Magnesium carbonate from magnesium chloride
Answer:
When magnesium carbonate is reacted with hydrochloric acid it forms magnesium chloride.
MgCO₃ + 2 HCl → MgCl₂ + H₂O + CO₂

Question 20.
Which of the following reactions are considered as neutralization reactions? Why?
1) NaOH + HCl → NaCl + H₂O
2) CaO + 2 HCl → CaCl₂ + H₂O
3) CO₂ + 2 NaOH → Na₂CO₃ + H₂O
4) SiO₂ + CaO → CaSiO₃
Answer:
All of them are considered as neutralization reactions.

1. An acid (HCl) reacts with base (NaOH) and forms salt and water. So it is a neutralization reaction.
2. Here metallic oxide which is basic in nature reacts with acid and forms salt and water. So it is also a neutralization reaction.
3. In third case non-metallic oxide (acidic oxide) reacts with base (NaOH) and forms salt and water. So it is also a neutralization reaction.
4. In fourth case a metallic oxide (CaO) reacts with non-metallic oxide (SiO₂) and forms salt. So it is also a neutralization reaction in the absence 6f water.

Question 21.
Which metals produce hydrogen gas when they are reacted with bases like NaOH and KOH? Write the chemical equations for the reactions.
Answer:
Zinc, aluminium and lead react with bases like NaOH and KOH and produce hydrogen gas.

Question 22.
i) A solution has a pH of 7. How would you increase its pH and decrease its pH? Explain.
Answer;
We can increase the pH of a solution by adding base because we know that bases have pH > 7. We can decrease the pH of a solution by adding an acid because acidic solution have pH < 7.

ii) If a solution changes the colour of litmus from red to blue, then what can you say about its pH?
Answer:
Bases can change red litmus into blue. So the pH of the solution is greater than 7.

iii) What can you say about pH of a solution that liberates carbon dioxide from sodium carbonate?
Answer:
Acids react with carbonates and liberate hydrogen gas. So the pH of the solution is less than 7.

Question 23.
Write the pH values of some solutions.
Answer:

Question 24.
Fill the following table of results of reactions between some substances (acids, bases, neutral substances) and indicators.

Answer:

Question 25.
The pH values of six solutions A, B, C, D, E, F are given as 5,2,1,3,7 and 9 respectively. Which solution is
a) Neutral
b) Strongly alkaline
c) Strongly acidic
d) Weakly acidic?
Arrange the pH in increasing order of Hydrogen ion concentration.
Answer:
a) Solution E is neutral.
b) Solution F is Alkaline.
c) Solution C is strongly acidic.
d) Solution A is weakly acidic.
e) Solution B is strongly acidic.
f) Solution D is strongly acidic.
g) Ascending order of increase of Hydrogen ion concentration is F, E, A, D, B, C.

Question 26.
Collect information about various organic acids different occurring naturally and prepare a table.
Answer:

Question 27.
Complete the table.

Answer:

Question 28.
Fill the table.

Answer:

Question 29.
Draw a diagram to show the reaction of acids with metals.
Answer:

Reaction of Zinc granules with dil. HCl and testing hydrogen gas by a burning candle

Question 30.
Draw a diagram to show that all metal carbonates and react hydrogen carbonates
Answer:

Question 31.
What are the uses of Bleaching powder?
Answer:

• It is used for bleaching cotton and linen in the textile industry for bleaching wood pulp in paper industry and for bleaching washed clothes in laundry.
• Used as an oxidizing agent in many chemical industries.
• Used for disinfecting drinking water to make it free of germs.
• Used as a reagent in the preparation of chloroform.

Question 32.
What are the uses of Baking soda?
Answer:
1) Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid. When baking powder is heated or mixed in water, the following reaction takes place.

Carbon dioxide produced during the reaction causes bread or cake to rise making them soft and spongy.

2) Sodium hydrogen carbonate is also an ingredient in antacids. Being alkaline, it neutralizes excess acid in the stomach and provides relief.

3) It is also used in soda-acid, fire extinguishers.

4) It acts as mild antiseptic.

Question 33.
What are the uses of Washing soda?
Answer:

• Sodium carbonate (washing soda) is used in glass, soaps and paper industries.
• It is used in the manufacture of sodium compounds such as borax.
• Sodium carbonate can be used” as a cleaning agent for domestic purposes.
• It is used for removing permanent hardness of water.

Question 34.
Write the chemical formulae of the following :
i) Bleaching powder
ii) Sodium Chloride
iii) Slaked lime
iv) Baking Soda
v) Washing Soda
vi) Gypsum
vii) Plaster of Paris
viii) Acetic acid
ix) Sodium Hydroxide
x) Limestone
Answer:
i) Bleaching powder = CaOCl₂
ii) Sodium Chloride = NaCl (Common Salt)
iii) Slaked lime (or) lime water = Ca(OH)₂
iv) Baking Soda = NaHCO₃
v) Washing Soda = Na₂CO₃, 10H₂O
vi) Gypsum = CaSO₄ . 2H₂O
vii) Plaster of Paris = CaSO₄ . ½H₂O
viii) Acetic acid = CH₃COOH
ix) Sodium Hydroxide = NaOH
x) Limestone = CaCO₃

Question 35.
What are the various applications of neutralization?
Answer:

• The acidity of soil is reduced by adding slaked lime.
• The sting of yellow wasps contains alkalis. If acetic acid is rubbed on affected area, they are neutralized.
• Ants and bees have formic acid in their stings which can be neutralised by applying soap and some other alkali.
• Antacids tablets contain magnesium hydroxide, persons suffering from acidity are administered these tablets.
• The affect of nettle plant leaves is neutralized by leaves of dock plant.

Question 36.
Mention two situations where you use hydrated and unhydrated salts in your daily life.
Answer:

• NaCl is unhydrated salt. It flows freely when filled in a container.
• NaHCl (Baking soda) is unhydrated salt. It flows freely when filled in a container.
• NaCO₃ . 10H₂O (washing soda) is hydrated salt. It leaves wetness inside the container.
• CaSO₄ . 2H₂O (Gypsum) is also hydrated salt. On careful heating of gypsum it loses water molecules partially to become (CaSO₄ . ½H₂O) P.O.P. It is used in hospitals as plaster for supporting fractured bones in right position.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 12 Electromagnetism.

AP State Syllabus SSC 10th Class Physics Important Questions 12th Electromagnetism

10th Class Physics 12th Lesson Electromagnetism 1 Mark Important Questions and Answers

Question 1.
Draw the diagram showing the magnetic field lines of bar mannet. (TS June 2016)
Answer:

Question 2.
Correct the diagram according to Lenz law and draw it again. (TS March 2017)
Answer:

Question 3.
What is the use of slip – ring in AC motor? (TS June 2018)
Answer:
Uses of slip rings :
Slip rings are used to change the direction of current in the coil continuously.

Question 4.
Draw the direction of magnetic lines force, assuming that the current is flowing into the page. (AP SCERT: 2019-20 )
Answer:

Question 5.
What happens when a current carrying coil is placed in a uniform magnetic field? (TS June2019)
Answer:
The rectangular coil comes into rotation in clock – wise direction because of equal and opposite pair of forces acting on the two sides of the coil.

1. If the direction of the current in the coil unchanged it rotates half clock – wise and comes to half and rotates in anticlock – wise direction.
2. If the direction of the current in the coil changed after the first half rotation, the coil continuously rotates in a same direction.

Question 6.
Write the name of the device that converts mechanical energy into electrical energy. (AP June 2019)
Answer:
Generator.

Question 7.
Name some sources of direct current.
Answer:
Dry cell, lead-acid battery.

Question 8.
Which sources produce alternating current?
Answer:
A.C generator, thermal power station, hydroelectric stations.

Question 9.
What is the role of split ring in an electric motor?
Answer:
The split rings are used to change the direction of current flowing through the coil.

Question 10.
Write one method of inducing current in the coil.
ANswer:
By pushing or pulling a bar magnet into or away from the coil we can induce current. Name two safety measures commonly used in electric circuit, i) Fuse ii) Earthing

Question 11.
On what factors does the magnetic induction at the centre of the coil depend?
Answer:
It depends on current, number of turns and radius of the coil.

Question 12.
Which is more dangerous AC or DC?
Answer:
AC is more dangerous.

Question 13.
State two serious hazards of electricity.
Answer:

• If a person touches the live wire, he gets severe shock which may prove fatal.
• Short-circuiting can cause a spark which may lead to fire in a building.

Question 14.
Why is earthing of electrical appliances recommended?
Answer:
To protect the user from any accidental electrical shock caused due to leakage of current.

Question 15.
Why is a spark produced at the place of short circuit? Why is the spark in white colour?
The resistance of circuit decreases, and a sudden flow of large current heats up the live wire and vapourises the metal. This causes spark. The metal of wire becomes very hot and naturally emits white light.

Question 16.
What is electromagnetic induction?
Answer:
Mechanical energy can be converted into electrical energy by moving a magnet inside a coil.

Question 17.
What is Maxwell’s right hand screw rule?
Answer:
The direction of current is the direction in which the tip of the screw advances and direction of ration of the screw gives the direction of magnetic lines of force.

Question 18.
What type of energy transformation takes place in electric generator?
Answer:
Electrical energy from mechanical energy.

Question 19.
Where are the electromagnets used?
Answer:
In electric generators and televisions.

Question 20.
What is electromagnet?
Answer:
When current carrying conductor is wound over a magnetic material like soft iron it gets magnetized.

Question 21.
What are the different types of power stations?
Answer:
Electrical energy is produced in different power stations from mechanical energy of water, meat energy, and nuclear energy.

Question 22.
If the current in the coil is in anti-clockwise, then what would be the face of the coil?
Answer:
It behayes as north pole.

Question 23.
If the current in the ceil is in clockwise, then what would be the face of the coil?
Answer:
It behaves as south pole.

Question 24.
What is the frequency of the A.C. supplied in your house?
Answer:
It is approximately 50 Hz.

Question 25.
What type of current is generated in electric power station?
Answer:
Alternating current.

Question 26.
What is the shape of magnetic lines due to straight current carrying conductor?
Answer:
They are concentric circles.

Question 27.
What is a transformer?
Answer:
It is a device which increases or decreases the voltage.

Question 28.
State two ways by which speed or rotation of electric motor can be increased.
Answer:

1. By increasing strength of the current.
2. By increasing number of turns in the coil.

Question 29.
What happens if an iron piece is dropped between two poles of strong magnet?
Answer:
Eddy current is produced in it. These eddy currents oppose the motion of the piece of iron. So it falls as it is moving through a viscous liquid.

Question 30.
If a copper rod carries a direct current, then where will be the magnetic field in the conductor?
Answer:
It will be both inside and outside the rod.

Question 31.
In what form is the energy in a current carrying coil stored?
Answer:
It is stored in the form of magnetic field.

Question 32.
What is solenoid?
Answer:
A solenoid is a long wire wound in a close packed helix.

Question 33.
What is the pattern of field lines inside a solenoid around when current carrying solenoid?
Answer:
Parallel to each other.

Question 34.
List any two properties of magnetic field lines.
Answer:

• Inside the magnet they start from south pole and end at north pole whereas outside the magnet they start at north pole and end at south pole.
• Two magnetic lines of force never intersect each other.

Question 35.
Why does the picture appear distorted when the bar magnet is brought close to the screen of a television?
Answer:
Picture on a television screen is due to motion of the electrons reaching the screen. These electrons are affected by magnetic field of bar magnet.

Question 36.
What is meant by electromagnetic induction?
Answer:
Whenever there is continuous change of magnetic flux linked with a closed coil, current generated in the coil is called electromagnetic induction.

Question 37.
State the Lenz’s principle.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 38.
What is induced emf?
Answer:
Change in magnetic flux produces emf in the circuit called induced emf.

Question 39.
What do you mean by magnetic effect of current?
Answer:
Current carrying conductor produces a magnetic field around it. This is called magnetic effect of current.

Question 40.
What is the direction of magnetic field at the centre of the coil carrying current in (i) clockwise, (ii) anti-clockwise direction?
Answer:
i) Along the axis of coil inwards.
ii) Along the axis of coil outwards.

Question 41.
Why does a current carrying freely suspended solenoid rest along a particular direction?
Answer:
A current carrying solenoid behaves like a bar magnet.

Question 42.
What effect will there be on a magnetic compass when it is brought near a current carrying solenoid?
Answer:
The needle of the compass will rest in the direction of magnetic field due to solenoid at that point.

Question 43.
How is magnetic field due to solenoid carrying current affected, if a soft iron bar is introduced inside the solenoid?
Answer:
The magnetic field increases when iron bar is introduced inside the solenoid.

Question 44.
What happens to magnetic field if we reverse the current direction?
Answer:
The magnetic field also gets reversed.

Question 45.
How do magnetic field lines inside a current carrying solenoid appear?
Answer:
They form along the axis and parallel to each other.

Question 46.
In which of the following cases does the electromagnetic induction occur?
i) A current is started in a wire held near a loop of wire.
ii) The current is stopped in a wire held near a loop of wire.
iii) A magnet is moved through a loop of wire.
iv) A loop of wire is held near a magnet.
Answer:
In first three cases there is a change in magnetic flux. So electromagnetic induction occurs in first three cases.

Question 47.
Why must an induced current flow in such a direction so as to oppose the change producing it?
Answer:
So that the mechanical energy spent in producing the change, is transformed into the electrical energy in form of induced current.

Question 48.
What is the maximum force acting on the current (i) carrying conductor of length (l) in the presence of magnetic field (B)?
Answer:
F = Bil

Question 49.
A charged particle q is moving with a speed v perpendicular to the magnetic field of induction B?
Find the equation of time period of the particle.
Answer:
\(\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{Bq}}\)

Question 50.
Name two safety measures commonly used in electric circuit.
Answer:
a) Fuse
b) Earthing

Question 51.
On what factors the magnetic induction at the centre of coil depends?
Answer:
a) Number of turns
b) Radius of the coil.

Question 52.
Write the formula for magnetic flux passing through an Area A with an angle θ.
Answer:
Flux ΦV= BA cos θ

Question 53.
Write the Lenz’s law.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 54.
What is the difference between AC and DC generator?
Answer:
In AC generator, ends of coil are connected to two slip rings.
In DC generator ends of coil are connected to two half split rings.

Question 55.
What are the uses of electromagnet?
Answer:
It is used in electric bells, electric motors, telephone diaphragms, etc.

Question 56.
What is the principle of Electric motor?
Answer:
When a rectangular coil is placed in magnetic field and current is passed through it, two equal and opposite forces act on the coil which rotates it continuously.

Question 57.
What factors are influence the speed of rotation of the motor?
Answer:

1. Strength of current
2. Number of turns
3. Area of the coil
4. Strength of magnetic field

Question 58.
Which two physical quantities are interrelated in Oerstead experiment?
Answer:
Electricity and magnetism are interrelated.

Question 59.
Which property of proton can change while it moves freely in a magnetic field?
Answer:
When a proton (the charge) moves in a magnetic field, then magnetic force is acting on proton. So its momentum changes.

Question 60.
Which type of conductors are producing magnetic field?
Answer:

1. Long straight current carrying conductor.
2. Circular loop.
3. Solenoid.

Question 61.
How much force acting on a neutron particle is moving with velocity V in a mag¬netic field with induction B?
Answer:
Zero, because neutron is charge less particle.

Question 62.
What are the instruments used in A.C Generator?
Answer:
Rectangular coil, brushes, slip rings, and magnetic poles.

Question 63.
What are the ways to produce the induced current in a coil?
Answer:
When a magnet is moved towards or away from coil or there is a relative motion between coil and magnet a current is induced in the coil.

Question 64.
At the time of short circuit, what happens to the current?
Answer:
At the time of short circuit, the current in the circuit increases heavily becomes the resistance of the conductor becomes almost zero.

Question 65.
A wire with green insulation is usually the line wire of an electric supply. Is it true?
Answer:
It is false, the wire with green insulation is the earth wire, not the line wire.

Question 66.
Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
Yes. current will be induced in coil ‘B’, because flux linked the coil ‘B’ changes with respect to time.

Question 67.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Electric fuse and Miniature Circuit Breaker (MCB).

Question 68.
An attemating current has frequency of 50Hz. How many times does it change its direction in one second?
Answer:
Frequency of AC = 50 Hz ⇒ 50 cycles in one sec. So it reverses its direction 100 times in one second.

Question 69.
Under what orientation, the induced current produced in moving conductor in a magnetic field can be maximum?
Answer:
The current induced in a conductor is maximum when direction of motion of conductor is at right angle to the magnetic field.

Question 70.
How could you make the coil rotate continuously in motor?
Answer:
The direction of current through the coil is reversed every half rotation, the coil will rotate continuously and the same direction.

Question 71.
What is the formula for induced cmf when change the magnetic flux?
Answer:
Induced EMF = \(-\frac{\Delta \phi}{\Delta t}\)

Question 72.
The magnetic flux is varying with time. Which cases E.M.F is induced?

Answer:
In OA and BC cases, E.M.F is induced.

Question 73.
In above problem, how much EMF is induced in BC curve?
Answer:

Question 74.
What is the equation of motional E.M.F?
Answer:
Motional E.M.F = Blv;
where B = magnetic induction,
l = length of rod,
v = velocity of rod.

Question 75.
When a magnet and a coil are moving same direction with same speed. Then induced E.M.F in coil is zero. Why?
Answer:
E.M.F = \(\frac{-\Delta \phi}{\Delta \mathrm{t}}\), but both moving same direction, so change in flux linked with coil is zero i.e., ∆Φ = 0

Question 76.
What is shape of conductor is drawn when current is passing through conductor?
Answer:

Question 77.
Draw the symbols of North and South pole when depends on current.
Answer:

Question 78.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:

Question 79.
Draw the diagram of AC and DC current varying with time.
Answer:

Question 80.
Where are electric motors used?
Answer:
Electric fans, water-pumps, coolers.

Question 81.
Mention two uses of solenoid.
Answer:
It is used in electric bells, fans, and motors.

Question 82.
Mention applications of electromagnetic induction.
Answer:
It is used in devices which convert mechanical energy into electrical energy.

Question 83.
What are the advantages of an electromagnet over permanent magnet?
Answer:

1. An electromagnet can produce a strong magnetic field.
2. The strength of electromagnet can be changed.
3. The polarity of electromagnet can be changed.

10th Class Physics 12th Lesson Electromagnetism 2 Marks Important Questions and Answers

Question 1.
Anand appreciated the law behind the making of ‘generator’. Name the law and state it. (AP June 2017)
Answer:
1) The law behind the making of ‘generator’ is Faraday’s law.

2) Faraday’s Law :
“Whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Question 2.
Explain Oersted experiment to show that Electricity and Magnetism were related phenomena. (AP June 2018)
Answer:

• Place a compass needle underneath a wire and then turn on electric current.
• Immediately the needle of compass shows the deflection. By this we can conclude that electricity and magnetism are related phenomena.

Question 3.
With the help of the given figure, the teacher explained that magnetic field lines are closed lines and not open lines. Write the questions which you will ask to rest whether the given statement is right or wrong. (TS June 2015)

Answer:

• Are there any magnetic field lines inside the magnet?
• If magnetic field lines are there inside the magnet, what is the direction of field lines inside the magnet?
• What is the direction of field lines outside the magnet?

Question 4.
State Right-hand rule with a labelled diagram. (TS March 2015)
Answer:

(OR)

• Right hand rule indicates the direction of magnetic force acting on a moving charge.
• It is used when velocity and field are perpendicular to each other. If the fore finger points towards the direction of velocity of charge or current (I), middle finger points to the direction of field (B), then thumb gives the direction of force (F).

Question 5.
A coil of insulated Copper wire is connected to a Galvanometer. (TS March 2015)
What happens, if a bar magnet is ………….
1) pushed into the coil?
2) withdrawn from inside the coil?
3) held stationary inside the coil?
Answer:

1. If a bar magnet is pushed into the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
2. If a bar magnet is withdrawn from inside the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
3. If a bar magnet is held stationary inside the coil, then the needle in Galvanometer does not get deflected. Because current is not generated in the coil.

Question 6.
Compare the magnetic field lines of force formed around a current carrying solenoid with the magnetic field lines of force of a bar magnet.
Answer:

Magnetic field lines of a bar magnet	Magnetic field lines of a solenoid
2) The direction of the field lines of the outside the magnet is from north to south pole.	2) The direction of the field lines of the outside the solenoid is from north to south pole.
3) The direction of the field lines of the inside of the magnet field looks like south to north pole.	3) The direction of the field lines of the inside of the solenoid is from south to north pole.
4) These lines are closed loops.	4) These lines are also closed loops.
5) We cannot find the field lines inside the bar magnet.	5) We can find the field lines inside the solenoid.
6) These field lines are same as field lines formed by a solenoid.	6) These field lines are also same as field lines formed by a bar magnet.
7) More field lines are found at poles.	7) More field lines are found at poles.

Question 7.
Which energy we get from an electric motor? Write two daily life applications of the electric motor. (TS June 2017)
Answer:

• We get mechanical energy from electric motor.
• In our daily life we use motor in
  i) Mixies
  ii) Grinders
  iii) Water Pumps
  iv) Fans / Coolers, etc.

Question 8.
List out the material required for Oersted experiment and mention the precautions to be taken in the experiment. (TS March 2019)
Answer:
Materials required for Oersted experiment are :

1. Thermocol sheet.
2. Wooden sticks.
3. Copper wire of 24 gauge.
4. Battery.
5. Key.
6. Magnetic compass.

Precautions to be taken are :

1. Copper Wire is made through the slits of the wooden sticks tightly.
2. Arrange/complete the circuit correctly.

Question 9.
What happens when magnetic flux passing through a coil changes continuously? Where does this process is used? (TS June 2019)
Answer:
1) When there is a continuous change of magnetic flux passing through a coil a current is generated in the coil.

2) This process is used in

1. Induction stoves
2. Security checking entrance/exit doors.
3. ATM cards scanners.

Question 10.
Distinguish between AC and DC.
Answer:

Question 11.
Distinguish between AC motor and DC motor.
Answer:

Question 12.
Distinguish between AC generator and DC generator.
Answer:

Question 13.
What is Faraday’s law of electromagnetic induction? Write its expression.
(OR)
State the Faraday’s law of electromagnetic induction. Write the equation of this law.
Answer:
The induced EMF generated in a closed loop is equal to the rate of change of magnetic flux passing through it.
Induced EMF = Change in flux / time
ε = ∆Φ / ∆t

Let Φ₀ be flux linked with single turn. If there are N turns of the coil, the flux linked with the coil is NΦ₀.
∴ ε = NAΦ₀ / ∆t

Question 14.
State the right-hand thumb rule. How does the rule help us?
Answer:
When you curl your right hand fingers in the direction of current thumb gives the direction of magnetic field.

It is useful to find the magnetic field direction as well as current direction.

Question 15.
A flat rectangular coil is rotated between the pole pieces of a horseshoe magnet. In which position of coil with respect to magnetic field, will the emf (i) be maximum, (ii) be zero, (iii) change its direction.
Answer:
i) The emf is maximum when the plane of coil is parallel to the magnetic field.
ii) The emf is zero when the plane of coil is normal to the magnetic field.
iii) The emf will change its direction when the plane of coil passes from the position normal to the magnetic field.

Question 16.
State two factors on which the magnitude of induced emf depend?
Answer:
The magnitude of induced emf depends on the following two factors.

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes i.e., the rate of change of magnetic flux.

More the change in magnetic flux, more is the emf induced. Further if more rapid the magnetic flux changes, more is the emf induced.

Question 17.
How do you increase the magnetic field of solenoid?
Answer:
The magnetic field of solenoid can be increased by the following two ways.

1. by increasing the number of turns of winding in the solenoid.
2. by increasing the current through the solenoid.

Question 18.
State the function of split ring in a DC motor.
Answer:

• The split ring acts as a commutator in a DC motor.
• With the split ring, the direction of current through the coil is reversed after every half rotation of coil.
• Thus the direction of couple rotating the coil remains unchanged and the coil continues to rotate in the same direction.

Question 19.
A DC motor is rotating in a clockwise direction. How can the direction of rotation be reversed?
Answer:
The direction of rotation of motor can be reversed by interchanging the connections at the terminals of the battery joined to the brushes of the motor.

Question 20.
State the effect of inserting a soft iron core within the coil of DC motor.
Answer:

• On inserting a soft iron core within the coil of a DC motor, the speed of rotation of coil increases.
• The reason is that the strength of magnetic field between the pole pieces of magnet increases due to which the deflecting couple on coil increases.

Question 21.
State condition when magnitude of force on a current-carrying conductor placed in a magnetic field is (a) zero, (b) maximum.
Answer:
a) When current in conductor is in direction of magnetic field,
b) When current in conductor is normal to the magnetic field.

Question 22.
A flat coil ABCD Is freely suspended between the pole pieces of a U – shaped permanent magnet with the plane of coil parallel to the magnetic field.
a) What happens when current is passed in the coil?
b) When will the coil come to rest?
c) Name the instrument which makes use of the principle stated above.
Answer:
a) The coil will experience a torque due to which it will rotate.
b) The coil will come to rest when its plane becomes normal to the magnetic field.
c) DC motor.

Question 23.
Why is it more difficult to move a magnet towards a coil which has a large number of turns?
Answer:

1. Emf induced in the coil becomes more when the number of turns in the coil are made large.
2. So it is difficult to move a magnet towards a coil which has a large number of turns.

Question 24.
A coil ABCD mounted on an axle is placed between the poles N and S of permanent magnet as shown in figure.

a) In which direction will the coil begin to rotate when the current is passed through the coil in direction ABCD by connecting a battery at the ends A and D of the coil.
b) Why is commutator necessary for continuous rotation of the coil?
Answer:
a) Anti-clockwise direction.
b) Commutator is needed to change the direction of current in the coil after each half rotation of coil.

Question 25.
Draw the magnetic field lines in uniform magnetic field.
Answer:

Question 26.
Draw a labelled diagram to make an electromagnet from soft iron bar AB. Mark the polarity at its ends. What precaution would you observe?
Answer:

The labelled diagram is shown in figure. The polarity at the end A where the current is clockwise, is south (S), while at the end B where the current is anti-clockwise is north (N).
Precaution :
The source of current must be the DC source.

Question 27.
State the principle of an electric motor. Name some appliances in which the Electric motor is used.
Answer:
Current carrying coil rotates when it is kept in a uniform magnetic field. It is the working principle of electric motor.
Appliances containing electric motor are :

1. Fans,
2. Mixies,
3. Grinders,
4. Machines, etc.

10th Class Physics 12th Lesson Electromagnetism 4 Marks Important Questions and Answers

Question 1.
How can you verify with experiment “The magnetic field lines are closed loops”? (AP March 2017)
Answer:

• Place a retort stand on the plank as shown in the figure.
• Pass a copper wire through hole of the plank and rubber knob of the retort stand in such a way that the wire be arranged in a vertical position and not touch the stand.
• Connect the two ends of the wire to a battery via switch. Allow the current flows through wire.
• By sprinkling the iron fillings around the wire, we can observe the magnetic field lines are in circular shape.

Conclusion :
Hence it is proved that “Magnetic field lines are closed loops”.

Question 2.
Name the device that converts electrical energy into mechanical energy. Draw its diagram and label the parts. (AP March 2018)
Answer:
1) The device that converts electrical energy into mechanical energy is motor.
2) Diagram of motor

Question 3.
List out the materials required for the Oersted experiment of electromagnetism. Write the procedure of the experiment. What do you understand by this experiment? (TS March 2016)
Answer:
List of material required for Oersted experiment:
A thermocol sheet, two small sticks, insulated copper wires, 9 V battery, switch, magnetic compass.

Procedure :

1. Take a thermocol sheet and fix two thin wooden sticks of height 1 cm which have small slit at the top of their ends.
2. Arrange a copper wire so that it passes through these slits and makes a circuit.
3. The circuit consists of a 9 V batten’, key, and copper wires connected in series.
4. Now keep a magnetic compass below the wire.
5. Now switch on the circuit and observe the compass needle.
6. Change the directions of current and observe the compass needle.

Observation :

1. When current is passed through circuit, we observe deflection of compass needle in one direction.
2. When the direction of current is changed, the compass needle deflects in another direction.
3. This shows that a current carrying conductor possesses magnetic field around it.

Question 4.
Write the experimental procedure and observations of the experiment that is to be performed to observe the magnetic field formed due to solenoid. (TS June 2017)
Answer:

• Fix a white paper on a wooden plank.
• Make two holes to the plank at appropriate distance.
• Make some more holes parallel to these two holes.
• Insert a copper wire through these holes. It looks as a coil.
• Connect the two ends of the coil to a battery and a switch in series.
• When swich is on current flows through the wire.
• Sprinkle some iron filings around the coil and tap the plank.

Observation :
An orderly pattern of iron filing is seen on the paper. These are magnetic field lines. The magnetic field lines set up by solenoid resemble those that of a bar magnet.

Question 5.
Why the current-carrying straight wire which is kept in a uniform magnetic field, perpendicularly to the direction of the field bends aside? Explain this process with a diagram showing the direction of forces acting on the wire. (TS March 2017)
Answer:

Uniform magnetic field (due to horse shoe magnet)

Magnetic field due to current carrying straight wire

Net field formed due to the above two fields

Explanation :
The net field in upper part is strong and in lower part it is weak. Hence a non-uniform field is created around wire. Therefore the wire tries to move to the weaker field region.

Question 6.
List out the apparatus and experimental procedure for the experiment to observe a current-carrying wire experiences a magnetic force when it is kept in uniform magnetic field. (TS June 2018)
Answer:
Required apparatus :
i) Horseshoe magnet,
ii) Conducting wire,
iii) Battery, switch

Experimental procedure :
1) Arrange the circuit :
Take a wooden plank and arrange two wooden sticks with slits and arrange a conductor through the sticks and make a circuits with switch, battery.

2) Put horse shoe magnet:
Arrange the horse shoe magnet on the conductor in such a way that the conductor should be in between the two poles of the magnet.

3) Deflections in conductor :
Allow the current to pass through the circuit. We can find that conductor deflects up wards or down wards.

4)

Question 7.

Answer the following questions by observing above diagram. (TS March 2018)
1. Which device function of working does the above figure gives?
Answer:
Motor

2. What is the angle made by AB and CD with magnetic field?
Answer:
90°

3. What are the directions of magnetic forces on sides AB and CD?
Answer:
By applying right hand rule to get the directions of magnetic force. At ‘AB’, the magnetic force acts inward perpendicular to field of the magnet and on ‘CD’, it acts outwards.

4. What is the net force acting on the rectangular coil?
Answer:
The net force on ‘AB’ is equal and opposite to the force on ‘CD’ due to external magnetic field because they carry equal currents in the opposite direction. Sum of these forces is zero. Similarly, the sum of the forces on sides ‘BC’ and ‘DA’ is also zero.

So, net force on the rectangular coil is zero.

Question 8.
Explain the working process of induction stove. (TS March 2019)
Answer:

• An induction stove works on the principle of electromagnetic induction.
• A metal coil is kept just beneath the cooling surface.
• It carries alternating current (AC) so that AC produce an alternating magnetic field.
• When you keep a metal pan with water on it, the varying magnetic field beneath it crosses the bottom surface of the pan, and EMF is induced in it.
• Induced EMF produces induced current in the metal pan.
• The pan has a finite resistance.
• The flow of induced current produces heat in it.
• That heat is conducted to the water. In this way, induction stove works and water will be heated.

Question 9.
Which device is used to convert mechanical energy into electrical energy? Draw a neat diagram and label the parts of this device. (TS March 2019)
Answer:
1) Dynamo is used to convert mechanical energy into electrical energy. They are also called Generators.

2) AC Generator (or) DC Generator

AC Generator

DC Generator

Question 10.
A coil is hung as shown in the figure. A bar magnet with north pole facing the coil is moved perpendicularly

a) How does the magnetic flux passing through the coil change?
b) State the direction of the flow of the current induced in the coil, keeping the direction of bar magnet in view.
c) Draw the diagrams showing the magnetic field formed due to bar magnet at the surface of the coil and the magnetic field formed due to induced current.
d) Explain the reason for induced current.
Answer:
a) A bar magnet with north pole facing the coil is moved perpendicularly, the magnetic flex increases when passing through the coil.
b) The direction of the flow of the. current induced in the coil, keeping the direction of bar magnet is anti-clockwise due to north pole. KT
c) Φ = 0

Plane of coil is parallel to ‘B’.
d) Electromagnetic induction is the reason for induced current.

Question 11.
Conductor of length T moves perpendicular to its length with the speed V. Length of the conductor is perpendicular to the magnetic field of the conductor. Let us assume that electrons could move freely in the conductor and the charge of an electron is ‘e’.

a) What is the magnetic force acting on electron in the conductor?
b) In which direction does the above force act?
c) What effect does this force have on motion of electrons?
Answer:
a) Magnetic field acting on the electron inside the conductor is = F_m = e (\(\bar{V} \times \bar{B}\)) = BeV
This field acts from P to Q.

b) Consider in the field P and Q are ends of a conductor. ‘Q’ will act as negative end and ‘P’ will act as positive end then flow passes from P to Q means downwards.

c) The force on electrons shows an effect creates a potential difference at the ends of the rods.
∴ BeV = eE ⇒ E = BY

Question 12.
A charged particle q is moving with a speed V perpendicular to the magnetic field induction B. Find the radius of the path and time period of the particle.
Answer:

1. Let us assume that the field is directed into the page.
2. Then the force experienced by the particle F = qvB.
3. We know that this force is always directed perpendicular to velocity.
4. Hence the particle moves along a circular path and the magnetic force on a charged particle acts like a centripetal force.
5. Let r be the radius of the circular path.

Question 13.
Explain different ways to induced current in a coil.
Answer:

• Moving a north pole of a magnet into a coil.
• Withdrawing north pole from a coil.
• Moving a south pole of magnet into a coil.
• Withdrawing a south pole of a magnet from a coil.
• Moving a coil towards a magnet.

Question 14.
What are the similarities between a current-carrying solenoid and a bar magnet?
Answer:
1) The magnetic field lines due to current carrying solenoid are identical to those of bar magnet. Thus a current-carrying solenoid behaves just like a bar magnet with fixed polarities at the ends. The end at which the direction of current is clockwise behaves like a south pole and the end at which current is anti-clockwise behaves like a north pole.

2) A current-carrying solenoid when suspended freely, will set itself in the north south direction exactly in the same manner as a bar magnet does.

3) A current-carrying solenoid also acquires the- attractive property of magnet. If iron filings are brought near the solenoid, it attracts them when current flows through the solenoid.

Question 15.
What are the dissimilarities between a current-carrying solenoid and a bar magnet?
Answer:

• The magnetic field strength due to solenoid can be altered by altering current in it, while the magnetic field strength of a bar magnet cannot be changed.
• The direction of magnetic field due to solenoid can be reversed by reversing the direction of current in it, but the direction of magnetic field of the bar magnet cannot be changed.

Question 16.
Compare electromagnet with a permanent magnet.
Answer:

Question 17.
What are the characteristics of magnetic field lines due to current in a loop (or circular coil)?
Answer:

• The magnetic lines are nearly circular in the vicinity of coil.
• Within the space enclosed by the wire the magnetic field lines are in the same direction.
• Near the centre of loop, the magnetic field lines are nearly parallel and the magnetic field may be assumed to be uniform in a small space near the centre.
• At the centre, the magnetic lines are along the axis of the loop and at right angles to the plane of the loop.
• The magnetic field lines become denser if the strength of current in the loop is increased and there are more number of turns in the loop.

Question 18.
A straight conductor passes vertically through a cardboard having some iron filings sprinkled on it.
a) Show the setting of iron filings when current is passed in the downward direction and then the cardboard is gently tapped. Draw arrows to represent the direction of magnetic field lines.
b) What changes occur if
i) current is increased ?
ii) the single conductor is replaced by several parallel conductors each carrying same current flowing in the same direction?
c) Name the law used by you to find the direction of magnetic field lines.
Answer:

a) The figure shoy/s, the pattern in which iron filings will set themselves. The arrows show the direction of magnetic field lines.
b) i) The arrangement of iron filings remains unchanged, but they become denser and Cardboar get arranged up to a larger distance from the conductor when the strength of current is increased and it is effective up to a larger distance from the conductor.
ii) The magnetic field at a point due to each conductor will be in same direction, so they will be added up. Thus the magnetic field strength is increased and it is effective up to a large distance so the magnetic field lines come closer and iron filings get arranged up to a larger distance.
c) Right hand thumb rule.

Question 19.
In figure A and B represent two straight wires carrying equal currents in a direction normal to the plane of paper inwards.
a) Sketch separately the magnetic field lines produced by each current.
b) Give a reason why the magnetic field at K (mid point of the line joining A and B) will be zero.
c) What will be the effect on the magnetic field at the point K if the current in wire B is reversed?

Answer:
a) Figure shows the sketch of magnetic field A and B.

b) The point K is equidistant from the wires A and B, and the wires A and B carry equal currents. So the magnetic fields at K due to wires A and B are equal in magnitude, but opposite in direction. Due to the wire A, it is downwards in the plane of paper, while due to the wire B, it is upwards in the plane of paper. So the net magnetic field at the point K is zero as the two fields cancel each other.
c) On reversing the direction of current in the wire B, the direction of magnetic field due to current is reversed at the point K, i.e. it becomes downwards in the plane of paper.

Question 20.
The diagram given below shows two coils X and Y. The coil X is connected to a battery S and a key K. The coil Y is connected to a galvanometer G.

When the key K is closed state the polarity.
i) At the end B of the coil X.
ii) At the end C of the coil Y.
iii) At the end C of the coil Y if the coil Y is (a) moved towards the coil X, (b) moved away from the coil X.
Answer:
i) On closing the key K, the current at the end B of the coil X is anti-clockwise, therefore at this end there is a north pole.

ii) While closing the key, polarity at the end C of the coil Y will be north. But there will be no polarity at the end C of the coil Y when the current becomes steady in the coil X

iii) a) With the key K closed, while the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.
b) With the key K closed, while the coil Y is moved away from the coil X, the polarity at the end C of the coil Y is south.

Question 21.
How do you increase the speed of rotation of coil in a DC motor?
Answer:
The speed of rotation of coil can be increased by

1. Increasing the strength of current.
2. Increasing the number of turns in coil.
3. Increasing the strength of magnetic field. To increase the strength of magnetic field a soft iron core can be inserted within the coil.

Question 22.
Explain Lenz’s law with an activity.
Answer:
Lenz law :
The induced current will appear in such a direction that it opposes the change in the flux in the coil.

Explanation :

1. We know that when a bar magnet is pushed towards a coil with its north pole facing the coil an induced current is set up in the coil.
2. Let the direction be clockwise then current carrying loop behaves like a magnet with its south pole facing the north pole of bar magnet.
3. In such a case, the bar magnet attracts the coil. Then it gains kinetic energy. This is contradictory to conservation of energy.
4. Hence our assumption is wrong. So correct induced current direction is anti-clockwise.
5. Let us see a case where the bar magnet is pulled away from the coil with the north pole facing the coil. In such case, the coil opposes the motion of bar magnet to balance the conversion of mechanical energy into electric energy.
6. It happens only when the north pole of the magnet faces the south pole of the coil.
7. So, the direction of induced current in the coil must be in anti-clockwise direction.
8. In simple terms, when flux increases through coil, the coil opposes the increase in the flux and when flux decreases through coil, it opposes the decrease in the flux. This is Lenz law.

Question 23.
What are the uses of electromagnet?
Answer:
Electromagnets are mainly used for the following purposes.

1. For lifting and transporting the large masses of scrap, girders, plates, etc. especially to the places where it is not convenient to take the help of human labour.
2. For loading furnaces with iron.
3. For separating the magnetic substances such as iron from other debris.
4. For removing pieces of iron from wounds.
5. In several electrical devices such as electric bell, telegraph, electric thumb, electric motor, relay, microphone, loudspeaker, etc.
6. In scientific research, to study the magnetic properties of a substance in a magnetic field.

Question 24.
Write the differences between AC generator and DC motor.
Answer:

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 8 Structure of Atom.

AP State Syllabus SSC 10th Class Chemistry Important Questions 8th Structure of Atom

10th Class Chemistry 8th Lesson Structure of Atom 1 Mark Important Questions and Answers

Question 1.
Write the electronic configuration of chromium. (AP June 2016)
Answrr:
The electronic configuration of chromium is
1s²2s²2p⁶3s²3p⁶4s¹3d⁵ or [Ar] 4s¹3d⁵

Question 2.
Out of 3d and 4s, which has more (n + l) value? Explain. (AP June 2017)
Answer:
1) 3d ⇒ n + l ⇒ 3 + 2 ⇒ 5 (energy)
4s ⇒ n + l ⇒ 4 + 0 ⇒ 4 (energy)

2) Hence, ‘3d’ has more (n + l) value than ‘4s’.

Question 3.
Prepare a question on nl^x method. (AP SA-I:2018-19)
Answer:

• How is nl^x method useful.
• Explain the nl^x method with an example.

Question 4.
Which colours do you observe when an iron rod is gradually heated to higher tem-peratures? (TS June 2015)
Answer:
First iron turns into red (lower energy corresponding to higher wavelength) and as the temperature rises it glows and turns into orange, yellow, blue or even white respectively (higher energy and lower wavelength).

Question 5.
Which principle is not followed in writing the electronic configuration of 1s² 2s¹ 2p⁴? Give reasons. (TS June 2015)
Answer:
1) Principle :
Aufbau principle is not followed in writing the electronic configuration of 1s² 2s¹ 2p⁴.

2) Reasons :
i) According to Aufbau principle electron enters into orbital of lowest energy.
ii) Between 2s and 2p, 2s has least energy. So 2s must be filled before the electron has to enter 2p.

Question 6.
Write the symbol of the outermost shell of magnesium (Z = 12) atom. How many electrons are present in the outermost shell of magnesium? (TS June 2017)
Answer:
Symbol of the outermost shell of magnesium (3rd shell) = M
No. of electrons in outermost shell of Magnesium = 2.

Question 7.
The four quantum number values of the 21st electrons of scandium (Sc) are given in the following table. (TS March 2017)

Write the values of the four quantum numbers for the 20th electron of scandium (Sc) in the form of the table.
Answer:

Question 8.
If n = 3, mention the orbitals present in the shell and write maximum number of electrons in the shell. (TS March 2018)
(OR)
Write the maximum number of electrons and number of orbitals in the shell, when n = 3.
Answer:

1. When n = 3, number of subshells = 3 (3s₁ 3p₁ 3d)
2. Number of orbitals = 9 (3s(1) 3p(3) 3d(5))
3. Maximum number of electrons (3s² 3p⁶ 3d¹⁰) = 18

Question 9.
What is dispersion?
Answer:
The splitting of light into different colours is called dispersion.

Question 10.
What is an electromagnetic wave?
Answer:
When electric field and magnetic fields are perpendicular to each other and at right angles to the direction of propagation of wave is formed. Such a wave is called electromagnetic wave.

Question 11.
What is a Zeeman effect?
Answer:
The splitting of spectral lines in the presence of magnetic field is called Zeeman effect.

Question 12.
What is a spectrum?
Answer:
Group of wavelengths is called spectrum (or) A collection of dispersed light giving its wavelength composition is called a spectrum.

Question 13.
What is speed of electromagnetic wave?
Answer:
It is equal to speed of light, i.e. 3 x 10⁸ ms^-1

Question 14.
Which colour has highest wavelength and which colour has least wavelength on visible spectrum?
Answer:
The colour that has highest wavelength in visible spectrum is red and least wavelength is violet.

Question 15.
If n = 5, then what is the maximum value for l?
Answer:
The maximum value for l is 4.

Question 16.
If l = 4, what is the number erf values for m_l?
Answer:
m_l = 2l + 1 = 2(4) +1=9.

Question 17.
What are the values of m_s?
Answer:
½ or – ½

Question 18.
What is electronic configuration?
Answer:
Distribution of electrons in shells, sub-shells and orbital in an atom is known as electronic configuration.

Question 19.
What is Heisenberg’s principle of uncertainty?
Answer:
It is not possible to find the exact position and velocity of electron simultaneously.

Question 20.
Give ascending order of various atomic orbitals according to Moeller diagram.
Answer:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4d < 5d < 6p < 7s < 5f < 6d < 7p < 8s.

Question 21.
What is Hund’s Rule?
Answer:
Electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

Question 22.
What is Planck’s equation?
Answer:
Planck’s equation is E = hv.
E = Energy of the radiation
h = Planck’s constant = 6.625 × 10^-34 J
v = Frequency of radiation.

Question 23.
What is electromagnetic spectrum?
Answer:
Electromagnetic waves can have a wide variety of wavelengths. The entire range of wavelengths is known as the electromagnetic spectrum.

Question 24.
What is wavelength?
Answer:
The distance from one wave peak to the next is called wavelength (λ).

Question 25.
What is frequency?
Answer:
The number of wave peaks that pass by a given point per unit time is called frequency.

Question 26.
When are electromagnetic waves produced?
AnElectromagnetic waves are produced when an electric charge vibrates.

Question 27.
Which is the example for line spectrum?
Answer:
The atomic spectrum of hydrogen atom.

Question 28.
Which model explains fine spectrum of atom?
Answer:
Bohr – Sommerfeld model.

Question 29.
How are wavelength and velocity of light related?
Answer:
c = vλ
where
c = velocity of light,
v = frequency of light,
λ = wavelength of light.

Question 30.
Give the equation which gives electromagnetic energy (light) that can have only Certain discrete energy values.
Answer:
E = hv
E = Energy of light
h = Planck’s constant = 6.62 5 × 10^-27 erg sec or 6.625 × 10^-34 Joule-sec
v = Frequency of radiation
This equation is called Planck’s equation.

Question 31.
Which group elements are called Noble gases?
Answer:
VIII A group or 18th group elements are called inert gases (or) Noble gases.

Question 32.
Which elements are highly stable?
Answer:
Noble gases are highly stable.

Question 33.
Write the set of quantum numbers for the electrons in a 3p_z orbital.
Answer:

Question 34.
What is the difference between an orbit and orbital?
Answer:
An orbit is a well defined path of electron that revolves around the nucleus.

An orbital is the space around the nucleus, where the probability of finding electrons is maximum.

Question 35.
What are the factors which influence electromagnetic energy?
Answer:
Electromagnetic energy depends on two factors

1. wavelength
2. frequency.

Question 36.
What is a wave?
Answer:
The disturbance occurred in a medium is called wave.

Question 37.
When cupric chloride is kept in non-luminous flame then what is the colour of flame?
Answer:
Green colour.

Question 38.
If the colours gradually changes there are no sharp boundaries in between them, then what is the name given to that type of spectrum?
Answer:
Continuous spectrum of emission.

Question 39.
What is the information given by magnetic orbital quantum numbers?
Answer:
Orientation of orbitals in space.

Question 40.
How many orbitals are present in a sub-shell?
Answer:
The number of orbitals are present in a sub-shell is n² (where n is principal quantum number).

Question 41.
What happens when an object is suitably excited by heating?
Answer:
Light is emitted by the object.

Question 42.
What is meant by Aufbau?
Answer:
The German word Aufbau means building up.

Question 43.
Which elements are examples for Noble gases?
Answer:
Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn) are examples for Noble gases.

Question 44.
Which element has duplet configuration?
Answer:
Helium (1s²)

Question 45.
Write the set of quantum number for the added electron of oxygen atom.
Answer:
Configuration of oxygen is 1s² 2s² 2p⁴.
The added electron is 4th in the 2p.
The set of quantum numbers (2, 1, -1, -½)

Question 46.
Can we apply c = vλ, to sound waves?
Answer:
Yes. It is a universal relationship and applies to all waves.

Question 47.
What is the value of Planck’s constant?
Answer:
The value of Planck’s constant is 6.626 × 10^-34 Js.

Question 48.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:’
The light emitted by different kind of atoms is different because the excited states electrons will go are different.

Question 49.
How many values can ‘l’ have for n = 4?
Answer:
If n = 4, l can take values 0, 1, 2, 3. So there are four values.

Question 50.
Write the four quantum numbers for the differentiating electrons of lithium (Li) atom.
Answer:
The electronic configuration of lithium is 1s² 2s¹. So differentiating electron enters into 2s. The values of four quantum numbers are as given below.

Question 51.
Write four quantum numbers for 2p¹ electrons.
Answer:
The four quantum numbers for 2p are

Question 52.
Which rule is violated in the following electronic configuration?

Answer:
The rule violated is Hund’s rule.

Question 53.
How many maximum number of electrons that can be accommodated in N principle energy shell?
Answer:
For N shell n = 4.
The maximum number of electrons accommodated in a shell is 2n².
∴ A maximum of 32 electrons can be filled in N shell.

Question 54.
How many maximum number of electrons that can be accommodated in a ‘l’ sub-shell?
Answer:
l sub-shell has 3 orbitals. Each orbital accommodates 2 electrons. So 6 electrons can be filled in l sub-shell.

Question 55.
How many maximum number of electrons can be accommodated in ‘d’ sub-shell?
Answer:
d sub-shell has 5 orbitals. So 10 electrons can be filled in d sub-shell.

Question 56.
How many sub-shells present in a ‘M’ principal energy shell?
Answer:
For M shell n = 3
The number of sub-shells in M shell is 3.

Question 57.
How many spin orientations are possible for an electron in s-orbital?
Answer:
The spin quantum number values for electrons are ½ or -½. So 2 spin orientations are possible.

Question 58.
Write valence electronic configuration of element which has the following set of quantum numbers.
Answer:

n = 3 indicates 3rd orbit and l = 1 indicates p orbital and there is one electron in p orbital. So the valence electron configuration is 3p¹.

Question 59.
How many unpaired electrons are present in chromium?
Answer:
The electronic configuration of chromium 1s² 2s² 2p⁶ 3s² 3p6 4s¹ 3d⁵.

The number of unpaired electrons = 6.

Question 60.
Find the four quantum number values of 3^rd and 4^th electrons of Beryllium.
The electronic configuration of Beryllium is 1s² 2s².

Question 61.
What is the n + l value of 4f orbital?
Answer:
For 4f orbital n = 4 and f orbital l = 3.
∴ n + l = 4 + 3 = 7

Question 62.
When you heat iron rod first it turns red. Why?
Answer:
Iron turns into red because red has higher wavelength. So it has lower energy which is emitted by iron.

Question 63.
What is the significance of Planck’s proposal?
Answer:
Electromagnetic energy can be gained or lost in discrete values and not in a continuous manner.

Question 64.
When do you see an emission line?
The energy emitted by an electron is seen in the form of an electromagnetic energy and when the wavelength is. in the visible region it is visible as an emission line.

Question 65.
How many elliptical orbits are there in 4^th orbit of Sommerfeld?
Answer:
The number of elliptical orbits in 4^th orbit of Sommerfeld is 3.

Question 66.
Why is spin quantum number introduced?
Answer:
When we observe spectrum of yellow light by using high resolution spectroscope it has very closely spaced doublet. Similar patterns are shown by Alkali and Alkaline earth metals. In order to account this spin quantum number is introduced.

Question 67.
Which of the following magnetic quantum number values is not possible for 3d orbital?
a) – 2
b) – 1
c) 0
d) 4
Answer:
For d orbital the possible m; values – 2, – 1, 0, 1, 2. So the value 4 is not possible.

Question 68.
If an element has 11 electrons in its M shell, then what is the name of element and its atomic number?
Answer:
The electronic configuration of element is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³.
[M shell electrons = 2 + 6 + 3=11]

So the element is Titanium.

Question 69.
The wave length of a wave is 100 nm. Find its frequency.
Answer:

10th Class Chemistry 8th Lesson Structure of Atom 2 Marks Important Questions and Answers

Question 1.
Explain Hund’s Rule with an example. (TS March 2016) (AP SA-1:2018-19)
Answer:
Hund’s Rule :
According to this rule electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

(1) Ex :

1. The configuration of carbon atom (Z = 6) is 1s² 2s² 2p².
2. The first four electrons go into the 1s and 2s orbitals.
3. The next two electrons go into 2p_x and 2p_y orbitals.
   
4. But, they do not pair in 2p_x orbital.

(2) Ex :

1. In oxygen atom (₈O), distribution of electrons is given below.
2. 
3. Here, pairing of electrons in 2px orbital takes place after, 2p_x, 2p_y and 2p_z orbitals are filled with a single electron.

Question 2.
The electronic configuration of Sodium is 1s² 2s² 2p⁶ 3s¹. (AP March 2017)
What information that it gives?
Answer:

1. Its atomic number is 11
2. It is s-block element
3. It is in 3rd period ’
4. It is in 1st group
5. Its valency is 1
6. Number of valency electrons are 1
7. It can form uni positive ion
8. It can form ionic bond, 9. It is metal.

Question 3.
Explain the principle which describes the arrangement of electrons in degenerate orbitals.
Answer:
According to Hund’s rule the degenerate orbitals are occupied with one electron each before pairing of electron starts.
Ex : Electronic configuration of carbon is 1s² 2s² 2p²

the last two electrons will enter into separate 2p orbitals.

Question 4.
Name the principle, which says an orbital can hold only 2 electrons and explain. (AP March 2018)
Answer:

• Name of the principle : Pauli’s exclusion principle.
• No, two electrons in an orbital can have all four quantum numbers same.
• It says there is a chance to hold only 2 electrons, one rotates in clockwise direction (+½)other rotates in anticlock wise direction (-½).

Question 5.
For a better understanding about the electronic configuration in an atom, the teacher wrote shorthand notation nl^x on the blackboard.
Looking at this notation, what could be the probable questions that generate in the student’s mind? Write any two of them. (TS March 2015)
Answer:

1. What n, l, x indicates related to atoms?
2. How nl^x indicates the position of the electrons in the atom?

Question 6.
Write the ‘Octet Rule’. How does Mg (12) get stability while reacting with chlorine as per this rule? (TS June 2017)
Answer:
Octet Rule :
The atoms of elements tend to undergo chemical changes that help to leave their atoms with eight outer – shell electrons.

Magnesium atom looses 2 electrons and get 8 electrons in its outermost shell as Neon. So that it gets stability.

Question 7.
Write the electronic configuration of the atom of an element having atomic number 11. Write the names of the rules and the laws followed by you in writing this electronic configuration. (TS March 2017)
Answer:
1s²2s²2p⁶3s¹.
(OR)
Principles followed :

1. Aufbau principle.
2. Hund’s Rule.
3. Pauli Exclusion Principle

Question 8.
The electron enters into 4s orbital after filling 3p orbital but not into 3d. Explain the reason (TS March 2018)
Answer:
Based on (n+l) values energy value of 3d orbitals is 3 + 2 = 5, energy value of 4s orbitals is 4 + 0 = 4
The energy level of 4s orbital is less than the 3d orbital according to the Aufbau principle electron enters into lower energy orbital first.
Thats why electrons enters into 4s orbital after filling 3p, but not into 3d.

Question 9.
Write the electronic configuration of Na⁺ and Cl^–.
Answer:
Electronic Configuration of Na⁺ is 1s² 2s² 2p⁶ and Cl^– is 1s² 2s² 2p⁶ 3s² 3p⁶.

Question 10.
Observe the given table and answer the following questions. (TS March 2019)

1) Mention the divalent element name.
2) Name the element belongs to 3^rd period and VA Group.
Answer:

1. Name of the divalent element in the table is Calcium.
2. Name of the element which belongs to 3^rd period and VA Group is Phosphorous.

Question 11.
Your friend is unable to understand nl^x. What questions will you ask him to understand nl^x method? (AP SCERT: 2019-20)
Answer:

1. What is nl^x method?
2. Where does it use ?
3. What is meant by ‘n’, 7′ and ‘x’?
4. How can we use nl^x method in the writing of electronic configuration?

Question 12.
Why do valency electrons involve in bond formation, than electrons of inner shells? (AP SCERT: 2019-20)
Answer:

1. When two atoms come sufficiently close together the valence electrons of each atom experience the attractive force of the nucleus in the other atom.
2. The nucleus and the electrons in the inner shell remain unaffected when atoms come close together.
3. The electrons in outer most shell of an atom get affected.
4. Thus electrons in valence shell are responsible for the formation of bond between atoms.

Question 13.
Explain Pauli’s exclusion principle with an example. (AP SA-I:2019-20)
Answer:
Pauli’s exclusion principle :
No two electrons of the same atom can have all four quantum numbers the same. If n, l and m_l are same for two electrons, then m_s must be different.
Suppose take the example of Helium atom.
The four quantum numbers for two electrons in the Helium atom given below.

We can observe from the table that three of the quantum numbers are same but fourth quantum number is different. The electronic configuration of Helium atom is² ↑↓. So the maximum number of electrons filled in an orbital is 2.

Question 14.
Explain Aufbau principle. (AP SA-I:2019-20)
Answer:
Aufbau principle :

1. In the ground state the electronic configuration can be built up by placing electrons in the lowest available orbitals until the total number of electrons added is equal to atomic number.
2. Thus orbitals are filled in the order of increasing energy.
3. Electrons are assigned to orbitals in order of increasing value of n + l.
4. For sub-shells with the same value of n + l, electrons are assigned first to the subshell with lower n.

Question 15.
The electronic configuration of an atom is as follows 1s² 2s² 2p².
a) Which element’s atom is it?
b) Which orbital is the last electron in?
c) When excited what could be the number of lone / single electrons in this atom?
d) What is the value of principal quantum numbers of two electrons in the first box?
Answer:
Given electronic configuration of atom is 1s² 2s² 2p².
a) The element is carbon.
b) The valence electron enters 2p orbital.
c) In excited state the electron in 2s orbital enters 2p orbital. So it has 4 unpaired electrons.
d) The value of principal quantum number is 1.

Question 16.
Draw the table which gives the information about the quantum numbers and the number of the quantum states.
Answer:

Question 17.
Explain briefly about spin quantum number.
Answer:

• This gives spin of the electrons about their own axes. It is denoted by ms.
• This quantum number refers to the two possible orientations of the spin of an electron, one clockwise and the other anti-clockwise spin.
• These are represented by +½ -½ and .

Question 18.
Write electronic configurations of following elements,
a) Hydrogen
b) Helium
c) Lithium
d) Beryllium
e) Boron
Answer:

Question 19.
What does a line spectrum tell us about the structure of an atom?
Answer:
The electrons in ground state i.e. lowest, energy state absorb energy and move into excited state where they are unable to stay for long periods so lose the energy and come back to the ground state. The emitted radiation appears as line in line spectrum.

Question 20.
What are the spins of electrons in Helium atom?
Answer:
The quantum numbers for two electrons of Helium are given below as per Pauli’s exclusive principle.

Three of quantum numbers are same. So fourth must be different so the two electrons have anti-parallel spins.

Question 21.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ is the electronic configuration of Cu (Z = 29). Which rule is violated while writing this configuration? What might be the reason for writing this configuration?
Answer:
The rule violated is Aufbau principle. The elements which have half filled or completely filled orbitals have greater stability. So copper can get stability by transferring one electron from 4s to 3d (their energies are close to each other).

So the electronic configuration of copper is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹, not 1s² 2s² 2p 3s² 3p⁶ 3d⁹ 4s².

Question 22.
Why are chromium and copper exceptions to electronic configuration?
Answer:
Elements which have half-filled or completely filled orbitals have greater stability. So in chromium and copper the electrons in 4s and 3d redistribute their energies to attain stability by acquiring half-filled and completely filled d orbitals.

Hence the actual electronic configurations of chromium and copper are as follows.

Question 23.
Distinguish between emission and absorption spectrum.
Answer:

Question 24.
Draw a table which will give the relationship between l values and number of orbitals and name of sub-shell and maximum number of electrons.
Answer:

Question 25.
Distinguish between line and band spectrum.
Answer:

Question 26.
What is Dispersion of light? Explain a natural example of dispersion of light.
Answer:
Dispersion:
Splitting of white light into colours (VIBGYOR) is called Dispersion of light.

The natural example for dispersion of light is formation of Rainbow. It is caused by dispersion of sunlight by tiny water droplets present in atmosphere which act as small prisms.

Question 27.
Wien is an electromagnetic wave produced? Write about characteristics of electromagnetic wave.
Answer:
Electromagnetic wave is produced when an electric charge vibrates (moves back and fortn).

Characteristics of electromagnetic waves :

1. Efectric field and magnetic fields are perpendicular to each other and at right angles to direction of propagation of wave.
2. It travels with speed of light i.e., 3 × 10⁸ ms^-1.
3. Electromagnetic energy is characterized by wavelength (λ) and frequency (v). The relation is given by c = vλ.

Question 28.
The valence electron configuration of element is given as 4s¹. Then give the following information.
1) What is the name of that element?
2) What is the outermost orbit of element?
3) What is ‘l’ value of outermost sub-shell?
4) What is the atomic number of element?
Answer:

1. Potassium.
2. N
3. The outermost sub-shell is 4s its l value is ‘O’.
4. Its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. So its atomic number is 19.

Question 29.
Write all the quantum numbers for valance electron of sodium.
Answer:

1. The electronic configuration of sodium is 1s² 2s² 2p⁶ 3s¹.
2. The valance orbital is 3s.
3. The quantum numbers for this orbital is
   

Question 30.
Give electronic configurations of following elements.
a) Sodium
b) Phosphorous
Answer:
Sodium – 1s² 2s² 2p⁶ 3s¹
Phosphorous – 1s² 2s² 2p⁶ 3s² 3p³

Question 31.
Why does nitrogen has more chemical stability when compared with oxygen?
Answer:
The electronic configurations of Nitrogen and Oxygen are as follows.

Nitrogen has half filled 2p³ configuration. So it has greater chemical stability when compared with Oxygen.

Question 32.
Ramu gave electronic configuration of potassium as 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ whereas Ravi expressed the configuration as 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. Who gave the correct configuration? Why?
Answer:
Ravi gave the correct configuration because according to Aufbau principle after completion of 3p orbital electron may enter either 4s or 3d.
Their n + l values are given below.

So 4s orbital has lower n + l value when compared with 3d orbital. So electron enters into 4s.

Question 33.
Given the valence electron configuration of an element is 4s¹. Then what are its quantum number values. Which element does it represent?
Answer:
Given valence electronic configuration is 4s¹. So n = 4 for ‘s’ sub shell l = 0 and if l is ‘O’ then m_l is also zero. m_s takes only two values that is +½ or -½ for convenience we can take m_s as +½.
∴ The quantum number values are like this

The element is potassium.

Question 34.
The atomic number of an element is 17, then calculate the total number of electrons present in its s and p orbitals.
Answer:
The element with atomic number 17 is chlorine.
Its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
So, the total number of electrons present in s orbitals = 2 + 2 + 2 = 6.
The total number of electrons present in p orbitals = 6 + 5 = 11.

Question 35.
Based on Aufbau’s principle, in which of the three 4d, 5p and 5s orbitals the electrons will be filled first? Why?
Answer:
According to Aufbau’s principle, the electron enters the orbital having lower n + l value. If both orbitals have same n + l values, electron enters the orbital with lower ‘n’ value.

So, 3s has least n + l value. Therefore the electron enters 5s orbital first.

Question 36.
Find the following.
1) Number of orbitals present in M orbit.
Answer:
Number of orbitals present in an orbit = n² For M orbit n = 3.
∴ So the number of orbitals = 3² = 9.

2) The maximum and minimum possible m_l values for 4f orbital.
Answer:
For f orbital ‘l’ value is 3. If l is 3, then m_l takes values from – 3 to + 3.
So the minimum value for m_l is – 3 and maximum value is + 3.

3) The possible values of l if n = 4.
Answer:
If n = 4, then l take values from 0 to 3 i.e., 0, 1, 2, 3.

4) The maximum number of electrons that can be filled in ‘N’ energy level.
Answer:
For N orbit n = 4. The maximum number of electrons present in a shell = n².
∴ The maximum number of electrons filled in N shell = 4² = 16.

Question 37.
How do the vibrating electric and magnetic fields around the charge become a wave that travel through space?
Answer:

• A vibrating electric charge creates a change in the electric field.
• The changing electric field creates a changing magnetic field.
• This process continues, with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.

Question 38.
What information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nuclear where the electrons are found and also their energies.

Question 39.
An electron in an atom has the following set of four quantum numbers. Which orbital does it belong to?

Answer:
The quantum numbers of an atom is given below.

By using nl^x method n = 3, if l = 0 then the sub-shell is s.
So the electron belongs to 3s.

Question 40.
Look at the following table and answer the following questions.

a) What is the law that is voilated in the above table? State that law.
b) Write the correct table using that law.
Answer:
a) The law violated is Pauli’s Exclusive Principle. Pauli’s law states that no two electrons of the same atom can have all the four quantum numbers same.
b)

10th Class Chemistry 8th Lesson Structure of Atom 4 Marks Important Questions and Answers

Question 1.
Draw Moeller chart of filling order of atomic orbitals.
(OR)
Draw a diagram showing the increasing value of (n + l) of orbitals. (AP June 2017)
Answer:

The filling order of atomic orbitals (Moeller Chart)

Question 2.
Complete the following table based on quantum numbers related to atomic orbitals and electron of an atom. (AP SA-I-2018-19)

Answer:

Question 3.
Based on the information given in the table, answer the questions given below.

i) For the 4th main shell, how many values are there for m_l? What are they? (TS June 2016)
Answer:
1) There are 16 m( values in the 4th main shell.
2) They are

ii) For sub-shell with n = 3, l = 1, write the m, values.
Answer:
For sub – shell with n = 3, l = 1 the m, values are -1, 0, 1.

iii) Write the principal quantum number value for ‘N’ shell. How many sub-shells are there in the main shell?
Answer:

1. The principal quantum number value for ‘N’ shell is 4.
2. The number of sub-shells is 4. They are 4s, 4p, 4d, 4f.

iv) In the above table m_l and l values are given. Write a formula that gives the relationship between m_l and l based on those values.
Answer:
m_l (No. of values) = 2l + 1.

Question 4.
Observe the information provided in the table about quantum numbers. Then answer the questions given below it. (TS June 2017)

i) Write the ‘l’ value and symbol of the spherical shaped sub-shell.
ii) How many values that ‘m_l‘ takes for 1 = 2? What are they?
iii) Write the symbols of the orbitals for l = 1 sub-shell.
iv) What is the shape of the sub-shell for l = 2? What is the maximum number of electrons that can occupy this sub-shell?
Answer:
i) Spherical shaped sub-shell “l” value is zero and symbol is ‘s’.
ii) Number of m; values for 1 = 2 is 5, those are -2, -1, 0, 1, 2.
iii) Symbols of the orbitals for l = 1 sub-shell are p_x, p_y, p_z.
iv) Shape of the subshell l = 2 is double dumbel.
The maximum number of electrons that can occupy in this sub-shell is 10.

Question 5.
Write postulates and limitations of Bohr Hydrogen atomic model. (TS June 2018)
Answer:
Postulates :

1. Niels Bohr proposed that electrons in an atom occupy ‘stationary’ orbits of fixed energy at different distances from the nucleus.
2. When an electron jumps from a lower energy state (ground state) to higher energy state it absorbs energy or emits energy when such a jump occurs from a higher energy state.
3. The energies of an electron in an atom can have only certain values E₁, E₂, E₅…. that is, the energy is quantized.

Limitations :

1. Bohr’s model failed to account for splitting of line spectra of hydrogen atom into liner lines.
2. Bohr theory could not explain the Zeeman and Stark effect.

Question 6.
Explain Bohr’s model of hydrogen atom and its limitations. (March 2019)
Answer:
Niels Bohr proposed that,
a) electrons in an atom occupy stationary orbits of fixed energy (K, L, M, N,…) at different distances from the nucleus.

b) when an electron jumps from a lower energy state to higher energy state, it ab¬sorbs energy or emits energy when such a jump occurs from a higher energy state to lower energy state.

c) the energies of an electron in an atom can have only certain values E₁, E₂, E₃, ……… i.e. the energy is quantized. The states corresponding to these energies are called stationary states and the possible values of the energy are called energy levels.

d) the angular momentum of electron is multiple integral of \(\frac{L}{2 \pi}\).
∴ L = mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
m = mass of electron ;
v = velocity of electron ;
r = radius of circular path;
h = plank constant

Limitations :

1. Bohr’s model failed to account for splitting of line spectra (Zeeman effect).
2. This model failed to account for the atomic spectra of atoms of more than one electron.
3. Bohr theory was not explained the quantisation of angular momentum of an electron.
4. It was not explained the formation of chemical bonds.

Question 7.
Explain four quantum numbers with an example. (AP SA-I; 2019-20)
Answer:
Each electron iii an atom is described by a set of three numbers called Quantum numbers.

1) Principal quantum number (n) :
It is used to know the size and energy of the main shell The values of ‘n’ are 1, 2, 3 …..
energy of the shell, n = 1 < energy of the Shell n = 2.

2) Angular momentum of quantum number (l) :
It is used to know the shape of a particular sub shell.
The values of ‘l’ are 0, 1, 2, 3
l = 0 = s orbital = spherical in shape
l = 1 = p orbital = dumbel in shape
l = 2 = d orbital = double dumbel Shape

3) i) Magnetic quantum number (m_l) :
It is used to describe the orientation of the orbital in space relative to the other orbitals in the atom.
The values of mt are – ‘l’ to ’+l’ including zero,

ii) Spin quantum number (m_s) :
It is used to know the orientation of the spins of electrons.
The values of ms are +½ and – ½.

Question 8.
How does Hund’s rule helps in writing electronic configuration of an atom? Explain with a suitable example. (TS June 2019)
Answer:
Hund’s rule :
According to this rule electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

This Hund’s rule helps in writing of electronic configuration of an element.

Example :

1. The electronic configuration of carbon (C) atom (Z = 6) is 1s² 2s² 2p⁶.
2. The first four electrons go into the Is and 2s orbitals.
3. The next two electrons go into separate 2p orbitals, with both electrons having the same spin.
   

Another example :

1. The electronic configuration of oxygen (₈O) is 1s² 2s² 2p⁴.
2. The first four electrons go into the 1s, 2s orbitals.
3. The next four electrons go into 2p orbits as 2 in 2p_x, 1 in 2p_y and 1 in 2p_z orbital.
   
4. Here pairing of electron in 2px starts after filling of electron in each 2p_x, 2p_y, 2p_z orbitals.
5. But electrons do not occupy like this
   

Question 9.
Your father asked you to go to the market and purchase an electric lamp. The shop-keeper displayed two lamps – one is violet and another is red. Which coloured lamp do you purchase to put in your bedroom? Support your choice of solution.
Answer:
Red has highest wavelength and violet has lowest wavelength. We know that relationship between energy as follows.
E = hv = h\(\frac{c}{\lambda}\)

h and c are remains constant.
∴ E ∝ \(\frac{1}{\lambda}\)

Energy is inversely proportional to wavelength. Red coloured light has lower energy and least intensity. So red coloured lamp is preferable as bed light.

Question 10.
Heisenberg contradicts Neils Bohr. Explain in what way he contradicts.
Answer:

• According to Bohr, electrons revolve around nucleus in definite paths or orbits. So the exact position of the electron at various times will be known to us.
• In order to explain Bohr’s postulate we have to know the velocity and exact position of electron.
• In order to find the position of electron we have to take the help of suitable light to find the position. As the electrons are very small, light of very short wavelength is required for this task.
• This short wavelength light interacts with the electron and disturbs the motion of the electron.
• Hence, it is not possible to find the exact position and velocity of electron simultaneously. This was stated by Heisenberg which is called Heisenberg’s principle of uncertainty.
• In this way Heisenberg contradicts Neils Bohr.

Question 11.
Explain Bohr-Sommerfeld model of an atom. What is the merit of this model? What are its limitations?
Answer:

• In an attempt to account for the structure of line spectra, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.
• While retaining the first of Bohr’s circular orbit as such, he added one elliptical orbit to Bohr’s second orbit, two elliptical orbits to Bohr’s third orbit, etc.
• Nucleus of the atom is one of the principal foci of these elliptical orbits because periodic motion under the influence of a central force will lead to elliptical orbits with the force situated at one of the foci.

Merit:
Bohr-Sommerfeld model is successful in accounting for the fine line structure of hydrogen atomic spectra.

Limitations :

1. This model failed to account for the atomic spectra of atoms of more than one electron.
2. It did not explain Zeeman and Stark effects.

Question 12.
In an atom the number of electrons in N-shell is equal to the number of electrons in K, L and M shells. Answer the following questions.
i) Which is the outermost shell?
Answer:
The outermost shell is p(n = 6).

ii) How many electrons are there in its outermost shell?
Answer:
Two electrons are there in outermost shell.

iii) What is the atomic number?
Answer:
Its atomic number is 56.

iv) Write the electronic configuration of the elements.
Answer:
The electronic configuration of element is
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s².

Question 13.
Explain the following electron configurations by using nl^x method.
a) 2p¹
b) 3d⁵
c) 4f⁹
d) 6s²
Answer:
In nl^x method n is the principle quantum number and l is the angular momentum quantum number and x is number of electrons. Now let us explain the following configurations
a) 2p¹ – It indicates that there is one electron in ‘p’ sub-shell of second orbit or shell.
b) 3d⁵ – It indicates that there are five electrons in ‘d’ sub-shell of third orbit or shell.
c) 4f⁹ – It indicates that there are nine electrons in ‘f’ sub-shell of fourth orbit or shell.
d) 6s² – It indicates that there are two electrons in ‘s’ sub-shell of sixth orbit or shell.

Question 14.
In an atom the number of electrons in L shell is equal to three times of K shell. Then answer the following.
1) Which is the outermost shell?
2) How many electrons are there in outermost shell?
3) What is the atomic number of element?
4) Write electronic configuration of element.
5) Write name of element.
Given that the number of electrons in L shell is three times of K shell.
Answer:
We know that number of electrons in K shell is 2.
Therefore number of electrons in L shell = 3 × 2 = 6

1. So the outermost shell is L.
2. The number of electrons in outermost shell is 6.
3. The atomic number of element is 8.
4. The electronic configuration of element is 1s² 2s² 2p⁴.
5. The element is oxygen.

Question 15.
We know that the electron configuration of copper is [Ar] 4s¹ 3d¹⁰. Is it against to Aufbau principle or not. If so, why is the configuration violated?
Answer:

• The atomic number of copper is 29. So its electron configuration should be [Ar] 4s² 3d⁴.
• But if one electron from 4s orbital jumps into 3d orbital, then copper gets half filled ‘d’ orbitals which gives stability to the atom.
• The energy difference between 4s and 3d is very less. So one electron can easily jump from 4s to 3d which gives half-filled 3d⁵ configuration.
• So, in order to get additional stability, Aufbau principle is violated i.e., electron enters the orbital of higher energy before the completion ortrital of lower energy.

Question 16.
Electronic configurations of following elements are written wrongly. Correct those configurations with proper explanation.

Answer:
1) The electron configuration of oxygen is 1s² 2s² 2p⁴ because the maximum number of electrons that can be filled in s orbital is 2 and so the extra electron should be entered in 2p.

2) The correct electron configuration of nitrogen is

The reason is that the pairing of electrons does not take place until each degenerate orbital is filled with one electron each (Hund’s principle).

3) The correct electronic configuration of scandium is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ because after completion of 3p orbital electron enters in 4s because the energy of 4s orbital is less than 3d (Aufbau principle).

4) The correct electronic configuration of chromium is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵. Because atoms having half filled or completely filled orbitals are more stable. So by transferring one electron from 4s to 3d the atom gets extra stability.

Question 17.
Here is set of quantum numbers. Do they form correct values of quantum numbers or not. If not, give reason.

The values l, m_l, m_s are not correct values for given ‘n’.
Answer:

• The maximum value for l is n – 1. If n = 3, then l takes values from 0 to 2. That is 0, 1, 2.
• m_l values depend on ‘l’. m_l take values from – l to + l including zero. So, the possible values for m7 may be from – 2 to + 2.
• Spin quantum number has only two values, i.e. and –\(\frac{1}{2}\) and \(\frac{1}{2}\). So \(\frac{1}{4}\) value is not possible.

Question 18.
Answer the following questions.
a) If n = 4, then what energy level does it represent?
Answer:
If n = 4, then it represents N energy level.

b) If n = 5, then what is the maximum value for l and why?
Answer:
If n = 5, then the maximum value of l for 4 because the maximum value for l is n- 1.

c) If l = 3, then what are the maximum possible values for m_l?
Answer:
Given l = 3.
Then possible values for m_l is 2l + 1.
∴ Maximum possible values = 2(3) +1 = 7

d) What is the number of electrons present in M energy level?
Answer:
For M energy level n = 3.
The maximum number of electrons in an orbit = 2n² = 2 × 3² = 2 × 9 = 18

Question 19.
Draw the shapes of s, p and d orbitals.
Answer:
s – orbital (Spherical) :

p – orbital (Dumbell) :

All P:

d – orbital (Double Dumbell):

Question 20.
Draw electromagnetic wave and label its parts.
Answer:

Question 21.
Draw the diagram of electromagnetic spectrum.
Answer:

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.1

Question 1.
a) Write any three rational numbers.
Solution:
\(\frac{3}{4}, \frac{5}{9}, \frac{2}{7}\)

b) Explain rational number is in your own words.
Solution:
A number which can be expressed in algebraic form i.e., in \(\frac { p }{ q }\) form is called a rational number.
E.g.: \(\frac { 3 }{ 5 }\), \(\frac { -4 }{ 9 }\) etc.

Question 2.
Give one example each to the following statements.
i) A number which is rational but not an integer.
Solution:
7/11

ii) A whole number which is not a natural number.
Solution:
‘0’ (Zero)

iii) An integer which is not a whole number.
Solution:
-8

iv) A number which is natural number, whole number, integer and rational number.
Solution:
5

v) A number which is an integer but not a natural number.
Solution:
-4

Question 3.
Find five rational numbers between 1 and 2.
Solution:

Question 4.
Five rational numbers between \(\frac { 2 }{ 3 }\) and \(\frac { 3 }{ 5 }\)
Solution:

Question 5.
Represent \(\frac { 8 }{ 5 }\) and \(\frac { -8 }{ 5 }\) on a number line.
Solution:

Step – 1 : Draw a number line.
Step – 2 : Divide each unit into 5 equal parts.
Step – 3 : Take 8 – equal parts from ‘0’ on its right side and mark it as \(\frac { 8 }{ 5 }\) (similarly) on left side \(\frac { -8 }{ 5 }\) .

Question 6.
Express the following rational numbers as decimal numbers.
Solution:
I. i) \(\frac { 242 }{ 1000 }\) .
ii) \(\frac { 354 }{ 500 }\) .
iii) \(\frac { 2 }{ 5 }\) .
iv) \(\frac{115}{4}\)
Solution:
i) \(\frac { 242 }{ 1000 }\) = 0.242
ii) \(\frac{354}{500}\)
\(=\frac{354 \times 2}{500 \times 2}\)
\(=\frac{708}{1000}\)
\(=0.708\)
iii) \(\frac{2}{5}\)
\(=\frac{2 \times 2}{5 \times 2}\)
\(=\frac{4}{10}\)
\(=0.4\)
iv)

II. i) \(\frac{2}{3}\)
Solution:

ii) \(\frac{-25}{36}\)
Solution:

iii) \(\frac{22}{7}\)
Solution:

iv) \(\frac{11}{9}\)
Solution:

Question 7.
Express each of the following decimals in \(\frac{p}{q}\) form where q ≠ 0 and p, q are integers.
i) 0.36
Solution:
0.36 = \(\frac{36}{100}=\frac{9}{25}\)

ii) 15.4
Solution:
15.4 = \(\frac{154}{10}=\frac{77}{5}\)

iii) 10.25
Solution:
10.25 = \(\frac{1025}{100}=\frac{41}{4}\)

iv) 3.25
Solution:
3.25 = \(\frac{325}{100}=\frac{13}{4}\)

Question 8.
Express each of the following decimal number in the \(\frac { p }{ q }\) form.
i) \(0 . \overline{5}\)
Solution:
Let x = \(0 . \overline{5}\) = 0.5555
Multiplying both sides by 10

ii) \(3 . \overline{8}\)
Solution:
Let x = \(3 . \overline{8}\)
(i.e) x = 3.888 ………..
Multiplying both sides by 10

iii) \(0 . \overline{36}\)
Solution:
Let x \(0 . \overline{36}\)
(i.e) x = 0.363636 ………..
Multiplying by 100 on both sides

iv) \(3.12 \overline{7}\)
Solution:
Let x = \(3.12 \overline{7}\)
x = 0.12777
Multiplying by 10 on both sides

Question 9.
Without actually dividing find which of the following are terminating
decimals.
i) \(\frac { 3 }{ 25 }\)
Solution:
Check the denominator, if it consists of 2’s or 5’s or combination of both then only it reduces to a terminating decimal.
25 = 5 x 5
Hence \(\frac { 3 }{ 25 }\) is a terminating decimal.

ii) \(\frac { 11 }{ 18 }\)
Solution:
Denominator 18 = 2 × 3 × 3,
hence \(\frac { 11 }{ 18 }\) is a non-terminating decimal 13

iii) \(\frac { 13 }{ 20 }\)
Denominator 20 = 2 × 2 × 5,
hence \(\frac { 13 }{ 20 }\) is a terminating decimal.

iv) \(\frac { 41 }{ 42 }\)
Solution:
Denominator 42 = 2 × 3 × 7,
hence \(\frac { 41 }{ 42 }\) is a non-terminating decimal.

AP State Board Syllabus AP SSC 10th Class Telugu Solutions 10th Class Telugu Grammar Sandhulu సంధులు Notes, Questions and Answers.

AP State Syllabus SSC 10th Class Telugu Grammar Sandhulu సంధులు

తెలుగు సంధులు

నా చిన్నప్పుడు చేసిన పనులు గుర్తుకు వచ్చాయి.

గమనిక :
పై వాక్యంలో “చిన్నప్పుడు” అనే పదం, చిన్న + అప్పుడు అనే రెండు పదాలు కలవడం వల్ల వచ్చింది. దీనినే “సంధి పదం” అంటారు. ఉచ్చరించడంలో సౌలభ్యం కోసం రెండు పదాలను వెంట వెంటనే కలిపి మాట్లాడవలసినప్పుడు, లేదా రాయవలసినప్పుడు “సంధి పదం” ఏర్పడుతుంది.

సంధి :
వ్యాకరణ పరిభాషలో రెండు స్వరాల (అచ్చుల) కలయికను “సంధి” అని పిలుస్తారు.

తెలుగు సంధులు :
రెండు తెలుగుపదాల మధ్య జరిగే సంధులను “తెలుగు సంధులు” అంటారు.

సంధి కార్యం : రెండు అచ్చుల మధ్య జరిగే మార్పును “సంధి కార్యం” అని పిలుస్తారు.

పూర్వ స్వరం :
సంధి జరిగే మొదటి పదం చివరి అక్షరంలోని అచ్చును (స్వరాన్ని) “పూర్వ స్వరం” అని పిలుస్తారు.

పర స్వరం :
సంధి జరిగే రెండవ పదం మొదటి అక్షరంలోని అచ్చును (స్వరాన్ని) “పర స్వరం” అని పిలుస్తారు.

ఉదా :
రామ + అయ్య : ‘రామ’ లోని ‘మ’ లో ‘అ’ పూర్వ స్వరం, ‘అయ్య’ లోని ‘అ’ పర స్వరం.

1. అత్వ సంధి సూత్రం
అత్తునకు సంధి బహుళంగా వస్తుంది.

ఈ కింది పదాలను విడదీయండి.

ఉదా : మేనల్లుడు = మేన + ‘అల్లుడు – (న్ +) అ + అ = అ) = (అత్వ సంధి)
1) ఒకప్పుడు : ఒక అప్పుడు = (అత్వ సంధి)
2) వచ్చినందుకు – వచ్చిన అందుకు : (అత్వ సంధి)
3) రాకుంటే ఉంటే (అత్వ సంధి)
4) లేకేమి = లేక + ఏమి = (అ + ఏ = (అత్వ సంధి)
5) పోవుటెట్లు : పోవుట + ఎట్లు = (అ + ఎ (అత్వ సంధి)
6) కొండంత = కొండ + అంత = (అ + ఎ (అత్వ సంధి)

గమనిక :
పై సంధి పదాలలోని పూర్వ స్వరం ‘అ’. అది పర స్వరంలోని అచ్చుతో కలిస్తే, పూర్వ స్వరం ‘అ’ లోపిస్తుంది. పై ఉదాహరణలలో ‘అ’ లోపించింది కాబట్టి ఇది ‘అత్వ సంధి’.

దీనిని అత్వ సంధి లేక ‘అకార సంధి’ అంటారు. పొట్టి ‘అ’ అనే అక్షరానికి, అచ్చు పరమైతే ‘అత్వ సంధి’ వస్తుంది.

2. ఇత్వ సంధి సూత్రం
ఏమ్యాదుల ఇత్తునకు సంధి వైకల్పికంగా వస్తుంది.
ఏమ్యాదులు = ఏమి మొదలగునవి.

ఏమి, మణి, కి (షష్ఠి, అది, అవి, ఇది, ఇవి, ఏది, ఏవి మొదలైనవి ఏమ్యాదులు. (కి షష్ఠి అంటే ‘కిన్’)

ఈ కింది పదాలను విడదీయండి.
ఉదా :
అ) ఏమంటివి = ఏమి + అంటివి = (ఇ + అ = అ) = (ఇత్వ సంధి)
సంధి జరుగనప్పుడు “య” కారం ఆగమంగా వస్తుంది. దానినే ‘యడాగమం’ అని పిలుస్తారు.

ఆ) ఏమియంటివి = ఏమి + య్ + అంటివి = ఇ + (య్ + అ) = య (ఇకార సంధి రాని యడాగమ రూపం)

గమనిక :
ప్రథమ, ఉత్తమ పురుష బహువచన క్రియల ఇకారానికి, సంధి వైకల్పికంగా జరుగుతుంది.

వచ్చిరిపుడు = వచ్చిరి + ఇపుడు – (ఇ + ఇ + ఇ) = (ఇత్వ సంధి)
వచ్చిరియిపుడు = వచ్చిరి + య్ + ఇపుడు – (ఇ + ఇ + యి) (యడాగమం వచ్చిన రూపం)

గమనిక :
పై ఉదాహరణలలో హ్రస్వ ‘ఇ’ కారానికి అచ్చు కలిసినపుడు సంధి జరిగింది. దీనిని “ఇత్వ సంధి” అంటారు. ఇత్వ సంధి తప్పక జరగాలన్న నియమం లేదు.

వైకల్పికం :
సంధి జరుగవచ్చు లేక జరుగకపోవచ్చు. వ్యాకరణంలో ఈ పరిస్థితిని “వైకల్పికం” అని పిలుస్తారు.

అభ్యాసం :
ఉదా : 1) ఏమంటివి = ఏమి + అంటివి : (మ్ + ఇ + అ = మ) : ఇత్వ సంధి
2) పైకెత్తినారు : పైకి + ఎత్తినారు : (ఇ + ఎ = ఎ) : ఇత్వ సంధి
3) మనిషన్నవాడు = మనిషి + అన్నవాడు – (ఇ + అ = అ) – ఇత్వ సంధి

3. ఉత్వ సంధి సూత్రం
ఉత్తునకు అచ్చు పరమైనపుడు సంధి నిత్యంగా వస్తుంది.

ఈ కింది పదాలను విడదీయండి.
ఉదా : రాముడతడు = రాముడు + అతడు = (డ్ + ఉ + అ = డ) : (ఉత్వ సంధి)
1) అతడెక్కడ = అతడు + ఎక్కడ = (A + ఎ = ఎ) : (ఉత్వ సంధి)
2) మనమున్నాము = మనము + ఉన్నాము : (ఉ + ఉ = ఉ) : (ఉత్వ సంధి)
3) మనసైన . మనసు + ఐన = (A + ఐ = ఐ) : . (ఉత్వ సంధి)

గమనిక :
హ్రస్వ ఉ కారానికి, అనగా ఉత్తునకు, అచ్చు కలిసినప్పుడు, పూర్వ స్వరం ‘ఉ’ కారం లోపించి, పర స్వరం కనిపిస్తుంది. లోపించిన పూర్వ స్వరం ‘ఉ’ కాబట్టి, ఇది “ఉత్వ సంధి” అని పిలువబడుతుంది.

నిత్యం :
నిత్యం అంటే, తప్పక సంధికార్యం జరుగుతుందని అర్థం.

4. యడాగమ సంధి సూత్రం
సంధి లేనిచోట అచ్చుల మధ్య “య్” వచ్చి చేరడాన్ని “యడాగమం” అని పిలుస్తారు.

ఈ కింది పదాలను విడదీయండి.
ఉదా :
అ) మాయమ్మ = మా + అమ్మ – మాయమ్మ
ఆ) మాయిల్లు = మా + ఇల్లు = మాయిల్లు
ఇ) హరియతడు = హరి + అతడు = హరియతడు
గమనిక :
పై ఉదాహరణలలో సంధి జరుగలేదు. కాని కొత్తగా ‘య్’ వచ్చి చేరింది. అలా చేరడం వల్ల ఈ కింది విధంగా మార్పు జరిగింది.

అ) మా + య్ + అమ్మ : మా ‘య’ మ్మ
ఆ) మా + య్ . + ఇల్లు : మా ‘ఋ’ ల్లు
ఇ) హరి + య్ + అతడు = హరి ‘య’ తడు

5. ఆమ్రేడిత సంధి సూత్రం
అచ్చునకు ఆమ్రేడితం పరమైతే సంధి తరచుగా వస్తుంది.

ఆమ్రేడితం :
ఒక పదాన్ని రెండుసార్లు ఉచ్చరిస్తే, రెండోసారి ఉచ్చరింపబడిన పదాన్ని ‘ఆమ్రేడితం’ అంటారు.
ఉదా :
1) ఆహాహా = ఆహా + ఆహా – ‘ఆహా’ అనే పదం రెండుసార్లు వచ్చింది. అందులో రెండవ ‘ఆహా’ అనే దాన్ని ఆమ్రేడితం అని పిలవాలి.
2) అరెరె : అరె అరె : రెండవసారి వచ్చిన ‘అరె’ ఆమ్రేడితం.
3) ఔరౌర – ఔర + . ఔర = రెండవసారి వచ్చిన ‘ఔర’ ఆమ్రేడితం.

గమనిక :
పై ఉదాహరణలలో ఒక్కొక్క పదం రెండుసార్లు వచ్చింది. రెండవసారి వచ్చిన పదాన్ని ‘ఆమ్రేడితం’ అంటారు.

ఆమ్రేడిత సంధికి ఉదాహరణములు :
ఔర + ఔర = (ఔర్ + అ)
ఆహా + ఆహా = (ఆహ్ + ఆ)
ఓహో + ఓహో = (ఓహ్ + ఓ)

గమనిక :
పై ఉదాహరణలలో పూర్వ పదం అనగా మొదటి పదం చివర, అ, ఆ, ఓ వంటి అచ్చులున్నాయి. ఈ అచ్చులకు ఆమ్రేడితం పరమైతే, సంధి వస్తుంది.
ఔర + ఔర = ఔరౌర = (అ + ఔ = ఔ)
ఆహా + ఆహా = ఆహాహా = (ఆ + ఆ = ఆ)
ఓహో + ఓహో = ఓహోహో = (ఓ + ఓ = ఓ)
ఏమి + ఏమి = ఏమేమి = (ఇ + ఏ = ఏ)
ఎట్లు + ఎట్లు = ఎట్లెట్లు = (ఉ + ఎ = ఎ)
ఏమిటి + ఏమిటి = ఏమిటేమిటి = (ఇ + ఏ = ఏ)
అరె + అరె = అరెరె = (ఎ + అ = ఎ)

గమనిక :
ఆమ్రేడిత సంధి, ఈ కింది ఉదాహరణలలో వికల్పంగా జరుగుతుంది. వీటిని గమనిస్తే, సంధి జరిగిన రూపం ఒకటి, సంధిరాని రూపము మరొకటి కనబడతాయి.
ఉదా :
ఏమి + ఏమి = ఏమేమి, ఏమియేమి (సంధి వైకల్పికం)
ఎట్లు + ఎట్లు – ఎబ్లెట్లు, ఎట్లుయెట్లు (సంధి వైకల్పికం)
ఎంత + ఎంత = ఎంతెంత, ఎంతయెంత (సంధి వైకల్పికం)

6. ఆమ్రేడిత ద్విరుక్తటకారాదేశ సంధి సూత్రం
ఆమ్రేడితం పరమైనపుడు, కడాదుల తొలి యచ్చు మీది వర్ణాల కెల్ల అదంతమైన ద్విరుక్తటకారం వస్తుంది.
కడాదులు (కడ + ఆదులు) = కడ, ఎదురు, కొన, చివర, తుద, తెన్ను, తెరువు, నడుము, పగలు, పిడుగు, బయలు, మొదలు, మొదలైనవి కడాదులు.

కింది ఉదాహరణలను గమనించండి.
1) పగలు + పగలు = పట్టపగలు
2) చివర + చివర = చిట్టచివర
3) కడ + కడ = కట్టకడ

గమనిక :
1) పగలు + పగలు – పట్టపగలు అవుతోంది. అంటే ‘ప’ తర్వాత ఉన్న ‘గలు’ అన్న అక్షరాలకు బదులుగా, ‘ఋ’ వచ్చింది. ‘మీ’ వచ్చి, ‘పట్టపగలు’ అయింది.
2) చివర + చివర అన్నప్పుడు ‘చి’ తర్వాత రెండక్షరాల మీద ‘జ’ వచ్చి, ‘చిట్టచివర’ అయింది.
3) కడ + కడ అన్నప్పుడు ‘డ’ స్థానంలో ‘జ’ వచ్చి ‘కట్టకడ’ అయింది. ఇప్పుడు కిందివాటిని కలిపి రాయండి.
ఎదురు + ఎదురు = ఎట్టయెదురు
కొన + కొన = కొట్టకొన
మొదట + మొదట = మొట్టమొదట
బయలు + బయలు = బట్టబయలు
తుద + తుద : తుట్టతుద

గమనిక :
ఆమ్రేడితం పరంగా ఉంటే, కడ మొదలైన శబ్దాల మొదటి అచ్చు మీద ఉన్న అన్ని అక్షరాలకు, ‘ఋ’ (ద్విరుక్తటకారం) వస్తుండడం గమనించాం.

7. ద్రుతప్రకృతిక సంధి సరళాదేశ సంధి
ద్రుతప్రకృతికం మీది పరుషాలకు సరళాలు వస్తాయి.

ఈ కింది పదాలు చదివి, పదంలోని చివర అక్షరం కింద గీత గీయండి. 1) పూచెను 2) చూచెన్ 3) తినెను 4) చేసెన్ 5) ఉండెన్

గమనిక :
పై పదాలను గమనిస్తే పదాల చివర, ను, న్ లు కనిపిస్తాయి. అంటే పదాల చివర నకారం ఉంది. ఈ నకారాన్ని ‘ద్రుతం’ అంటారు. ద్రుతము చివర గల పదాలను, “ద్రుతప్రకృతికాలు” అంటారు.

గమనిక :
పూచెను, చూచెన్, తినెను, చేసెన్, ఉండెన్ – అనేవి ద్రుతప్రకృతికాలు

కింది ఉదాహరణములను గమనించండి.
ఉదా:
అ) పూచెన్ + కలువలు = పూచెన్ + గలువలు
ఆ) దెసన్ + చూచి = దెసన్ + జూచి
ఇ) చేసెన్ + టక్కు = చేసెన్ + డక్కు
ఈ) పాటిన్ + తప్ప : పాటిన్ + దప్ప
ఉ) వడిన్ + పట్టి = వడిన్ + బట్టి
ఊ) చేసెను + తల్లీ : ‘ ‘ చేసెను + దల్లీ
ఋ) దెసను + చూసి : దెసను + జూసి

గమనిక :
ద్రుతప్రకృతికానికి ‘క’ పరమైతే ‘గ’, ‘చ’ పరమైతే ‘జ’, ‘ట’ పరమైతే ‘డ’, ‘త’ పరమైతే ‘ద’, ‘ప’ పరమైతే ‘బ’ ఆదేశంగా వస్తాయి.
1) క – ‘గ’ గా
2) చ – ‘జ’ గా
3) ట – ‘డ’ గా
4) త – ‘ద’ గా
5) ప – ‘బి’ గా మార్పు వచ్చింది.

ఇందులో ‘క చ ట త ప’ లకు, ‘పరుషాలు’ అని పేరు, ‘గ జ డ ద బ’ లకు, ‘సరళాలు’ అని పేరు.

దీనిని బట్టి సరళాదేశ సంధి సూత్రం ఇలా ఉంటుంది.

ద్రుత ప్రకృతిక సంధి సూత్రం (1):
ద్రుత ప్రకృతికం మీది పరుషాలకు, సరళాలు వస్తాయి.

గమనిక :
ఇప్పుడు పై ఉదాహరణలలో మార్పు గమనించండి.
ఉదా :
పూచెఁ గలువలు (ద్రుతం అరసున్నగా మారింది)
1) పూచెను + కలువలు (పూచెం గలువలు (ద్రుతం సున్నగా మారింది)
2) పూచెనలువలు (ద్రుతం మీద హల్లుతో కలిసి సంశ్లేష రూపం అయ్యింది)
3) పూచెను గలువలు. (ద్రుతం మార్పు చెందలేదు) దీనికి సూత్రం చెపితే సూత్రం ఇలా ఉంటుంది.

ద్రుత ప్రకృతిక సంధి సూత్రం (2) : ఆదేశ సరళాలకు ముందున్న ద్రుతానికి, బిందు, సంశ్లేషలు విభాషగా వస్తాయి.
గమనిక :
అంటే ఒక్కోసారి బిందువు వస్తుంది. ఒక్కోసారి సంశ్లేష వస్తుంది.

8. గసడదవాదేశ సంధి సూత్రం
ప్రథమ మీది పరుషాలకు గ,స,డ,ద,వ లు బహుళంబుగా వస్తాయి.

కింది పదాలను ఎలా విడదీశారో గమనించండి.
1) గొప్పవాడు = గదా – గొప్పవాడు + కదా (డు + క)
2) కొలువుసేసి = కొలువు + చేసి (వు + చే)
3) వాడు డక్కరి = వాడు + టక్కరి (డు + ట)
4) నిజము దెలిసి = నిజము + తెలిసి (ము + తె)
5) పాలువోయక = పాలు + పోయక (లు + పో)

గమనిక : పై ఉదాహరణలలో పూర్వపదం చివర, ప్రథమా విభక్తి ప్రత్యయాలు ఉన్నాయి. పరపదం మొదట క, చ, ట, త, ప, లు ఉన్నాయి. ఈ విధంగా ప్రథమావిభక్తి మీద ప్రత్యయాలు, క, చ, ట, త, ప లు పరమైతే, వాటి స్థానంలో గ, స, డ, ద, వ లు ఆదేశంగా వస్తాయి. అంటే
1) క – గ గా మారుతుంది.
2) చ – స గా మారుతుంది.
3) ట – డ గా మారుతుంది.
4) త – ద గా మారుతుంది.
5) ప – వ గా మారుతుంది.

అంటే క, చ, ట, త, ప లకు గ, స, డ, ద, వ లు ఆదేశంగా వస్తాయి.

ద్వంద్వ సమాసంలో : గసడదవాదేశ సంధి

కింది పదాలను గమనించండి.

కూరగాయలు = కూర + కాయ + లు
కాలుసేతులు = కాలు + చేయి + లు
టక్కుడెక్కులు = టక్కు + టెక్కు + లు
తల్లిదండ్రులు = తల్లి + తండ్రి + లు
ఊరువల్లెలు : ఊరు + పల్లె + లు

గమనిక :
పై ఉదాహరణలు ద్వంద్వ సమాసపదాలు. పై ఉదాహరణలలో కూడా క చట త ప లకు, గసడదవ లు వచ్చాయి. దీన్నే గసడదవా దేశం అంటారు.

గసడదవాదేశ సంధి సూత్రం:
ద్వంద్వ సమాసంలో మొదటి పదం మీద ఉన్న క చట త ప లకు గసడదవలు క్రమంగా వస్తాయి.

కింది పదాలను కలపండి.
1) అక్క , చెల్లి : అక్కసెల్లెండ్రు
2) అన్న తమ్ముడు – అన్నదమ్ములు

9. టుగాగమ సంధి సూత్రం
కర్మధారయములందు ఉత్తునకు, అచ్చుపరమైతే టుగాగమం వస్తుంది.

ఈ కింది పదాలను పరిశీలించండి.
నిలువు + అద్దం = నిలువుటద్దం
తేనె + ఈగ = తేనెటీగ
పల్లె + ఊరు = పల్లెటూరు

గమనిక :
వీటిలో సంధి జరిగినపుడు ‘ట్’ అదనంగా చేరింది. ఇలా ‘ట్’ వర్ణం అదనంగా వచ్చే సంధిని, ‘టుగాగమ సంధి’ అంటారు. అలాగే కింది పదాలు కూడా గమనించండి. తేనె, పల్లె అనే పదాల చివర ‘ఉ’ లేక పోయినా, టుగాగమం వచ్చింది.
1) చిగురు + ఆకు = చిగురుటాకు / చిగురాకు
2) పొదరు + ఇల్లు : పొదరుటిల్లు / పొదరిల్లు

గమనిక :
వీటిలో ‘ట్’ అనే వర్ణం సంధి జరిగినపుడు రావచ్చు. ‘ట్’ వస్తే “టుగాగమం” అవుతుంది. ‘ట్’ రాకుంటే ‘ఉత్వ సంధి’ అవుతుంది.

సూత్రం :
కర్మధారయమునందు పేర్వాది శబ్దాలకు అచ్చు పరమైనపుడు, టుగాగమం విభాషగా వస్తుంది.
ఉదా :
1) పేరు + ఉరము = పేరు టురము / పేరురము
2) చిగురు + ఆకు = చిగురుటాకు / చిగురాకు
3) పొదరు + ఇల్లు = పొదరుటిల్లు / పొదరిల్లు

10. లులన సంధి సూత్రం
లు, ల, న లు పరమైనపుడు ఒక్కొక్కప్పుడు ‘ము’ గాగమానికి లోపమూ, దాని పూర్వ స్వరానికి దీర్ఘమూ విభాషగా వస్తాయి.

ఈ కింది ఉదాహరణములు గమనించండి.
1) పుస్తకములు – పుస్తకాలు
2) దేశముల – దేశాల
3) జీవితమున – జీవితాన
4) గ్రంథములు – గ్రంథాలు
5) రాష్ట్రముల – రాష్ట్రాల
6) వక్షమున – వృక్షాన

పై పదాల్లో మార్పును గమనించండి.

పుస్తకములు, గ్రంథములు, దేశములు, రాష్ట్రములు, జీవితమున, వృక్షమున – వీటినే మనం పుస్తకాలు, గ్రంథాలు, దేశాలు, రాష్ట్రాలు, జీవితాన, వృక్షాన అని కూడా అంటాం.

గమనిక :
ఈ మార్పులో, లు,ల, న అనే అక్షరాల ముందున్న ‘ము’ పోయింది. ‘ము’ కంటే ముందున్న అక్షరానికి దీర్ఘం వచ్చింది.

11. పడ్వాది సూత్రం
పడ్వాదులు పరమైనపుడు ‘ము’ వర్ణకానికి లోప పూర్ణబిందువులు విభాషగా వస్తాయి.

ఈ కింది ఉదాహరణలు గమనించండి.
1) భయము + పడు = భయంపడు, భయపడు

విడదీసిన పదాలకూ, కలిపిన పదాలకూ తేడా గమనించండి. కలిపిన పదంలో ‘ము’ కు బదులుగా సున్న(0) వచ్చింది. మరో దానిలో ‘ము’ లోపించింది.

గమనిక :
పడ్వాదులు = పడు, పట్టె, పాటు అనేవి.

12. త్రిక సంధి సూత్రం
1. ఆ, ఈ, ఏ అనే సర్వనామాలు త్రికం అనబడతాయి.
2. త్రికం మీది అసంయుక్త హల్లుకు ద్వితం బహుళంగా వస్తుంది.
3. ద్విరుక్తమైన హల్లు పరమైనపుడు, ఆచ్చికమైన దీర్ఘానికి హ్రస్వం వస్తుంది.

ఈ కింది ఉదాహరణ చూడండి.
అక్కొమరుండు = ఆ + కొమరుండు
ఆ + కొమరుండు = అనే దానిలో, ‘ఆ’, త్రికంలో ఒకటి. ఇది ‘అ’ గా మారింది. సంయుక్తాక్షరం కాని హల్లు ‘కొ’ ద్విత్వంగా ‘క్కొ’ గా మారింది.

అలాగే ఈ, ఏ లు అనే త్రికములు కూడా, ఇ, ఎ లుగా మారతాయి.
ఉదా :
ఈ + కాలము = ఇక్కాలము
ఏ + వాడు = ఎవ్వాడు

త్రిక సంధి సూత్రం (1):
త్రికము మీది అసంయుక్త హల్లుకు ద్విత్వం బహుళంగా వస్తుంది.
ఉదా :
ఈ + క్కాలము
ఏ + వ్వాడు

త్రిక సంధి సూత్రం (2) : ద్విరుక్తమైన హల్లు పరమైనపుడు అచ్ఛిక దీరానికి హ్రస్వం అవుతుంది.
ఉదా : 1) ఇక్కాలము 2) ఎవ్వాడు

13. రుగాగమ సంధి
సూత్రం : పేదాది శబ్దాలకు ‘ఆలు’ శబ్దం పరమైతే, కర్మధారయంలో రుగాగమం వస్తుంది.
పేదాదులు = (పేద + ఆదులు) పేద మొదలైనవి.

పేద, బీద, ముద్ద, బాలెంత, కొమ, జవ, అయిదువ, మనుమ, గొడ్డు మొదలైనవి పేదాదులు.
ఉదా : పేద + ఆలు = పేద + ర్ + ఆలు = పేదరాలు

పై రెండు పదాలకు మధ్య ‘5’ అనేది వచ్చి, ప్రక్కనున్న ‘ఆ’ అనే అచ్చుతో కలిస్తే ‘రా’ అయింది. అదెలా వస్తుందంటే, పేద, బీద, బాలెంత ఇలాంటి పదాలకు ‘ఆలు’ అనే శబ్దం పరమైతే, ఇలా ‘రుగాగమం’ అంటే ‘5’ వస్తుంది. ఆగమం : రెండు పదాలలో ఏ అక్షరాన్ని కొట్టివేయకుండా, కొత్తగా అక్షరం వస్తే “ఆగమం” అంటారు.

మనుమ + ఆలు = మనుమరాలు
బాలెంత + ఆలు = బాలెంతరాలు

రుగాగమ సంధి సూత్రం (2) :
కర్మధారయంలో తత్సమ పదాలకు, ఆలు శబ్దం పరమైతే, పూర్వ పదం చివరనున్న అత్వానికి ఉత్వమూ, రుగాగమమూ వస్తాయి.
ఉదా :
ధీరురాలు = ధీర + ఆలు
గుణవంతురాలు = గుణవంత + ఆలు
విద్యావంతురాలు = విద్యావంత + ఆలు

14. పుంప్వాదేశ సంధి సూత్రం
కర్మధారయ సమాసాల్లో “ము” వర్ణకానికి బదులు “పుంపులు” ఆదేశంగా వస్తాయి.

గమనిక :
“ము” అనే వర్ణకానికి బదులుగా “పు” కాని, “ఎపు” కాని వస్తుంది. దీన్ని వ్యాకరణ దృష్టిలో “ఆదేశం” అని పిలుస్తారు. కింది పదాలు విడదీసి చూడండి. మార్పును గమనించండి.
ఉదా :
అచ్చపు పూలతోట = అచ్చము + పూలతోట

అ) నీలపుఁగండ్లు = నీలము + కండ్లు
ఆ) ముత్తెపుసరులు = ముత్తెము + సరులు
ఇ) సరసపుమాట = సరసము + మాట

గమనిక :
పైన పేర్కొన్న ఉదాహరణలలో రెండు మార్పులను మనం గమనించవచ్చు.

అ) మొదటి పదాల్లో ‘ము’ వర్ణకం లోపించింది.
ఆ) ప్రతి సంధి పదంలోనూ మొదటి పదం విశేషణాన్ని తెలుపుతుంది.

గమనిక :
సమాసంలో మొదటి పదం విశేషణం, రెండవ పదం విశేష్యం అయితే, ఆ సమాసాలను “విశేషణ పూర్వపద కర్మధారయ సమాసం” అంటారని మనకు తెలుసు. అంటే ఈ పుంప్వాదేశ సంధి, కర్మధారయ సమాసాల్లో ఏర్పడుతుందని గ్రహించాలి.

అభ్యాసం :
కింది పదాలను విడదీసి, సంధి సూత్రాన్ని సరిచూడండి.

అ) సింగపు కొదమ
జవాబు:
సింగపుకొదమ : సింగము + ‘కొదమ = పుంప్వాదేశ సంధి

గమనిక :
ఇక్కడ సింగము అనే మొదటి పదం చివర ఉన్న “ము” వర్ణకం పోయి, “పు” ఆదేశంగా వచ్చింది. కాబట్టి ఇది “పుంప్వాదేశ సంధి.”

ఆ) ముత్యపుచిప్ప
జవాబు:
ముత్యపుచిప్ప = ముత్యము + చిప్ప + పుంప్వాదేశ సంధి
గమనిక :
ఇక్కడ ముత్యము అనే మొదటి పదం చివర ఉన్న “ము” వర్ణకం పోయి, “పు” ఆదేశంగా వచ్చింది. కాబట్టి ఇది “పుంప్వాదేశ సంధి.”

ఇ) కొంచెపు నరుడు
జవాబు:
కొంచెపునరుడు = కొంచెము + నరుడు = పుంప్వాదేశ సంధి
గమనిక :
ఇక్కడ కొంచెము అనే పూర్వపదం చివర ఉన్న “ము” వర్ణకం పోయి, “పు” ఆదేశంగా వచ్చింది. కాబట్టి ఇది పుంప్వాదేశ సంధి.

15. ప్రాతాది సంధి సూత్రం
సమాసాలలో ప్రాతాదుల తొలి అచ్చు మీది వర్ణాలకెల్ల లోపం బహుళంగా వస్తుంది.

కింద గీత గీసిన పదాలను విడదీయండి. మార్పులు గమనించండి.

అ. పూరెమ్మ అందంగా ఉన్నది.
జవాబు:
పూరెమ్మ : పూవు + రెమ్మ – ప్రాతాది సంధి

ఆ. గురుశిష్యులు పూదోటలో విహరిస్తున్నారు.జవాబు:
పూదోట : పూవు + తోట = ప్రాతాది సంధి

ఇ) కొలనులో కెందామరలు కొత్త శోభను వెదజల్లుతున్నాయి.
జవాబు:
కెందామరలు – కెంపు + తామరలు = ప్రాతాది సంధి

ఈ) ఆ ముసలివానిలాగే అతని ప్రాయిల్లు జీర్ణమైయున్నది.
జవాబు:
ప్రాయిల్లు : పాత + ఇల్లు = ప్రాతాది సంధి

పై సంధి పదాల విభజనను సరిచూడండి. వచ్చిన మార్పు గమనించండి.
అ) పూవు + రెమ్మ = పూరెమ్మ = ప్రాతాది సంధి
ఆ) పూవు + తోట = పూఁదోట = ప్రాతాది సంధి
ఇ) కెంపు + తామరలు = కెందామరలు = ప్రాతాది సంధి
ఈ) ప్రాత + ఇల్లు = ప్రాయిల్లు = ప్రాతాది సంధి
ఉ) మీదు + కడ = మీఁగడ = ప్రాతాది సంధి

గమనిక :
1) ఈ ఐదు సందర్భాలలోనూ మొదటి పదంలోని మొదటి అక్షరం తరువాత ఉన్న వర్ణాలన్నీ లోపిస్తాయి.
2) రెండవ పదం మొదట ఉన్న పరుషా (త, క) లు, సరళా (ద, గ) లుగా మారాయి.

పై మార్పులను బట్టి, ప్రాతాది సంధి నియమాలను ఈ విధంగా అర్థం చేసుకోవచ్చు.

ప్రాతాది సంధి

ప్రాతాది సంధి సూత్రము (1):
సమాసాలలో ప్రాతాదుల తొలి అచ్చు మీది వర్ణాలకెల్ల లోపం బహుళంగా వస్తుంది.
1) పూవు + తోట = పూ + తోట
2) కెంపు + తామర = కెల + తామర
3) మీదు + కడ = మీ + కడ

ప్రాతాది సంధి సూత్రము (2) :
లోపింపగా మిగిలిన (సంధిలోని) మొదటి అక్షరానికి పరుషాలు పరమైతే నుగాగమం (‘న్’ ఆగమంగా) వస్తుంది.
1) పూ + న్ + తోట = పూదోట
2) కె +న్ + తామర = కెందామర
3) మీ + 5 + కడ = మీగడ

గమనిక :
నుగాగమంలోని ‘స్’ అనేది, పూర్ణబిందువుగా గాని, అర్థబిందువుగా గాని, సరళాదేశ సంధి వల్ల మారుతుంది.
ఉదా : 1) పూఁదోట
2) కెందామర
3) మీఁగడ

గమనిక :
ప్రాతాదులు అంటే ‘పాత’ మొదలయిన కొన్ని మాటలు. అవి
1) ప్రాత
2) లేత
3) క్రొత్త
4) క్రిందు
5) కెంపు
6) చెన్ను మొ||నవి.

16. నుగాగమ సంధి సూత్రం
ఉదంతమగు తద్ధర్మార్థక విశేషణానికి అచ్చు పరమైనపుడు నుగాగమం వస్తుంది.

అభ్యాసం :
కింది పదాలను విడదీయండి. మార్పును గమనించండి.

అ) నుగాగమ సంధి:

అ) చేయునతడు = చేయు + అతడు
ఆ) వచ్చునపుడు = వచ్చు + అపుడు

గమనిక :
చేయు, వచ్చు వంటి క్రియలు చివర ‘ఉత్తు’ను, అనగా హ్రస్వ ఉకారాన్ని కల్గి ఉంటాయి. కాబట్టి వీటిని ‘ఉదంతాలు’ అంటారు. ఇవి తద్ధర్మార్థక క్రియలు. అంటే ఏ కాలానికైనా వర్తించే క్రియలు. ఈ ఉదంత, తద్ధర్మార్థక క్రియలకు అచ్చు కలిసింది. అప్పుడు ‘న్’ అనే అక్షరం కొత్తగా వస్తుంది. ‘న్’ ఆగమంగా అనగా ఉన్న అక్షరాలను కొట్టి వేయకుండా వస్తుంది. కాబట్టి ఇది “నుగాగమ సంధి’.
ఉదా :
చేయు + అతడు = చేయు +న్ + అతడు = చేయునతడు
వచ్చు + అప్పుడు = వచ్చు + న్ + అప్పుడు = వచ్చునప్పుడు

గమనిక :
అతడు, అప్పుడు అనే పదాలలోని ‘అ’ అనే అచ్చు పరంకాగా, కొత్తగా ‘న్’ ఆగమంగా వచ్చి, నుగాగమం అయింది.

ఉదా :
1) వచ్చునప్పుడు
2) చేయునతడు

ఆ. నుగాగమ సంధి:

గమనిక :
ఈ పై సందర్భాలలోనే కాక, మరికొన్ని చోట్ల కూడా నుగాగమం వస్తుంది. కింది పదాలు విడదీసి, పరిశీలించండి.
అ) తళుకుం గజ్జెలు
ఆ) సింగపుం గొదమ

పై సంధి పదాలను విడదీస్తే
అ) తళుకుం గజ్జెలు – తళుకు + గజ్జెలు
ఆ) సింగపుం గొదమ = సింగము + కొదమ

గమనిక :
‘తళుకు గజ్జెలు’ అనే సంధి పదంలో ‘తళుకు’ అనే పదం, ఉత్తు చివర గల స్త్రీ సమశబ్దం. ఇటువంటి ఉదంత స్త్రీ సమపదాలకు, పరుషాలుగాని, సరళాలుగాని పరమైతే నుగాగమం వస్తుంది.
ఉదా :
తళుకు + 5 + గజ్జెలు = ద్రుతానికి సరళ స్థిరాలు పరమైతే పూర్ణబిందువు వస్తుంది. తళుకుం గజ్జెలు అవుతుంది.

అలాగే పుంపులకు, పరుష సరళాలు పరమైతే, నుగాగమం వస్తుంది.

గమనిక :
సింగపు + కొదమ అనే చోట ‘సింగపు’ అన్న చోట చివర ‘పు’ ఉంది. దానికి ‘కొదమ’ లో మొదటి అక్షరం ‘అ’ అనేది పరుషం పరమైంది. పుంపులకు పరుషం పరమైంది. కాబట్టి ‘నుగాగమం’ అనగా ‘5’ వస్తుంది.
ఉదా :
సింగపు + 5 + కొదమ సరళాదేశం రాగా సింగపుఁగొదమ అవుతుంది.

అభ్యాసం :
కింది ఉదాహరణలు పరిశీలించి, లక్షణ సమన్వయం చేయండి.
ఉదా :
తళుకుంగయిదువు – తళుకు + 5 + కయిదువు = సరళాదేశ సంధి రాగా, తళుకుం గయిదువు అవుతుంది.

గమనిక :
తళుకు అన్నది (ఉదంత స్త్రీ సమం); ‘కయిదువు’ పదంలో మొదట ‘క’ అనే పరుషము ఉంది. కాబట్టి నుగాగమం వస్తుంది.

1. చిగురుం గయిదువు
జవాబు:
చిగురుం గయిదువు = చిగురు + కయిదువు
గమనిక :
‘చిగురు’ అనేది ఉదంత స్త్రీ సమశబ్దం. దానికి కయిదువు అనే పదం పరమయ్యింది. కయిదువులో ‘క’ అనే పరుషం పరమైంది.

ఉదంత స్త్రీ సమశబ్దాలకు పరుషం పరమయింది కాబట్టి ‘నుగాగమం’ వచ్చింది.
ఉదా :
చిగురు + న్ + కయిదువు , తరువాత సరళాదేశం రాగా చిగురుఁగయిదువు అవుతుంది.

2. సరసపుఁదనము
జవాబు:
సరసము + తనము
గమనిక :
‘సరసముతనము’ అనేది కర్మధారయ సమాసం. అందువల్ల కర్మధారయాలలోని ‘ము’ వర్ణకానికి పు, ంపులు వస్తాయి.
ఉదా : సరసపు + తనము

ఇక్కడ పుంపులకు పరుష సరళాలు పరమైతే నుగాగమం వస్తుంది. ‘సరసపు’ లోని పుంపులకు ‘తనము’లోని ‘త’ పరుషం పరమైంది. కాబట్టి నుగాగమం (న్) వచ్చింది.
ఉదా :
సరసపు + న్ + తనము – చివరకు సరళాదేశం రాగా ‘సరసపుఁదనము’ అవుతుంది.

ఇ) నుగాగమ సంధి:
గమనిక :
ఇంకా మరికొన్ని సందర్భాల్లోనూ నుగాగమం వస్తుంది. కింది పదాలను విడదీయండి.

అ) విధాతృనానతి = విధాతృ + ఆనతి
ఆ) రాజునాజ్ఞ = రాజు + ఆజ్ఞ
విధాతృ + ఆనతి = విధాత యొక్క ఆనతి
రాజు + ఆజ్ఞ = రాజు యొక్క ఆజ్ఞ

విగ్రహవాక్యాలను అనుసరించి, పై సంధి పదాలు షష్ఠీ తత్పురుష సమాసానికి చెందినవి.

పూర్వపదం చివర స్వరాలు అనగా అచ్చులు “ఋ” కారం, “ఉత్తు” ఉన్నాయి.

గమనిక :
ఈ విధంగా షష్ఠీ తత్పురుష సమాసపదాల్లో, “A” కార, “ఋ” కారములకు అచ్చుపరమైతే నుగాగమం (5) వస్తుంది.
ఉదా :
1) విధాతృ + ఆనతి (విధాత యొక్క ఆనతి) = విధాతృ + న్ + ఆనతి
2) రాజు + ఆజ్ఞ (రాజు యొక్క ఆజ్ఞ = రాజు + న్ + ఆజ్ఞ

ఈ) నుగాగమ సంధి సూత్రం :
షష్ఠీ తత్పురుష సమాసంలో “ఉ” కార, “బు” కారాలకు అచ్చుపరమైనపుడు నుగాగమం వస్తుంది.
ఉదా :
విధాతృ + న్ + ఆనతి = విధాతృనానతి
రాము + న్ + ఆజ్ఞ = రామునాజ్ఞ

షష్ఠీ తత్పురుషంలో నుగాగమ సంధి సూత్రం :
షష్ఠీ సమాసమందు “ఉ” కార, ‘ఋ” కారాలకు అచ్చుపరమైతే నుగాగమం వస్తుంది.

పాఠ్యపుస్తకంలోని ముఖ్యమైన తెలుగు సంధులు

సంధి పదము విడదీత సంధి పేరు
1) వంటాముదము = వంట + ఆముదము = అత్వ సంధి
2) ఏమనిరి = ఏమి + అనిరి = ఇత్వ సంధి
3) అవ్విధంబున = ఆ + విధంబున = త్రిక సంధి
4) సింగపుకొదమ = సింగము + కొదమ = పుంప్వాదేశ సంధి
5) ముత్యపుచిప్ప = ముత్యము + చిప్ప = పుంప్వాదేశ సంధి
6) కొంచేపునరుడు = కొంచెము + నరుడు = పుంప్వాదేశ సంధి
7) బంధమూడ్చి = బంధము + ఊడ్చి = ఉత్వ సంధి
8) అవ్వారల = ఆ + వారల = త్రిక సంధి
9) భక్తురాలు = భక్త + ఆలు = రుగాగమ సంధి
10) బాలెంతరాలు = బాలెంత + ఆలు = రుగాగమ సంధి
11) గుణవంతురాలు = గుణవంత + ఆలు = రుగాగమ సంధి
12) దేశాలలో = దేశము + లలో = లులన సంధి
13) పుస్తకాలు = పుస్తకము + లు = లులన సంధి
14) సమయాన = సమయము + న = లులన సంధి
15) చిగురుంగయిదువు = చిగురు + కయిదువు = నుగాగమ సంధి
16) సరసపుఁదనము = సరసము + తనము = నుగాగమ సంధి
17) ఆనందాన్నిచ్చిన = ఆనందాన్ని + ఇచ్చిన = ఇత్వ సంధి
18) మేమంత = మేము + అంత = ఉత్వ సంధి
19) ఇవ్వీటిమీద = ఈ + వీటిమీద = త్రిక సంధి
20) భిక్షయిడదయ్యే = భిక్ష + ఇడదయ్యె = యడాగమ సంధి
21) ఆగ్రహముదగునె = ఆగ్రహము + తగునె = గసడదవాదేశ సంధి
22) ఉన్నయూరు = ఉన్న + ఊరు = యడాగమ సంధి
23) అందుఁ జూడాకర్ణుడు = అందున్ + చూడాకర్ణుడు = సరళాదేశ సంధి
24) చూడాకర్ణుడను = చూడాకర్ణుడు + అను = ఉత్వ సంధి
25) పరివ్రాజకుడు గలడు = పరివ్రాజకుడు + కలడు = గసడదవాదేశ సంధి

సంస్కృత సంధులు

1. సవర్ణదీర్ఘ సంధి సూత్రం
అ, ఇ, ఉ, ఋ లకు అవే అచ్చులు పరమైనపుడు వాని దీర్ఘాలు ఏకాదేశంగా వస్తాయి.

గమనిక :
‘అ’ వర్ణానికి – ‘అ’, ‘ఆ’ లు సవర్ణాలు
‘ఇ’ వర్గానికి – ‘ఇ’, ‘ఈ’ లు సవర్ణాలు
‘ఉ’ వర్గానికి – ‘ఉ’, ‘ఊ’ లు సవర్ణాలు
‘ఋ’ వర్గానికి – ‘ఋ’, ‘బూ’ లు సవర్ణాలు

1. ఉదా :
1) రామానుజుడు = రామ + అనుజుడు = (అ + అ = ఆ) = సవర్ణదీర్ఘ సంధి
2) రామాలయం = రామ + ఆలయం = (అ + ఆ = ఆ) = సవర్ణదీర్ఘ సంధి
3) గంగాంబ = గంగ + అంబ = (అ + అ = ఆ) = సవర్ణదీర్ఘ సంధి

2. ఉదా :
4) కవీంద్రుడు = కవి + ఇంద్రుడు – (ఇ’ + ఇ = ఈ), = సవర్ణదీర్ఘ సంధి
5) శ్రీకాళహస్తీశ్వరా = శ్రీకాళహస్తి + ఈశ్వర – (ఇ + ఈ = ఈ) = సవర్ణదీర్ఘ సంధి

3. ఉదా :
6) భానూదయం . = భాను + ఉదయం = (ఉ + ఉ = ఊ) = సవర్ణదీర్ఘ సంధి
7) వధూపేతుడు : వధూ + ఉపేతుడు = (ఊ + ఉ = ఊ) – సవర్ణదీర్ఘ సంధి

4. ఉదా :
8) పిత్రణం : పితృ + ఋణం = (ఋ + ఋ = ఋ) = సవర్ణదీర్ఘ సంధి
9) మాతణం = మాతృ + ఋణం = (బ + ఋ – ఋ) = సవర్ణదీర్ఘ సంధి

2. గుణ సంధి సూత్రం
అకారానికి ఇ, ఉ, ఋ లు పరమైతే, ఏ, ఓ, అర్ లు క్రమంగా ఏకాదేశంగా వస్తాయి.
1. ఉదా :
రాజేంద్రుడు = రాజ + ఇంద్రుడు = (అ + ఇ = ఏ) = గుణ సంధి
మహేంద్రుడు = మహా + ఇంద్రుడు = (ఆ + ఇ = ఏ) = గుణ సంధి
నరేంద్రుడు – నర + ఇంద్రుడు = (అ + ఇ = ఏ) = గుణ సంధి
రామేశ్వర = రామ + ఈశ్వర (అ + ఇ = ఏ) = గుణ సంధి

2. ఉదా :
పరోపకారం = పర + ఉపకారం = (అ + ఉ = ఓ) = గుణ సంధి
మహోన్నతి = మహా + ఉన్నతి = (అ + ఉ = ఓ) = గుణ సంధి
దేశోన్నతి = దేశ + ఉన్నతి = (అ + ఉ = ఓ) = గుణ సంధి
గృహోపకరణం = గృహ + ఉపకరణం = (అ + ఉ = ఓ) = గుణ సంధి

3. ఉదా :
రాజర్షి = రాజ + ఋషి = (అ + ఋ = అర్) = గుణ సంధి
మహర్షి = మహా + ఋషి = (అ + ఋ = అర్) = గుణ సంధి

పై ఉదాహరణలు పరిశీలిస్తే
1) అ, ఆ లకు ఇ, ఈ లు కలిసి ‘ఏ’ గా మారడం
2) అ, ఆ లకు ఉ, ఊ లు కలిసి ‘ఓ’ గా మారడం
3) అ, ఆ లకు ఋ, ౠలు కలిసి ‘అర్’ గా మారడం – గమనించగలం.

పై మూడు సందర్భాల్లోనూ, పూర్వ స్వరం అంటే, సంధి వీడదీసినపుడు, మొదటి పదం చివరి అచ్చులు, అ, ఆ లుగానూ, పర స్వరం, అంటే విడదీసిన రెండవ పదంలో మొదటి అచ్చులు ఇ, ఉ, ఋ – లుగానూ ఉన్నాయి.

గమనిక :
1) అ, ఆ లకు – ‘ఇ’ కలిస్తే ‘ఏ’ గా మారుతుంది.
2) అ, ఆ లకు – ‘ఉ’ కలిస్తే ‘ఓ’ గా మారుతుంది.
3) అ, ఆ లకు – ‘ఋ’ కలిస్తే ‘అర్’ గా మారుతుంది.
ఏ, ఓ, అర్ అనే వాటిని గుణాలు అంటారు. ఇలా గుణాలు వచ్చే సంధిని “గుణ సంధి” అంటారు.

3. యణాదేశ సంధి సూత్రం
ఇ, ఉ, ఋ లకు అసవర్ణాచ్చులు పరమైతే, య, వ, రలు ఆదేశంగా వస్తాయి.
ఈ కింది పదాలను విడదీయండి. మార్పును గమనించండి.
1. ఉదా :
అ) అత్యానందం = అతి + ఆనందం – (త్ + ఇ + ఆ = యా) = యణాదేశ సంధి
1) అత్యంతం = అతి + అంతం = (త్ + ఇ + అ = య) = యణాదేశ సంధి
2) అభ్యంతరం = అ + అంతరం = (త్ + ఇ + అ = య) = యణాదేశ సంధి

2. ఉదా :
ఆ) అణ్వస్త్రం = అస్త్రం (ణ్ + ఉ + అ = వ) = యణాదేశ సంధి
2) గుర్వాజ్ఞ = (ర్ + ఉ + ఆ = వ) = యణాదేశ సంధి

3. ఉదా :
ఇ) పిత్రాజ్ఞ = పితృ + ఆజ్ఞ = (ఋ + ఆ = రా) = యణాదేశ సంధి
3) మాత్రంశ = మాతృ + అంశ = (ఋ + అ = ర) = యణాదేశ సంధి

గమనిక :
ఇ, ఉ, ఋ లకు అసవర్ణాచ్చులు (వేరే అచ్చులు) పక్కన వచ్చినపుడు, క్రమంగా వాటికి య – వ -రలు వచ్చాయి. యవరలను ‘యణులు’ అంటారు. యజ్ఞులు చేరితే వచ్చే సంధిని ‘యణాదేశ సంధి అంటారు. యణాదేశ సంధిలో ‘ఇ’ కి బదులుగా “య్”, ‘ఉ’ కి బదులుగా ‘ప్’, ‘ఋ’ కి బదులుగా ‘5’ వచ్చాయి.

4. వృద్ధి సంధి సూత్రం
అకారానికి ఏ, ఐలు పరమైతే ‘ఐ’ కారమూ, ఓ, ఔ లు పరమైతే ‘ఔ’ కారమూ వస్తాయి.

ఈ కింది పదాలను విడదీయండి.
1. ఉదా :
వసుధైక = వసుధా + ఏక = = (ఆ + ఏ = ఐ) = వృద్ధి సంధి
అ) రసైక = రస + ఏక = (అ + ఏ = ఐ) = వృద్ధి సంధి
ఆ) సురైక = సుర + ఏక = (అ + ఏ = ఐ) = వృద్ధి సంధి

2. ఉదా :
సమైక్యం = సమ + ఐక్యం = (అ + ఐ = ఐ) = వృద్ధి సంధి
ఇ) అప్లైశ్వర్యం = అష్ట + ఐశ్వర్యం = (అ + ఐ = ఐ) = వృద్ధి సంధి
ఈ) దేవైశ్వర్యం =దేవ + ఐశ్వర్యం = (అ + ఐ = ఐ) = వృద్ధి సంధి

3. ఉదా :
పాపౌఘము = పాప + ఓఘము = (అ + ఓ = ఔ) = వృద్ధి సంధి
ఉ) వనౌకసులు = వన + ఓకసులు = (అ + ఓ = ఔ) = వృద్ధి సంధి
ఊ) వనౌషధి = వన + ఓషధి = (అ + ఓ = ఔ) = వృద్ధి సంధి

4. ఉదా :
రనౌచిత్యం = రస + ఔచిత్యం = (అ + ఔ – ఔ) = వృద్ధి సంధి
ఋ) దివ్యౌషధం = దివ్య + ఔషధం = (అ + ఔ – ఔ) = వృద్ధి సంధి
ఋ) దేశోన్నత్యం = దేశ + ఔన్నత్యం = (అ + ఔ – ఔ) = వృద్ధి సంధి

గమనిక :
పైన పేర్కొన్న పదాలను విడదీసినపుడు మీరు గమనింపదగిన విషయం ఇది
1. వృద్ధి సంధి ఏర్పడేటప్పుడు ప్రతిసారీ పూర్వ స్వరంగా ‘అ’.వచ్చింది.
2. పర స్వరం స్థానంలో వరుసగా “ఏ, ఏ, ఐ, ఔలు ఉన్నాయి.
3. అకారానికి ఏ, ఐలు కలిపినపుడు ‘బి’ వచ్చింది.
4. అకారానికి ఓ, ఔ లు కలిపినపుడు ‘&’ వచ్చింది.

వృద్ధులు = ఐ, ఔలను ‘వృద్ధులు’ అంటారు.

5. జశ్వ సంధి సూత్రం
పరుషాలకు వర్గ ప్రథమ ద్వితీయాక్షరాలు, శష స లు తప్ప, మిగిలిన హల్లులు కానీ, అచ్చులు కానీ, పరమైతే వరుసగా సరళాలు ఆదేశమవుతాయి.
ఉదా :
సత్ + భక్తులు = సద్ + భక్తులు = సద్భక్తులు

పై సంధి పదాలను పరిశీలించండి. మొదట విడదీసిన పదాలలోని ‘త’ కార స్థానములో, ‘ద’ కారం ఆదేశంగా వచ్చి, ‘సద్భక్తులు’ అనే రూపం వచ్చింది.

గమనిక :
ఈ విధంగా మొదటి పదం చివర, క, చ, ట, త, ప (పరుషాలు),లలో ఏదైనా ఒక అక్షరం ఉండి, రెండవ పదం మొదట క ఖ, చ ఛ, ట ఠ, త థ, ప ఫ, లు మరియు శ ష స లు తప్ప, మిగిలిన హల్లులూ, అచ్చులలో ఏ అక్షరం ఉన్నా ‘గ, జ, డ, ద, బ’ లు వరుసగా ఆదేశం అవుతాయి.

కింది పదాలను విడదీయండి.
1) దిగంతము = దిక్ + అంతము = జశ్వ సంధి (క్ – గ్ గా మారింది)
2) మృదటము = మృత్ + ఘటము = జశ్వ సంధి (త్ -ద్ గా మారింది)
3) ఉదంచద్భక్తి = ఉదంచత్ + భక్తి = జత్త్వ సంధి (త్ -ద్ గా మారింది)
4) వాగీశుడు = వాక్ ఈశుడు = జత్త్వ సంధి (క్ – గ్ గా మారింది)
5) వాగ్యుద్ధం = వాక్ + యుద్ధం = జశ్వ సంధి (క్ – గ్ గా మారింది)
6) వాగ్వాదం = వాక్ + వాదం = జశ్వ సంధి (క్ – గ్ గా మారింది)
7) తద్విధం = తత్ + విధం = జశ్వ సంధి (త్ -ద్ గా మారింది)

6. అనునాసిక సంధి సూత్రం
వర్గ ప్రథమాక్షరాలకు (కటతలకు) ‘న’ గాని, ‘మ’ గాని పరమైనప్పుడు అనునాసికములు ఆదేశంగా వస్తాయి.

ఈ కింది పదాలను విడదీయండి.
అ) వాజ్మయం = వాక్ + మయం = అనునాసిక సంధి
ఆ) రాణ్మహేంద్రవరం = రాట్ + మహేంద్రవరం = అనునాసిక సంధి
ఇ) జగన్నాథుడు = జగత్ + నాథుడు = అనునాసిక సంధి

గమనిక :
పై సంధులు జరిగిన తీరు గమనించండి.
అ) వాక్ + మయం = వాజ్మయం = ‘క్’ స్థానంలో ‘జ’ వచ్చింది.
ఆ) రాట్ + మణి = రాణ్మణి = ‘ట్’ స్థానంలో ‘ణ’ వచ్చింది.
ఇ) జగత్ + నాథుడు = జగన్నాథుడు = ‘త్’ స్థానంలో ‘న’ వచ్చింది.

గమనిక :
1) పై మూడు సంధి పదాలలోనూ మొదటి పదం చివర వరుసగా క, ట, త, లు ఉన్నాయి.
2) వాటికి ‘మ’ గాని, ‘న’ గాని పరమయినాయి. అంటే తరువాత కలిశాయి.
3) 1) అప్పుడు పూర్వపదం చివర గల ‘క’ కారం, ‘క’ వర్గకు అనునాసికమైన ‘జ’ గా మారుతుంది. (క, ఖ, గ, ఘ, ఙ)
2) అప్పుడు పూర్వపదం చివర గల ‘ట’ కారం, ‘ట’ వర్గకు అనునాసికమైన ‘ణ’ గా మారింది. (ట, ఠ, డ, ఢ, ణ)
3) అప్పుడు పూర్వపదం చివర గల ‘త’ కారం, దాని అనునాసికమైన ‘న’ (త థ ద ధ న) గా ఆదేశం అవుతాయి.
దీనినే ‘అనునాసిక సంధి’ అంటారు.

అభ్యాసము :
1) తన్మయము = తత్ + మయము = అనునాసిక సంధి
2) రాణ్మణి = రాట్ + మణి , – అనునాసిక సంధి
3) వాజ్మయము = వాక్ + మయము = అనునాసిక సంధి
4) మరున్నందనుడు = మరుత్ + నందనుడు = అనునాసిక సంధి

7.అ) విసర్గ సంధి సూత్రం
అకారాంత పదాల మీద ఉన్న విసర్గకు, వర్గ ప్రథమ ద్వితీయాక్షరాలు (క, చ, ట, త, ప, ఖ, ఛ, ఠ, థ, ఫ, శ, ష, సలు కాక, మిగతా అక్షరాలు) కలిసినపుడు, విసర్గ లోపించి “అ” కారం ‘ఓ’ కారంగా మారుతుంది.

ఉదాహరణలు చూడండి :
అ) నమోనమః = నమః + నమః
ఆ) మనోహరం : మనః + హరం
ఇ) పయోనిధి : పయః + నిధి
ఈ) వచోనియమం = వచః + నియమం

గమనిక :
ఈ నాలుగు ఉదాహరణలలో అకారాంత పదాల మీద ఉన్న విసర్గ లోపించి, ‘అ’ కారం ‘ఓ’ కారంగా మారింది.

ఆ) విసర్గ సంధి సూత్రం
విసర్గకు శ,ష,సలు కలిసినపుడు, విసర్గ శ,ష,స,లుగా మారి శ,ష,సలు ద్విత్వాలుగా మారుతాయి.
ఉదాహరణలు :
అ) మనశ్శాంతి : మనః + శాంతి
ఆ) చతుషష్టి : చతుః + షష్టి
ఇ) నభస్సుమం : నభః + సుమం

గమనిక :
విసర్గము, ప్రక్కనున్న శ, ష, స లుగా మారి, ద్విత్వాలుగా అయ్యింది. ఆయా పదాలను కలుపగా, వరుసగా మనశ్శాంతి, చతుషష్టి, నభస్సుమం అనే రూపాలు ఏర్పడ్డాయి.

ఇ) విసర్గ సంధి సూత్రం
విసర్గకు క,ఖ,ప,ఫ,లు కలిస్తే విసర్గమారదు. (సంధి ఏర్పడదు.)
ఉదాహరణలు :
అ) ప్రాతఃకాలము = ప్రాతః + కాలము = విసర్గ సంధి
ఆ) తపఃసలము = తపః + ఫలము = విసర్గ సంధి

గమనిక :
పై ఉదాహరణలలో విసర్గకు క, ఫ లు పరం అయ్యాయి. కాబట్టి విసర్గ మారకుండా యథాప్రకారంగానే ఉంది.

ఈ) విసర్గ సంధి సూత్రం
అంతః, దుః, చతుః, ఆశీః, పునః మొదలయిన పదాల తరువాత ఉండే విసర్గ, రేఫ (‘ర్’)గా మారుతుంది.
ఉదా :
అంతః + ఆత్మ : అంతర్ – + ఆత్మ = అంతరాత్మ
అ) దుః + అభిమానం = దుర్ + అభిమానం = దురభిమానం
ఆ) చతుః + దిశలు = చతుర్ + దిశలు = చతుర్దశలు
ఇ) ఆశీః + వాదము = ఆశీర్ + వాదము = ఆశీర్వాదము
ఈ) పునః + ఆగమనం = పునర్ + ఆగమనం = పునరాగమనం
ఉ) అంతః + మథనం = అంతర్ + మథనం = అంతర్మథనం

ఉ) విసర్గ సంధి సూత్రం
ఇస్, ఉర్ల విసర్గకు, క, ఖ, ప, ఫ, లు కలిస్తే, విసర్గ ‘ష’ కారంగా మారుతుంది.\
ఉదా :
ధనుష్కోటి : ధనుస్ట్ + కోటి = ధనుష్ + కోటి = ధనుష్కోటి
అ) నిష్ఫలము = నిస్ + ఫలము = నిష్, + ఫలము = నిష్ఫలము
ఆ) దుష్కరము = దుస్ + కరము – దుష్ – + కరము = దుష్కరము

గమనిక :
ఇస్ (ఇజి), ఉస్ (43) ల విసర్గలకు క,ఖ,ప,ఫ లు కలిసినపుడు, విసర్గ అనగా ‘స్’ కారము ‘ష’ కారంగా మారుతుంది.

ఊ) విసర్గ సంధి సూత్రం
విసర్గకు (అనగా ‘స్’ కు) చ ఛ లు పరమైతే ‘శ’ కారం, ట ఠలు పరమైతే ‘ష’ కారం, త థ
లు పరమైతే ‘స’ కారం వస్తాయి.
ఉదా :
అ) దుశ్చేష్టితము = దుః + చేష్టితము (విసర్గము – శ్ గా మారింది)
ఆ) ధనుష్టంకారం = ధనుః + టంకారము (విసర్గము ష్ గా మారింది)
ఇ) మనస్తాపము = మనః + తాపము (విసర్గము ‘స’గా మారింది)
ఈ) నిస్తేజము = నిః + తేజము (విసర్గము ‘స’గా మారింది)

గమనిక :
పై ఉదాహరణలలో విసర్గకు చ ఛలు పరమైతే ‘శ’ కారం, ట ఠలు పరమైతే ‘ష’ కారం త థలు పరమైతే ‘స’ కారం వస్తుంది.

పై ఉదాహరణలు గమనిస్తే విసర్గ సంధి ఆరు విధాలుగా ఏర్పడుతోందని తెలుస్తోంది.

8. శ్చుత్వ సంధి సూత్రం
‘స’ కార, ‘త’ వర్గాలకు, ‘శ’ కార ‘చ’ వర్గా (చ ఛ జ ఝ) లు పరమైతే, ‘శ’ కార ‘చ’ వర్గాలే వస్తాయి.

కింది ఉదాహరణలను పరిశీలించండి.
అ) తప + శక్తి → తపస్ + శక్తి → తపశ్శక్తి
ఆ) నిః + శంక → నిస్ + శంక → నిశ్శంక
ఇ) మనః + శాంతి → మనస్ + శాంతి → మనశ్శాంతి

గమనిక :
పై ఉదాహరణలలో పూర్వపదాలలో ఉన్న విసర్గ సంధి కార్యంలో విసర్గను ‘స’ గా తీసుకుంటున్నాం. విసర్గకు ‘స్’ కారం వస్తుంది. అలా విసర్గ స కారం కాగా, ఆ ‘స’ కారానికి ‘శ’ వర్ణం పరం అవుతుంది. ఇలా పరం అయినపుడు ఆ ‘స’ కారం, ‘శ’ కారంగా మారుతుంది. అనగా ‘శవర్ణ ద్విత్వం వస్తుంది.
ఆ) నిస్ + చింత → నిశ్చింత
సత్ + ఛాత్రుడు → సచ్ఛాత్రుడు
శరత్ + చంద్రికలు → శరచ్చంద్రికలు
జగత్ + జనని → జగజ్జనని
శార్జిన్ + జయః → శారిఞ్జయః

గమనిక :
పై పదాల్లో ‘స’ కార, ‘త’ వర్గాలు పూర్వపదాంతంగా ఉన్నాయి. ‘శ’ కార, ‘చ’ వర్గాలు (త, న) పరమైనాయి. అలా పరమైనప్పుడు ‘శ’ కార, చ వర్గాలుగా మారుతాయి.

అనగా

1) స్ + చి = శ్చి
2) త్ + జ = జ్జ
3) త్ + శా = చ్చా
4) న్ + జ = ఞ్జ
5) త్ + చ = చ్చ

ఈ విధంగా ‘స’ కార ‘త’ వర్గాలకు (తథదధన) లకు, ‘శ’ కార, ‘చ’ వర్గాలు వస్తే అది “శ్చుత్వ సంధి” అవుతుంది.

పాఠ్యపుస్తకంలోని ముఖ్యమైన సంస్కృత సంధులు

1) అత్యంత = అతి + అంత = యణాదేశ సంధి
2) పుణ్యవాసము = పుణ్య + ఆవాసము = సవర్ణదీర్ఘ సంధి
3) స్నిగ్గాంబుదము = స్నిగ్ధ + అంబుదము = సవర్ణదీర్ఘ సంధి
4) పురాతనాపాదితము = పురాతన + ఆపాదితము = సవర్ణదీర్ఘ సంధి
5) సహస్రాబ్దం = సహస్ర + అబ్దం = సవర్ణదీర్ఘ సంధి
6) వేదోక్తము = వేద + ఉక్తము = గుణ సంధి
7) మదోన్మాదము = మద + ఉన్మాదము = గుణ సంధి
8) సరభసోత్సాహం = సరభస + ఉత్సాహం = గుణ సంధి
9) జీవనోపాధి = జీవన + ఉపాధి = గుణ సంధి
10) మహోపకారం = మహా + ఉపకారం = గుణ సంధి
11) గుణౌద్ధత్యం = గుణ + ఔద్ధత్యం = వృద్ధి సంధి
12) రసైకస్థితి = రస + ఏకస్థితి = వృద్ధి సంధి
13) తన్మయము = తత్ + మయము = అనునాసిక సంధి
14) వాజ్మయము = వాక్ + మయము = అనునాసిక సంధి
15) రాణ్మణి = రాట్ + మణి = అనునాసిక సంధి
16) మరున్నందనుడు = మరుత్ + నందనుడు = అనునాసిక సంధి
17) రాణ్మహేంద్రపురం = రాట్ + మహేంద్రపురం = అనునాసిక సంధి
18) జగన్నాథుడు = జగత్ + నాథుడు = అనునాసిక సంధి
19) నమోనమః = నమః . + నమః = విసర్గ సంధి
20) మనోహరం = మనః . + హరం = విసర్గ సంధి
21) పయోనిధి = పయః + నిధి = విసర్గ సంధి
22) వచోనిచయం = వచః + నిచయం = విసర్గ సంధి
23) ప్రాతఃకాలము = ప్రాతః + కాలము = విసర్గ సంధి
24) తపఃఫలము = తపః + ఫలము = విసర్గ సంధి
25) నిష్ఫలము = నిస్ + ఫలము = విసర్గ సంధి
26) దుష్కరము = దుస్ + కరము = విసర్గ సంధి
27) ధనుష్టంకారము = ధనుః + టంకారము = విసర్గ సంధి
28) మనస్తాపము = మనః + తాపము = విసర్గ సంధి
29) దురభిమానం = దుః + అభిమానం = విసర్గ సంధి
30) నిరాడంబరం = ఆడంబరం = విసర్గ సంధి
31) దుర్భేద్యము = దుః + భేద్యము = విసర్గ సంధి
32) తపోధనుడు = తపః + ధనుడు = విసర్గ సంధి
33) నిరాశ = నిస్ + ఆశ = విసర్గ సంధి
34) దుశ్చేష్టితము = దుస్ + చేష్టితము = శ్చుత్వ సంధి
35) నిశ్చింత = నిస్ + చింత = శ్చుత్వ సంధి
36) సచ్ఛాత్రుడు = సత్ + ఛాత్రుడు = శ్చుత్వ సంధి
37) శరచ్చంద్రికలు = శరత్ + చంద్రికలు = శ్చుత్వ సంధి
38) జగజ్జనని = జగత్ + జనని = శ్చుత్వ సంధి
39) శారిజ్జయః = శార్జిన్ + జయః = శ్చుత్వ సంధి

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 4 Movement of Materials Across the Cell Membrane Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 4th Lesson Movement of Materials Across the Cell Membrane

9th Class Biology 4th Lesson Movement of Materials Across the Cell Membrane Textbook Questions and Answers

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Question 1.
The structure which controls the entry and exit of the materials through the cell is
A) Cell wall
B) Cell membrane
C) Both
D) None of them
Answer:
Cell membrane.

Question 2.
Fill in the blanks.
a) The smell of flowers reaches us through the process of …………………..
Answer:
Diffusion

b) The MIC gas of Bhopal tragedy was spread throughout the city through the process of …………………
Answer:
Diffusion

c) Water enters the potato osmometer due to a process called ………………
Answer:
Osmosis

d) The fresh grape wrinkles, if kept in salt water because of …………………
Answer:
Osmosis

Question 3.
What do you mean by permeability of membrane? Explain with suitable example.
Answer:
Allowing only certain materials to pass through the membrane is called permeability.

Example :

1. The cell membrane is very much permeable to gases such as carbondioxide, oxygen, nitrogen and fat solvent compounds such as alcohol, ether and chloroform.
2. It is impermeable to polysaccharides, phospholipids and proteins.

Question 4.
If the dried vegetables are kept in water they become fresh. What is the reason?
Answer:

1. The dried vegetables have less water content and high salt concentration in cells.
2. When they are kept in water they absorb water and become fresh.
3. The water enter into the vegetables by a process known as osmosis.

Question 5.
Name the process by which we can get fresh water from sea water.
Answer:
Reverse Osmosis.

Question 6.
What will happen to a marine fish if kept in fresh water aquarium? Support your answer with reasons.
Answer:
The marine fish dies.
Reasons:

1. Usually marine fishes have high concentration of salts in their body.
2. When they are kept in fresh water, the water from the fresh water aquarium enters the body of fishes due to osmosis.
3. More amount of fresh water enters the cells of fish. This results in bursting of cells and fish dies.

Question 7.
Why do the doctors administer saline (salt solution) only, but not the distilled water?
Answer:

• Distilled water causes cells to lyse, so injecting distilled water into a vein will cause some degree of haemolysis.
• Haemolysis is the rupture of red blood cells.
• Large amount of distilled water would cause much more damage not just limited to haemolysis and also cause brain damage or cardiac arrest and death.
• That is why fluids are administered to patients as saline (which include appropriate amount of salt)

Question 8.
What will happen if 50% glucose solution (dextrose) is injected intravenously (into vein)?
Answer:

• 50% glucose solution (dextrose) is used for reduction of increased cerebrospinal pressure and cerebral edema.
• If 50% glucose solution is injected intravenously it may produce allergic reactions in sensitive persons.
• The allergic reactions include nervous excitement infection at the joint site, tissues necrosis, venous thrombosis extending from the site of injection etc.
• Hence concentrated dextrose (glucose) should be administered via central vein only after suitable dilution.

Question 9.
What will happen if cells do not have ability of permeability?
Answer:

1. If the cells do not have ability of permeability they would not be able to carryout any of their fundamental life functions.
2. Oxygen, glucose, fats, proteins and vitamins are needed by cells to perform life process.
3. Mature cells become impermeable to any molecules or atoms it would die of toxicity and it would not be able to remove its wastes.

Question 10.
Draw the flow chart showing different stages in doing the experiment with egg.
Answer:

Question 11.
You have purchased a coconut in the market. By shaking it you found there is less water in coconut. Can you fill the coconut with water without making a hole to the coconut?
Answer:

• No, it is not possible to fill the coconut with water without making a hole.
• The husk of coconut is mostly made up of sclerenchymatous cells which are dead.
• Osmosis do not takes place in dead cells.
• It is not possible to fill the coconut with water without making a hole.

Question 12.
What are your observations in experiments to know about diffusion?
Answer:
Observations in experiments to know about diffusion are :

1. Materials kept in medium (water/air) get dissolves in the medium.
2. These dissolved molecules gradually move randomly in all directions. (from center to periphery)
3. They move from higher concentration to lower concentration.
4. This movement occures till these molecules spread equally throughout the medium.

Question 13.
Discuss with your friends and write the list of incidences where diffusion occurs.
Answer:

• A sugar cube in a glass of milk/water diffuses throughout it and make it sweet.
• The smell of cookies diffuses through the house as they bake.
• Tea leaf pigments diffuse through the tea bag into the water to give it colour and taste.
• Air freshner/deodorent molecules diffuse into the air when put on so we can smell it.
• If the cooking gas is leaked it spreads all over the house through diffusion.
• CO₂ bubbles in soft drink diffuses out of soda leaving the soda flat.
• Robbin Blue drops diffuses in water, making the water blue.
• Agarbatti, mosquito repellents work on the principle of diffusion.

Question 14.
How diffusion is useful in everyday life?
Answer:

• A wilted carrot made firm again by soaking in water.
• Cigarette smoke. It diffuses into air and spreads through the room.
• A sugar cube in a glass of water that is not stirred will dissolve slowly and the sugar molecules will distribute over the water by diffusion.
• The smell of cookies diffuses through the house as they bake.
• Tea leaf pigments diffuse through the tea bag to give the water its colour and taste of tea.
• Air freshner / deodorant molecules diffuse into the air when put on. So we can smell it.
• If the cooking gas is leaked, it spreads all over the house through diffusion.
• CO₂ bubbles in soft drink diffuses out of our soda leaving our soda flat.
• Air freshners, agarbatti, mosquito repellents work on the principle of diffusion.

Question 15.
Give examples of three daily life activities in which osmosis is involved?
Answer:

• Water enters into the roots through osmosis.
• In our body waste materials are filtered from the blood.
• Osmosis helps in the opening and closing of stomata.

9th Class Biology 4th Lesson Movement of Materials Across the Cell Membrane Activities

Activity – 1

Question 1.
Look at the substances in the table identify the (✓) substances that should go into the cell and should go out of the cell?
Answer:

Procedure :

• Keep the raw eggs in dil HCl / toilet cleaning acid for 4 to 5 hours.
• Take out the egg with the help of table spoon.
• Wash the eggs under tap water.

• Measure the circumference of each egg with long strip of paper, as its widest place, and mark on the paper with pen or pencil.
• Prepare a concentrated salt solution in a beaker.
• Place one egg in the beaker with tap water and place the other in the salt water.

• Leave the beakers for 2 to 4 hours.
• Take the eggs out, wipe them and measure the circumference with the same strip of paper. Mark on the paper with pen or pencil.

Observation :
The egg placed in salt water shrinks, the egg placed in the tap water swells.

Result:

• Shrinking of egg placed in the salt water is due to exosmosis in which water molecules leave the cell.
• Swelling of egg placed in the tap water is due to endosmosis in which water molecules enter the cell.

Lab Activity – 3

Question 2.
Prepare semi-permeable membranes and conduct an experiment to prove osmosis with it.
Answer:
Preparing semi-permeable membranes.

• Take two raw eggs.
• Keep the two eggs in dil. HCl for 4 to 5 hours.
• The shells which are made of calcium carbonate (CaCO₃) are dissolved.
• Wash the eggs under tap water.
• Carefully pierce a pencil sized hole in the egg membrane and drain the contents.
• Wash the membrane with fresh water. Now the semi-permeable membrane is ready for use.