Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.2

10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.
Evaluate the following.
i) sin 45° + cos 45°
Answer:
sin 45° + cos 45°
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= \(\frac{1+1}{\sqrt{2}}\)
= \(\frac{2}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}\)
= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Answer:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) – \(\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)
= \(\frac{8+3-3}{4}\)
= \(\frac{8}{4}\)
= 2

v)

Answer:

Question 2.
Choose the right option and justify your choice.
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) sin 60°
b) cos 60°
c) tan 30°
d) sin 30°
Answer:

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) tan 90°
b) 1
c) sin 45°
d) 0
Answer:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-(1)^{2}}{1+(1)^{2}}\)
= \(\frac{0}{1+1}\) = \(\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\)
a) cos 60°
b) sin 60°
c) tan 60°
d) sin 30°
Answer:

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?
Answer:
Take sin 60°.cos 30° + sin 30°.cos 60°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{(\sqrt{3})^{2}}{4}\) + \(\frac{1}{4}\)
= \(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{3+1}{4}\)
= \(\frac{4}{4}\) = 1 …… (1)
Now take sin (60° + 30°)
= sin 90° = 1 …….. (2)
From equations (1) and (2), I conclude that
sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?
Answer:
L.H.S. = cos (60° + 30°)
cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
∴ L.H.S = R.H.S
Yes, it is right to say
cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.
i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.
In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.
Answer:
Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = \(\frac{\text { Opposite side to } \angle P}{\text { Adjacent side to } \angle P}\)
√3 = \(\frac{RQ}{6}\)
which gives RQ = 6√3 cm ……. (1)
To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.
In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.
Answer:
Note: In the problem take
YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.
By Pythagoras theorem
XZ² = XY² + YZ²
(2x)² = (x)² + YZ²
4x² = x² + YZ²
YZ² = 4x² – x² = 3x²
YZ = \(\sqrt{3 x^{2}}\) = √3x
Now, from the △XYZ
tan X = \(\frac{XZ}{XY}\) = \(\frac{\sqrt{3} x}{x}\)
tan X = √3 = tan 60°
∴ Angle YXZ is 60°.
tan Z = \(\frac{XY}{YZ}\) = \(\frac{x}{\sqrt{3} x}\)
tan Z = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle YZX is 30°.
Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.
Is it right to say that
sin (A + B) = sin A + sin B? Justify your answer.
Answer:
Let A = 30° and B = 60°
L.H.S = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
R.H.S = sin 30° + sin 60°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
Hence L.H.S ≠ R.H.S
So, it is not right to say that sin (A + B) = sin A + sin B

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.3

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.3 Textbook Questions and Answers

Question 1.
Find the area of the triangle whose vertices are
i) (2, 3), (-1, 0), (2,-4)
Answer:
Given: A (2, 3), B (- 1, 0) and C (2, – 4) are the vertices of a △ABC.
Area of the triangle ABC = \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|2(0+4)-1(-4-3)+2(3-0)|\)
= \(\frac{1}{2}|8+7+6|\)
= \(\frac{21}{2}\)
= 10\(\frac{1}{2}\) sq.units

ii) (-5, -1), (3, -5), (5, 2)
Answer:
Given: A (- 5, – 1), B (3, – 5) and C (5, 2) are the vertices of △ABC.
Area of the △ABC
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|-5(-5-2)+3(2+1)+5(-1+5)|\)
= \(\frac{1}{2}|35+9+20|\)
= \(\frac{64}{2}\)
= 32 sq.units

iii) (0, 0), (3, 0), (0, 2)
Answer:
Given: O (0, 0), A (3, 0) and B (0, 2) are the vertices of a triangle, △AOB.
Area of the △AOB
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(0-2)+3(2-0)+0(0-0)|\)
= \(\frac{1}{2}|6|\)
= 3 sq.units

Or

△AOB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 3 × 2
= 3 sq.units

Question 2.
Find the value of ‘K’ for which the points are collinear.
i) (7, -2), (5, 1), (3, K)
Answer:
Given: A (7, – 2), B (5, 1) and C (3, K) are collinear.
∴ Area of △ABC = 0
But area of triangle
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
⇒ \(\frac{1}{2} \mid 7(1-\mathrm{K})+5(\mathrm{~K}+2)+3(-2-1)\) = 0
⇒ \(|7-7 K+5 K+10-9|\) = 0
⇒ \(|-2 \mathrm{~K}+8|\) = 0
⇒ -2K + 8 = 0
⇒ -2K = -8
⇒ K = \(\frac{8}{2}\)
i.e., K = 4

ii) (8, 1), (K,-4), (2,-5)
Answer:
Given: A (8, 1), B(K, – 4) and C (2, – 5) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|8(-4+5)+\mathrm{K}(-5-1)+2(1+4)|\) = 0
⇒ \(|8-6 \mathrm{~K}+10|\) = 0
⇒ \(|18-6 \mathrm{~K}|\) = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
i.e., K = 3

iii) (K,K), (2, 3), and (4,-1)
Answer:
A (K, K), B (2, 3) and C (4, – 1) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|\mathrm{~K}(3+1)+2(-1-\mathrm{K})+4(\mathrm{~K}-3)|\) = 0
⇒ \(|4K-2-2K+4K-12|\) = 0
⇒ \(|6 \mathrm{~K}-14|\) = 0
⇒ 6K – 14 = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\) = \(\frac{7}{3}\)
∴ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer:

Given: A (0, – 1), B (2, 1) and C (0, 3) are the vertices of △ABC.
Let D, E and F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}\).

Area of a triangle ABC =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|\)
= \(\frac{1}{2}|8|\)
= 4 sq.units
Area of △DEF = \(\frac{1}{2}|1(2-1)+1(1-0)+0(0-2)|\)
= \(\frac{1}{2}|1+1|\)
= \(\frac{2}{2}\)
= 1 sq.units
Ratio of areas = △ABC : △DEF = 4 : 1.
△ADF ≅ △BED ≅ △DEF ≅ △CEF
∴ △ABC : △DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices taken inorder are (-4, -2), (-3, -5),(3, -2) and (2, 3).
Answer:

Given: A (- 4, – 2), B (- 3, – 5), C (3, – 2) and D (2, 3) are the vertices of the quadrilateral ▱ ABCD.
Area of ▱ ABCD = △ABC + △ACD.
Area of a triangle =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)

Question 5.
Find the area of the triangle formed by the points by using Heron’s formula.
i) (1, 1), (1, 4) and (5, 1)
ii) (2, 3), (-1,3) and (2, -1)
Answer:
i) (1, 1) (1, 4) and (5, 1)
let A (1, 1) B(l, 4) and C(5, 1) are the vertices then length of sides can be calculated using the formula
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
now

now formula for area of triangle using Heron’s formula = △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
∴ s = \(\frac{3+4+5}{2}\) = \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

ii) let the vertices of given triangle A (2, 3), B (-l, 3) and C (2, -1)

∴ a = 5, b = 4, c = 3 units
now from using Heron’s formula area of triangle
= △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
= \(\frac{5+4+3}{2}\)
= \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.1

10th Class Maths 11th Lesson Trigonometry Ex 11.1 Textbook Questions and Answers

Question 1.
In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.
Answer:
Given that
△ABC is a right angle triangle and Lengths of AB, BC and CA are 8 cm, 15 cm and 17 cm respectively.
Among the given lengths CA is longest.
Hence CA is the hypotenuse in △ABC and its opposite vertex having right angle.
i.e., ∠B = 90°.
With reference to ∠A, we have opposite side = BC = 15 cm
adjacent side = AB = 8 cm
and hypotenuse = AC = 17

sin A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{BC}{AC}\) = \(\frac{15}{17}\)
cos A = \(\frac{\text { Adjacent side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{8}{17}\)
tan A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Adjacent side of } \angle \mathrm{A}}\) = \(\frac{BC}{AB}\) = \(\frac{15}{8}\)
∴ sin A = \(\frac{15}{17}\);
cos A = \(\frac{8}{17}\)
tan A = \(\frac{15}{8}\)

Question 2.
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find, tan Q – tan R.
Answer:

Given that △PQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.
By Pythagoras theorem QR² = PQ² + PR²
(25)² = (7)² + PR²
PR² = (25)² – (7)² = 625 – 49 = 576
PR = √576 = 24 cm
tan Q = \(\frac{PR}{PQ}\) = \(\frac{24}{7}\);
tan R = \(\frac{PQ}{PR}\) = \(\frac{7}{24}\)
∴ tan Q – tan R = \(\frac{24}{7}\) – \(\frac{7}{24}\)
= \(\frac{(24)^{2}-(7)^{2}}{168}\)
= \(\frac{576-49}{168}\)
= \(\frac{527}{168}\)

Question 3.
In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Answer:
Given that ABC is a right angle triangle with right angle at B, and BC = a = 24 units, CA = b = 25 units and ∠BAC = θ.

By Pythagoras theorem
AC² = AB² + BC²
(25)² = AB² + (24)²
AB² = 25² – 24² = 625 – 576
AB² = 49
AB = √49 = 1
With reference to ∠BAC = θ, we have
Opposite side to θ = BC = 24 units.
Adjacent side to θ = AB = 7 units.
Hypotenuse = AC = 25 units.
Now
cos θ = \(\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{7}{25}\)
tan θ = \(\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}\) = \(\frac{BC}{AB}\) = \(\frac{24}{7}\)
Hence cos θ = \(\frac{7}{25}\) and tan θ = \(\frac{24}{7}\)

Question 4.
If cos A = \(\frac{12}{13}\), then find sin A and tan A.
Answer:
From the identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\left(\frac{12}{13}\right)^{2}\)
= 1 – \(\frac{144}{169}\)
= \(\frac{169-144}{169}\)
= \(\frac{25}{169}\)
∴ sin A = \(\sqrt{\frac{25}{169}}\) = \(\frac{5}{13}\)

∴ sin A = \(\frac{5}{13}\); tan A = \(\frac{5}{12}\)

Question 5.
If 3 tan A = 4, then find sin A and cos A.
Answer:
Given 3 tan A = 4
⇒ tan A = \(\frac{4}{3}\)
From the identify sec² A – tan² A = 1
⇒ 1 + tan² A = sec² A

If cos A = \(\frac{3}{5}\) then from
sin² A + cos² A = 1
We can write sin²A = 1 – cos²A
= 1 – \(\left(\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}\)
⇒ sin² A = \(\frac{16}{25}\)
⇒ sin A = \(\frac{4}{5}\)
∴ sin A = \(\frac{4}{5}\); cos A = \(\frac{3}{5}\)

Question 6.
In △ABC and △XYZ, if ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.
Answer:

In the given triangle, cos A = cos X
⇒ \(\frac{AC}{AX}\) = \(\frac{XC}{AX}\)
⇒ AC = XC
⇒ ∠A = ∠X (∵ Angles opposite to equal sides are also equal)

Question 7.
Given cot θ = \(\frac{7}{8}\), then evaluate

Answer:

cot² θ = (cot θ)²
= \(\left(\frac{7}{8}\right)^{2}\) = \(\frac{49}{64}\) …… (1)

= sec θ + tan θ
So cot θ = \(\frac{7}{8}\)
⇒ tan θ = \(\frac{8}{7}\)
⇒ tan² θ = \(\left(\frac{8}{7}\right)^{2}\) = \(\frac{64}{49}\)
From sec² θ – tan² θ = 1
⇒ 1 + tan² θ = sec² θ

Question 8.
In a right angle triangle ABC, right angle is at B, if tan A = √3, then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Answer:
Given, tan A = \(\frac{\sqrt{3}}{1}\)
Hence \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{\sqrt{3}}{1}\)
Let opposite side = √3k and adjacent side = 1k

In right angled △ABC,
AC² = AB² + BC²
(By Pythagoras theorem)
⇒ AC² = (1k)² + (√3k)²
⇒ AC² = 1k² + 3k²
⇒ AC² = 4k²
∴ AC = \(\sqrt{4 k^{2}}\) = 2k

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)

Sl.No.	Electron Configuration
1.	1s2 2s2 2p6 3s2 3p3
2.	1s2 2s2 2p6 3s2 3p6 4s2
3.	1s2 2s2 2p6 3s2 3p6

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:

Emission spectrum	Absorption spectrum
1) The spectrum produced by emitted radiation is called emission spectrum.	1) The spectrum produced by absorption of radiation is called absorption spectrum.
2) The emission spectrum contains bright lines on the dark back-ground.	2) The absorption spectrum contains dark lines on the bright back-ground.
3) The emission spectrum corresponds the radiation emitted when an excited electron comes back to the ground state.	3) The absorption spectrum corresponds the radiation absorbed in exciting an electron from lower to higher energy levels.

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:

Line spectrum	Band spectrum
1) The spectrum has sharp distinct lines.	1) The spectrum has many closed, spaced lines.
2) The spectrum is characteristic of atoms and is also called atomic spectrum.	2) The spectrum is characteristic of molecules is also called molecular spectrum.
3) The spectrum is given by inert gases, metal vapours and atomised non-metals.	3) The spectrum is given by hot metals and molecular non-metals.

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.

Orbital	n + l Values
4s	4 + 0 = 4
3d	3 + 2 = 5

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.

Oprbital	n + l value
4d	4 + 3 = 7
5p	5 + 1 = 6
5s	5 + 0 = 5

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?

n	l	ml	ms
3	0	0	+½

Answer:
The quantum numbers of an atom is given below.

n	l	ml	ms
3	0	0	+½

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

Orbital	ml values	Total
4s	0	1
4p	-1, 0, + 1	3
4d	-2, -1, 0, + 1, +2	5
4f	-3, -2, -1, 0, +1, +2, +3	7
Total		16

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)

n	l	ml
1	0	0
2	0	0
	1	-1, 0, +1
3	0	0
	1	-1, 0, +1
	2	-2, -1, 0, +1, +2

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 Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.4

10th Class Maths 11th Lesson Trigonometry Ex 11.4 Textbook Questions and Answers

Question 1.
Evaluate the following:
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
Answer:
Given (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

ii) (sin θ + cos θ)² + (sin θ – cos θ)²
Answer:
Given (sin θ + cos θ)² + (sin θ – cos θ)²
= (sin² θ + cos² θ + 2 sin θ cos θ) + (sin² θ + cos² θ – 2 sin θ cos θ) [∵ (a + b)² = a² + b² + 2ab
(a – b)² = a² + b² – 2ab]
= 1 + 2 sin θ cos θ + 1 – 2 sin θ cos θ [∵ sin² θ + cos² θ = 1]
= 1 + 1
= 2

iii) (sec² θ – 1) (cosec² θ – 1)
Answer:
Given (sec² θ – 1) (cosec² θ – 1)
= tan² θ × cot² θ [∵ sec² θ – tan² θ = 1; cosec² θ – cot² θ = 1]
= tan² θ × \(\frac{1}{\tan ^{2} \theta}\) = 1

Question 2.
Show that (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:

Question 3.
Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Answer:
Given that L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)
Rationalise the denominator, rational factor of 1 – sin A is 1 + sin A.

[∵ (a + b)(a + b) = (a + b)²]; (a – b)(a + b) = a² — b²]
= \(\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)
= \(\frac{1+\sin A}{\cos A}\)
= \(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A = R.H.S.

Question 4.
Show that \(\frac{1-\tan ^{2} A}{\cot ^{2} A-1}\) = tan² A
Answer:

Question 5.
Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ – sin θ.
Answer:

Question 6.
Simplify sec A (1 – sin A) (sec A + tan A)
Answer:
L.H.S. = sec A (1 – sin A) (sec A + tan A)
= (sec A – sec A . sin A) (sec A + tan A)
= (sec A – \(\frac{1}{\cos A}\) . sin A) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec² A – tan² A [∵ sec² A – tan² A = 1]
= 1

Question 7.
Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Answer:
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A . cosec A) + (cos² A – sec² A + 2 cos A . sec A) [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + cosec² A + 2 sin A . \(\frac{1}{\sin A}\) + sec² A + 2 cos A . \(\frac{1}{\cos A}\)
[∵ \(\frac{1}{\sin A}\) = cosec A; \(\frac{1}{\cos A}\) = sec A]
= 1 +(1 + cot² A) + 2 + (1 + tan² A) + 2
[∵ sin² A + cos² A = 1; cosec² A = 1 + cot² A; sec² A = 1 + tan² A]
= 7 + tan² A + cot² A
= R.H.S.

Question 8.
Simplify (1 – cos θ) (1 + cos θ) (1 + cot² θ)
Answer:
Given that
(1 – cos θ) (1 + cos θ) (1 + cot² θ)
= (1 – cos² θ) (1 + cot² θ)
[∵ (a – b) (a + b) = a² – b²]
= sin² θ. cosec² θ [∵ 1 – cos² θ = sin² θ; 1 + cot² θ = cosec² θ]
= sin² θ . \(\frac{1}{\sin ^{2} \theta}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

Question 9.
If sec θ + tan θ = p, then what is the value of sec θ – tan θ?
Answer:
Given that sec θ + tan θ = p ,
We know that sec² θ – tan² θ = 1
sec² θ – tan² θ = (sec θ + tan θ) (sec θ – tan θ)
= p (sec θ – tan θ)
= 1 (from given)
⇒ sec θ – tan θ = \(\frac{1}{p}\)

Question 10.
If cosec θ + cot θ = k, then prove that cos θ = \(\frac{k^{2}-1}{k^{2}+1}\)
Answer:
Method-I:
Given that cosec θ + cot θ = k
R.H.S. = \(\frac{k^{2}-1}{k^{2}+1}\)

Method – II:
Given that cosec θ + cot θ = k ……..(1)
We know that cosec² θ – cot² θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1 [∵ a² – b² = (a -b)(a + b)]
⇒ k (cosec θ – cot θ) = 1
⇒ (cosec θ – cot θ) = \(\frac{1}{k}\)
By solving (1) and (2)

According to identity cos² θ + sin² θ = 1

Hence proved.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1

10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers

Question 1.
In △PQR, ST is a line such that \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and also ∠PST = ∠PRQ.
Prove that △PQR is an isosceles triangle.
Answer:
Given : In △PQR,
\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST= ∠PRQ.

R.T.P: △PQR is isosceles.
Proof: \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)
Hence, ST || QR (Converse of Basic proportionality theorem)
∠PST = ∠PQR …….. (1)
(Corresponding angles for the lines ST || QR)
Also, ∠PST = ∠PRQ ……… (2) given
From (1) and (2),
∠PQR = ∠PRQ
i.e., PR = PQ
[∵ In a triangle sides opposite to equal angles are equal]
Hence, APQR is an isosceles triangle.

Question 2.
In the given figure, LM || CM and LN || CD. Prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Answer:
Given : LM || CB and LN || CD In △ABC, LM || BC (given) Hence,

Adding ‘1’ on both sides.

From (1) and (2)
∴ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Question 3.
In the given figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\).

Answer:
In △ABC, DE || AC
Hence \(\frac{BE}{EC}\) = \(\frac{BD}{DA}\) ………. (1)
[∵ A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]
Also in △ABE, DF || AE
Hence \(\frac{BF}{FE}\) = \(\frac{BD}{DA}\) ………. (2)
From (1) and (2), \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\) Hence proved.

Question 4.
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).
Answer:

Given: In △ABC; D is the mid-point of AB.
A line ‘l’ through D, parallel to BC, meeting AC at E.
R.T.P: E is the midpoint of AC.
Proof:
DE || BC (Given)
then
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)(From Basic Proportional theorem)
Also given ‘D’ is mid point of AB.
Then AD = DB.
⇒ \(\frac{AD}{DB}\) = \(\frac{DB}{DB}\) = \(\frac{AE}{EC}\) = 1
⇒ AE = EC
∴ ‘E’ is mid point of AC
∴ The line bisects the third side \(\overline{\mathrm{AC}}\).
Hence proved.

Question 5.
Prove that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)
Answer:
Given: △ABC, D is the midpoint of AB and E is the midpoint of AC.
R.T.P : DE || BC.
Proof:

Since D is the midpoint of AB, we have AD = DB ⇒ \(\frac{AD}{DB}\) = 1 ……. (1)
also ‘E’ is the midpoint of AC, we have AE = EC ⇒ \(\frac{AE}{EC}\) = 1 ……. (2)
From (1) and (2)
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.
∴ DE || BC by Basic proportionality theorem.
Hence proved.

Question 6.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Answer:
Given: △PQR, DE || OQ; DF || OR
R.T.P: EF || QR
Proof:
In △POQ;
\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ……. (1)
[∵ ED || QO, Basic proportionality theorem]
In △POR; \(\frac{PF}{FR}\) = \(\frac{PD}{DO}\) ……. (2) [∵ DF || OR, Basic Proportionality Theorem]
From (1) and (2),
\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)
Thus the line \(\overline{\mathrm{EF}}\) divides the two sides PQ and PR of △PQR in the same ratio.
Hence, EF || QR. [∵ Converse of Basic proportionality theorem]

Question 7.
In the given figure A, B and C are points on OP, OQ and OR respec¬tively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:
Given:
In △PQR, AB || PQ; AC || PR
R.T.P : BC || QR
Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……… (1)
(∵ Basic Proportional theorem)
and in △OPR, Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……… (2)
From (1) and (2), we can write
\(\frac{OB}{BQ}\) = \(\frac{OC}{CR}\)
Then consider above condition in △OQR then from (3) it is clear.
∴ BC || QR [∵ from converse of Basic Proportionality Theorem]
Hence proved.

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that\(\frac{AO}{BO}\) = \(\frac{CO}{DO}\).
Answer:
Given: In trapezium □ ABCD, AB || CD. Diagonals AC, BD intersect at O.
R.T.P: \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\)

Construction:
Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In △ACD, EO || CD
∴ \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) …….. (1)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In △ABD, EO || AB
Hence, \(\frac{DE}{AE}\) = \(\frac{DO}{BO}\)
[∵ Basic proportionality theorem]
\(\frac{BO}{DO}\) = \(\frac{AE}{ED}\) …….. (2) [∵ Invertendo]
From (1) and (2),
\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)
⇒ \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) [∵ Alternendo]

Question 9.
Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
Answer:

Steps of construction:

1. Draw a line segment \(\overline{\mathrm{AB}}\) of length 7.2 cm.
2. Construct an acute angle ∠BAX at A.
3. Mark off 5 + 3 = 8 equal parts (A₁, A₂, …., A₈) on \(\stackrel{\leftrightarrow}{\mathrm{AX}}\) with same radius.
4. Join A₈ and B.
5. Draw a line parallel to \(\stackrel{\leftrightarrow}{\mathrm{A}_{8} \mathrm{~B}}\) at A₅ meeting AB at C.
6. Now the point C divides AB in the ratio 5:3.
7. Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.

 

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.1

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers

Question 1.
Fill in the blanks.
i) A tangent to a circle intersects it in ——— point(s). (one)
ii) A line intersecting a circle in two points is called a ———. (secant)
iii) The number of tangents drawn at the end of the diameter is ———. (two)
iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)
v) We can draw ——— tangents to a given circle. (infinite)

Question 2.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.
Answer:
Given: A circle with centre O and radius OP = 5 cm
\(\overline{\mathrm{PQ}}\) is a tangent and OQ = 12 cm
We know that ∠OPQ = 90°
Hence in △OPQ
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
12² = 5² + PQ²
∴ PQ² = 144 – 25 .
PQ² = 119
PQ = √119

Question 3.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer:
Steps:

1. Draw a circle with some radius.
2. Draw a chord of the circle.
3. Draw a line parallel to the chord intersecting the circle at two distinct points.
4. This is secant of the circle (l).
5. Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.

Question 4.
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.
Answer:
Given: A circle with radius OP = 9 cm
A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm
Now in △POQ, ∠P = 90°
OP² + PQ² – OQ²
9² + PQ² = 15²
PQ² = 15² – 9²
PQ² = 144
∴ PQ = √144 = 12 cm.
Hence the length of the tangent =12 cm.

Question 5.
Prove that the tangents to a circle at the end points of a diameter are parallel.
Answer:
A circle with a diameter AB.
PQ is a tangent drawn at A and RS is a tangent drawn at B.
R.T.P: PQ || RS.
Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.
∴ OA ⊥ PQ ……….(1)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
Similarly, OB ⊥ RS ……….(2)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
But, OA and OB are the parts of AB.
i.e., AB ⊥ PQ and AB ⊥ RS.
∴ PQ || RS.
O is the centre, PQ is a tangent drawn at A.
∠OAQ = 90°
Similarly, ∠OBS = 90°
∠OAQ + ∠OBS = 90° + 90° = 180°
∴ PQ || RS.
[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers

Question 1.
A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road.
Answer:

Let the height of the tower = h mts say
Width of the road be = x m.
Distance between two points of observation = 10 cm.
Angles of elevation from the two points = 60° and 30°.
From the figure
tan 60° = \(\frac{h}{x}\)
√3 = \(\frac{h}{x}\)
⇒ h = √3x …….(1)
Also tan 30° = \(\frac{h}{10+x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{10+x}\)
⇒ h = \(\frac{10+x}{\sqrt{3}}\) ………(2)
From equations (1) and (2) h
h = √3x = \(\frac{10+x}{\sqrt{3}}\)
∴ √3x = \(\frac{10+x}{\sqrt{3}}\)
√3 × √3x = 10 + x
⇒ 3x – x = 10
⇒ 2x = 10
⇒ x = \(\frac{10}{2}\) = 5m
∴ Width of the road = 5 m
Now Height of the tower = √3x = 5√3 m.

Question 2.
A 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.
Answer:

Height of the temple = 30 m
Height of the man = 1.5 m
Initial distance between the man and temple = d m. say
Let the distance walked = x m.
From the figure
tan 30° = \(\frac{30-1.5}{d}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{d}\)
∴ d = 28.5 × √3m ………(1)
Also tan 60° = \(\frac{28.5}{d-x}\)
⇒ √3 = \(\frac{28.5}{d-x}\)
⇒ √3(d-x) = 28.5
⇒ √3(28.5 × √3-x) = 28.5
⇒ 28.5 × 3 – √3x = 28.5
⇒ √3x = 3 × 28.5-28.5
⇒ √3x = 2 × 28.5 = 57
∴ x = \(\frac{57}{\sqrt{3}}=\frac{19 \times 3}{\sqrt{3}}\) = 19√3
= 19 × 1.732
= 32.908 m.
∴ Distance walked = 32.908 m.

Question 3.
A statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.
Answer:

Height of the pedestal = 2 m.
Let the height of the statue = h m. Angle of elevation of top of the statue = 60°.
Angle of elevation of top of the pedestal = 45°.
Let the distance between the point of observation and foot of the pedestal = x m.
From the figure
tan 45° = \(\frac{2}{x}\)
1 = \(\frac{2}{x}\)
∴ x = 2 m.
Also tan 60° = \(\frac{2+h}{x}\)
⇒ √3 = \(\frac{2+h}{x}\)
⇒ 2√3 = 2 + h
⇒ h = 2√3 – 2
= 2(√3-1)
= 2(1.732 – 1)
= 2 × 0.732
= 1.464 m.

Question 4.
From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower.
Answer:

Angle of elevation of the top of the tower = 60°.
Angle of depression to the foot of the tower = 45°.
Distance between tower and building = 7 m.
Let the height of the building = x m and tower = y m.
From the figure
tan 45° = \(\frac{x}{7}\)
1 = \(\frac{x}{7}\)
∴ x = 7 m.
Also tan 60° = \(\frac{y-x}{7}\)
⇒ √3 = \(\frac{y-x}{7}\)
⇒ 7√3 = y – 7
∴ y = 7 + 7√3
= 7 (√3 + 1)
= 7(1.732 + 1)
= 2.732 × 7
= 19.124 m.

Question 5.
A wire of length 18 m had been tied with electric pole at an angle of eleva¬tion 30° with the ground. As it is covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ?
Answer:

Length of the wire = 18 m
Let the length of the wire removed = x
Height of the pole be = h
From the figure
sin 30° = \(\frac{h}{18}\)
⇒ \(\frac{1}{2}\) = \(\frac{h}{18}\)
⇒ h = \(\frac{18}{2}\) = 9 m
Also sin 60° = \(\frac{h}{18-x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{9}{18-x}\)
√3(18-x) = 9 × 2
18√3 – √3x = 18
√3x = 18√3 – 18
√3x = 18(√3-1)
x = \(\frac{18(\sqrt{3}-1)}{\sqrt{3}}\)
= \(\frac{6 \times 3(\sqrt{3}-1)}{\sqrt{3}}\)
= 6√3(√3-1)
= 6(3-√3)
= 18 – 6√3
= 18 – 10.392
= 7.608 m.

Question 6.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.
Answer:

Height of the tower = 30 m
Angle of elevation of the top of the tower = 60°.
Angle of elevation of the top of the building = 30°.
Let the distance between the foot of the tower and foot of the building be d m and height of the building be x m.
From the figure
tan 60° = \(\frac{30}{d}\)
√3 = \(\frac{30}{d}\)
⇒ d = \(\frac{30}{\sqrt{3}}=\frac{10 \times 3}{\sqrt{3}}\) = 10√3m
Also tan 30° = \(\frac{x}{d}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{x}{10 \sqrt{3}}\)
⇒ x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10 m
∴ Height of the building = 10 m

Question 7.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answer:

Width of the road = 120 f.
Angle of elevation of the top of the 1st tower = 60°.
Angle of elevation of the top of the 2 tower = 30°.
Let the distance of the point from the 1st pole = x.
Then the distance of the point from
the 2nd pole = 120 – x.
and height of each pole = h say.
From the figure
tan 60° = \(\frac{h}{x}\)
⇒ √3 = \(\frac{h}{x}\)
⇒ h = √3x ……..(1)
Also tan 30° = \(\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ h = \(\frac{120-x}{\sqrt{3}}\)
From (1) and (2)
√3x = \(\frac{120-x}{\sqrt{3}}\)
⇒ √3.√3x = 120-x
⇒ 3x = 120 – x
⇒ 3x + x = 120
⇒ 4x = 120
⇒ x = \(\frac{120}{4}\) = 30 ft
Now h = √3x = √3 × 30 = 1.732 x 30 = 51.960 feet
∴ Distances of the poles = 30 ft. and 120 – 30 fts = 90 ft.
Height of each pole = 51.96 ft.

Question 8.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.
Answer:

Let the height of the tower = h m.
Angles of elevation of the top of the tower from two points = x° and (90° – x)
From the figure
tan x = \(\frac{h}{4}\) ……. (1)
Also tan (90° – x) = \(\frac{h}{9}\)
⇒ cot x = \(\frac{h}{9}\)
⇒ \(\frac{1}{\tan x}\) = \(\frac{h}{9}\)
∴ tan x = \(\frac{9}{h}\) …….. (2)
From (1) and (2)
tan x = \(\frac{h}{4}\) = \(\frac{9}{h}\)
∴ \(\frac{h}{4}\) = \(\frac{9}{h}\)
h × h = 9 × 4
⇒ h² = 36
⇒ h = 6 m

Question 9.
The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)
Answer:

Height of the plane from the ground PM = RN = 1500√3 m.
Angle of elevation are 30° and 60°.
From the figure
tan 60° = \(\frac{PM}{QM}\)
√3 = \(\frac{1500 \sqrt{3}}{\mathrm{QM}}\)
QM = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m
Also tan 30° = \(\frac{RN}{QN}\)
\(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{\mathrm{QM}+\mathrm{MN}}\)
QM + MN = 1500√3 × √3
1500 + MN = 1500 × 3
MN = 4500 – 1500
MN = 3000 mts.
∴ Distance travelled in 15 seconds = 3000 mts.
∴ Speed of the jet plane = \(\frac{\text { distance }}{\text { time }}=\frac{3000}{15}\) = 200 m/s
= 200 × \(\frac{18}{5}\) kmph
= 720 kmph
Speed = 200 m/sec. or 720 kmph.

AP SSC 10th Class Maths Textbook Solutions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.2

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers

Question 1.
Choose the correct answer and give justification for each.
(i) The angle between a tangent to a circle and the radius drawn at the point of contact is
a) 60°
b) 30°
c) 45°
d) 90°
Answer: [ d ]
If radius is not perpendicular to the tangent, the tangent must be a secant i.e., 90°.

(ii) From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is
a) 7 cm
b) 12 cm
c) 15 cm
d) 24.5 cm
Answer: [ a ]

O – centre of the circle
OP – a circle radius = ?
OQ = 25 cm
PQ = 24 cm
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
25² = OP² + 24²
OP² = 625 – 576
OP² = 49
OP = √49 = 7 cm.

iii) If AP and AQ are the two tangents a circle with centre O, so that ∠POQ = 110°. Then ∠PAQ is equal to

a) 60°
b) 70°
c) 80°
d) 90°
Answer: [ b ]
In □ OPAQ,
∠OPA = ∠OQA = 90°
∠POQ = 110°
∴ ∠O + ∠P + ∠A + ∠Q = 360°
⇒ 90° + 90° + 110° + ∠PAQ – 360°
⇒ ∠PAQ = 360° – 290° = 70°

iv) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
a) 50°
b) 60°
c) 70°
d) 80°
Answer: [None]

If ∠APB = 80°
then ∠AOB = 180° – 80° = 100°
[∴ ∠A + ∠B = 90° + 90° = 180°]

v) In the figure XY and XV are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XV at B then ∠AOB =
a) 80°
b) 100°
c) 90°
d) 60°

Answer: [ c ]

Question 2.
Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Given: Two circles of radii 3 cm and 5 cm with common centre.

Let AB be a tangent to the inner/small circle and chord to the larger circle.
Let ‘P’ be the point of contact.
Construction: Join OP and OB.
In △OPB ;
∠OPB = 90°
[radius is perpendicular to the tangent]
OP = 3cm OB = 5 cm
Now, OB² = OP² + PB²
[hypotenuse² = Adj. side² + Opp. side², Pythagoras theorem]
5² = 3² + PB²
PB² = 25 – 9 = 16
∴ PB = √l6 = 4cm.
Now, AB = 2 × PB
[∵ The perpendicular drawn from the centre of the circle to a chord, bisects it]
AB = 2 × 4 = 8 cm.
∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Question 3.
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:

Given: A circle with centre ‘O’.
A parallelogram ABCD, circumscribing the given circle.
Let P, Q, R, S be the points of contact.
Required to prove: □ ABCD is a rhombus.
Proof: AP = AS …….. (1)
[∵ tangents drawn from an external point to a circle are equal]
BP = BQ ……. (2)
CR = CQ ……. (3)
DR = DS ……. (4)
Adding (1), (2), (3) and (4) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + DC = AD + BC
AB + AB = AD + AD
[∵ Opposite sides of a parallelogram are equal]
2AB = 2AD
AB = AD
Hence, AB = CD and AD = BC [∵ Opposite sides of a parallelogram]
∴ AB = BC = CD = AD
Thus □ ABCD is a rhombus (Q.E.D.)

Question 4.
A triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively (See below figure). Find the sides AB and AC.

Answer:
The given figure can also be drawn as

Given: Let △ABC be the given triangle circumscribing the given circle with centre ‘O’ and radius 3 cm.
i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.
It is given that BD = 9 cm
CD = 3 cm

∵ Lengths of two tangents drawn from an external point to a circle are equal.
∴ BF = BD = 9 cm
CD = CE = 3 cm
AF = AE = x cm say
∴ The sides of die triangle are
12 cm, (9 + x) cm, (3 + x) cm
Perimeter = 2S = 12 + 9 + x + 3 + x
⇒ 2S = 24 + 2x
or S = 12 + x
S – a = 12 + x – 12 = x
S – b = 12 + x – 3 – x = 9
S – c = 12 + x – 9 – x = 3
∴ Area of the triangle

Squaring on both sides we get,
27 (x² + 12x) = (36 + 3x)²
27x² + 324x = 1296 + 9x² + 216x
⇒ 18x² + 108x- 1296 = 0
⇒ x² + 6x – 72 = 0
⇒ x² + 12x – 6x – 72 = 0
⇒ x (x + 12) – 6 (x + 12) = 0
⇒ (x – 6) (x + 12) = 0
⇒ x = 6 or – 12
But ‘x’ can’t be negative hence, x = 6
∴ AB = 9 + 6 = 15 cm
AC = 3 + 6 = 9 cm.

Question 5.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythagoras Theorem.
Answer:
Steps of construction:

1. Draw a circle with centre ‘O’ and radius 6 cm.
2. Take a point P outside the circle such that OP =10 cm. Join OP.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM or MO as radius draw a circle. Let the circle intersects the given circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tan¬gents of lengths 8 cm each.

Proof: In △OAP
OA² + AP² = 6² + 8²
= 36 + 64 = 100
OP² = 10² = 100
∴ OA² + AP² = OP²
Hence AP is a tangent.
Similarly BP is a tangent.

Question 6.
Construct, a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer:
Steps of construction:

1. Draw two concentric circles with centre ‘O’ and radii 4 cm and 6 cm.
2. Take a point ‘P’ on larger circle and join O, P.
3. Draw the perpendicular bisector of OP which intersects it at M.
4. Taking M as centre and PM or MO as radius draw a circle which intersects smaller circle at Q.
5. Join PQ, which is a tangent to the smaller circle.

Question 7.
Draw a circle with the help of a bangle, take a point outside the circle. Con-struct the pair of tangents from this point to the circle measure them. Write conclusion.
Answer:
Steps of construction:

1. Draw a circle with the help of a bangle.
2. Draw two chords AB and AC. Perpendicular bisectors of AB and AC meets at ‘O’ which is the centre of the circle.
3. Taking an outside point P, join OP.
4. Let M be the midpoint of OP. Taking M as centre OM as radius, draw a circle which intersects the given circle at R and S. Join PR, PS which are the required tangents.

Conclusion: Tangents drawn from an external point to a circle are equal.

Question 8.
In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.
Answer:
Let ABC be a right triangle right angled at P.
Consider a circle with diametere AB.
From the figure, the tangent to the circle at B meets BC in Q.
Now QB and QP are two tangents to the circle from the same point P.
QB = QP …….. (1)
Also, ∠QPC = ∠QCP
∴ PQ = QC (2)
From (1) and (2);
QB = QC Hence proved.

Question 9.
Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point? [Hint: The distance of two points to the point of contact is the same.
Answer:
Only two tangents can be drawn from a given point outside the circle.

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 Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 6 Refraction of Light at Curved Surfaces.

AP State Syllabus SSC 10th Class Physics Important Questions 6th Lesson Refraction of Light at Curved Surfaces

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 1 Mark Important Questions and Answers

Question 1.
Suppose you are inside the water in a swimming pool. Your friend is standing on the edge. Do you find your friend taller or shorter than his actual height? Why? (AP June 2018)
Answer:
Friend is seemed to be taller. Because of refraction of light.

Question 2.
What happens to the image, if a convex lens is made up of two different transparent materials as shown in figure? (TS March 2016)

Answer:
The convex lens is made up of two different materials. So the refractive i these two materials will be different. Hence two images will be formed.

Question 3.
Write the list of materials required for the experiment to find the focal length of a convex lens. (TS June 2017)
Answer:
Convex Lens, Scale, Piece of paper, Sunrays.
(OR)
Convex Lens, V-Stand, Candle, Match box, Screen, Scale.

Question 4.
Complete the following ray diagram. (TS March 2019)

Answer:

The parallel rays coming with some angle to principal axis meet on focal plane.

Question 5.
If the object is placed between the focal point and the optical centre of a convex lens, what will be the characteristics of the image formed? (AP SA-1:2019-20)
Answer:
Object is placed between F₂ and optic centre P :

Nature :
Virtual, erect and magnified.

Position :
Same side of the lens where object is placed.

Question 6.
For a concave lens, what type of image will be formed if the object is placed at the centre of curvature? (AP SA-1:2019-20)
Answer:
Same size of object, inverted and real image will be formed.

Question 7.
Write lens formula. (AP SA-I:2019-20)
Answer:
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Question 8.
What is a lens? (or) Define lens.
Answer:
A lens is formed when a transparent material is bounded by two spherical surfaces.

Question 9.
What is a double convex lens?
The lens having two spherical surfaces bulging outwards is called double convex lens.

Question 10.
What about the thickness of double convex lens?
Answer:
It is thick at the middle as compared to edges.

Question 11.
What is a double concave lens?
Answer:
The lens having two spherical surfaces curved inward is called a double concave lens.

Question 12.
Write about the thickness of concave lens.
Answer:
It is thin at the middle and thicker at the edges.

Question 13.
What is centre of curvature?
Answer:
The centre of sphere which contains the part of curved surface is called centre of curvature.

Question 14.
What is radius of curvature?
Answer:
The distance between the centre of curvature and curved surface is called radius of curvature.

Question 15.
What is the mid point of lens called?
Answer:
The mid point of lens is called pole (or) optical centre.

Question 16.
What is a focus?
Answer:
The point where rays converge or the point from which rays seem emanate is called focal point (or) focus.

Question 17.
What is the distance between pole and focal point called?
Answer:
Focal length.

Question 18.
What happens if the ray passes through principal axis?
Answer:
It will be undeviated.

Question 19.
Where do light rays travelling to principal axis converge?
Answer:
They converge at focus.

Question 20.
What happens to light rays passing through focus?
Answer:
The path of the rays is parallel to principal axis after refraction.

Question 21.
What is a focal plane?
Answer:
A plane which is perpendicular to principal axis at the focus is called focal plane.

Question 22.
What is lens formula?
Answer:
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Question 23.
On what factor does focal length of a lens depend?
Answer:
It depends on refractive index of the medium, object distance and image distance.

Question 24.
What is lens maker formula?
Answer:
\(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)

Question 25.
What happens to be image formed by a convex lens if its lower part is blackened?
Answer:
Every part of a lens forms a complete image. If the lower part of the lens is blackened the complete image will be formed but its intensity will be decreased.

Question 26.
From which point of lens are all the distances are measured?
Answer:
The optical centre of lens.

Question 27.
Is it possible for a lens to Act as a convergent lens in are medium and a divergent lens in another?
Answer:
Yes. A convergent lens is placed in a higher refractive index of medium the nature of the lens changes i.e., it acts as divergent lens.

Question 28.
What are paraxial rays?
Answer:
The rays which move very close to the principal axis which can be treated as parallel are called paraxial rays.

Question 29.
What is absolute refractive index?
Answer:
It is the ratio of speed of light in air to speed of light in any medium.

Question 30.
Give mathematic expression for power lens and explain the terms in the formula.
Answer:
Power (P) = \(\frac{1}{f}\)
where f is focal length of lens.

Question 31.
If the size of image is same as object through a convex lens, then where is the object placed?
Answer:
The object is placed at centre of curvature.

Question 32.
How will you identify a concave lens by touching it?
Answer:
A concave lens is thinner at centre and thicker at edges.

Question 33.
How will you identify a convex lens by touching it?
Answer:
A convex lens is thicker at centre and thinner at edges.

Question 34.
Give the sign conventions for lenses with regard to the object and image distance.
Answer:
The distance measured in the direction of incident ray is taken as positive.
The distance measured against the direction of incident ray is taken as negative.

Question 35.
Give the sign conventions for lenses with regard to the height of objects and images.
Answer:
All the heights of objects and images above principal axis are positive and below the axis are negative.

Question 36.
When light of two colours A and B passes through a plane boundary, A is bent more than B. Which colour travels more slowly in the second medium?
Answer:
Colour A travels slowly.

Question 37.
What type of lens behaviour will an air bubble inside water show?
Answer:
It will act as a concave lens.

Question 38.
Is it possible for a lens to act as a convergent lens in one medium and a divergent lens in another?
Answer:
Yes. A lens is placed in a medium of a high refractive index than that of the lens then nature of lens changes (M_L > M_g).

Question 39.
The image formed by a lens is always erect and diminished. What is the nature of lens?
Answer:
Given that the lens is forming an image which is always erect and diminished. So it is virtual also. Such type of image is formed by concave lens.

Question 40.
If a student observed an image of same size with a convex lens of focal length 20 cm, then where should he keep the object in front of lens?
Answer:
Because the student got image of same size the object should be placed at a distance of twice the focal length, i.e. 40 cm.

Question 41.
For an object placed at a distance of 20 cm in front of convex lens, the image formed is at a distance of 20 cm behind the lens. Find the focal length of lens.
Answer:
The object distance and image distance are same. So the object is kept at twice the focal length. So the focal length of the convex lens is 10 cm.

Question 42.
A doctor suggested spectacles for a student which has negative focal length. Which type of lens is that?
Answer:
Focal length negative indicates that it is a concave lens.

Question 43.
What happens to a light ray which passes through optical centre?
Answer:
The light ray which passes through optical centre does not deviate.

Question 44.
When do you get image at infinity with a convex lens?
Answer:
When the object is at the focal point.

Question 45.
When do you get a virtual image with a convex lens?
Answer:
When the object is placed between focus and pole.

Question 46.
Is focal length of a lens zero? If not, why?
Answer:
No, focal length of lens never equals to zero because it is the distance between focal point and optical centre.

Question 47.
A thin lens has a focal length of 12 cm. Is it a convex leps or a concave lens?
Answer:
It is a convex lens, because f is positive.

Question 48.
Name the different apparatus where we are using the convex and concave lenses.
Answer:
The magnifying lense, telescope, microscope.

Question 49.
Draw the given diagram in your answer book and complete it for the path of ray of light beyond the lens.

Answer:

Question 50.
The diagram below shows two incident rays P and Q which emerge as parallel rays R and S respectively. The appropriate device used in the box is ……..

Answer:
The rays are diverging and they produced the parallel from the device after refraction. So the device is concave lens.

Question 51.
The following figure shows the incident and refracted rays pass through a lens kept in the box. Draw the lens and complete the path of rays.

Answer:
The incident rays 1 and 2 have converged after refraction. So the lens is convex.

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 2 Marks Important Questions and Answers

Question 1.
A convex lens is made of five different materials as shown in the figure. How many images does it form? Why? (AP March 2017)

Answer:
The given convex lens is made up of five different materials.
So they have different refractive indices / different focal lengths.
Hence they form five different images.

Question 2.
The focal length of a converging lens is 20cm. An object is 60cm from the lens. Where will be image be formed and write characteristics of the image. (AP March 2018)
Answer:
Focal length = f = 20 cm (+ 20cm)
Object distance = u = 60 cm (- 60cm)
Image distance = v = ?

Here, (-) indicates inverted images.
m = \(\frac{1}{2}\) < 1 indicates diminished image.
Image forms between F₁ and 2F₁.
Characteristics of the image :

1. real
2. inverted
3. diminished.

Question 3.
When a light rays enters a medium with refractive index n2 from a medium with refractive index n, at curved interface with radius of curvature R is given by
\(\frac{\mathbf{n}_{2}}{\mathbf{v}}-\frac{\mathbf{n}_{1}}{\mathbf{u}}=\frac{\mathbf{n}_{2}-\mathbf{n}_{1}}{\mathbf{R}}\)
Now assume that the surface is plane and rewrite the formula with suitable changes.
Answer:
Assume that the interface is plane surface
Then R becomes infinity
R = ∞ (or) R = 1/0
Substitute the above value in the given equation

Question 4.
Two convex lenses of same focal length are fixed in a PVC pipe at a distance double to their focal length. What happens if a boy sees the moon with that arrangement? (TS March 2017)
Answer:

• The rays coming from moon are parallel. The first lens converges the rays at focus.
• The converging point is the focus of second lens. So the second lens convert the diverging rays into parallel.
• Hence, in the rays of moon, there will be no change when we see moon with this arrangement or without this arrangement.

(OR)

This arrangement does not make any difference in the rays coming from moon.
The moon appears same if we see directly or with this arrangement.

Question 5.
Focal length of the lens depends on its surrounding medium. What happens, if we use a liquid as surrounding media of refractive index, equal to the refractive index of lens? (TS June 2018)
Answer:

• When the refractive index of surrounding media is equal to the refractive index of lens, the lens looses its characteristics.
• Lens do not diverge or converge the light.
• Light do not get refracted when it passes through that lens.

Question 6.
Complete the ray diagram given below (TS March 2018)

Answer:

Question 7.
The refractive index of convex lens material is 1.46. The refractive index of Benzene and water is 1.5 and 1.0 respectively. How does the lens behaves when it is kept in Benzene and water? Given and write. (TS March 2018)
Answer:

• When the convex lens with refractive index 1.46 is kept in Benzene with refractive index 1.5, then the lens acts as a diverging lens.
• If the same lens is kept in water whose refractive index is 1, then it acts as a converging lens.

Question 8.
Write the applications of lenses in day to day life. (AP SCERT: 2019-20)
Answer:
Uses of lenses in day to day life :

• Lenses are used for correcting eye defects.
• They are used as magnifying lenses.
• They are used in microscopes, telescopes, binoculars.
• They are used in cinema projectors and cameras.

Question 9.
Water lens is made of double convex lens of radius of curvature “R”. Write lens makers formula for water lens. (AP SA-I: 2019-20)
Answer:
1) Radius of curvatures of water lens are R₁ = R₂ = R and n = 1.5.
2) Sign conversion R₁ = + R₁, R₂ = -R₂.
3) Lens makers formula

Question 10.
Find the focal length of plane convex dens if its radius of curvature is R and its refractive index is n.
Answer:
Given lens is plano-convex lens; radius of curvature = R
Refractive index = n

Question 11.
In a classroom, four friends found out the focal length of a lens by conducting an experiment. The value came out to be 12.1cm, 12.2cm, 12.05 cm, 12.3 cm. The friends discussed the reasons for the differences or defects. Mention those reasons.
Answer:
Students got different focal lengths.

1. By observing the values they got all positive values. This indicates they are given by convex lens.
2. All the students got exact interger but different decimal value.

Reasons :
The difference in values is due to least count errors, parallax errors, random errors and systematic errors, etc.

Question 12.
How will you decide whether a given piece of glass is a convex lens, concave lens or a plane plate?
Answer:
Hold the given piece of glass over some printed matter.

1. If the letters appear magnified, then the given piece of glass is convex lens.
2. If the letters appear diminished, then the given piece of glass is concave lens.
3. If the letters appear to be same size, then it is a plane glass piece.

Question 13.
State the type of lens used as a magnifying glass. Draw a labelled ray diagram to show that the image of the object is magnified.
Answer:

A single convex lens is used as a magnifying glass, i.e. for seeing small object magnified. When the object to be seen in between the focus and optical centre of the lens, a vertical, erect magnified image of the object is formed as shown and convex lens is and to act as magnifying glass AB’ is the magnified image of AB.

Question 14.
Give conventions used in lenses.
(OR)
Write the signs of convex and concave lens using in drawing ray diagrams.
Answer:

• All distances are measured from optical centre of the lens.
• Distances measured along the direction of the incident light are taken as positive.
• The distances against the incident light are taken as negative.
• The heights measured vertically above the axis, are taken as positive.
• The heights measured vertically down the axis, are taken as negative.

Question 15.
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Answer:
Yes, the statement is correct.
For magnified virtual image :
The object should be placed between optic centre (C) and focus (F) (< 20)

For magnified real image:
Placed between focus (F) and centre of curvature (2F) (20 – 40)

Question 16.
Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building ? What is the approximate focal length of this lens?
Answer:
As the image is real, therefore the lens use is convex lens. The distance of the real image formed by a convex lens from the lens decreases as the object distance from the lens increases. Hence, the screen has to be moved towards the lens to obtain the sharp image of the building. Approximate focal length of the lens =15 cm as the rays of light coming the window pane are considered to come from infinity. These rays of light are focussed by the convex lens at its focus, (i.e. on the screen).

Question 17.
What do you see when your friend brings a sheet of paper on which arrow was drawn behind the empty cylindrical shaped transparent vessel? Why do you see a diminished image?
Answer:
We will see a diminished image of the arrow.

When the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass and enters air, then it again undergoes refraction curved surface of the vessel and comes out into the air. In this way, light travels in two media, comes out of the vessel and forms a diminished image.

Question 18.
Using the formula of refraction at curved surfaces, write the formula for plane surfaces.
Answer:
For curved surfaces the formula for refraction is \(\frac{\mathrm{n}_{2}}{\mathrm{v}}-\frac{\mathrm{n}_{1}}{\mathrm{u}}=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)}{\mathrm{R}}\)

For plane surface, the radius of curvature (R) approaches infinity. Hence 1/R becomes zero.
\(\frac{\mathrm{n}_{2}}{\mathrm{v}}-\frac{\mathrm{n}_{1}}{\mathrm{u}}=0 \Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{v}}=\frac{\mathrm{n}_{1}}{\mathrm{u}}\)

Question 19.
Explain how a convex lens behaves on converging lens and diverging lens.
Answer:
The convex lens behaves as a converging lens, if it is kept in a medium with refractive index less than the refractive index of the lens. It behaves like a diverging lens when it is kept in transparent medium with greater refractive index than that of lens.
e.g. : Air bubble in water behaves like a diverging lens.

Question 20.
When does Snell’s law fail?
Answer:

• Snell’s law fails when light is incident normally on the surface of refracting medium.
• Both media have same refractive index.

Question 21.
If on applying sign convention for lens the image distance obtained is negative, state the significance of negative sign.
Answer:

• Negative sign of image distance means the image is virtual and erect.
• It is formed on the same side of the object with respect to lens.

Question 22.
Magnification of lens is found to be +2. What type of lens is that?
Answer:
Magnification +ve indicates the image is erect and virtual.
Magnification 2 indicates it is magnified.
Magnified virtual image is formed by only convex lens.

Question 23.
For same angle of incidence in media A, B and C the angle of refractions are 30°, 25° and 20° respectively. In which medium will the velocity of light be minimum?
Answer:
In medium R the velocity of light is minimum because it has greater refractive index. Refractive index and velocity of light in a medium are inversely proportional. So in medium R the velocity is minimum.

Question 24.
The radius of curvature (twice the focal length) of a convex lens is 40 cm. A student wants to get various images of following types (a) enlarged virtual image, (b) enlarged real image, (c) image of same size, (d) diminished image.
In order to get these images where should the object should be kept in front of convex lens?
Answer:
a) The object should be kept in less than 20 cm (i.e. less than focal length).
b) In order to get enlarged real image, object should be kept between 20 cm to 40 cm in front of lens.
c) In order to get image of same size object should be kept at a distance of 40 cm from the lens.
d) In order to get diminished image the object should be kept beyond 40 cm from the lens.

Question 25.
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images.
Answer:

• Yes, the statement is correct.
• A convex lens of focal length 20 cm will produce a magnified virtual image if object is placed at a distance less than 20 cm from the lens.
• A convex lens of focal length 20 cm will produce magnified real image if the object is placed at a distance more than 20 cm and less than 40 cm.

Question 26.
A concave mirror and a convex lensare held in water. What changes, if any, do you expect in their focal length?
Answer:
The focal length of a concave mirror independent of the medium and A convex lens depends on medium when they are placed in water.
The focal length of the mirror – Does not change.
The focal length of the convex lens – Changes (means increases 4 times).

Question 27.
When does a convex lens behave like a diverging lens? Given example.
Answer:
A Convex lens behaves like a diverging lens when it is kept in a tranparent medium with greater refractive index than that of the lens.
Eg : An air bubble in water behaves like a diverging lens.

Question 28.
A pond appears to be shallower than it really is when viewed obliquely. Why?
Answer:

• Suppose two rays are originated from the bottom of the pond.
• As these rays get refracted into air, they bend away from the normal.
• When these two refracted rays produced backwards they seem to meet at a point higher than the bottom of the pond.
• This point gives the apparent position of the bottom of the pond.
• Thus the pond appears to be shallower.

Question 29.
What happens to the image formed by a convex lens if its lower part is blackened?
Answer:

1. Every part of lens forms complete image.
2. If lower part of the lens is blackened, the complete image will be formed.
3. But its intensity will decrease.

Question 30.
Is it possible for a lens to act as a convergent lens in one medium and a divergent lens in another?
Answer:

• Yes, the type of lens changes, if it is placed in medium having higher refractive index that of lens.
• For example, convex lens acts as converging lens when it is placed in a medium of lower refractive index otherwise it behaves like a diverging lens.

Question 31.
Draw the different types of convex and concave lens.
Answer:

Question 32.
Complete the ray diagram and give reason.

Answer:

The light ray which passes through optical centre does not undergo refraction. So it goes straight.

Question 33.
How do you appreciate the refraction at plane surfaces and at curved surfaces?
Answer:
The refraction of curved surfaces are used in various aspects such as

1. In microscope to enlarge microscopic objects.
2. In telescopes to see celestial objects.
3. To correct eye defects like myopia, hypermetropia and presbyopia.

So refraction at curved surfaces is thoroughly appreciated.

Question 34.
An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.
Answer:

Image distance is negative that indicates it is a virtual and erect image.

Question 35.
A bird is flying at the height 3m above the river surface while a fish is 4 m below the surface. At what depth would the fish appear to the bird ? At what height the bird would appear to the fish? (given a/w = 4/3)
Answer:
Given that refractive index of air / water = \(\frac{4}{3}\)
The height the bird would appear to fish = 4 + \(\frac{4}{3}\) × 3 = 4 + 4 = 8m

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 4 Marks Important Questions and Answers

Question 1.
Complete the ray diagram when an object is placed between F₂ and 2F₂. (AP June 2017)

Answer:

Question 2.
An object is placed at the following distances from a convex lens of focal length 10 cm.
(a) 8 cm.
(b) 15 cm.
(c) 20 cm.
(d) 25 cm.
Which position of the object will produce……. (TS March 2015)
(i) a diminished, real and inverted image?
(ii) a magnified, real and inverted image?
(iii) a magnified, virtual and erect image?
(iv) an image of same size as the object?
Justify your answer in each case.
Answer:
i) d (or) 25 cm
Reason : Object placed between centre of curvature and focal point.

ii) b (or) 15 cm
Reason : Object placed beyond centre of curvature.

iii) a (or) 8cm
Reason : Object placed between focal point and optic centre.

iv) c (or) 20 cm
Reason :Object placed at centre of curvature.

Question 3.
The ray diagrams showing the image formed by a convex lens are given in the following table. From these diagrams complete the table. (TS June 2016)

Answer:

Question 4.
Explain the behaviour of light rays in any four situations of their incidence on a convex lens. (TS March 2016)
Answer:
1) A ray passing along the principal axis is undeviated.

2) Any ray passing through optic centre is also undeviated.

3) The rays passing parallel to principal axis converge at focus.

4) The rays passing through the focus will take a path parallel to principal axis.

Question 5.
Draw the ray diagrams to find the images when an object is placed in front of the lens (i) at a distance of 8 cm, and (ii) at a distance of 10 cm on the principal axis of a convex lens whose focal length is 4 cm. Write the characteristics of images in both the cases. (TS June 2017)
Answer:
(i) Ray diagram :

Characteristics of Image :
i) Size of the image equal to the size of the object,
ii) Inverted image,
iii) Real image,
iv) Image formed at C.

(ii) Ray diagram

Characteristics of Image :
i) Image size is less than that of object size,
ii) Inverted image,
iii) Real image,
iv) Image is formed between F & C.

Question 6.
A double concave lens with the refractive index (n) = 1.5 is kept in the air. Its two spherical surfaces have radii R₁ = 20 cm and R₂ = 60 cm. Find the focal length of the lens. Write the characteristics of the lens. (TS March 2017)
Answer:

Hence f = – 30 cm (Here minus indicates that the lense is divergent)
Characteristics of the biconcave lens :

1. It is a diverging lens.
2. It is thin at the middle and thicker at the edges.

Question 7.
Draw ray diagrams for a double concave lens of focal length 4 cm, when objects are placed at 3 cm and 5 cm on principal axis. Write characteristics of images. (TS June 2018)
Answer:
i)

ii)

Characteristics of images :

1. Image formed between P and F
2. Diminished image
3. Errected image
4. Virtual image

Question 8.
Write the characteristics of the images which are formed when objects are placed at 50cm and 75cm on the principle axis of a convex lens with focal length of 25 cm. (TS March 2018)
Answer:
i) Object is placed at 50cm.

Characteristics of the image :

1. Image forms at 2Fp
2. Image is real,
3. Image is inverted,
4. Image is same size

ii) Object is placed at 75cm
f = +25cm; u = -75cm; v = ?

Characteristics of the image :

1. Image forms between F₁ and 2F₁
2. Image is real
3. Image is inverted
4. Image is diminished.

Question 9.
Write the role of lenses in our daily life. (AP March 2019)
Answer:
The role of lenses in our daily life :

1. Used for correcting eye defects.
2. Used as magnifying lens.
3. Used in Microscopes.
4. Used in Telescopes.
5. Used in Binoculars.
6. Used in Cinema Projectors.
7. Used in Cameras.

Question 10.
Draw the ray diagrams for the following positions of objects in front of a convex lens mention the characteristics of the image. (AP SCERT: 2019-20)
a) Object is placed beyond 2F₂.
b) Object is placed between focal point and opint center.
Answer:
a)

b)

a) Characteristics of the Image :

1. real
2. inverted
3. diminished.

b) Characteristics of the Image :
If we place an object between focus and optic centre, we will get an image which is virtual, erect and magnified.

Question 11.
The focal length of a convex lens is 2 cm. Draw the ray diagram of an object placed on principal axis at the ‘C’ of lens and at a distance of 3 cm from its optic centre.
Answer:
1) Object is placed on principal axis of a convex lens at ‘C’.

2) Object is placedon principal axis at a distance of 3 cm from its optic center.

Question 12.
Using biconvex lens, a point image is made on its principal axis S. Let us assume that we know optical centre P and its focus F. We also know PF > PS. Draw the ray diagram to identify the point source and give reasons.
Answer:

Given lens is biconvex lens and given condition is PF > PS’ means image is formed between optic centre (P) and Focus (F).

According to Snell’s law this condition is possible when the object is also placed between P and F. Because reflected rays are divergent.

Question 13.
Write about the behaviour of light rays when they incident on a lens.
Answer:
1) Situation I:
Ray passing through the principal axis.
⇒ It is not deviated.

2) Situation II:
Ray passing through the pole.
⇒ It is also undeviated.

3) Situation III:
Rays travelling parallel to the principal axis.
⇒ They converge at focus or diverge from the focus.

4) Situation IV :
Ray passing through focus.
⇒ It will take a path parallel to principal axis after refraction.

5) Situation V :
Parallel rays fall on a lens making some angle with principal axis.
⇒ They converge at a point or diverge from a point lying on a focal plane.

Question 14.
Write characteristics of image formed due to convex lens at various distances.
Answer:

Question 15.
Write characteristics of image formed by a concave lens at various distances.
Answer:

Question 16.
You are given a convex lens of focal length 10 cm. Where would you place an object to get a real inverted and highly enlarged image of the object? Draw a ray diagram.
Answer:
If an object is placed at the focus of the lens it forms real, inverted and highly enlarged image. Thus, the distance of the object from the optical centre of the lens is equal to the focal length of the lens =10 cm.
The ray diagram is as shown

Question 17.
Derive the formula of image formation in refraction at curved surfaces.
Answer:
1) Object at infinity :
The rays coming from the object at infinity are parallel to principal axis and converge to the focal point after refraction. So, a point-sized image is formed at the focal point.

2) Object placed beyond the centre of curvature on the principal axis :
When an object is placed beyond the centre of curvature 2F₂, a real, inverted and diminished image is formed on the principal axis between F₁ and 2F₁

3) Object placed at the centre of curvature :
When an object is placed at the centre of curvature (2F₂) on the principal axis, a real, inverted image is formed at 2F₁ which is same size as that of the object.

4) Object placed between the centre of curvature and focal point:
When an object is placed between centre of curvature (2F₂) and focus (F2), we will get an image which is real, inverted and magnified. This image will form beyond 2F₁.

5) Object located at focal point:
When an object is placed at focus (F₂), the image will be at infinity.

6) Object placed between focal point and optic centre :
If we place an object between focus and optic centre, we will get an image which in virtual, erect and magnified.

Question 18.
Distinguish between convex lens and concave lens.
Answer:

Convex lens	Concave lens
1. Objects appear to be big in the lens.	1. Objects appear to be shrink in the lens.
2. It generally forms real image, (except object is placed between optical centre and focal point)	2. It always forms virtual image.
3. Light rays tend to converge after refraction from lens.	3. Light rays tend to diverge from lens after refraction.
4. The image due to lens may be enlarged or same size or diminished.	4. The image is always diminished.
5. The image due to lens may be inverted or erect.	5. The image is always erect.
6. It is used to correct hypermetropia.	6. It is used to correct myopia.

Question 19.
The ray diagram given below illustrates the experimental set up for the determination of the focal length of a converging lens using a plane mirror.

1) State the magnification of the image formed.
2) Write the characteristics of the ipiage formed.
3) What is the name given to the distance between the object and optical centre of the lens in the following diagram?
Answer:

1. The magnification of the image formed is unity (or 1).
2. The image is a) real and b) inverted is at the same position of the object.
3. The distance between the object and the optical centre of the lens is called object distance.

Question 20.
A concave lens made of a material of refractive index n₁ is kept in medium of refractive index n₂. A parallel beam of light incident on the lens. Complete the path of rays of light emerging from the concave lens if
i) n₁ > n₂
ii) n₁ = n₂
iii) n₁ < n₂.
Answer:
i) When n₁ > n₂, light goes from rarer to denser medium. Therefore, in passing through a concave lens it diverges.

ii) When n₁ = n₂, there is no change in medium. Therefore no bending or refraction occurs.

iii) When n₁ < n2, light goes from a denser to rarer medium. Therefore, in passing through a concave lens it converges.

Question 21.
One half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:

• Every part of a lens forms an image.
• For formation of image we require only two light rays to converge.
• Therefore, if the lower half of the lens is covered, it will still form a complete image.
• However the intensity of the image will be reduced.
• This can be verified experimentally by observing the image of distant object like tree on a screen, when lower half of the lens covered with a black paper.

Question 22.
Complete the following table if the object is placed at various positions in front of a convex lens.

Position of object	Position of image	Nature of image
1. At infinity
2.	Between F1 and 2F1
3.		Same size, real and inverted
4.	Seen in the lens

Answer:

Position of object	Position of image	Nature of image
1. At infinity	On Fj (focal point)	Highly diminished, real and inverted
2. Beyond 2F2	Between F1 and 2F1	Diminished, real and inverted
3. On 2F1	On 2F2	Same size, real and inverted
4. Between focus and optical centre	Seen in the lens	Enlarged, virtual and erect

Question 23.
A student focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observations.

Distance of flame from the lens (cm)	Distance of the screen from the lens (cm)
1. 60	20
2. 40	24
3. 30	30
4. 24	40
5. 15	70

a) From the above table, find the focal length of lens without using lens formula.
b) Which set of observations is incorrect and why?
c) In which case the size of the object and image will be same? Give reason for your answer.
Answer:
a) From the observations, it is clear that for u = 30, v = 30 cm. This means this value must be equal to twice the focal length of the convex lens.
∴ Focal length of convex lens = 30/2 = 15 cm

b) The observation (5) is not correct because if u = 15 cm i.e., the object is kept at focus so the image should be at infinity and not at 70 cm.

c) For twice the focal length we know size of object = size of image. So when object is kept at 30 cm the size of object and image are same.

Question 24.
Draw the ray diagrams when incident ray striking a convex surface or a concave surface moving from one medium to another medium.
Answer:

Question 25.
Draw ray diagrams of image formed by a convex lens at various distances.
Answer:
1) Object at infinity.

2) Object placed beyond the centre of curvature (2F₂).

3) Object placed at the centre of curvature.

4) Object placed between 2F₂ and F₂.

5) Object at the focal point.

6) Object placed between F and P

Question 26.
Write about the focal length of the lens with diagram.
Answer:

• A parallel beam of light incident on a lens converges to a point as shown in figure (a) or seems to emanate from a point on the principal axis as shown in figure (b).
• The point of convergence (or) the point from which rays seem to emanate is called focal point or focus (F).
• Every lens has two focal points.
• The distance between the focal point and optic centre is called the focal length of lens denoted by ‘f’
• To draw ray diagram for lenses, we need two more points in addition to focal points F₁ and F₂.
• These points are equidistant from centre of the lens and also equal to double the focal length. So we call them 2F₁ and 2F₂.
• For drawing ray diagrams related to lenses, we represent convex lens with a symbol £ and concave lens with J as shown in the figure c and d.

Question 27.
The diagram shows an object OA and its image IB formed by a lens. The image is same size as the object.
a) Complete the ray diagram and locate the focus of lens by labelling it as F.
b) State whether the lens is convex or concave.
Show it in the diagram.

Answer:

a)

1. Optical centre goes undeviated, therefore to find the optical centre, join A to B to meet the line 01 at the point P which gives the position of optical centre of the lens.
2. Draw a line CP through P perpendicular to the line 01 to represent lens.
3. Draw another ray AC from the point A parallel to the principal axis 01 to meet the lens line CP at a point C.
4. This ray AC will reach the image point B while passing through the focus, therefore join C to B to meet line 01 at a point F which is the focus of the lens. The completed ray diagram is shown above.

b) Since the image is real and inverted, the lens is convex.

c) Since the size of object and image of equal, the object must be at a distance of twice the focal length, i.e., at 2F₂.

Question 28.
The diagram shows an object AB placed on the principal axis of the lens L. The two foci of lens are F₁ and F₂. The image formed by the lens is erect, virtual, and diminished.

i) Draw the outline of the lens L used and name it.
ii) Draw a ray of light starting from B and passing through ‘O’. Show the same ray after refraction by the lens.
iii) Draw another ray from B which is incident and parallel to the principal axis. Show how it emerges after refraction from the lens.
iv) Locate the final image formed.
Answer:
i) Since the image formed by the lens is erect, virtual, and diminished, the lens is concave.
ii) The light ray BO incident at the optical centre ‘O’ of the lens, passes undeviated as OC after refraction by the lens.
iii) The light ray BP is incident and parallel to the principal axis. It emerges as PQ after refraction which appears to diverge from the second focus F₂ of the lens.
iv) The refracted rays OC and PQ do not actually meet, but they appear to diverge from a point B’ (i.e. they meet at a point B’ when they are produced backwards).
v) Thus, B’ is the complete image A’B’ is obtained by drawing perpendicular from B’ on the line F₂OF₁. The image is formed between the optical centre O and focus F₂ of the lens.

Question 29.
Figure below shows the refracted ray BC through a concave lens and its foci marked as F₁ and F₂. Complete the diagram by drawing the corresponding incident ray and also give reason.

Answer:

1. Figure shows the refracted ray parallel to the principal axis. Therefore, the incident ray must be travelling towards the focus F₂.
2. Thus, to find incident ray, F₂ is joined to the starting point B of the refracted ray and produced backward as BA.
3. Then AB is the required incident ray.
4. The completed diagram is shown below.

Question 30.
State the type of lens used as a simple magnifying glass. Draw a labelled diagram to show that the image of the object is magnified.
Answer:

• A single convex lens is used as a magnifying glass, i.e. for seeing small object magnified.
• When the object to be seen between the focus and optical centre of the lens, a virtual, erect and enlarged image is formed.
• So a convex lens acts as magnifying glass for object AB as shown in the figure.

Question 31.
Radii of biconvex lens are equal. Let us keep an object at one of the centres of curvature. Refractive index of lens is ‘n’. Assume lens is in the air. Let us take R as the radius of the curvature.
a) How much is the focal length of the lens?
b) What is the image distance ?
c) Discuss the nature of the image.
Answer:
Radii of curvatures (R) of biconvex lens are equal, so R₁ = R₂ = R


c) The nature of the image is inverted and v < u.

Question 32.
Refractive index of a lens is 1.5. When an object is placed at 30 cm, image is formed at 20 cm. Find its focal length. Which lens is it? If the radii of curvature are equal, then what is its value?
Answer:
Object distance = u = – 30 cm (Infornt of the lens)
Image distance = v = 20 cm

Question 33.
A convex lens of focal length 10 cm is placed at a distance of 12 cm from a wall. How far from the lens should an object be placed so as to form its real image on the wall?
Answer:

Question 34.
A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. If image distance is thrice the focal length, find object distance, image distance and nature of image.
Answer:
Given that
Focal length of lens = + 20 cm
Object distance = -u
Image distance (v) = + 3u

So the object is between F₂ and 2F₂. So the image beyond 2F₁ it is real, inverted, and magnified.

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

• The objective lens of a telescope, camera, slide projector, etc. is a convex lens which forms real and inverted image of object.
• Our eye lens is also a convex lens. The eye lens forms the inverted image of the object on the retina.
• The eye defects are corrected by lenses.
• A magnifying glass is nothing but a convex lens of short focal length fitted in a steel (or plastic) flame provided with a handle.
• In spectroscope, convex lenses are used for obtaining a pure spectrum.
• A concave lens is used as eye lens in a Galilean telescope to obtain an erect final image of the object.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 5 Refraction of Light at Plane Surfaces.

AP State Syllabus SSC 10th Class Physics Important Questions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces 1 Mark Important Questions and Answers

Question 1.
Take a bright metal ball and make it black with soot on a candle flame. Immerse it in the water. Mention one observation. (AP June 2015)
Answer:
1) The ball shines.
2) The ball appears to raise up in water.

Question 2.
What is critical angle? (AP March 2015)
Answer:
The angle of incidence at which the light ray propagates from denser to rarer graze along interface is called critical angle of denser medium.

Question 3.
If refractive index of glass is \(\frac{3}{2}\), then what is speed of light in glass? (AP June 2016)
(OR)
Find the speed of light in a transparent medium, whose refractive index is 3/2.
Answer:
The refractive index of glass or transparent medium = \(\frac{3}{2}\)

Question 4.
Write any two questions about the ‘formation of mirages’. (AP June 2017)
Answer:

1. When does a mirage form?
2. How does a mirage form?

Question 5.
Optical Fibre Cable (OFC) are oftenly used in tele-communications. What is the working principle behind the OFC? (AP March 2017)
Answer:
Total Internal Reflection.

Question 6.
Among objects made of glass and diamond, which one shines more? Why? (AP June 2015)
Answer:
Diamond shines more because of low conical angle of 24.4°.

Question 7.
Suggest reasons for the phenomenon associated with the following : Twinkling of stars. (TS March 2015)
Answer:
Refraction of light is the reason for the twinkling of stars.

Question 8.
Draw the diagram showing the path of the ray when it travels from denser medium to rarer medium when the incident angle is more than the critical angle. (TS June 2016)
Answer:

Question 9.
Why does the light ray deviate in refraction? (AP SA-1:2019-20)
Answer:
Light ray always chooses the path of least time to travel. Hence speed of light changes at interface of two media. So, the light ray deviate in refraction.

Question 10.
Name the phenomenon involved in the function of optical fibre. (AP SCERT : 2019-20)
Answer:
Total Internal Reflection.

Question 11.
What is Fermat’s principle?
Answer:
The light ray always travels in a path which needs shortest possible time to cover the distance between the two given points.

Question 12.
What happens when light travels from one medium to another medium?
Answer:
It bends towards or away from normal.

Question 13.
When does speed of light decrease?
Answer:
When it travels from rarer to denser medium.

Question 14.
What do you mean by denser medium?
Answer:
The medium which has more optical density.

Question 15.
What is refraction?
Answer:
The process of changing speed when light travels from one medium to another is called refraction of light.

Question 16.
Which quantity will compare the refractive indices of two media?
Answer:
Relative refractive index.

Question 17.
What is relative refractive index?
Answer:
It is the ratio of refractive index of second medium to refractive index of first medium.

Question 18.
When does a light ray bend away from normal?
Answer:
When a light ray moves from denser to rarer medium it bends away from normal.

Question 19.
When will angle of refraction be equal to 90°?
Answer:
When angle of incidence is equal to critical angle then angle of refraction will be equal to 90°.

Question 20.
When angle of incidence of light ray is greater than critical angle, what happens?
Answer:
Light ray undergoes total internal reflection.

Question 21.
What happens to refractive index of air with height?
Answer:
Refractive index of air increases with height.

Question 22.
Which has greater refractive index between these?
1) cool air at the top
2) hotter air just above the road
Answer:
Cooler air has greater refractive index due to more density.

Question 23.
What is mirage?
Answer:
The virtual images of distant high objects cause the optical illusion called mirage.

Question 24.
What happens to a light ray when it falls perpendicular to one side of the slab surface?
Answer:
It comes out without any duration.

Question 25.
What are the conditions for total internal reflection?
Answer:

1. The rays of light must travel from denser to rarer medium.
2. The angle of incidence of denser medium must be greater than critical angle.

Question 26.
What is meant by a vertical shift?
Answer:
When a ray emerges out of a glass slab, it is parallel to the incident ray but is displaced laterally relative to incident ray. This shift of emergent ray is called vertical shift.

Question 27.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
It bends towards the normal. This is because it travels from an optically rarer to , optically denser medium.

Question 28.
When does Snell’s law fail?
Answer:
Snell’s law fails when light is incident normally on the surface of a refracting medium.

Question 29.
Why does a ray of light bend when it travels into another medium?
Answer:
It bends because its velocity changes when it moves from one medium to the other.

Question 30.
A pencil when dipped in a glass tumbler containing water appears to be bent at the interface of air and water. Explain why.
Answer:

• When light travels obliquely from one transparent medium to another, the direction of propagation of light changes due to refraction of light.
• In this case, light travels from denser medium to rarer medium, hence it bends away from the normal and the pencil appears to be bent.

Question 31.
Why does light travel in vacuum?
Answer:
Light travels in vacuum because it does not require medium for its propagation.

Question 32.
What are the factors on which refractive index depends?
Answer:

1. Nature of material
2. Wavelength of light used.

Question 33.
What does the ratio of sine of angle of incidence and sine of angle of refraction give?
Answer:
The ratio of sine of angle of incidence and sine of angle of refraction gives refractive index.

Question 34.
What is the relationship between critical angle and refractive index?
Answer:
The relationship between critical angle and refractive index

Question 35.
What is the relationship between angle of incidence and shift?
Answer:
As the angle of incidence increases, the shift also increases.

Question 36.
Why does a coin placed in a water appear to be raised?
Answer:
It is due to refraction of light.

Question 37.
Can you guess what happens when light travels from denser medium to rarer medium?
Answer:
The light ray bends away from normal.

Question 38.
A ray of light falls normally on a face of a glass slab. What are the values of angle of incidence and angle of refraction of this ray?
Answer:
Both angles are zero.

Question 39.
When does light ray from slab not undergo any deviation?
Answer:
The light ray that incidents perpendicular to one side of the slab surface comes out without any deviation.

Question 40.
What is the factor on which refraction depends?
Answer:
Refraction depends on optical density.

Question 41.
What is absolute refractive index?
Answer:
It is the ratio of speed of light in vacuum to speed of light in medium.
\(\mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}}\)

Question 42.
In water filled vessel, the coin of the bottom can be seen at a height. Give reasons.
Answer:
Rising of coin when water is poured in a cylindrical transparent vessel.

Question 43.
Write one activity in showing the process of ‘total internal refraction’.
Answer:
Due to refraction of light speed of light changes when it travels from one medium to another medium.

Question 44.
Define “Glass Slab”.
Answer:
A thin glass slab is formed when a medium is isolated from its surroundings by two plane surfaces parallel to each other.

Question 45.
A ray of light is incident normally on a plane glass slab. What will be the angle of refraction and angle of deviation for the ray?
Answer:
The ray is incident normally on a plane glass slab. So there is no deviation of light ray. Therefore the angle of refraction and angle of deviation both have 0° values.

Question 46.
A light ray in passing from water to a medium (a) speeds up, (b) slows down. In each case get one example of the medium.
Answer:
a) Air, because its optical density is less than water,
b) Glass, because its optical density is more than water.

Question 47.
If an angle of refraction is 90°, what is the corresponding angle of incidence called?
Answer:
The angle of incidence is called critical angle.

Question 48.
If the angle of incidence is more than critical angle, what happens to light ray if the light ray travels from denser to rarer medium?
Answer:
The light ray undergoes total internal reflection.

Question 49.
The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?
Answer:
It means that light travels 2.42 times faster in vacuum than in diamond.

Question 50.
For the same angle of incidence 45°, the angle of refraction in two transparent media, I and II is 20° and 30° respectively. Out of I and II, which medium is optically denser and why?
Answer:
Medium I is optically dense as angle of refraction is lesser in it, hence light bends towards normal.

Question 51.
For which colour of white light is the refractive index maximum and for which colour of white light is the refractive index minimum?
Answer:
The refractive index is maximum for violet because its wavelength is least.
The refractive index is minimum for red because its wavelength is maximum.

Question 52.
Correct the statement. “If the angle of incidence is greater than the critical angle the light is refracted when it falls on the surface from a denser medium to rarer medium”.
Answer:
If the angle df incidence is greater than the critical angle the light undergoes total internal reflection when it falls on the surface from a denser medium to rarer medium.

Question 53.
A light ray passes from medium 1 to medium 2. Which of the following quantities of refracted ray will differ from that of the incident ray?
Speed, intensity, frequency, wavelength.
Answer:
Speed, intensity and wavelength will differ from that of incident ray.

Question 54.
The refractive indices of alcohol and turpentine oil with respect to air are 1.36 and 1.47 respectively. Find the refractive index of turpentine oil with respect to alcohol. Which one of these permits the light to travel faster?
Answer:
The refractive index of turpentine oil with respect to alcohol = \(\frac{1.47}{1.36}\) = 1.08.

The refractive index increases when the speed of light decreases. So light travels faster in alcohol as its refractive index is less.

Question 55.
Light enters from air to diamond which has refractive index of 2.42. Calculate the speed of light in diamond, if speed of light in air 3 × 10⁸ ms^-1.
Answer:
Absolute refractive index = \(\frac{c}{v}\)
2.42 = \(\frac{3 \times 10^{8}}{\mathrm{v}}\) ⇒ v = 1.24 × 10⁸ ms^-1.

Question 56.
A glass block 3.0 cm thick is placed over a stamp. Calculate the height through which image of stamp is raised. Refractive index of glass is 1.54.
Answer:

Question 57.
The refractive index of water is 4/3. Calculate the critical angle for water – air interface (sin 49 = 3/4).
Answer:

10th Class Physics 5th Lesson Reflection of Light by Different Surfaces 2 Marks Important Questions and Answers

Question 1.
A ray of light enters from air to a medium X. The speed of light in the medium is 1.5 × 10⁸ m/s and the speed of light in air is 3 × 10⁸ m/s.
Find the refractive index of the medium X. (TS March 2015)
Answer:

Question 2.
What are the applications of optical fibres?
(OR)
Write two uses of fibre optics in daily life. (TS June 2016)
Answer:
Applications oruses of optical fibres :

1. Light pipes using optical fibres may be used to see places which are difficult to reach things such as inside of a human body.
2. The other important application of fibre optics is to transmit communication signals through light pipes.

Question 3.
Focal length of the lens depends on its surrounding medium. What happens, if we use a liquid as surrounding media of refractive index, equal to the refractive index of lens? (TS June 2018)
Answer:

• When the refractive index of surrounding media is equal to the refractive index of lens, the lens looses its characteristics.
• Lens do not diverge or converge the light.
• Light do not get refracted when it passes through that lens.

Question 4.
Why does the light ray travel slowly in diamond when compared to vacuum? (AP SA-I : 2019-20)
Answer:

• Refractive index of diamond (2.42) is greater than that of vacuum (1).
• Speed of light is inversely proportional to refractive index of substances.
• Hence, light ray travel slowly in diamond when compared to vacuum.

Question 5.
Write about laws of refraction.
(OR)
Write the laws of refraction.
Answer:
Laws of refraction :

1. The incident ray, the refractive ray and the normal to interface of two transparent media at a point of incidence lie in the same plane.
2. During refraction light follows Snell’s law, i.e., the ratio of sine of angled of incidence to sine of angle of refraction is constant.
   n₁ sin i = n₂ sin r (OR) \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) = constant.

Question 6.
What is total internal reflection ? What are the applications of total internal reflection?
Answer:
When the angle of incidence is greater than critical angle, the light ray is reflected into denser medium at interface i.e., light never enters rarer medium. This phenomenon is called total internal reflection.
1) Brilliance of diamonds :
Total internal reflection is the main cause for brilliance of diamonds. The critical angle of diamonds is very low (24.4°). So if a light ray enters a diamond it is very likely to get total internal reflection which makes the diamond shine brilliant.

2) Optical fibres:
Total internal reflection is the basic principle for working of optical fibre.

Question 7.
What are optical fibres? How do they work?
Answer:

• An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer.
• A bunch of such thin fibres forms a light pipe.

Working :

1. Because of the small radius of the fibre, light going into it makes a nearly glancing incidence on the wall.
2. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
3. The light is thus transmitted along the fibre.

Question 8.
How can a patient’s stomach be viewed by using optical fibres?
(OR)
How do you observe patient’s stomach by using a light pipe?
Answer:

• The patient’s stomach can be viewed by inserting one end of a light pipe into the stomach through the mouth.
• Light is sent down through one set of fibres in the pipe.
• This illuminates the inside of the stomach.
• The light from the inside travels back through another set of fibres in the pipe and the viewer gets the image at the outer end.

Question 9.
State four differences between reflection and total internal reflection.
Answer:

Reflection	Total internal reflection
1) Smooth polished surface is required for reflection.	1) No smooth polished surface is required for total internal reflection.
2) It takes place for all angles of incidence.	2) It takes place only, when angle of incidence is greater than critical angle.
3) It takes place when the rays of light travel from rarer to denser medium to an opaque medium.	3) It takes place when rays of light travel from denser to rarer medium.
4) Some amount of light is absorbed by reflecting surface.	4) No light is absorbed by reflecting surface.

Question 10.
The figure shows refraction and emergence of a ray of light incident on a rectangular glass slab. Copy the diagram and mark the lateral displacement of the incident ray. Name the two factors on which the lateral displacement depends.

Answer:
The lateral displacement depends on

1. The angle of incidence of the incident ray PQ, on the slab and
2. The thickness of the glass slab.

The perpendicular distance between the emergent forward.

Question 11.
1) What happens to a ray of light when it travels from one equal refractive indices?
2) State the cause of refraction of light.
Answer:
1) No refraction or bending would take place. The light will travel in a straight line.

2) The refraction occurs due to change in speed of light as it enters from one medium to another.

Question 12.
A coin placed at the bottom of a tank appears to be raised when water is poured into it. Explain.
Answer:

• It happens due to the phenomenon of refraction of light.
• When the rays of light from the coin, in the denser medium fall on the interface separating the two media, the rays of light move away from the normal after refraction.
• The point from which the refracted rays appear to come gives the apparent position of the coin.
• As the rays appear to come from a point above the coin, therefore, the coin seems to be raised.

Question 13.
Define refractive index. Explain the relationship between the refractive index of the medium and to the speed of light in the medium.
Answer:
The ratio of speed of light in vacuum to the speed of light in that medium is defined as refractive index ‘n’ with respect to the vacuum. It is also called absolute refractive index.

When refractive index of a medium is high, then the speed of light is low and vice-versa.

Question 14.
Explain lateral shift and vertical shift.
Answer:
Lateral shift:
The distance between incident and emergent ray is called lateral shift.

Vertical shift :
The perpendicular distance between object and its image is called vertical shift.

Question 15.
During refraction of light, which of the following quantities does not change.
(1) velocity,
(2) wavelength,
(3) frequency,
(4) amplitude.
Answer:
During refraction of light velocity of light changes and also wavelength and amplitude. Frequency does not change during refraction.

Question 16.
The upper surface of water contained in a beaker and held above the eye level appears silvery. Why?
Answer:
Critical angle for water is 48°. The rays of light entering in water from below, suffer refraction. If these rays strike the water-air surface at an angle which is greater than 48°, they get totally internally reflected. These rays on emerging out of water, appear to come from the upper surface of water, which in turn appear silvery.

Question 17.
Why don’t the planets twinkle?
Answer:

• The planets are much closer to the earth, and are thus seen as extended sources.
• We can consider a planet as a collection of a large number of point-sized sources of light.
• The total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying twinkling effect

Question 18.
Why did an empty test tube placed obliquely in water, appears filled with mercury, when seen from above?
Answer:
When the rays of light travelling through water they strike the water glass interface of test tube at an angle, which is more than critical angle for water, they suffer total internal reflection. When these totally reflected rays reach eye, then to the eye they appear as they come from surface of test tube, which in turn appears filled with mercury.

Question 19.
Why are the bubbles rising up the fish tank appear silvery?
Answer:
When the rays of light travelling through water they strike the water air interface of the bubble at an angle, which is greater than critical angle for water, they get totally internally reflected. These reflected rays on reaching the eye appear to come from air bubble, which in turn appears silvery.

Question 20.
Why does a crack in a window pane appear silvery?
Answer:
There is always some amount of air present in the crack. When the rays of light travelling through glass, strike the glass, the glass air interface at an angle, greater than critical angle of glass, they are totally internally reflected. When these reflected rays reach eye, then to the eye they appear to come from the crack, which in turn appears silvery.

Question 21.
Explain why a straight stick appears to be bent when dipped in water.
Answer:

• Suppose two rays originate from the end of the stick in water.
• As these rays get refracted into the air, they bend away from the normal.
• When these two refracted rays are produced backwards they seem to meet at a point higher than the end of stick.
• This point gives the apparent position of the end of the stick. Thus, the stick appears to be bent.

Question 22.
A pond appears to be shallower than it really is when viewed obliquely. Why?
Answer:

• Suppose two rays originate from the bottom of the pond. As these rays get refracted into the air, they bend away from the normal.
• When these two refracted rays are produced backwards they seem to meet at a point higher than the bottom of the pond.
• This point gives the apparent position of the bottom of the pond.
• Thus, the pond appears to be shallower.
• This effect is absent if the pond is viewed normally.

Question 23.
Frame some questions to know about the formation of mirages.
Answer:

1. What are mirages?
2. What is the principle involved in mirages?
3. Can mirages be photographed?
4. Where does the water on the road go?

Question 24.
A glass slab is placed over a piece of paper on which VIBGYOR is printed with each letter into corresponding colours.
1) Will the image of all the letters be in the same place?
2) The letter of which colour appears to be raised maximum and which colour minimum? Explain your answer.
Answer:

1. The image of all letters will not be in the same place.
2. The letter of violet colour appears to be raised maximum, while the letter of red colour appears to be raised minimum because refractive index of glass is most for the violet light while least for the red light, therefore the apparent depth is least for violet and most for red.

Question 25.
Why does sun appear bigger during the sunset or the sunrise?
Answer:

1. We already know that the apparent position of sun is higher than actual position in the horizon.
2. Moreover, due to refraction, the apparent image of sun is closer to eye than the actual position. Since during sunset or sunrise, the rays of light travel through maximum length of atmosphere therefore the refraction is also maximum.
3. Hence apparent image of sun is very much closer to eye. Thus it appears bigger.

Question 26.
Write the material required in finding out the relation between angle of incidence and angle of refraction.
Answer:
Material required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2cm, pencil and laser light.

Question 27.
Write the aim and apparatus experiment in finding the refractive index of the glass slab.
Answer:
Aim :
Finding the refractive index of the glass slab.

Apparatus :
Glass slab, white chart, pin.

Question 28.
Observe the following table and answer the following question.

Questions :
1) Find out from the table the medium having highest optical density and the medium with lowest optical density.
2) You are given kerosene, turpentine oil and water. In which of these does the light travel fast? Use the information given in the table.
3) The refractive index of diamond is 2.42. What is the meaning of this statement?
4) When light travels from water to crown glass, what happens?
5) When light travels from diamond to air, what happens?
Answer:
1) The medium with highest optical density is diamond as its refractive index is maximum, i.e. 2.42.
The medium with lowest optical density is air, as its refractive index is minimum, i.e. 1.0003.

2) The refractive index of medium is given by the expression, n = \(\frac{c}{v}\) or v = \(\frac{c}{n}\)
This expression shows that light travels faster in the medium whose refractive index is minimum. From the table, we can find that water has the minimum value of refractive index. Therefore light travels faster in water.

3) This statement means that light travels 2.42 times faster in vacuum than in diamond.

4) The light bends towards normal.

5) The light bends away from the normal.

Question 29.
A ray of light enters from a medium A into a slab made up of a transparent substance B. Refractive indices of medium A and B are 2.42 and 1.65 respectively. Complete the path of ray of light till it emerges out of slab.

Answer:

Question 30.
A glass slab made of material of refractive index n₁ is kept in medium of refractive index n₂. A light ray is incident on the slab. Complete the path of rays of light emerging from glass slab, if a) n₁ > n₂ b) n₁ = n₂ c) n₁ < n₂.
Answer:
a) n₁ > n₂

b) n₁ = n₂
There is no deviation of light ray

c) n₁ < n₂

Question 31.
How do you appreciate the process of total internal reflection in nature?
Answer:

1. Total internal reflection is responsible for brilliance of diamond.
2. Total internal reflection is basic pruxiplo behind working of optical fibres which are used in getting the images ol internal ouaiis and also used in telecommunications. So the role of total internal reflection is thoroughly appreciated.

Question 32.
Write the application of optical fibres in communication.
Answer:

• Optical fibres are used to transmit communication signals through light pipes.
• For example, about 2000 telephone signals, approximately mixed with lightwaves, may be simultaneously transmitted through a typical optical fibre.
• The clarity of the signals transmitted in this way is much better than other conventional methods.

Question 33.
Write the applications of total internal reflection.
Answer:
Application of total internal reflection :

1. Brilliance of diamonds,
2. Optical fibres.

Question 34.
A monochromatic ray of light strikes the surface of transparent medium at an angle of incidence 60° and gets refracted into the medium at an angle of refraction 45°. What is the refractive index of the medium?
Answer:

Question 35.
A light ray enters a liquid at an angle of incidence 45° and it gets refracted on liquid at angle of refraction 30°. Calculate the refractive index of the liquid.
Answer:

Question 36.
Refractive index of water is 4/3. Calculate the speed of light in water.
Answer:

Question 37.
A postage stamp placed under glass appears raised by 8 mm. If refractive index of glass is 1.5, calculate the actual thickness of glass slab.
Answer:
Let real thickness of glass = x.
Vertical shift = 8 mm.

Question 38.
Refractive index of glass is 1.5. Find its critical angle.
Answer:

Question 39.
What is the advantage of using prism in place of plane mirror in periscope or binocular?
Answer:

• When total internal reflection occurs from a prism, the entire incident light is reflected back into the denser medium.
• Whereas in ordinary reflection from a plane mirror, some light is refracted and absorbed. So the reflection is partial.
• This is the reasons why total reflecting prism is used in place of a plane mirror to deviate the light ray by 90° in a periscope and 180° in a binocular.

10th Class Physics 5th Lesson Reflection of Light by Different Surfaces 4 Marks Important Questions and Answers

Question 1.
What is the angle of deviation produced by the glass slab? Explain with ray diagram. (AP June 2015)
(OR)
Which angle of deviation is produced by glass slab? Write your explanation with a ray diagram.
Answer:

1. Angle of deviation is the angle between incident ray and emergent ray.
2. The angle of deviation produced by a glass slab is ‘O’, because the incident ray and emergent ray are parallel to each other that can be seen in the figure.

Question 2.
Explain the phenomenon of total internal reflection with two examples. (AP June 2018)
(OR)
What is total internal reflection? Explain with examples. (AP SA-I:2019-20)
Answer:

• When the angle of incidence is greater than the critical angle, the light ray is reflected into denser medium at interface. This phenomenon is called total internal reflection.
• Total internal reflection is the main reason for brilliance of diamonds. The critical angle of a diamond is very low. So if a light ray enters a diamond it is very likely to undergo total internal reflection which makes the diamond shine.
• Total internal reflection is the basic principle behind working of optical fibre. Because of the small radius of the fibre light going into it makes a nearly glancing incidence is greater than the critical angle and hence total internal reflection takes place. The light is thus transmitted along the fibre.

Question 3.
Explain the relation between angle of incidence and angle of refraction with an experiment. (AP March 2018)
Answer:
Aim :
To verify the relation between angle of incidence and angle of refraction.

Material required :
A plank, white chart, protractor, semicircular glass disc, pencil and leser light.

Procedure :

1. Take a drawing sheet on a cardboard and mark different angles (on both side of MM line)
2. Place a semi circular glass disc, so that its diameter coincides with the line “MM”.
3. Send a laser light along a line with makes 15° with NN.
4. Let it is incident angle.
5. Measure its corresponding angle of refraction by observing light coming from outside of the glass slab.
6. Repeat this experiment with various values of angle of incidence, refraction and not in the table.

7.

8. From the above table we observe that \(\frac{\sin \mathrm{i}}{\sin r}\) = constant.

Question 4.
Give some daily life consequences of refraction of light.
Answer:

• A star appears twinkling in the sky.
• The sun is seen a few minutes before it rises above the horizon in the morning and in the evening few minutes longer after it sets.
• A coin kept in a vessel not visible when seen from just below the edge of the vessel, can be viewed from the same position when water is poured into the vessel.
• A print appears to be raised when a glass block placed over it.
• A piece of paper stuck at the bottom of a glass block appears to be raised when seen from above.
• A tank appears shallow than its actual depth.
• A person’s legs appear to be short when standing in a tank.
• An object placed in a denser medium when viewed from a rarer medium appears to be at a lesser depth.
• An object in a rarer medium, when viewed from a denser medium, appears to be at a greater distance than its real distance.

Question 5.
What are the factors which influence refractive index of material?
Answer:

• Nature of medium, i.e. its optical density. Smaller the speed of light in a medium relative to air, higher is the refractive index of the medium.
• Physical condition such as temperature. With rise in temperature the speed of light in medium increases, so the refractive index of medium decreases.
• The colour or wavelength of light (refractive index increases with decrease in wavelength, eg : µ_v > µ_R).

Question 6.
What is the advantage of total internal reflection over reflection?
Answer:

• In the process of total internal reflection, 100% energy is reflected back.
• No other device such as plane mirror, etc. produces 100% reflection due to absorption and refraction of some part of light.
• Due to this property the phenomenon total internal reflection is of great practical application in the construction of periscope, binocular and certain type of camera.

Question 7.
The diagram below shows a glass block suspended in a liquid. A beam of light of single colour is incident from liquid on one side of block.
1) Draw diagrams to show how light bends when it travels from liquid to glass and then to liquid if (i) the light slows down in glass (ii) the light speeds up in glass.
2) State two conditions under which the light ray moving from liquid to glass passes straight without bending. Will the glass be visible them?
Answer:
1) If light slows down in going from liquid to glass (i.e., µ_glass > µ_liquid), it will bend towards the normal at the point of incidence in passing from liquid to glass at the first surface, while it is bent away from normal at the second surface in passing from glass to liquid. In the ray diagram, the light beam suffers lateral shift.

2) If light speeds up in going from liquid to glass (i.e., µ_glass < µ_liquid). It will bend away from the normal at the point of incidence on the first surface in passing from liquid to glass, while it bends towards the normal at the second surface in passing from glass to liquid. The light beam suffers lateral shift in direction opposite to that
Note that in both cases, the emergent ray is parallel to the incident ray.

Question 8.
A ray of light is incident on a rectangular glass block PQRS, which is silvered at the surface RS. The ray is partly reflected and partly refracted.

1) Trace the path of reflected and refracted rays.
2) Show at least two rays emerging from the surface PQ after reflection from the surface RS.
3) How many images are formed in the above case? Which image is the brightest?
Answer:
1) In the figure OB is reflected ray and OC is the refracted ray for the incident ray AO.

2) Two rays emerging from surface PQ after reflections for the surface RS are labelled as 1 and 2.

3) Multiple (or infinite) images are formed. The second image formed due to first reflection at C at the silvered surface RS is the brightest. It is seen in the direction of ray 1.

Question 9.
What are the factors which affect critical angle? The critical angle for a given pair of media depends on their refractive index which is affected by the following factors.
Answer:
1. Effect of colour of light:
The refractive index of transparent medium is more for violet light and less for red light, therefore the critical angle for pair of media is less for the violet light and more for the red light. Thus critical angle increases with increase in wavelength of light.

2. Effect of temperature :
On increasing the temperature of medium, its refractive index decreases, so the critical angle for that pair of media increases. Thus critical angle increases with increase in temperature.

Question 10.
The table shows the refractive index of some material media.

Material Medium	Refractive Index
Air	1.0003
Ice	1.31
Water	1.33
Kerosene	1.44
Fused quartz	1.46
Turpentine oil	1.47
Crown glass	1.52
Benzene	1.50

Answer the following questions with the help of the above table.
1) Find the speed of light in Benzene.
2) Write the relationship between mass density and optical density of kerosene and water.
3) What are the factors that refractive index depends on?
4) Write the relative refractive index of kerosene with water.
Answer:
1)

Speed of light in benzene = 2 × 10⁸ m/s

2) Optical denser medium may not possess greater mass density. Kerosene with high refractive index is optically denser than water although its mass density is lesser than water.

3) Refractive index depends on
1) nature of material,
2) wavelength of light used.

4) Relative refractive index of kerosene with water = \(\frac{1.44}{1.33}\) = 1.08

Question 11.
Red light of wavelength 6600A travelling in air gets refracted in water. If the speed of light in air is 3 × 108 ms-1 and refractive index of water is 4/3, find the
(i) frequency of light in air,
(ii) the speed of light in water,
(iii) the wavelength of light in water.
Answer:

Question 12.
Draw the ray diagram which shows the ray takes curved path because of total internal reflection.
Answer:

Question 13.
Give some daily life consequences of total internal reflection?
Answer:

• On a hot sunny day, a driver may see a pool of water on the road before him. It is the phenomenon of mirage which is often observed in desert.
• An empty test tube placed in a beaker with mouth outside the water surface shines like a mirror.
• A crack in a glass vessel often shines like a mirror.
• A piece of diamond sparkles when viewed from certain directions.
• An optical fibre is used to transmit a light signal over a long distance with negligible loss of energy.

Question 14.
Light travels from air to water, then the refraction index of water is 1.33. Hence find the refractive index when light travels from water to air.
Answer:
Refractive index of water (n₂₁) = 1.33
Refractive index of air (n₁₂) = \(\frac{1}{1.33}\) = 0.75

Question 15.
The refractive index of diamond is 2.42 and the refractive index of glass is 1.5; compare the critical angle between them. (Diamond 24°, glass 42°)
Answer:
Refractive index of diamond (µ₁) = 2.42

Question 16.
A ray of light travels from an optically denser to rarer medium. The critical angle of the two media is ‘C’. What is the maximum possible deviation of the ray?
Answer:
The relation between angle of deviation and angle of incidence, angle of emergence and angle of prism is given by
Angle of deviation = i₁ + i₂ – A
For maximum deviation, Angle of incidence (i₁) = 90°
Angle of emergence (i₂) = 90°

∴ Maximum deviation = i₁ + i₂ – A = 90 + 90 – 2C = 180 – 2C = n – 2C.

Question 17.
A ray of light strikes a glass slab 5 cm thick making an angle of incidence equal to 30°.
a) Construct the ray diagram showing emergent ray and refracted ray through the glass block. The refractive index of glass is 1.5.
b) Measure the lateral shift of the ray.
Answer: