Waves & Thermodynamics

Solutions of

Waves & Thermodynamics Lesson 14th to 19th

By DC Pandey

14. Wave Motion Introductory Exercise 14.1 1. A function, f can represent wave equation, if it satisfy 2 ¶2 f 2 ¶ f = v ¶t 2 ¶x 2 For, y = a sin wt, ¶2 y = - w2 a sin wt = - w2 y 2 ¶t ¶2 y but, =0 ¶x 2 So, y do not represent wave equation. 2

2

2. y ( x, t) = ae - (bx - et) = ae - ( kx - wt) w c Þ k = b and w = e Þ v = = k b 1 3. y ( x, t) = represent 1 + ( 4 x + wt) 2

the

given pulse, where, 1 1 y( x, 0) = = 2 2 1+ k x 1 + x2 Þ

k =1

1 1 = 2 1 + ( x - 2w) 1 + ( x - 1) 2 1 Þ w= 2 w 1/ 2 \ v= = = 0.5 m/s k 1 10 a 4. y = = 2 5 + ( x + 2t) b + ( kx + wt) 2 a 10 Amplitude, y max = = = 2m b 5 and k = 1; w = 2 w v = = 2 m/s and is travelling in (–) x k direction. 10 5. y = ( kx - wt) 2 + 2 10 10 y( x, 0) = 2 2 = 2 Þk =1 k x +2 x +2 and y( x, z) =

w = vk = 2 m/s ´ 1 m -1 = 2 rad/s 10 Þ y= ( x - 2t) 2 + 2

Introductory Exercise 14.2 x t ö 1. y( x, t) = 0.02 sin æç + ÷m è 0.05 0.01 ø = A sin ( kx + wt) m 1 1 -1 Þ A = 0.02 m, k = s m -1 , w = 0.05 001 . w 0.05 (a) v = = m/s = 5 m/s k 0.01 ¶y (b) v p = = Awcos ( kx + wt) ¶t 1 v p (0.2, 0.3) = 0.02 ´ 0.01

0.2 0.3 ö cos æç + ÷ è 0.5 0.01 ø = 2 cos ( 4 + 30) = 2 cos 34 = 2( - 0.85) = - 1.7 m/s w 2. Yes, (v p ) max = Aw = Ak × = ( Ak)v k 3. l = 4 cm, v = 40 cm/s (given) v 40 cm /s (a) n = = = 10 Hz l 4 cm

2 | Waves & Motion

(b) D f =

2p l =

2p

´ 2.5 cm =

5p

4 cm 4 T 1 (c) D t = Df = Df 2p 2 pn 1 p = ´ 2p ´ 10 3 1 s = 60 (d) v p = (v p ) max = - Aw = - 2pAn = - 2p ´ 2 cm ´ 10 s -1 = - 40 p cm/s = - 1.26 cm/s 4. (a)

y = A sin ( wt - kx) 2p ö æ 2p = A sin ç v × tx÷ l l ø è 2p 2p ö æ = 0.05 sin ç 12 ´ tx÷ 0.4 0.4 ø è

Dx rad

= 0.05 sin (60 pt - 5px) (b) y(0.25, 0.15) = 0.05 sin (60 p ´ 0.15 - 5p ´ 0.25) = 0.05 sin (9 p - 1.25 p) = 0.05 sin (7.75 p) = 0.05 sin (1.75 p) = - 0.0354 m = - 3.54 cm Df 0.25 p T (c) Dt = Df = = 2p w 60 p 1 s = 4.2 ms = 240

y

x

Introductory Exercise 14.3 1.

v= =

T = m

T = m/ l

500 ´ 2 0.06

=

Tl m

100 5 = 129.1 m/s 3

2.

v= =

T T = m r× A 0.98 = 10 m/s 9.8 ´ 10 3 ´ 10 -6

Introductory Exercise 14.4 1W P 1 = = W /m 2 2 2 4p 4 pr 4 p ´ (1 m) r 2. For line source, I = 2prl 1.

I =

1 and as I µ A 2 r 1 Þ Aµ r Þ I µ

Waves & Motion

| 3

AIEEE Corner ¢

Subjective Questions (Level 1) 1.

y( x, t) =6.50 mm cos 2p æ ö p t ç ÷ ç 28.0 cm - 0.0360s ÷ è ø x tö æ = A cos 2p ç - ÷ èl Tø

A = 6.50 mm, l = 28.0 cm, 1 1 n= = s -1 = 27.78 Hz T 0.036 v = nl = 28.0 cm ´ 27.78 s -1 = 778 cm/s = 7.78 m/s The wave is travelling along ( +)ve x-axis. x ö 2. y = 5 sin 30p æç t ÷ 240 ø è p = 5 sin æç 30pt - x ö÷ = A sin ( wt - kx) 8 ø è p (a) y (2, 0) = 5 sin æç 3 p ´ 0 - ´ 2 ö÷ 8 è ø p 5 = - 5 sin = = - 3.535 cm 4 2 2p 2 p (b) l = = = 16 cm k p/ 8 w 30p (c) v = = = 240 cm/s k p/ 8 w 30 p (d) n = = = 15 Hz 2p 2p Þ

3. y = 3 cm sin (3.14 cm -1 x - 314 s -1 t) = 3 cm sin ( p cm -1 x = 100 ps -1 t) = A sin ( kx - wt) (a) (v p ) max = Aw = 3 cm ´ 100p s -1 = 300 p cm/s = 3 p m/s = 9.4 m/s (b) a = - w2 y = - (100 ps -1 ) 2 ´ 3 cm sin (6p - 111p) = - 300p sin ( -105p) = 0 v/n l 350 p 4. (a) Dx = Df = Df = ´ 2p 2p 500 ´ 2p 3

7p 7 = m = 0.166 m 60p 50 2p (b) Df = Dt = 2pn Dt = 2p ´ 500 ´ 10 -3 T = p = 180° 6 5. y ( x, t) = ( kx + wt) 2 + 3 6 6 y ( x, 0) = 2 2 = 2 k x +3 x +3 =

Þ k = 1 m -1 Þ w = vk = 4.5 m/s ´ 1 m -1 = 4.5 rad/s 6 Þ y ( x, t) = ( x - 4.5t) 2 + 3 x t ö 6. y = 1.0 sin p æç ÷ è 2.0 0.01 ø x t ö = 1.0 sin 2p æç ÷ è 4.0 0.02 ø x t = A sin 2p æç - ö÷ èl Tø (a) A = 1.0 mm, l = 4.0 cm, T = 0.02 s ¶y x t (b) v p = = - wA cos 2p æç - ö÷ ¶t èl Tø 2pA x t =cos 2p æç - ö÷ T èl Tø æ x 2p ´ 1.0 mm t ö ÷ =cos 2p çç ÷ 0.02 s è 4.0 0.02 s ø æ x p t ö ÷ = - m/s cos p çç ÷ 10 è 2.0 cm 0.01 s ø v p (1.0 cm, 0.01s) = p 1 0.01 ö - m/s cos p æç ÷ 10 è 2 0.01 ø p p = - m/s cos = 0 m/s 10 2 (c) v p (3.0, 0.01) p 3 =cos p æç - 1 ö÷ = 0 m/s 10 è2 ø

4 | Waves & Motion

v p (5.0 cm, 0.01 s) = -

p 5 ms cos p æç - 1 ö÷ 10 è2 ø

= 0 m/s p 7 v p (7.0 cm, 0.01s) = - m/s cos p æç - 1 ö÷ 10 è2 ø = 0 m/s (d) v p (1.0 cm, 0.011s) p = - m/s 10 1 0.011 ö cos p æç ÷ è 2 0.01 ø p 1 = - cos p æç - 1.1 ö÷ 10 è 12 ø p p 3p = - cos 0.6p = - cos = 9.7 cm/s 10 10 5 v p (1.0 cm, 0.012s) p 1 0.012 ö = - m/s cos æç ÷ 10 è 2 0.01 ø p = - cos p (0.5 - 1.2) 10 p = - cos 0.7 p = 18.5 cm/s 10 p v p (1.0 cm, 0.013 s) = - m/s 10 1 0.013 ö p æ cos p ç ÷ = - cos 0.8 p 10 è 2 0.01 ø = 25.4 cm/s 2p 2p p 7. (a) k = = = cm -1 l 40 cm 20 = 0.157 rad/cm 1 1 T = = s = 0.125 s n 8 w = 2pn = 16p rad/s = 50.26 rad/s v = nl = 8 s -1 ´ 40 cm = 320 cm/s (b) y ( x, t) = A cos ( kx - wt) = 15.0 cm cos (0.157 x - 50.3 t) 8. A = 0.06m and 2.5l = 20 cm 20 Þ l= cm = 8 cm 2.5 v 300 m/s n= = = 3750 Hz l 8 cm y = A sin ( kx - wt) = 0.06m

2p sin æç x - 2p ´ 3750 t ö÷ è 0.08 ø = 0.06 m sin (78.5 m -1 x - 23561.9 s -1 t) v 8.00 m/s 9. (a) n = = = 25 Hz l 0.32 m 1 1 T= = s = 0.043 Hz n 15 2p 2p k= = = 19.63 rad/m l 0.32 m x t (b) y = A cos ( kx + wt) = A cos 2p æç + ö÷ èl Tø æ x t ö ÷ = 0.07 m cos 2p çç + ÷ è 0.32 m 0.04 s ø 0.36 0.15 ö (c) y = 0.07 m cos 2p æç + ÷ è 0.32 0.04 ø 9 30 ö = 0.07 m cos 2p æç + ÷ 8 ø è8 39 = 0.07 m cos p 4 p = 0.07 m cos æç 10p - ö÷ 4ø è p = 0.07 m cos = 0.0495 m 4 Df p + p/4 T (d) Dt = Df = = 2p 2pn 2p ´ 25 3 = s = 0.015 s 200 Mg T T 10. v = = = m rA rA =

=

2 ´ 9.8 8920 ´ 3.14 ´ (1.2 ´ 10 -3 ) 2 2 ´ 9.8 ´ 104 89.2 ´ 3.14 ´ 1.44

= 22 m/s

11. l µ n µ T µ M Þ

l2 = l1 =

Þ

M2 M1 8 = 4 = 2. 2

l2 = 2l1 = 0.12 m.

Waves & Motion

T( x) m

12. T( x) = m ( L - x) g, v( x) = g ( L - x) dx = dt ; g ( L - x)

=

x

Þ

t=

2 ´ 2 l0 g

\

t=

8l0 g

15.

Let, L - x = y dx = - dy 0 - dy òL g y = t

2

Þ M = ò dm = ò kx dx = 0

2M L2 T v( x) = = m

13. (a) dm w2 R = 2T sin dq

dq

1 2 kL 2

Þ k= 0

T

dm = kx dx

m=

1 - y L t= =2 g g 1/2 1

\

| 5

\ dq T

R

mR 2dq w2 R = 2T dq T Þ w2 R2 = m T \ Wave speed, v = = w2 R2 = Rw m (b) Kink remains stationary when rope and kink moves in opposite sence i. e., if rope is rotating anticlockwise then kink has to move clockwise. 14. x is being measured from lover end of the string x 1 \ m( x) = ò dm = ò m 0 x dx = m 0 x2 0 2 m( x) g T( x) \ v( x) = = m m 1 m 0 x2 g 1 2 = = gx m0x 2 l dx t Þ ò 0 1 = ò 0 dt gx 2

T TL2 dx = = kx 2Mx dt

t = ò dt = ò

L

0

2Mx dx = TL2

1 +1 2 L

2M TL2 1 + 1 2

2 2ML3 2 2ML = 3 TL2 3 T Mg T 16. (a) v = = m m =

= (b) l =

1.5 ´ 9.8 0.055 16.3 m/s

= 16.3 m/s

v = = 0.136 m n 120 / s

(c) l µ v µ T µ M i. e., if M is doubled both speed and wavelength increases by a factor of 2 . 17. E = I At = 2p2 n2 a2 r vAt = 2p2 n2 a2 (rA) (v. t) = 2p2 n2 a2 m. l = 2p2 n2 a2 m = 2 ´ (3.14) 2 ´ (120) 2 ´ (0.16 ´ 10 -3 ) 2 ´ 80 ´ 10 -3 = 582 ´ 10 -6 J = 582 mJ = 0.58 mJ E 18. P = = IA = 2p2 n2 a2 rnA = 2p2 n2 a2 mv t = 2p2 n2 a2 Tm = 2 ´ (3.14) 2 ´ (60) 2 ´ (6 ´ 10 -2 ) 2 80 ´ 5 ´ 10 -2 = 4 (3.14 ´ 60 ´ 0.06) 2 = 511.6 W

6 | Waves & Motion

19. P = IA = 2p2 n2 a2 Tm = 2 ´ (3.14) 2 ´ (200) 2 ´ 10 -6 60 ´ 6 ´ 10 -3 = 8 ´ (3.14) 2 ´ 10 -2 ´ 6 ´ 10 -1 W = 0.474 W l E = Pt = P × v 0.474 ´ 2 0.474 ´ 2 = = J = 9.48 mJ 100 60 6 ´ 10 -3 ¢

20. P = 2p2 v2 a2 rvA = 2p2 v2 a2 mv ; v = m=

T m

T v2

T T × v = 2p2 n2 a2 2 v v 2 ´ (3.14) 2 ´ (100) 2 (0.5 ´ 10 -3 ) 2 ´ 100

= 2p2 n2 a2 =

100 = 2 ´ (3.14) ´ 10 ´ 0.25 ´ 10 -6 = 4.93 ´ 10 -2 W = 49 mW 2

4

Objective Questions (Level 1) 1. w = q=

150 ´ 2p p 4

60

= 5p rad/s, A = 0.04 m and

p \ y = A sin ( wt + q) = 0.04 sin æç 5pt + ö÷ 4ø è w 2. w = 600p, v = 300 Þ k = = 2p v

p 2p = ´ 0.04 6 l \ l = 12 ´ 0.04 = 0.48 m. v 300 m/s 5. l = = = 12 m n 25 Hz 2p 2p Df = Dx = (16 - 10) m = p l 12 m Þ

6. y = 0.02 sin ( x + 30t) = A sin ( kx + wt) Þ y = A sin ( wt - kx) = 0.04 sin (600pt - 2px) Þ k = 1, w = 30 3 y (0.75, 0.01) = 0.04 sin æç 600p ´ 0.01 - 2p ´ ö÷ w T v = = 30 m/s = 4ø è k m 3p ö = 0.04 sin æç 6p ÷ 2 4 2 ø è Þ T = m v = 1.3 ´ 10 ´ 900 = 0.117 N pö æ ¶y ¶y ¶x ¶y = 0.04 sin ç 4 p + ÷ = 0.04 m 7. v p = = × =v = slope ´ v 2ø è ¶t ¶x ¶t ¶x 1 up 3. y ( x, t) = In transverse wave they are 2 2 + 3 ( kx - wt) v 1 1 y ( x, 0) = = 2 + 3 k2 x2 2 + 3 x2 p perpendicular i. e., . In longitudinal Þ k =1 2 1 1 y ( x, 2) = = wave , they are either at 0 or up 2 + 3 ( x - 2w) 2 2 + 3 ( x - 2) 2 v p w p so, 0, and p are the possible angles Þ w = 1 \ v = = 1 m/s 2 k A 4. y = A sin ( wt - kx) = between v p and v. 2 p 8. w = 2p n = 200 p rad/s, Þ wt - kx = 6 3.5 ´ 10 -3 m w 2p T p k = = w = 200 p \ × - = kx v T 35 T 6 6

Waves & Motion

= 2p rad/m y = A cos ( wt - kx) = A cos (200pt - 2px) ¶y = 2p A sin (200pt - 2px) ¶x When, y = 0 Þ sin (200pt - 2px) = 0 Þ sin (200pt - 2px) = 1 p 1 \ 2pA = ÞA= = 0.025 m 20 40 \ y = 0.025 cos (200pt - 2px) 2p 2p 9. w = = = 8 p rad/s; T 0.25 w 8p p k= = = rad/cm v 48 6 y = A sin ( wt - kx) p = A sin æç 8 p ´ 1 - ´ 67 ö÷ 6 è ø p A = A sin = A sin 30° = = 3 cm 6 2 Þ A = 6 cm 10.

vA vB

= =

TA rA A dB dA

×

rA B

TA

TB =

TA p d2B

=

TB p d2A

dB

TB / 2

TB dB / 2 1 =2´ = 2 2 11. E µ A 2 n2

TB

for E to constant, An =

constant A A n A = A Bn B Þ A A 4n B = A Bn B Þ AB = 4 A A 12. k = 1 rad/m; v = 4 m/s w = vk = 4 rad/s 6 6 \ y= = 2 ( kx - wt) + 3 ( x - 4 t) 2 + 3 Þ

13. v l = Þ

Y and v t = r

Y

Dl l =v l r

Dl l

v l Dl 1 = l = 10 \ = Dl v t l 100

| 7

Dl Dl E =E = l l 100 p p p 14. A = 4 m, w = , k = , q = 5 9 6 w p/5 9 m/s \ v= = = k p/ 9 5 2p 2p l= = = 18 m k p/ 9 w p/ 5 1 n= = = Hz 2 p 2 p 10 Stress = Y

15. w = 10p and k = 01 . p 2p 2p Þ l= = = 20 m k 01 . p 2p 2p \ Df = Dx = ´ 10 = p l 20 2 16. y = (2x - 6.2t) 2 + 20 2 Þ A= = 0.1 m, k = 2 rad/m 20 and w = 6.2 rad/s w 6.2 \ v= = = 3.1 m /s k 2 w 6.2 n= = = 1 Hz 2 p 2 ´ 3.1 2p 2p l= = = pm k 2 1 17. I = 2p2 n2 A 2 rv = w2 A 2 rv 2 2 2 2 E IST 2p n A rv St u= = = V V V 1 = 2p2 n2 A 2 r = rw2 A 2 2 E 2 2 2 P = = I . S = 2p n A rv × S t 1 2 2 = rw A v × S 2 E P E = Pt Þ P = = IS Þ I = t S 18. y = A sin ( px + pt) y ( x, 0) = A sin ( px) Þ y = 0 for x =0 and 1 a = - w2 y = - w2 A sin ( px) 1 3 Þ a = ± w2 A at x = and 2 2

8 | Waves & Motion

1 3 and 2 2 So all the above options are correct. 2p 19. y = A sin ( x - bt) = A sin ( kx - wt) a 2p 2pb k= , w= a a w 2pb/ a Þv= = =b k 2p/ a 2p 2p l= = =a k 2p / a v P = pA cos ( px) Þ v p = 0 for x =

x t x t 20. y = A sin 2p æç - ö÷ = A sin 2p æç - ö÷ è a bø èl Tø Þ Þ 21.

l = a, T = b l a v = nl = = T b

y

a

x Þ

as

x

a = - w2 y

JEE Corner ¢

Assertion and Reason 1. For propagation of transverse waves medium require tension which is possible due to modulus of rigidity. And in gases there is no such Young’s modulus or surface tension. So the reason given is correct explanation. 2. Surface tension of water plays the role of modulus of rigidity and that is why transverse waves can travel on liquid surface. 3. Both the waves are travelling in same direction with a phase difference of p. So reason is false. 4. v = fl is constant for a particular medium so if frequency is doubled wavelength becomes half, and speed remains constant. Thus assertion is false. 5. Sound is mechanical wave which requires material medium for propagation and as on moon there is no atmosphere, sound cannot travel. 2p 6. Angular wave number, n = while l 1 wave number, k = which is defined as l the number of waves per unit length.

7. Electromagnetic wave are nonmechanical, they travel depending upon electric and magnetic properties of medium. They can travel in medium as well as an vacuum. So reason is false. T 1 8. As speed, v = in second Þ vµ m m string m is more (by looking) so v will be less. Thus reason is true explanation of assertion. 9. At point A both v p and Dl is zero ie, K.E. and P.E. are minimum while at B both v p and Dl are maximum i. e., both K.E. and P.E. are maximum. Thus both assertion and reason are true but not correct explanation. 12.

y P x

If P is moving downword then it shows that the wave is travelling in (-) ve x direction. So assertion is false.

Waves & Motion

2p 360° = = 120° 3 3 Assertion is true but the reason is false.

Df , for A = a 2 Df 1 p cos = = cos 2 2 3

11. A = 2a cos Þ

¢

| 9

\

Df =

Match the Columns 1. y = a sin ( bt - cx) = A sin ( wt - kx) (a)

v=

w b = k c

r

(b)

(v p ) max = Aw = ab

s

(c)

n=

w b = 2 p 2p 2p 2p l= = k c

p

(d)

s

(d)

aB = 0 ve y0 = 0®

r

2 2 2 E IST 2 p n A rvst = = V V V 1 2 2 2 2 2 = 2p n A r = rw A 2 [ML2 T -2 ] [u] = = [ML-1 T -2 ] ® s [L3 ] E ISt (b) P = = = IS = 2p2 n2 A 2 rvs t t 1 1 = w2 A 2 rvs = rw2 A 2 s v ® q 2 2 E [ML2 T -2 ] [ P] = = = [ML2 T -3 ] ® p t [T] E [ML2 T -2 ] (c) I = = St [L2 T]

s

v p = 4 p cm /s cos ( pt + 2px) a = - 4 p2 cm /s 2 sin ( pt + 2px) (a)

v p (0, t) = 4 p cm s cos pt = ± 4p for cos pt = ± 1 or pt = np Þ t = n = 0, 1, 2, 3, ®

q, r

(b)

a(0, t) = - 4 p2 cm /s2 sin pt = ± 4 p2 for sin pt = ± 1 p or pt = (2n + 1) 2 1 Þ t = n + = 0.5, 2. 5 ® 2

p, s

v p (05 . , t) = 4 p cm /s cos ( pt + p) = ± 4p for pt + p = np or t = n - 1 = 0, 1, 2, 3, ®

q, r

a (0.5, t) = - 4 p2 cm /s2 p = ± 4 p2Þ pt + p = (2n + 1) 2 1 or t = n - = 0.5, 15 . , 25 . ® 2

p

(d)

vB = ± Aw ®

4. (a) u =

2. y = 4 cm sin ( pt + 2px)

(c)

(c)

= [MT -3 ] = [ML0 T -3 ] ® s 1 (d) = [L-1 ] = [M 0 L-1 T 0 ] ® s l 5.

(a)

y = A sin ( wt - kx) ® v p = wA cos ( wt - kx) ®

p r

a = - w2 A sin ( wt - kx) (b)

y = A sin ( kx - wt) ®

p

v p = - wA cos ( kx - wt) a = - w2 A sin ( kx - wt) (c)

3. y = A sin ( wt ± kx) at t = 0

y = - A cos ( wt + kx) ®

q

v p = wA sin ( wt + kx)

Þ y = ± A sin kx v p = ± w A cos kx and a = - w2 y

a = w2 A cos ( wt + kx) ® (d) y = - A cos ( kx - wt) ®

(a)

v p = ± wA cos kx ®

s

v p = - A w sin ( kx - wt)

(b)

a A = ( +) ve as y A is negative ®

p

a = w2 A cos ( kx - wt) ®

s p ds

15. Superposition of Waves Introductory Exercise 15.1 1. Þ

When displacement of all the particles is momentarily zero, then there is no elastic potential energy stored in the string and as the speed is maximum at mean position, so entire energy is purely kinetic. T 2. (a) v = m v2 m1 Þ = v1 m2 m1 1 = 0.25 m 1 0.25 1 = =2 05 . Þ v2 = 2 v1 = 20 cm/s 2 v2 (b) a t = ai v1 + v2 2 ´ 20 4 = ai = ai 10 + 20 3 v2 - v1 and a r = ai v2 + v1 20 - 10 1 = ai = ai 20 + 10 3 =

3. (a) For fixed end, a phase change of p takes place in reflected wave and direction becomes opposite. as Yi = 0.3 cos (2x - 40 t) Þ Yr = 0.3 cos (2x + 40t + p)

(b) For free end, there is no change in phase for reflected wave and direction becomes opposite. as Yi = 0.3 cos (2x - 40 t) Þ Yr = 0.3 cos (2x + 40 t) w 50 4. v1 = = = 25 m/s and v2 = 50 m/s k1 2 2 ´ 50 2v2 Þ at = ai = ai v1 + v2 25 + 50 4 8 = ´ 2 ´ 10 -3 m = mm. 3 3 v2 - v1 50 - 25 ar = ai = ai v2 + v1 50 + 25 1 2 = ´ 2 ´ 10 -3 m = mm. 3 3 as v2 > v1 Þ the boundary is rearer and there is no phase change. w k2 = v2 50 p = =p 50 2 \ y r = ´ 10 -3 cos p (0.2 x + 50 t) 3 8 and y t = ´ 10 -3 cos p (1.0 x - 50 t) 3 2 ´ 40 cm 5. t1 = = 8 s, inverted 1 cm/s Þ 4cm

Þ

6cm

4cm

Superposition of Waves

4 + 10 + 6 1 = 20 s upright

y

t2 =

1

y

2

2

4

6

8 10 12 14

t = 2s

x

4

6

8 10 12 14

x

1 2

4

t = 2s

4 3

3 1

y 4 3

3

y

6.

2

4

6

8 10 12 14

x

t = 3s

6

8 10 12 14

y

4 3

3

1 2

4

6

8 10 12 14

x

1 2

4

t = 2s

6

8 10 12 14

t = 5s

Introductory Exercise 15.2 px cos 40pt = 2a sin kx cos w t 3 5 p a = = 2.5 cm, k = cm -1 , w = 40 ps -1 2 3 w 40p v= = = 120 cm/s k p/ 3 l 1 2p p p cm = 3 cm Dx = = × = = 2 2 k k p/ 9 dy px vP = = - 200 p sin sin 40 p t dt 3 9 p 3 v P æç 1.5, ö÷ = - 200 p sin æç × ö÷ 8 3 è ø è 2ø 9 sin æç 40 p ´ ö÷ 8ø è pö æ = - 200 p sin ç ÷ sin ( 45 p) è2ø

1. y = 5 sin

= - 200 p ´ 1 ´ 0 = 0 cm/s 2. Two waves with different amplitudes can produce partial stationary waves with amplitude of antinodes being a1 + a2 and amplitude of nodes being a1 ~ a2 . As here node is not stationary that is why energy is also transported through nodes. l 3. (a) = 2 m Þ l = 4 m, 2

T 100 102 = = = 50 m/s m 2 4 ´ 10 -2 v 50 n= = = 125 . Hz l 4 and is fundamental tone or first harmonic. 2p y = 0.1 sin x sin 2 pnt l 2p = 0.1 sin x sin 2p ´ 12.5 t 4 p = 0.1 sin x sin 25 p t 2 l 4 (b) 3 = 2 m Þ l = m and v = 50 m/s 2 3 v 50 n= = Hz = 37.5 Hz and is 2nd l 4 /3 overtone or 3rd harmonic. 2p y = 0.04 sin x sin 2p ´ 37.5 t 4 /3 3p = 0.04 sin x`sin 75 pt 2 400 ´ 4 F Fl 4. v = = = m m 160 ´ 10 -3 v=

= (a)

x

t = 3s

y

1

| 11

1600 = 102 = 100 m/s 16 ´ 10 -2

l0 = l Þ l 0 = 4 l = 16 m 4

x

12 | Superposition of Waves

3 l1 4 l 16 m = 5.33 m = l Þ l1 = = 4 3 3 5l 4 l 16 and 2 = l Þ l2 = m = 3.2 m = 4 5 5 v 100 (b) n 0 = = = 6.25 Hz l0 16 v 100 n1 = = = 18.75 Hz l1 16/ 3 v 100 n3 = = = 31.25 Hz l 3 16/ 5 l 0.54 5. l = n = n = 0.27 n 2 2 l¢ 0.48 and l = ( n + 1) = ( n + 1) 2 2 = 0.24 ( n + 1) Þ 0.27 n = 0.24 n + 0.24 Þ 0.03 n = 0.24 Þ n = 8 (a) These are 8th and 9th harmonic (b) l = 0.27 n = 0.27 ´ 8 = 2.16 m l (c) 0 = l Þ l 0 = 2l = 4.32 m 2 6. 5n 0 - 2n 0 = 54 Hz Þ 3n 0 = 54 Hz Þ n 0 = 18 Hz

7.

n0 = n2 = n1

Þ

=

Þ

M + 2.2

Þ 8.

2.2

1 2l

F m

F2 F1 M + 2.2 2.2

=

260 13 = 220 11

169 48 =1 + 121 121 M =1 + 2.2 48 ´ 2.2 9.6 M = = = 0.873 kg 121 11 =

nn 0 = 250 Hz and ( n + 1)n 0 = 300 Hz Þ n 0 = 50 Hz and n = 5 Þ So these are 5th and 6th harmonics. 1 F n0 = 2l m Þ F = 4 l 2 v20 m = 4 ´ 502 36 ´ 10 -3 ´ = 360 N 1

AIEEE Corner ¢

Subjective Question (Level 1) 1. A =

A12 + A12 + 2 A1 A1 cos 90°

= A1 2 = 4 2 cm = 5.66 cm 2. v2 = 2 v1 v - v1 v 1 Ar = 2 A= 1 A= A v2 + v1 3v1 3 2 v2 4v1 4 At = A= A= A v2 + v1 3v1 3 2

and

I r æ Ar ö 1 =ç ÷ = Ii è A ø 9 It 1 8 =1 - = Ii 9 9

A = 102 + 202 + 2 ´ 10 ´ 20 cos

p 3

= 100 + 400 + 200 = 700 = 10 7 = 26.46 units p 20 sin 3 tan q = p 10 + 20 cos 3 p 3 = sin = 3 2 -1 æ 3 ö Þ q = tan ç ÷ = 0.714 rad è 2 ø \ Phase = 5x + 25t + 0.714 rad. 4. y1 = 1 cm sin ( p cm -1 x - 50 ps -1 t)


p y2 = 1.5 cm sin æç cm -1 x - 100 ps -1 t ö÷ è2 ø Þ 250 p ö æ y1 ( 4.5, 5 ´ 10 -3 ) = 1 cm sin ç 4.5 p ÷ 1000 ø Þ è

| 13

Þ t=0s

t = 0.01 s

t = 0.02 s

9 p = 1 sin æç p - ö÷ 4ø è2 17 p ö æ = 1 sin ç ÷ è 4 ø p = 1 cm sin æç 4 p + ö÷ 4ø è p 1 cm and = 1 sin = 4 2

6. (a) 1 cm/3 Þ

1 cm

1cm 1cm 1cm t=1s

1 cm 1 cm 2 cm

t=0 1 cm/s Þ

1 cm

æ 9 p 500 p ö y2 ( 4.5, 5 ´ 10 ) = 1.5 cm sin ç ÷ 1000 ø è 4 -3

1 cm/s

1 cm

1cm 1cm t = 2s

Þ t = 3s

1 cm 1 cm

\

5. v =

9p p ö = 1.5 cm sin æç - ÷ 2ø è 4 5p ö æ = 1.5 sin ç ÷ è 4ø p = 1.5 sin æç p + ö÷ 4ø è pö æ = - 1.5 sin ç ÷ è4ø 1.5 =cm 2 1.5 1 y = y1 + y2 = 2 2 0.5 1 ==cm 2 2 2 16 N T = -3 m 0.4 ´ 10 ´ 102 kg/N =

16 ´ 102 4

= 20 m /s

(a) For same shape, time, 2 l 2 ´ 0.2 t= = s = 0.02 s v 20 (b)

1 cm t = 4s

(b) 1 cm

1 cm/s Þ

1 cm 1 cm 1 cm t=0 1 cm

1 cm/s

1 cm

1cm 1cm 1cm t = 1s

1 cm/s

2 cm

Þ 1cm 1cm t = 2s

1 cm t = 3s

1 cm 1cm 1cm t = 4s

7. y = 1.5 sin (0.4 x) cos (200t) = 2A sin kx cos wt 2p 2p l= = = 5p m = 15.7 m k 0.4 w 200 100 n= = = Hz = 31.8 Hz 2p 2p p w 200 v= = = 500 m /s k 0.4


8. y = y1 + y2 = 3 cm sin ( px + 0.6 pt) + 3 cm sin ( px - 0.6 p t) = 6 cm sin px cos 0.6 pt = R cos 0.6 pt where, R = 6 cm sin px. 1 (a) R (0.25) = 6 cm sin p ´ 4 6 = = 3 2 cm = 4.24 cm 2 1 (b) R (0.50) = 6 cm sin p ´ = 6 cm 2 3p (c) R (1.50) = 6 cm sin = - 6 cm 2 Þ |R| = 6 cm (d) For antinodes, R = ± 6 cm p Þ sin px = ± 1 Þ px = (2n + 1) 2 1 or x = n + = 0.5 cm, 1.5 cm, 2.5 cm 2 2p 2p 9. l = = = 4 cm 4 p/ 2 (a) Distance between successive l antinodes = = 2 cm 2 (b) R( x) = 2 A sin kx p = 2 ´ p cm sin ´ 0.5 2 p = 2p sin 4 2p = = 2 p cm 2 n+1 T n+1 20 10. n n = = 2l m 2 ´ 20 9 ´ 10 -3 =

n+1

´

100 2

=

5 2 5 2 = ( n + 1) 9 9

60 3 = 0.786 ( n + 1) = 0.786 Hz 1.57 Hz, 2.36 Hz, 3.14 Hz 11. (a) T = mv2 = mn2 l2 1.2 ´ 10 -3

´ (220) 2 ´ (1.4) 2 0.7 = 162.6 N (b) n2 = 3 n 0 = 3 ´ 220 Hz = 660 Hz =

n+1 T 50 = 2l m 2 ´ 0.6 0.01 50 2 = ( n + 1) = 58.93 ( n + 1) Hz 1.2 n n £ 20,000 Hz Þ n = 338 \ n 338 = 339 ´ 58.93 = 199758 . Hz = 19.976 kHz

12. n n =

n+1

13. nn 0 = 420 Hz and ( n + 1) n 0 = 490 Hz Þ n 0 = 70 Hz and n = 6 450 1 T 1 T . \n0 = Þl = = 0005 2l m 2n 0 m 2 ´ 70 300 = = 2.143 m 140 v 400 m /s 1 14. l = = = m, n 800 Hz 2 l l = 4 = 2l = 1 m 2 (a) 4 n 0 = 400 Hz Þ n 0 = 100 Hz (b) 7n 0 = 700 Hz 1 T 16. n 0 = 2l m 1 Þ n0 µ l n1 : n2 : n 3 = 1 : 2 : 3 1 1 1 = : : l1 l2 l 3 1 1 1 Þ l1 : l2 : l 3 = : : 1 2 3 = 6 : 3 : 2 = 6x : 3 x : 2x 6x + 3 x + 2x = 1 m 1 Þ x= m 11 6 \position of first bridge = 6x = m 11 and position of second bridge 9 m = 6x + 3 x = 9 x = 11 9 2 From the same end or 1 m = 11 11 from other end .


17. n 0 =

v 2l

Þ

n¢0 =

Þ

l¢ =

(b) 2A = 5.60 cm Þ A = 2.80 cm l 3 2p 3 p (c) l = 3 = × = 2 2 k k 3p = cm = 277.2 cm 0.0340 2p 2p (d) l = = cm = 184.8 cm k 0.0340 w 50 n= = = 7.96 Hz 2p 2p 1 1 T= = s = 0.216 s n 7.96 v = nl = 7.96 Hz ´ 184.8 cm = 1470 cm /s (e) (v p ) max = Rmax w = 2 Aw = 5.60 cm ´ 50 rad/s = 280 cm/s (f) for eight harmonic, l l 277 .2 8 = l Þ l¢ = = = 69.3 cm 2 4 4 2 p 2 ´ 3.14 k= = = 0.0907 rad/cm l¢ 69.3 8 8 v¢ = 8 v 0 = × v = ´ 7.96 Hz 3 3 = 21.22 Hz w¢ = 2pn = 133.4 rad/s Þ y = 5.60 cm sin (0.0907 rad/s × x) sin × (133 rad/s × t)

v 2 l¢

n0 n¢0

l=

124 ´ 90 cm = 60 cm 186

Thus length of the vibrating string has to be 60 cm. l 18. = 15 cm Þ l = 30 cm, 2 Rmax = 2 A = 0.85 cm, T = 0.075 s (a) y = 2 A sin kx sin wt æ 2p ö æ 2p ö = 0.85 cm sin çç x ÷÷ sin çç t ÷÷ è 0.3 m ø è 0.075 s ø w 2p/ 0.075 0.3 (b) v = = = = 4 m /s k 2p/ 0.3 0.075 l 30 (c) = = 7.5 cm 4 4 \ R (7.5 + 3) = 2 A sin kx 2p = 0.85 sin ´ 10.5 30 21p ö \ R(115 . cm) = 085 . sinæç ÷ è 30 ø = 0.85 sin (0.7 p) = 0.85 (126° ) = 0.688 cm 19. n 0 = and

v 48 = = 16 Hz 2 l 2 ´ 1.5

| 15

21. (a) v = nl = n 0 × 2 l = 60 ´ 2 ´ 0.8 = 96 m/s

l 0 = 2l = 3 m

v 48 n2 = 3n 0 = 48 Hz and l2 = = =1 m n2 48

(b) T = m v2 =

40 ´ 10 -3 80 ´ 10 -2

´ (96) 2

962 = 460.8 N 20 (c) (v p ) max = Rmax w = 0.3 cm ´ 2p ´ 60 rad/s = 113 cm /s = 1.13 m/s a max = w2 Rmax = (120 p) 2 ´ 0.3 cm /s 2 = 426.4 m /s 2

n 3 = 4 n 0 = 64 Hz v 48 3 and l 3 = = = = 0.75 m n 3 64 4

=

20. y = 5.60 cm sin (0.340 rad/cm x) sin (50.0 rad/s t) = 2 A sin ( kx) sin ( w t) (a) ¢

Objective Questions (Level 1)


1.

n2 = n1

T2 3 Þ = T1 2

T + 2.5 T

9 T = 4 ( T + 2.5) 5T = 10 Þ T = 2N n+1 T n+1 T 2. n = = 2l m 2l pr2 r n+1 T n+1 T = = = constant. 2 l r pr ld pr ld n+1µ T n1 + 1 l1 d1 T2 = × n2 + 1 l2 d2 T1 1 1 = ´ ´ 2 2 3 1 = =1:3 2 3 2 n +1 or 2 =3 2 n1 + 1 1 3. f µ ; l = l1 + l2 + l 3 l 1 1 1 1 Þ = + + f0 f1 f2 f3 Þ Þ

4. During overlapping the displacement of particles is zero while velocity is maximum. So the entire energy is purely kinetic. 5.

y ( x, y) = y1 + y2 = a cos ( kx + wt) + y2 = - 2a sin kx sin wt is necessary for a node at x = 0 . Thus, y2 = 2a sin kx sin wt - a cos ( kx + wt) = - 2a sin kx sin wt - a cos kx cos wt + a sin kx sin wt = - a[cos kx cos wt + sin kx sin wt] = - a cos ( kx - wt)

6. In transverse stationary wave, longitudinal strain is maximum at node. While in longitudinal stationary wave at displacement node pressure and density are maximum. So all are correct.

7. In stationary wave all particles errors the mean position simultaneously and are at their maximum displacement simultaneously at different instant at this time all of them are at rest. So all are correct. 8. Maximum displacement y max = 3 A - A + 2 A = 4 A Y Dl vt vt = = vl vl

= 10. f n =

n+1 2´ 1

l r Y r

=

Dl l

1 l h 1 = l h 100 = 50 ( n + 1) 0.01

= 50 Hz, 100 Hz, 150 Hz 2n + 1 100 nn = = 25 (2m + 1) 4 ´ 1 0.01 = 25 Hz, 75 Hz, 125 Hz f + f2 \ n2 = 75 Hz = 1 2 50 Hz + 100 Hz = = 75 Hz 2 11. In stationary waves all particles perform SHM such that they are at their positive and negative N N extremes N one time each in a time period, where they come to rest. Particles between two successive nodes are in phase while beside node are in opposite phase. So all the particles cannot be at positive extreme simultaneously. 12. The question is wrong, string has to be fix at one end and free at other. Then (2n + 1) n 0 = 90 Hz, (2n + 3) n 0 = 50 Hz and (2n + 5) n 0 = 210 Hz


Þ 2 n 0 = 60 Hz or n 0 = 30 Hz and n = 1 i. e., vibrations are 3rd, 5th and 7th harmonic. l 0 = 2l = 1.6 m \ v = n 0 l 0 = 30 Hz ´ 1.6 m = 48 m /s p 13. y = y1 + y2 + y 3 = 12 sin æç q - ö÷ 2ø è p æ + 6 sin ( q + 0) + 4 sin ç q + ö÷ 2ø è

17. R = 2 A sin Kx = 4 mm sin 2mm

= 4 mm sin

(2n + 3) n 0 = 175 Hz 2n 0 = 70 Hz n 0 = 35 Hz 1 1 1 1 1 1 15. l1 : l2 : l 3 = : : = : : n1 n2 n 3 1 3 4 = 12 : 4 : 3 \ 12x + 4 x + 3 x = 114 cm 114 Þ x= cm = 6 cm 19 \ l1 = 12x = 72 cm, l2 = 4 x = 24 cm, l 3 = 3 x = 18 cm 16. f µ T f /2 f

=

Vrg - Vs1 g

= 1-

s2 1 = r 2

Vrg s1 1 s 3 Þ 1= Þ 1 = r 4 r 4 f /3 Vrg - Vs2 g s 1 = = 1- 2 = f Vrg r 3 s s 1 8 Þ 1- 2 = Þ 2 = r q r 9 s1 3 / 4 27 s 32 \ = = Þ 2 = = 1.18 s2 8 / 9 32 s1 27 where s1 is density of water and s2 is density of the other liquid.

x

2

2px

2 mm = 4 mm sin

3m 2 px

3 p Þ = 3 3 Þ x = 0.5 m Thus points 1 and 2 are at 0.5 m from their nearest boundary. So separation between them is 1.5 m - 2 ´ 0.5 m = 0.5 m = 50 m 2 px

= 100 = 10 mm and Þ Þ

l

4mm 2mm

1

= 6 sin q - 12 cos q + 4 cos q = 6 sin q - 8 cos q p Þ R = 62 + 8 2 + 2 ´ 6 ´ 8 cos 2 14. (2n + 1) n 0 = 105 Hz

2p

| 17

18. y = - A sin ( wt - kx) 2pn ö = - A sin æç 2pnt x÷ v è ø = - A sin (6 pt - 2px) y(3, t) = + A = - A sin (6 pt - 6 p) = A sin (6p - 6p t) p 11 \ 6p - 6p t = Þ = 6t 2 2 11 Þ t= s 12 2p 3 2p 19. Df1 = Dx = × l =3p l l 2 p p 5p and Df2 = - Þ Df = 3 p - = 2 2 2 20.

nn 0 = 400 Hz , ( n + 1)n = 450 Hz

Þ n 0 = 50 Hz and n = 8 1 T 1 T n0 = Þl = 2l m 2n 0 m 1 490 70 = = = 0.7 m 2 ´ 50 0.1 100 l 2 21. 3 × = 1 m, l = m 2 3 2 v = nl = 300 Hz ´ m = 200 m/s 3 l1 2l2 3 l 3 22. l = , , 2 2 2


2l 2l , l3 = 2 3 2l 2l l1 : l2 : l 3 = 2l : : 2 3 1 1 =1: : 2 3 l1 = 2l, l2 =

Þ \

23. Df = = =

2p l 2p vT 2p

Dx Dx =

2p 300 ´ 0.04

´ (16 - 10)

´6=p

12 w 30 24. v = = = 30 m/s k 1 T T = = m rA \ T = rAv2 = 8000 ´ 10 -6 ´ 900 = 7.2 N 2 25. 5n 0 = 480 Hz, 2n 0 = ´ 480 Hz 5 = 192 Hz 2

26.

Ir æA ö = 0.64 = ç r ÷ Ii è Ai ø Ar Þ = 0.8 Þ A r = 0.8 A i Ai v - v1 4 Ar = 2 A i = 0.8 A i = A i v2 + v1 5 Þ

5 v2 - 5 v1 = ± ( 4v2 + 4v1 )

Þ

v2 = 9v1 ,

1 v1 9

For, v2 > v1 the boundary is rarer and there will not be any change in phase of reflected wave and for v2 < v1 a phase change of 180° takes place. \ Yr = 0.8 A sin ( kx + wt + 30°+ 180° ) 1 T 1 T 27. n A = = 2 2 l rpd ld pr

nB

4 1 = 2 ´ 2l

2T 2r p

4 d2 4

=

1 4 ld

T 1 = nA pr 4

\Third overtone of n B = 4n B = n A ¢

Passage

(Q 28 to 30)

I r = (100% - 36%) I i = 64% I i = 0.64 I i v - v1 Ar Ir \ = = 0.64 = 0.8 = ± 2 Ai Ii v2 + v1 Þ 0.8 v2 + 0.8 v1 = ± (v2 - v1 ) Þ - 0.2 v2 = 1.8 v1 Þ v2 = 9 v1 for rarer boundary 1 or 1.8 v2 = 0.8 v1 Þ v2 = v1 9 for danser boundary 28. A r = 0.8 A p 29. Y = A sin æç ax + bt + ö÷ + 0.8 2ø è p A sin æç ax - bt + + p ö÷ 2 è ø p = A sin æç ax + bt + ö÷ - 0.8 2ø è p A sin æç ax - bt + ö÷ 2ø è = A cos ( ax + bt) - 0.8 A cos ( ax - bt) = A cos ax cos bt - A sin ax sin bt - 0.8 A cos ax cos bt - 0.8 A sin ax sin bt = 0.2 A cos ax cos bt - 1.8 A sin ax sin bt = 0.2 A cos ax cos bt - 0.2 A sin ax sin bt - 1.6 A sin ax sin bt = 0.2 A cos ( ax + bt) - 1.6 A sin ( ax) sin ( bt) = cA cos ( ax + bt) - 1.6 A sin ax sin bt Þ e = 0.2 30. For antinodes, sin ax = ± 1 p Þ ax = (2n + 1) 2 p p 3 p 5p , x = (2n + 1) = , 2a 2 a 2a 2a


So for second antinode, x = 31.

n 0 + 15 n0 Þ

=

1 + 0.21 1

3p

| 19

constant, wavelength increases while frequency is constant, wavelength increases while phase do not change during change in medium.

2a

= 1×1

15 = 0.1 n 0 Þ n 0 = 150 Hz 1.21 n2 T2 = = = 11 . n1 T1 1

Þ n2 = 110% of v1 l 0 = 2l which do not change So, (a), (c) and (d) are correct. 32. For interference, sources must be coherent there frequency has to be equal and phase different has to be constant. So, (a) and (d) are correct. 33. Stationary waves are formed due to superposition (here use of the term ‘interference’ is literary and not scientific because interference is a different phenomenon than stationary waves) of waves having some amplitude, same frequency and travelling opposite direction. Here nodes are the points who always remain at rest. Total energy is always conserved. 34. A medium is said to be rarer if speed of wave in it is higher. And as frequency is

35. Y = A sin kx cos wt = 2a sin kx cos wt A a = , third overtone means fourth 2 harmonic and wire oscillate with four loops. 2p 4 p l l = 4 = 2l = 2 × = 2 k k and stationary wave do not propagate. 36. For stationary waves, frequency and amplitude has to be same and direction has to be opposite with constant phase difference. It is satisfied in (b) and (d) only. 37.

y = y1 + y2 = 2 A cos kx sin wt = R sin wt R = 2 A cos kx so at x = 0 there is antinode. \ cos kx = ± 1 np p 2p , Þ kx = np, x = = 0, , k 4 x are antinodes.

JEE Corner ¢

Assertion and Reason 1. y1 + y2 = A sin ( wt + kx) + A cos ( wt - kx) p = A sin ( wt + kx) + A sin æç - wt + kx ö÷ è2 ø p wt + kx + - wt + kx 2 = 2 A sin 2 p wt + kx - + cot - kx 2 cos 2

p p = 2 A sin æç kx + ö÷ cos æç wt - ö÷ 4ø 4ø è è p = R cos æç wt - ö÷ 4ø è p where, R = 2 A sin æç kx + ö÷; 4ø è p R(0) = 2 A sin = A 2 4 So, at x = 0, node is not present, i. e., Assertion is false. 2. In stationary waves only nodes are at rest and not other particles. It is so


p + A sin æç q + ö÷ 2ø è

called as energy is not transmitted, thus assertion is false.

= - A cos q + A sin q + A cos q = A sin q \ R = AÞ I f = Ii Assertion and reason are both true but reason do no explain assertion.

3. In rarer medium speed of wave is higher and as 2 v2 At = Ai v1 + v2 Þ At > Ai so reason is correct explanation to assertion. 4. In second overtone or third harmonic there are three loops or three antinodes or four nodes. And length of the string, l l = 3 so, assertion and reason are both 2 true.

10. For two coherent sources phase difference has to be constant and that constant be same at all points as Df ¹ Df ( t). Different light sources can never be coherent. So phase difference must be same, thus assertion is false. ¢

Match the Columns v T T and v2 = = 1 m 9m 3 v1 Þ =3 v2 v - v1 / 3 2/ 3 1 Ar = 1 Ai = Ai = Ai v1 + v1 / 3 4 /3 2 2 v2 and A t = Ai v1 + v2 2 v1 / 3 1 = Ai = Ai v1 + v1 / 3 2 A1 A r 1/ 2 A i (a) = = = 1®q A2 A t 1/ 2 A i v (b) 1 = 3 ® r v2

1. v1 = N

A

N

A

N

A

N

5. As speed of wave is constant in stretched wire, and v = fl, so with increase in frequency, wavelength decreases. So reason is correct explanation of assertion. 6. In stationary waves, amplitude of nodes is zero and it is possible only when superposing waves has same amplitude. But it is not the only condition, there has to be same frequency, opposite direction of propagation and constant phase difference. So assertion is not completely true. 7. Energy lying between conservative node and antinode is constant where it moves to and fro between node and antinode. 2

I 25 æ 5 ö2 æ A1 + A2 ö ÷÷ 8. max = = ç ÷ = çç I min 1 è1ø è A1 - A2 ø

Þ 5 ( A1 - A2 ) = A1 + A2 Þ 4 A1 = 6 A2 Þ A1 : A2 = 3 : 2. Thus reason is the correct explanation of assertion. p 9. y = A sin æç q - ö÷ + A sin q 2ø è

2

I æA ö 1 2 1 (c) r = ç r ÷ = æç ö÷ = and I i è Ai ø 4 è2ø It 1 3 =1 - = Ii 4 4 I 1 I r I r / I i 1/ 4 1 \ = = = = ®s I2 It I t / I i 3/ 4 3 (d) P = IS = 2p2 n2 A 2 r vr 1 T = w2 A 2 m 2 m 1 2 2 = w A Tm 2


1 2 2 w A1 Tm 1 P1 2 = P2 1 w2 A 2 Tm 2 2 2 A2 m 1 m1 = 12 = 9m 1 A2 m 2 1 = ®s 3 3 v n 3 2. (a) 2 = 2l = ® r 5 n4 v 5 2l (b) Number of nodes in 3rd harmonic is 4 and in Fifth harmonic 6, 4 2 so, = ® p 6 3 (c) Number of antnodes in 3rd harmonic is 3 and in fifth harmonic 3 5, 50, ® r 5 l2 n4 5 (d) = = ®s l4 n2 3 3. In danser medium speed of wave is lesser and in rarer medium it is greater. (a) When wave goes from denser to rarer medium its speed increases ® p (b) As frequency do not change with change in medium then with

| 21

increase in speed wavelength increases ® p (c) As v t > v i then A t > A i ® p (d) Frequency remains unchanged ® r 4.

A 2 + A 2 + 2 × A × A × cos q q = 2 A cos 2 60° (a) R(60° ) = 2 A cos 2 3 = 2 A cos 30° = 2 A × =A 3®s 2 (b) R(120° ) = 2 A cos 120°/ 2 = 2 A cos 60° 1 = 2A × = A ® s 2 (c) R (90° ) = 2 A cos 90°/ 2 = 2 A cos 45° = A 2 Þ I R = 2 A 2 = 2I i ® p (d) R(0° ) = 2 A cos (0°/ 2) = 2 A Þ I R = 4 A2 = 4I i ® r R=

5. n2 = 3n 0 = 210 Hz Þ n 0 = 70 Hz (a) n 0 (b) n2 (c) n 3 (d) n1

= 70 Hz ® s = 3n 0 = 210 Hz ® p = 4n 0 = 4 ´ 70 Hz = 280 Hz ® r = 2 n 0 = 140 Hz ® s

16. Sound Waves Introductory Exercise 16.1 p0 p0 = kB 2 p × rv2 l p0 vl p0 l p0 = = = 2 2 2p nrv 2p rv 2p nr v 10 m = 2 ´ 3.14 ´ 10 3 ´ 1.29 ´ 340

1. P0 = S0 kB B=

Þ

=

(b) S0 = P0 P l = 0 S0 k 2pS0 14 ´ 0.35 2 ´ 3.14 ´ 5.5 ´ 10 -6

= 1.4 ´ 105 N/m 2 1450 m /s v 2. l = Þ l max = = 72.5 m, n 20 Hz 1450 m /s l min = = 7.25 cm 20000 Hz 3. Pressure wave and displacement wave p has a phase difference of , so, 2 (a) When pressure is maximum, displacement is minimum i.e., zero.

= 3.63 ´ 10 -6 m P P0 P 4. S0 = 0 = = 0 kB 2p nr v wrv P0 k = 2 rw 12 ´ 8.18 = 129 . ´ (2700) 2 = 1.04 ´ 10 -5 m

Introductory Exercise 16.2 1.

v2 = v1

T2 = 2 Þ T2 = 4 T1 = 4 ´ 273 K T1

= 3 ´ 273 ° C = 819 ° C t ö1/ 2 t ö æ 2. v t = v 0 æç 1 + ÷ = v0 ç 1 + ÷ 273 ø 546 ø è è 30 3 ù v 30 - v 3 = v 0 é1 + -1 + êë 546 546 úû 33 ö = v 0 æç ÷ è 546 ø 33 = 332 ´ = 20.06 m/s 546

3.

v = nl = 250 ´ 8 = 2000 m/s

B = rv2 = 900 ´ (2000) 2 = 36 ´ 10 8 N/m = 3.6 ´ 10 9 Pa 7 ´ 8.314 ´ 273 gRt 4. v = = 5 M 32 ´ 10 -3 = 315 m/s

Sound Waves

| 23

Introductory Exercise 16.3 3 ö = 20 log æç ÷ = 20 dB è 0.3 ø

1. P0 = S0 kB = 2p nrv S0 = 2 ´ 3.14 ´ 300 ´ 1.2 ´ 344 ´ 6 ´ 10 -6 = 4.67 Pa P2 ( 4.67) 2 I = 0 = 2 rv 2 ´ 1.2 ´ 344 = 2.64 ´ 10 -2 W/m 2 2.64 ´ 10 -2 I L = 10 log = 10 log I0 10 -12 = 104 dB 2.

hI I 2L - L = 10 log - 10 log I0 I0

= 10 log (h) = 9 dB Þ log h = 0.9, h = 10 0. 9 = 7.9 1 k 3. I µ 2 Þ I = 2 r r I LF - LM = 10 log F IM ær = 10 log çç M è rF

4. (a) I =

P02 (28) 2 ; I max = 2rv 2 ´ 1.29 ´ 345

= 0.881 W/m 2 0.881 Lmax = 10 log = 119.45 dB 10 -12 (2 ´ 10 -5 ) 2 I min = 2 ´ 1.29 ´ 345 = 4.49 ´ 10 -13 W/m 2 4.49 ´ 10 -13 dB Lmin = 10 log 10 -12 = - 3.48 dB Po Po (b) S0 = = kB 2pnrv 28 ( S0 ) max = 2 ´ 3.14 ´ 500 ´ 1.29 ´ 345

2

ö ÷ ÷ ø

( S0 ) min

= 2 ´ 10 -5 m 2 ´ 10 -5 = 2 ´ 3.14 ´ 500 ´ 1.29 ´ 345 = 1.43 ´ 1011 m

Introductory Exercise 16.4 l = 12 cm 2 l and (2n + 1) = 36 cm 2 Þ l = 36 - 12 = 24 cm v 330 m /s n= = = 1375 Hz l 0.24 m l l p l 2. Dx = Df = × = 2p 2p 3 6 v 350 = = = 0.117 m = 11.7 cm 6 n 6 ´ 500 2p Df = Dt = 2p nD t = 2p ´ 500 ´ 10 -3 T = p rad = 180° 1.

(2n - 1)

3. Dx1 = 2 H 2 +

d2 - d = nl 4

and Dx2 = 2 ( H + h) 2 + Þ

d2 1 - d = æç n + ö÷ l 4 2ø è

l d2 d2 = 2 ( H + h) 2 + - 2 H2 + 2 4 4

or l = 4 ( H + h) 2 +

d2 -4 4

H2 +

d2 4

l = 2 4 ( H + h) 2 + d2 - 2 4 H 2 + d2 1 4. Dx p = d sin q = æç n + ö÷ l for minima 2ø è

24 | Sound Waves Y

Df =

P S1 q

d

=

í ì

l for first minima 2 l v ö q = sin -1 æç ö÷ = sin -1 æç ÷ è 2d ø è 2nd ø æ ö 340 ÷ = sin -1 çç ÷ è 2 ´ 600 ´ 2 ø

(a) \ d sin q =

= sin -1 (0.142) = 0.142 rad = 8.14 ° (b) For, first maxima d sin q = l l 340 ö Þ q = sin -1 æç ö÷ = sin -1 æç ÷ è dø è 1200 ø = 16. 46°

Þ

2 ´ 600

d l

= 3.53 340 n = 3 maxima. =

340

5. (a) For coherent speakers in phase, q I R = 4 I 0 cos 2 2 2p 2p l Df = Dx = × = p= q l l 2 p Þ I R = 4 I 0 cos = 0 2 (b) For incoherent sources, I R = I 1 + I 2 = I 0 + I 0 = 2I 0 (c) For coherent speakers with a phase difference 180°. Df¢ = 180° + Df = p + p = 2p 2p Þ I R¢ = 4 I 0 cos 2 = 4I 0 2 I0 6. 60 dB = 10 log 10 -12 Þ 10 6 ´ 10 -12 = I 0 Þ I 0 = 10 -6 W/m 2

2pn Dx v

´ (11 - 8) = 3 p = q

(a) \ I R = 4 I 0 cos 2

d sin q

(c) Dx max £ d Þ nl £ d, n £

Dx =

l 2 p ´ 170

X

ì

S2

v

2p

3p q = 4 I 0 cos 2 =0 2 2

(b) Df¢ = 3 p + p = 4 p Þ

4p = 4I 0 2 = 4 ´ 10 -6 W/m 2 4 ´ 10 -6

I R¢ = 4 I 0 cos

LR¢ = 10 log

10 -12 = 10 log 10 6 dB + 10 log 4 = 60 dB + 2 log 2 dB = 60 dB + 6 dB = 66 dB 2p ´ 85 2pn (e) Df¢¢ = × Dx = ´ (11 - 8) v 340 3p = =q 2 3p I R ¢¢ = 4 I 0 cos 2 4 p 2æ = 4 I 0 cos ç p - ö÷ = 2I 0 4ø è 2 ´ 10 -6 Þ LR ¢¢ = 10 log = 63 dB 10 -12 10 -3 10 -3 7. (a) I 1 = = 16 p 4 p ´ 22 = 19.9 ´ 10 -6 W/m 2 = 19.9 mW / m 2 10 -3 10 -3 I2 = = 36 p 4 p ´ 32 = 8.84 ´ 10 -6 W/m 2 = 8.84 mW / m 2 (b) ( I P ) max = ( I 1 + I 2 ) 2 = ( 4.46 + 2.97 ) 2 = 55.27 mW/m 2 (c) ( I P ) min = ( I 1 - I 2 ) 2 = ( 4.46 - 2.97) 2 = 2.22 mW/m 2 (d) I P = I 1 + I 2 = 28.7 mW/m 2

Sound Waves

| 25

Introductory Exercise 16.5 1. (a) n 0 =

345 m /s v v Þ lc = = 4lc 4n 0 4 ´ 220 Hz

harmonics are odd, which can be seen in closed organ pipe only. (b) These are 5th and 7th harmonic. v (c) n 0 = 4lc v 344 Þ lc = = = 1.075 m 4n 0 4 ´ 80

= 0.392 m 3v (b) = 5n 0 2l 0 3v 3 ´ 345 Þ l0 = = = 0.470 m 10 n 0 10 ´ 220

4. v = nl = 1000 ´ 2 ´ 6.77 ´ 10 -2 m /s

2. (a) N

AN Fundamental tone

N N A A ANAN A First overtone Second overtone

l 0.8 d A = l = 0.8 m d A = , l = m, 3 0.8 2.43 = m, m , 0.8 m 5 4

dA =

l 3l , ,l 5 5

=

8.314 ´ 400

As 1 < r < 2

(b) NA

A

N A N d A = 0,

Þ n = 2 Þ r = 0.7 ´ 2 = 1.4 =

N N AN A

A

2l 3 = 0,0.533 m

d A = 0n

3.

= 135.4 m /s gRT Mv2 v= Þ r= M RT n ´ 127 ´ 10 -3 ´ (135.4) 2

2l 4l , 5 5 = 0 m , 0.32 m, 0.64 m d A = 0,

5. n =

(2n + 1)v

=

= 0.7 n

7 diatonic 5

(2n + 3)v

4 l1 4 l2 2n + 3 l2 100 5 = = = 2n + 1 l1 60 3

2 400, 560 0 4 20, 28

n =1 4 l1 n 4 ´ 0.6 ´ 440 \v = = = 352 m/s 2n + 1 3 Þ

Þ HCF of the two shows, 80 and the values, 400 Hz and 560 Hz are odd multiples of 80. These conservative

Introductory Exercise 16.6 252Hz

256Hz

A1

260Hz

>4

<4

n A = 252 Hz n A = (256 ± 4) Hz and n A - n = (256 ± 6) Hz \ 256 ± 4 - n = 256 ± 6 ± 4 m 6 = nÞ n = - 4 + 6 = 2

n A = 256 - 4 = 252 Hz

\

A

2.

A'1

Þ

A'

B

381

384

A1

A 387

ý

B

ý

A'

ý

A'1

ý

1.

>3

<3

n A = 387 Hz n A = (384 ± 3) Hz and n A - n = 384 ± m, m < 3 \ 384 ± 3 - n = 385 ± m Þ

26 | Sound Waves

Þ Þ Þ Þ

and Þ

±3-n=±m ± 3 m m = n = ( +) ve n=+3-m n A = 384 + 3 = 387 Hz 1 TA 6 Hz = 600 Hz = 2l m 600 Hz = 606 = 600

= 1.02 TB v 4. 256 ± 4 = 2 ´ 0.25 v and 256 = 2 ´ (0.25 - x) 2 ´ 0.25 256 1 = = 252 2 (0.25 - x) 1 - 4 x

TB

1 2l

m

TA TB

TA

Þ

= 1.01

256 - 4 ´ 256 x = 252 4 = 4 ´ 256 x 1 100 x= m= cm = 0.4 cm 256 256

Þ

Introductory Exercise 16.7 1. When source is moving, v 1 n s¢ = n= n v v + vs 1m s v -1 v æ ö = ç1 m s ÷ n v ø è v ö u æ = ç 1 ± s ÷ n = æç 1 ± ö÷n v ø vø è è When observer is moving, n 0 =

v ± v0 v

v ö v æ = ç 1 ± 0 ÷ n = æç 1 ± ö÷ n v ø vø è è So, it can be seen that, n 0 and n s are equal if u << v. 340 2. l = = 1.7 m 200 80 (a) l¢ = l - uT = 1.7 m = 1.7 m 200 - 0.4 m = 1.3 m v 340 m /s (b) n¢ = = l¢ 1.3 m = 262 Hz

n

3. For doppler effect there has to be relative motion between source and receiver, but as they are at rest relative to each other that’s why there is no shift in wavelength and frequency. v 344 4. l = = = 0.688 m n 500 30 (a) l front = l - uT = 0.688 500 = 0.688 - 0.060 = 0.628 m (b) l behind = l + uT = 0.688 + 0.060 = 0.748 m 344 (c) n front = = 547.8 Hz 0.628 344 (d) n behind = = 459.9 Hz 0748 . v - w - v0 340 - 5 - 20 5. n¢ = n= ´ 300 Hz v - w + vs 340 - 5 + 10 315 = ´ 300 Hz = 273.9 Hz 345 6.

S

vs

P r

Sound Waves

| 27

AIEEE Corner ¢


t1 t +v 2 2 2 v 332 æ 3 5 ö = ( t1 + t2 ) = ç + ÷ 2 2 è 2 2ø

d = d1 + d2 = v

= 332 ´ 2 = 664 m The time for third eco is, 3 5 t = t1 + t2 = + = 4 s 2 2 7 ´ 8.314 ´ 300 gRT 2. v = = 5 M 2 ´ 10 -3 = 21 ´ 8.314 ´ 104 = 1321 m /s gp 3. v = r 5 ´ 76 ´ 10 -2 ´ 13.6 ´ 10 3 ´ 9.8 = 3 0.179 5 ´ 76 ´ 136 ´ 9.8 = = 971 m /s 3 ´ 0.179 4. (a) B = rv2 = rn2 l2 = 1300 ´ 16 ´ 104 ´ 64 = 1.33 ´ 1010 N/m 2 r l2 6400 ´ (15 . )2 (b) Y = rv2 = 2 = t (3.9 ´ 10 -4 ) 2 = 9.47 ´ 1010 Pa 2

5. v t =

Dl Dl Dl æç v t ö÷ F ; =Y vl Þ = l l l çè v l ÷ø A 2

æv ö 1 2 Y = Y çç t ÷÷ = Y æç ö÷ = 900 è 30 ø è tl ø 2 ´ 2 + 1 ´ 14 6. M mix = = 6 m/mole 2+1 v mix = vH 2

MH

2

M mix

=

2 1 = 6 3

T2 v 1 1 vH = v0 = 0 2 T1 3 3 3 v0 1300 = = = 787 m /s 2.73 2.73

v mix =

300 273

10 -6 = 60 log 10 = 60 dB 10 -12 10 -9 L2 = 10 log = 30 log 10 = 30 dB 10 -12 Þ L1 = 2L2 I 8. 100 dB = 10 log dB I0 7. L1 = 10 log

I = 1010 I 0 = 10 -2 W /m 2 P = 4 pr2 I = 4 p ´ ( 40) 2 ´ 10 -2 = 64 p W = 201 W I 9. (a) 60 dB = 10 log dB I0 Þ

Þ I = 10 6 I 0 = 10 -6 W/m 2 (b) P = AI = 120 ´ 10 -4 ´ 10 -6 W = 1.2 ´ 10 -8 watt I 10. (a) DL = 13 dB = 10 log 2 dB I1 Þ

I 2 = 101. 3 I 1 = 20 I 1

(b) As with doubling the intensity, loudness increases by 3 dB irrespective of the initial intensity. P 5 5 11. I = = = 4 pr2 4 p (20) 2 4 p ´ 400 1 = W/m 2 = 9.95 ´ 10 -4 W/m 2 320p 1 I (b) I = 2p2 n2 a2 rv Þ a = pn 2 r v =

1 300 p

1 320 p ´ 2 ´ 129 . ´ 330

=

1 300 p ´ 1012

= 1.15 ´ 10 -6 m

1 85.5

28 | Sound Waves

12. 60 dB = 10 log

I dB 10 -12

Þ I = 10 -6 W/m 2 and a = =

1 pn

T 2 rv

10 -6 = 13.6 ´ 10 -9 m 2 ´ 1 .29 ´ 330

1 800 p

13. 102 dB = 10 log Þ

(c) A a > A w ;

I dB I0

16.

I = 1010.2 I 0 = 1010.2 - 12 = 10 -1. 8 W/m 2 P = 4 p r2 I = 4 ´ 3.14 ´ (20) 2 ´ 10 -1. 8 = 80 W

14. I = 2p2 v2 a2 rv = 2 ´ (3.14) 2 ´ (300) 2 ´ (0.2 ´ 10 -3 ) 2 ´ 1.29 ´ 330 W/m 2 = 30 .25 W/m 2 I 30.25 L = 10 log dB = 10 log dB I0 10 -12 = 134.8 dB 15. (a) v w =

Aw

2.18 ´ 10 9 B = r 10 3

= 1.48 ´ 10 3 m /s I 1 = = 2 2 2p n rv pn =

1 3400 p

I 2rv

3 ´ 10 -6 2 ´ 10 3 ´ 1.48 ´ 10 3

= 9.44 ´ 10 -11 m v 1.48 ´ 10 3 lw = N = = 0.43 m n 3400 1.4 ´ 105 gp (b) v a = = = 341.6 m /s r 1.2 Aa =

1 3400 p

3 ´ 10 -6 2 ´ 1.2 ´ 3416 . -9

= 5.66 ´ 10 m 341.6 la = = 0.1 m 3400

5.66 ´ 10 -9 Aa = = 60 A w 9.44 ´ 10 -11

As bulk modulus of water is much larger than air, such that displacement of particles of medium becomes less. (6 ´ 10 -5 ) 2 p2 I = 0 = W/m 2 2 rv 2 ´ 1.29 ´ 343 = 4 ´ 10 - 12 W 4 ´ 10 -12 I \ L = 10 log = 10 log I0 10 -12

= 20 log 2 = 6 dB v 17. n o = = 594 Hz; 2l v n 0 594 nc = = = Hz = 297 Hz 4l 2 2 ( n + 1)v 344 18. n 0 = = ( n + 1) 2l 2 ´ 0.45 = ( n + 1) ´ 382.2 Hz = 3822 . Hz, 764.4 Hz, 1146.7 Hz, (2n + 1)v nt = 2l 344 = (2n + 1) 2 ´ 0.45 (2n + 1) ´ 191.1 Hz = 191.1 Hz, 573.3 Hz, 955.6 Hz v 19. n c = 4l Þ v = 4 l n c = 4 ´ 0.15 ´ 500 = 300 m /s v 300 no = = = 250 Hz 2l o 2 ´ 0.6 20. y = A cos kx cos wt 2p 330 t = A cos x cos 2p 1.6 1.6 = A cos 3.93 x cos 1296 t 2n + 1 2n + 3 21. n = v= v 4 ´ 0.5 4 ´ 0.84 2n + 3 84 Þ = = 1.68 2n + 1 50

Sound Waves

Þ 3 - 1.68 = 2n ´ 0.68 \ n = 0.97 = 1 as n is an integer 4l v 4 ´ 0.5 ´ 512 v= = m /s 2n + 1 3 = 341.3 m/s 2n + 5 n= n 4l 2n + 5 7 Þ l= v= ´ 341.3 4n 4 ´ 512 = 1.167 m = 116.7 cm v 340 22. n c = = = 85 Hz 4l 4 ´ 1 ns =

v 1 = l 0.4

F = 85 m

Þ F¢ = (85 ´ 0.4) 2 m = (34) 2 ´

4 ´ 10 -3 0. 4

= 11.65 N v 23. n c = Þ v = 4n ( l + l) 4( l + e) = 4n( l + 03 . d) = 4 ´ 480 (016 . + 03 . ´ 005 . ) = 336 m/s (2n + 1) n 24. (a) n e = 4l 5 ´ 330 Þ 440 = 4l 5 ´ 330 15 Þ l= = m 4 ´ 440 16 5l 15 A (b) N =l= 4 16 15 4 3 Þ l= ´ = m 16 5 4 2 p 15 Dp = Dp0 cos kx = Dp0 cos ´ 3 / 4 32 15 p 5p Dp0 = Dp0 cos = Dp0 cos = 12 4 2 (c) At open end there is pressure node, so, pmax = Dpmin = Dp0 (d) At closed end there is pressure antinode, such that, and pmax = p0 + Dp0 pmin = p0 - Dp0

25. (a) n c =

| 29

v 4lc

v 345 = = 0.392 m 4n c 4 ´ 220 5l 3l 3 4 (b) l 0 = , l 0 = = × lc 4 2 2 5 6 6 = l c = ´ 0.392 m = 0.47 m 5 5 vs v 26. n s = n c Þ = s 2 ´ 0.8 l c 4 l c v s 1.6 Þ = = 0.4 va 4 v 340 17 27. (a) l s = = = m = 113 . m n 300 15 v - vs (b) l a = l - v s T Þ n 340 - 30 31 = = = 1.03 m 300 30 v + vs l b = l + vs T = n 340 + 30 37 37 = = = = 1.23 m 300 30 30 1 F Dn 1 DF 28. n = Þ = 2l m n 2 F DF Dn 15 . 3 Þ =2 =2´ = = ± 0.68% F n 440 440 n¢ = n + Dn = 440 ± 1.5 = 438.5 Hz or 441.5 Hz Þ lc =

29. v = 0.32 m/s; l = vT = 0.32 ´ 1.6 m = 0.512 m. l - l¢ l¢ l¢a = l - v s T Þ v s = =v T T 0.12 = 0.32 = 0.245 m/s 1.6 l b ¢ ¢ = l + v s T = 0.512 m + 0.245 ´ 1.6 = 0.512 + 0.392 = 0.904 m. v - v0 340 + 18 30. (a) n a = n= ´ 262 Hz v - vs 240 - 30 358 = ´ 262 Hz = 302.5 Hz 310 v - v0 340 - 18 (b) n r = n= ´ 262 v + vs 340 + 30

30 | Sound Waves

322 ´ 262 Hz = 228 Hz 370 v v 31. Dn = nn v - vs v + vs 2 vv n 2v n vDn = 2 s2 ~ - s Þ v= v 2v s v - vs 340 ´ 4 \ n= = 680 Hz 2´ 1 v 32. n 0 = n c ± Dn = 110 ± 2.2; n c = 4lc v 330 Þ lc = = 4n c 4 ´ 110 2v 3 \ l c = m; n 0 = 4 2 l0 2 ´ 330 2v Þ l0 = = 2n 0 2 (330 ± 2.2) =

= 0.993 m or 1.007 m 7 33. n p = n Q ± and n P < n Q as beat 2 frequency increases waxing of P. v 332 nQ + 5 = nQ = nQ v - vs 332 - 5 332 5 = nQ Þ5 = nQ 327 327 Þ n Q = 327 Hz and 7 n P = 327 - = 323.5 Hz 2 When Q gives 5 beats with its own echo. OR 7 332 n P = n Q - = n¢ q - 5 = nQ - 5 2 327 7 5 Þ 5- = nQ 2 327 327 ´ 1.5 Þ nQ = = 98.1 Hz 5 Þ n P = 98.1 - 2.5 = 94.6 Hz When P gives 5 beats with the echo of Q. 2vv n 2v n v v 34. Dn = nnÞ 2 s 2 ~ - s v - vs v + vs v v - vs vn 340 ´ 2 Þ vs = = 2 n 2 ´ 680

1 \ v s = m/s 2 l 35. (2n + 1) = 11.5 cm 2 l (2n + 3) = 34.5 cm 2 2n + 3 34.5 Þ = = 3 Þ 4n = 0 Þ n = 0 2n + 1 11.5 l \ = 11.5 cm Þ l = 23 cm 2 v 331.2 m/s n= = = 1440 Hz l 0.23 m v 330 36. l = = = 1.5 m n 220 3 9 Dx = S2 P - S1 P = 3 - = m 4 4 3 3 3 l = × = l = (2n + 1) 2 2 2 2 3 1 Here, S1 P = = l 4 2 2p 2p l Þ f1 = S1 P = × =p l l 2 3 and S2 P = 3 = 2 × = 2l 2 2p Þ f2 = × 2l = 4 p l Destructive interference will take place at P. \ PP = Pmin = ( P1 - P2 ) 2 = ( 1.8 ´ 10 -3 - 1.2 ´ 10 -3 ) 2 = 0.6 ´ 10 -3 ( 3 - 2) 2 = 0.6 ´ 10 -3 ´ 0.1 = 6 ´ 10 -5 W x 2 37. Dx = 2 22 + æç ö÷ - x = nl = 1 ´ l è2ø 360 m/s = =1m 360 Hz x2 4 æ x2 4 çç 4 + 4 è

2 4+ or

=1 + x ö ÷ = 1 + 2x + x2 ÷ ø

16 - 1 = 2x x = 7.5 m

Sound Waves

Objective Questions (Level 1)

2. Longitudinal waves can travel through all mechanical mediums. gR ´ 288 gRT 3. = 32 28 32 8 T= ´ 288 K = ´ 288 K = 56° C 28 7

1 ´ 1.01 ´

0.01 v 1 1 11. n¢ = n = n = nn Þ n = = 0.5 v+v 2 2 12. I max = ( I + I ) 2 = 4 I = NI Þ N = 4 v 3v 13. n = = 4 ( l1 + e) 4 ( l2 + e)

4. Third overtone is 7th harmonic ie, there 4 nodes and 4 antinodes. A

l2 + e = 3 l1 + 3 e l - 3 l1 Þ e= 2 2 42 - 3 ´ 17 \ e= cm = 0.5 cm 2 v = 4n( l1 + e) = 4 ´ 500 (17 + 8.5) ´ 10 -2 = 20 ´ 17.5 = 350 m/s Þ

N N AN A N A

v Þ n µ T so with increase in l temperature, frequency increases.

5. n µ

6. For sound water is rarer medium and air is densor medium so, it bends towards normal while going from water to air. l v v 2 7. n c = = no = Þ c = =1:2 4lc 2l o lo 4 8.

n2 = n1

F2 F1 2

æn Þ F2 = çç 2 è n1

ö ÷÷ F1 ø

æn Þ M 2 = çç 2 è n1

ö 256 ö2 ÷÷ M 1 = æç ÷ ´ 10 kg è 320 ø ø

2

= 6.4 kg \ OM = M 2 - M 1 = 6.4 - 10 = - 3.6 kg i.e., Mass has to be decreased by 3.6 kg v v 9. n direct = n and n reflected = n v - vs u - vs as n D = n R so there will be no beats i. e., beat frequency will be zero. ( l - l2 ) v v 10. Dn = n2 - n1 = =v 1 l2 l1 l1 l2 l1 l2 Dn Þ v= Dl

v=

\

10 3 = 337 m/s

14. At the moment when velocity of source is perpendicular to the line joining source and observer then there is no Doppler effect i.e., n + n1 = n Þ n1 = 0 ( n + 1)v 340 15. n = = (2n + 1) = 85 (2n + 1) 4l 4´1 = 85, 255, 425, 595, 765, 935 \ 6 frequencies below 1 kHz. æ v - v0 v - v0 v + v0 16. Dn = nn = n çç 1 v - vs v + vs v + vs è A o

S

vs + v0

10 ´ 180 = 5 Hz 360 v v 17. Dn = n = n1 - n2 = l1 l2 v( l2 - l1 ) nl1 l2 = Þv = l1 l2 l1 - l2 = n×

18.

v + vs

=

A

A' 345Hz

B

250Hz ý

1. Sound cannot travel in vacuum, as it is mechanical wave.

ý

¢

| 31

Dn<5

Dn>5

355Hz

ö ÷ ÷ ø

32 | Sound Waves

As beat frequency between A and B decreases on loading A. i. e., n B < n A Þ n B = 345 Hz B

C' 341Hz

C

245 Hz

349Hz

After loading A, n A ¢ = 345 + 2 = 247 Hz and n A ¢ - n c = ± 6 Þ n c = l A ¢ m 6 = 347 m 6 = 341 or 353 Hz. As possible frequency of C are 341 Hz and 249 Hz then only 341 Hz is justified. l - 3 l1 122 - 3 ´ 40 19. e = 2 = cm = 1 cm 2 2 5v v So, = l ( l1 + e) 4( l1 + e) l 3 = 5l1 + 4 l = 5 ´ 40 + 4 ´ 1 = 204 cm Dn 1 DF 1 DF 20. = Þ Dn = n n 2 F 2 F 1 1 = ´ 200 ´ = 1 Hz 2 100 2n + 1 2n + 1 21. n = v Þ l= v 4l 4n 2n + 1 340 = (2n + 1) = m 4 ´ 340 4 1 3 5 = m, m, m. 4 4 4 As, l max = 120 cm Þ l = 25 cm 75 cm. \Height of water column = 120 cm - 75 cm = 45 cm 105 ´ 4 l 22. 7 = 105 cm Þ l = = 60 cm 4 7 Þ

N

N

N N

l 60 Þ = = 15 cm 4 4 l l l l So, nodes are at, , 3 , 5 and 7 from 4 4 4 4 closed end i. e., they are at, 15 cm, 45 cm, 75 cm and 105 cm.

v v v = 512 Hz, n o = =2 4l 2l 4l = 2 n c = 2 ´ 512 Hz = 1024 Hz 1 ´ 32 + 1 ´ 2 24. M min = = 17 1+1 23. n c =

v min = vH 2

MH

2

=

M min

2 17

25. n a > f and n r < f but n a = constant and n r = cosntant. So, curve in (b) represents correctly. (2n + 1) v ( m + 1) v 26. u = 4l 2l (2n + 1) v ( m + 1) v How, 4 ´ 2l 4 ´ 2l 4 = = 2 beat/s 2 v v 27. Dn = n a - n r = nn v - v1 v + v1 2 vv1n = (v - v1 ) (v + v1 ) 2 ´ 320 ´ 4 ´ 243 = = 6 Hz 316 ´ 324 28. n c =

(2n + 1) v 4lc

=

320 (2n + 1) 4´1

= (2n + 1) ´ 80 Hz = 80 Hz, 240 Hz, 400 Hz,... ( n + 1)v 320 n0 = = ( n + 1) 2l 0 2 ´ 1.6 = ( n + 1) ´ 100 Hz = 100 Hz, 200 Hz, 300 Hz, 400 Hz,... \ n c = n o = 400 Hz 29. I max = 4 I 0 and

I max ¢ = 4 I max = 16I 0 L¢ = 10 dB + 10 log 16 = 10 dB + 40 log 2 dB = 22 dB 2p 2p 30. l = = = 4m k p/2 l 4 l =5 =5´ m =5m 4 4

Sound Waves

31. d = (2n + 1)

l (2n + 1) 330 = × m 4 4 660

=

= 33.33 Hz and n ¢A ¹ n B¢ So both (a) and (b) options are wrong.

v + v0

v ö æ f = ç 1 + 0 ÷ f and v v ø è v æ ö fr = ç1 - 0 ÷ f v ø è f a v + v0 = fr v - v0

34. f a =

330 (2n + 1) cm = (2n + 1) ´ 13.75 cm. 24 = 13.75 cm, 41.25 cm, 68.75 cm, 96.25 cm etc. v ( l2 - l1 ) v v 32. Dn = = l1 l2 l1 l2 332 ´ 1 ´ 10 -2 = = 13.15 Hz 0.49 ´ 0.5 300 33. Dn B = n A - n B = ´ 300 - 300 300 - 30

| 33

Þ Þ

( f a - f r )v = ( f a + f r )v 0 f + fr v . = a v0 fa - fr

and fa - fr = Þ

2v0

æ f - fr ö ÷f f = 2 çç a ÷ v è fa + fr ø f + fr f = a . 2

JEE Corner ¢

Assertion and Reason v v 3v 5v = , , ,... 4l 4l 4l 4l ( n + 1)v v 2 v 3v 4v while, n 0 = = , , , ,... 2l 2l 2l 2l 2l

5. With increase in intensity sound level increases in lograthmic order so assertion is false.

it can be seen that n c ¹ n o at all situation 1 and n c = n o so assertion is true but 2 reason is false.

6. Speed of sound v =

1. n c = (2n + 1)

gp , with increase in r

only pressure density increases such that

p/r remains constant. Again gRT so both assertion and reason M

2. Apparent frequency is constant for constant relative velocity so assertion is false.

v=

3. At a point of minimum displacement pressure amplitude is maximum i. e., pressure difference is maximum not pressure. So assertion is false.

explanation of assertion.

4. The deriver receiver two sounds one v+u direct, n 0 = n and other n R = n such v -u that be detects beats. So reason is true explanation of assertion.

are true but reason is not correct 7. n A = n B + 4 when A is loaded with little wax then n A sightly decreases and then beat frequency decreases, but if it is heavily loaded with wax then its frequency goes much below n B such that beat frequency increases. So, assertion and reason are both true but reason is not correct explanation of assertion.

34 | Sound Waves

8. 150 450, 7 50, 3,

5

The frequencies are odd harmonics then the pipe is closed and fundamental frequency is also 150 Hz. So assertion and reason are both true but reason is not correct explanation of assertion.

¢

9. n µ

1 with increase in diameter end l+e

correction, e, increases and n decreases. So reason is correct explanation of assertion. 10. With increasing length of air column, number of overtone increases and not the wavelength so assertion is false.

Objective Questions (Level 2) 1. At the boundary between two mediums, one part of incident wave gets reflected and other part gets transmitted or refracted. 3.9 p 3l 2. = 3.9 p Þ l = 2 1.5 2p 2p 3 Þ k= = = r 0 = S0 kB 3.9 p l 3.9 1.5 10 -2 ´ 105 r0 Þ S0 = = 3 kB ´ 1.3 ´ (200) 2 3.9 3.9 ´ 10 -1 = = 0.025 m = 2.5 cm 12 ´ 1.3 2 v2 2 ´ 100 At 2 = = = A i v1 + v2 200 + 100 3 v v 4. Dn = nn v + v2 v + v1 vn (v1 - v2 ) ~ n (v1 - v2 ) = (v + v2 ) (v + v1 ) v vDn 340 ´ 10 \ v1 - v2 = = = 2 m/s n 1700

6. 7

l 2L =L Þl= 2 7

A = a cos kx = a cos

7. For maxima, nl = 3 3 v nv Þ l= ;n= = = 110 n. n l 3 \n = 110, 220, 330 Hz, ..etc. maxima will be formed so maximum will not be formed at 120 Hz and 100 Hz. 20 m/s

8. W 60° S

3.

5. v s = gt = 10 m/s v + v0 v - v0 Dn = nn v - vs v + vs æ 300 + 2 300 - 2 ö ÷ ´ 150 Hz = çç ÷ è 300 - 10 300 + 10 ø 302 298 ö = æç ÷ ´ 150 = 12 Hz è 290 310 ø

2p L × = a cos p = - a L 7 2 7

n¢ =

20 m/s

v + wcos 60°

v + w cos 60° - v s 300 + 10 = ´ 500 Hz 300 + 10 - 20 310 = ´ 500 = 534 Hz 290 év + 20 v + 20 ù 9. Dn = n R - n 0 = ê ú ´ 500 ë v - 10 v + 10 û 360 360 ö = æç ÷ ´ 500 Hz = 31 Hz è 300 350 ø 404 p 400p 10. Dn = = 202 - 200 = 2 Hz 2p 2p 2 I max æ 2 + 1 ö ç ÷ = =9 :1 I min çè 2 - 1 ÷ø

Sound Waves

l 4 = 34 cm Þ l = ´ 34 cm 4 3 v 136 Þ n = Þ v51 = nl = n l 3 273 + 16 v16 289 1 1 = = = = v51 273 + 51 324 1.121 1.1 nl51 Þ nl16 = 1.1 136 \ l16 = = 41.21 cm 3 ´ 1.1 v -v v+v 12. 176 ´ = 165 ´ v - 22 v

y

11. 3

176 v (v - v) = 165 (v + v) (v - 22) Þ \ 176 ´ 330 (330 - v) = 165(330 + v) (330 - 22) or 1.143 (330 - v) = 330 + v or 0.143 ´ 330 = 2.143 v Þ v = 22 m/s 2 ´ 32 + 3 ´ 48 13. M min = = 41.6 2+3 n2 v2 m1 32 = = = = 0.77 n1 v1 m2 41.6 = 0.875 = 175 Hz 14.

v 0 = gt = 30 m/s v + 30 1100 = ´ 1000, 1.1v = v + 30 v 0.1 v = 30 Þ v = 300 m/s

Passage (Q 5 to 17) v m + v p = 8 m/s, 50v m = 150v p Þ v m = 3v p , 4v p = 8 m/s v p = 2 m/s and v m = 6 m/s v+2 332 15. n¢ = f0 = f 0 = constant v -6 324 v -2 328 16. n¢ ¢ = f0 = f 0 = constant v+6 336 17. n¢ ¢ < f 0 < v¢ and graph is (a) 18.

| 35

D A

B

C

E

x

Both (a) and b are correct. More Than One (2n + 1)v 19. n = 4l v 330 Þl = (2n + 1) = (2n + 1) m 4n 4 ´ 264 = (2n + 1) ´ 31.25 cm = 31.25 cm, 93.75 cm, 156.25 cm 20. (a) v µ p0 , (b) v µ T Þ v2 µ T, where T is absolute temperature. 1 (c) v µ F (d) n µ l \(c) and (d) are correct. Dp Dp 21. P0 = BA k; B = Dp DV p V Dpr r \ Dr = = BA k = rA k B B r µ p; Pressure and density equations are in p opposite phase i. e., Df = and not p. 2 So, (a), (b) and (c) are correct. 5v 3v l 125 . 2 2 8 22. = Þ = Þ o = = . 4 l c 2l o lc lo l c 1.25 5 2v v v (a) n c = = = 5 4lc 4 ´ l 5 lo o 8 4 v 4 = × = no Þ nc < no 5 2l o 5 3v 3v 12 v (b) n c = = = × 4lc 4 ´ 5 l 5 2l 0 0 8 6 2v 6 = × = × no Þ nc > no. 5 2 lo 5

36 | Sound Waves

15 v 6v v = = 12 5 4 lc 4 ´ l lo 2l o o 8 = 12 v 0 twelbth harmonic.

(c) n c =

15 v

=

(d) Closed organ pipe cannot have tenth harmonic it only has odd harmonics. gRT v 1 23. f = = 4( l + e) 4( l + e) M

(b) increase in T Þ increase in v Þ increase in f (c) increase in M Þ decrease in v Þ decrease in f (d) increase in P Þ increase in r Þ no change in v Þ no change in f v v 24. f a = and f r = f f are v - vs v + vs constants during approach and received.

(a) increase in r Þ increase in e Þ decrease in f

¢

Match the Columns v =f 2l f v (a) n c = = = 0.25 f ® s 4 ´ 2l 4 5v 5 (b) n c2 = = × f = 1.25 f ® p 4 ´ 2l 4 3v 3 (c) n c1 = = f = 0.75 f ® r 4 ´ 2l 4 3v (d) n c1 = = 0.75 f ® r 4 ´ 2l

1. n o =

2 vv f æ v v ö÷ 2. Dn1 = çç f = 2 f ÷ v - v2s è v - vs v + vs ø v 2v 4 f = 16 ´ 1 f = 8 f = 15 2 15 v2 v2 16 æ v + vs ö 2vs Dn2 = çç - 1 ÷÷ f = f v - vs è v - vs ø 2 v/4 2 = f = f v - v/4 3 æ v v ö÷ Dn 3 = çç ÷f =0 è v - v5 v - v5 ø 8 (a) ® Dn1 = f®q 15 2 (b) ® Dn2 = f ® p 3 (c) Dn 3 = 0 ® s (d)Dn ¢3 = 0 ® s

3. f = f T - f S (a) If tuning fork is loaded f T decreases such that beat frequency may increase or decrease depending upon amount of wax ® r, s (b) If prongs are filed, beat frequency must increase ® p (c) If tension is increased beat frequency may increase or decrease depending upon the amount of change in tension. ® r, s (d) If tension is decreased, beat frequency must increase ® p 1 1 4. (a) For point source, I µ , and A µ ® r r r (b) ® q 1 (c) For line source, I µ and r 1 Aµ ®q r (d) ® p 2 p 2p 5. l = = =2m k p l 5 l = 5 = m = 2.5 m 4 2 (a) l = 2.5 m ® s (b) l = 2 m ® r l 2l (c) , = 1m, 2 m ® p, r 2 2 l l (d) , 3 = 0.5 m, 1.5 m ® q 4 4

17. Thermometry, Thermal Expansion & Kinetic Theory of Gases Introductory Exercise 17.1 C F - 32 5 for F = 0, C = - ´ 32 = 5 9 9 = - 17.8 ° C K - 273.15 F - 32 (b) for K = 0, = 5 9 9 F = - ´ 273.15 + 32 = - 459.67 ° F 5 x 2x - 32 10x 2. (a) = Þx= - 17.8 5 9 9 10 Þ 17.8 = æç - 1 ö÷ x è9 ø

C F - 32 9 = Þ x = x - 32 5 9 5 4 Þ x = - 32 5 5 Þ x = - ´ 32 = - 40 ° C 4 1 6. Dt = at Dq 2 1 = ´ 1.2 ´ 10 -5 ´ 86400 ´ 30 2 = 1.5 ´ 1.2 ´ 8.64 s = 15.55 s given.

1. (a)

5.

Þ x = 17.8 ´ 9 = 160.2 ° C x x / 2 - 32 5 (b) = Þ x= x - 17.8 5 9 18 13 Þ 17.8 = x Þ x = - 24.65 °C 18 C -5 F - 32 3. = 99 - 5 212 - 32 C - 5 F - 32 Þ = 94 180 52 - 5 F - 32 Þ = 94 180 180 Þ F = 32 + ´ 47 = 122 ° F 94 K - 273.15 F - 32 4. = 5 9 5 Þ x - 273.15 = x - 17.8 9 4 Þ x = 255.35 Þ x = 574.54 9

7. As from 0°C to 4°C, density of water increases so the volume of wooden block above water level increases and as from 4°C to 10°C density of water decreases so the volume of block above water decreases. 8. V1r 1 g = V2¢ s1 g and V2 r 2 g = V2¢ s2 g DV1 V¢ s =1 - 1 = -1 - 1 V1 V1 r1 DV2 s and =1 - 2 V2 r2 DV2 DV1 \ V2 V1 æ s ö æ s ö s s = çç 1 - 2 ÷÷ - çç 1 - 1 ÷÷ = 1 - 2 r2 ø è r1 ø r1 r2 è s1 s1 (1 - g 2 DT) = r1 r 1 (1 - g 1 T) Þ

38 | Thermometry, Thermal Expansion & Kinetic Theory of Gases DT

=

s1 æ g 2 - g 1 ö ç ÷ r 1 çè 1 - g 1 ÷ø T s1 (1 - g 2 DT)

=

s1 r1 r 1 (1 - g 1 T)

=

s1 æ g 2 - g 1 ö ç ÷ r 1 çè 1 - g 1 ÷ø T

DT

9. On cooling brass contracts more than iron ( a B > a Fe ) such that brass disk gets loosen from hole of iron. 10. V µ T Þ V = kT Þ ln V = ln k + ln T DV DT DV 1 Þ = Þ = =g V T VDT T

Introductory Exercise 17.2 1. For ideal gases, pV = nRT V VM Þ T= p= p nR mR VM Slope = mR 1 As slope µ Þ m2 < m1 m p T 2. pV = nRT Þ 2 = 2 p1 T1 360 6 = = 300 5 6 6 Þ p2 = p1 = ´ 10 atm = 12 atm 5 5 1 1 ´ 28 + ´ 44 7 + 11 4 3. M mix = 4 = = 36 1 1 1 + 4 4 2 m pV = nRT = RT M pM m Þ pM = RT = rRT Þ r = V RT 101 . ´ 105 ´ 36 ´ 10 -3 \ r= 8.31 ´ 290

=

101 . ´ 36 8.31 ´ 29 .

4. pV = nRT = Þ N=

=

= 15 . kg/m 3

N RT NA

pVN A

RT 10 -6 ´ 13.6 ´ 10 3 ´ 10 ´ 250 ´ 10 -6 ´ 6.02 ´ 1023

8.31 ´ 300 13.6 ´ 5 ´ 6.02 = ´ 1015 = 8.21 ´ 1015 8.31 ´ 6 5. pV = nRT Þ

Þ

nR ×T p 1 Slope µ r V =

p1 > p2

1 V Þ y = mx is a straight line passing through origin.

6. pV = nRT Þ p = ( nRT)

Introductory Exercise 17.3 1. Average velocity depends on the direction of motion of gas molecules and as container do not move such that their net effect becomes zero, due to the reason that some molecules are moving

in one direction while other are moving in opposite direction. But in case of average speed only magnitudes are in use which do not cancel each other.

Thermometry, Thermal Expansion & Kinetic Theory of Gases

2.

3 3 8.31 kT = ´ ´ 300 J 2 2 6 ´ 1023 3 = ´ 8.31 ´ 10 -21 J 4 = 6.21 ´ 10 -21 J

KE =

3. v rms = v He =

4 ´ 10 -3

= 1.37 ´ 10 3 m /s 3 ´ 8.31 ´ 300 v Ne = = 608.5 m/s 20.2 ´ 10 -3 3 KE = kT = 6.21 ´ 10 -21 J 2 3 RT 4. v rms = M Mv2rms 4 ´ 10 -3 ´ 10 6 ÞT = = 3R 3 ´ 8.31

5. r =

= 160.45 K n1r 1 + n2 r 2 n1 + n2

=

(1 - n2 ) r 1 + n2 r 2 1 - n2 + n2

= r 1 + n2 (r 2 - r 1 ) r - r1 1.293 - 1.429 Þ n2 = = r 2 - r 1 1.251 - 1.429 136 = = 0.764 = 76.4% by mass 178 ( p + hrg) V pT 6. 2 = 1 2 = 0 V1 p2 T1 p0 ´ 277 =

1 ´ 1023 ; 3 S = 4 pR2 = 4 ´ 3.14 ´ (6400 ´ 10 3 ´ 102 ) 2 =

\

3 RT , M 3 ´ 8.31 ´ 300

(1.01 ´ 105 + 40 ´ 10 3 ´ 10) ´ 293

1.01 ´ 105 ´ 277 5.01 ´ 293 = = 5.25 1.01 ´ 277 Þ V2 = 5.25 V1 = 105 cm 3 1 7. N = nN A = ´ 6 ´ 1023 18

| 39

= 5.14 ´ 1018 cm 2 N 1023 = S 3 ´ 5.14 ´ 1018

= 6.5 ´ 10 3 molecules/cm 2 3 (a) nCV = nR = 35 J/K 2 70 Þ n= = 2.8 mole 3R 3 (b) U = nRT = 35 J/K ´ 273 K = 9555 J 2 5 (c) C p = CV + R = R = 20.8 J/K mole 2 8. (a) n(C p - CV ) = nR = 29.1 J/K 29.1 Þn= mole = 3.5 mole 8.314 3 (b) CV = nc V = n R = 3.5 ´ 1.5 ´ 8.314 2 = 43.65 J/K 5 C p = nc p = n R = CV + nR 2 = 43.65 + 3.5 ´ 8.314 = 72.75 J/K 5 (c) CV ¢ = nc V = n ´ R = 72.75 J/K 3 7 C p¢ = nc p = n ´ R 2 = 72.75 + 3.5 ´ 8.314 = 101.85 J/K 3 RT 8 RT 10. v rms = and v av = M pM 8 Here 3 > Þ v rms > v av , p i. e., the statement is true.

40 | Thermometry, Thermal Expansion & Kinetic Theory of Gases

AIEEE Corner ¢


2.

3.

4.

5.

6.

C¢ 68 - 32 36 = = =4 5 9 9 K¢ - 273 68 - 32 Þ C¢ = 20° C ; = =4 5 9 Þ K¢ = 293 K C¢ 5 - 32 27 = == -3 5 9 9 K¢ - 273 5 - 32 Þ C¢ = - 15° C; = = -3 5 9 Þ K¢ = 258 K C¢ 176 - 32 144 = = = 16 5 9 9 K¢ - 273 Þ C¢ = 80° C; = 16 5 Þ K¢ = 353 K 30 F¢ - 32 = Þ F¢ = 54 + 32 = 86° F 5 9 = 546° R 5 F¢ - 32 = Þ F¢ = 9 + 32 = 41° F = 501° R 5 9 20 F¢ - 32 = 5 9 Þ F¢ = - 36 + 32 = - 41° F = 456° R x 32 x 9 4 = Þ 32 = x - x = - x 5 9 5 5 5 Þ x = - ´ 32 = - 40° 4 Þ - 40° C = - 40° F DC DF 9 9 = Þ DF = DC = ´ 40 = 72° 5 9 5 5 \ F2 = F1 + 72° = 140 .2° F 32 - 20 C¢ - 0 = 80 - 20 100 - 0 12 ´ 100 12 C¢ Þ = Þ C¢ = = 20° C 60 100 60 T2 p 160 = 2 = = 2 Þ T2 = 2T1 T1 p1 80

\ T2 = 2 ´ 273.15 K = 546.30 K 7. Rt = R0 (1 + a D q) Þ 3.50 = 250 . (1 + 100a) Þ 1 = 250 K 10 . or a = = 4 ´ 10 -3 /°C 250 \ 650 . = 250 . (1 + 4 ´ 10 -3 Dq) Þ 4 = 10 -2 ´ Dq Þ Dq = 400 Þ q2 = 400°C Þ Dq = 400 Þ q2 = 400°C i. e., boiling point of sulphur is 400°C. 75 + 45 120 3 T p 8. 2 = 2 = = = T1 p1 75 + 5 80 2 3 3 T2 = T1 = ´ 30015 . K 2 2 = 450225 . K = 177.08 ° C 9. Dg = g ( a Br - a Fe ) Dq Dg 1 Þ Dq = × g a Br - a Fe =

0.01 ´ 10 -3 6 ´ 10

-2

×

a Br

1 - a Fe

-3

= \

10 6 ( a Br - a Fe )

q2 = q1 +

= 30° C +

10 -3 6 ( a Br - a Fe )

10 -3 100 = 30° C + 6 ( a Br - a Fe ) 6 ´ 0.63

= 57.78 ° C. 10. (a) Dl = l aDq ~ - 88.42 ´ 2.4 ´ 10 -5 ´ 30 = 0.064 cm (b) Dl = l ( a Al - a St ) Dq = 88.42 (2.4 - 1.2) ´ 10 -5 ´ 30 = 0.032 cm l S = l + Dl = 88.42 + 0.032 cm = 88.45 cm


Dl ´ 100% = a D q ´ 100% l = - 1.2 ´ 10 -5 ´ 35 ´ 100% = - 0.042% Dl 12. F = YA = YA a Dq l = 2 ´ 1011 ´ 2 ´ 10 -6 ´ 1.2 ´ 10 -5 ´ 40 = 4 ´ 1.2 ´ 40 N = 160 ´ 1.2 N = 192 N 11.

13. Vsg = (50 - 45) ´ 10 -3 kg = 5 ´ 10 -3 kg V ¢ s¢ g = (50 - 451 . ) ´ 10 -3 kg = 4.9 ´ 10 -3 kg s V (1 + g s Dq) g = 4.9 ´ 10 -3 1 + g l Dq 1 + g s Dq 4.9 = 1 + g l Dq 5 Þ 5 + 5g s Dq = 4.9 + 4.9 g e Dq 0.1 + 5g s Dq 1 5 gl = = + gs 4.9 Dq 49 Dq 4.9 1 5 gs = + ´ 12 ´ 10 -6 49 ´ 75 4.9 = 272.1 ´ 10 -6 + 12.2 ´ 10 -6 = 2.84 ´ 10 -4 ° C 14. M = 14 + 3 = 17 g/mole = 17 ´ 10 -3 kg/mole 17 ´ 10 -3 kg/molecule ÞM = 6033 . ´ 10 -23 = 282 . ´ 10 -26 kg/molecule 6 -2 pV 1.52 ´ 10 ´ 10 15. n = = = 6.13 RT 8.314 ´ 298.15 r=

m nM 6.13 ´ 2 ´ 10 = = V V 10 -2

-3

= 1.23 kg/m 3 m¢ nM ¢ 16 nM r¢ = = = = 16 r V V V = 19.62 kg/m 3 V 76 16. p2 = p1 1 = 1 atm ´ V2 6 = 12.7 atm

| 41

p1 V1 T2 p T × = 1 × 2 × V1 T1 p2 p2 T1 1 270 = ´ ´ 500 m 3 = 900 m 3 0.5 300 pV p V 18. 1 1 = 2 2 T1 T2 æ mg + p ö A × h æ mg + p ö Ah ç ç 0 ÷ i 0 ÷ f A A ø ø Þè =è 293 273 373 373 Þ hf = hi = ´ 4 cm = 50.9 cm 293 293 25/ 28 40/ 4 n n 19. p1 = p2 Þ 1 = 2 = = V1 V2 L1 A L2 A L1 25 1 5 Þ = ´ = = 0.089 L2 28 10 56 n1 25/ 28 25 5 = = = = 0.089 n2 40/ 4 280 56 17. V2 =

20. n = n1 + n2 Þ p( V1 + V2 ) = p1 V1 + p2 V2 p V + p2 V2 Þ p= 1 1 V1 + V2 1.38 ´ 0.11 + 0.69 ´ 0.16 \p = MPa 0.11 + 0.16 0.1518 + 0.1104 0.2622 = = = 0.97 MPa 0.27 0.27 pV1 pV2 pV pV 21. + = 1 1 + 1 2 T T T1 T2 1 atm ´ 600 cm 3 293 K æ 400 cm 3 200 cm 3 ö ÷ = p1 ç + ç 373 K ÷ 273 K è ø 600/ 293 Þ p1 = atm 400 200 + 373 273 3 \ p1 = atm 2 1 ö æ 293 ç + ÷ è 373 273 ø 3 3 = = atm 1.57 + 1.07 2.64 = 1.136 atm


22. V =

nRT 1 ´ 8.314 ´ 273.15 3 = m p 1.013 ´ ´ 105

= 0.02242 m 3 = 22.42 litre pV T V T 23. p2 = 1 1 × 2 = p1 1 × 2 T1 V2 V2 T1 0.75 430 = 1.5 ´ 105 ´ ´ 0.48 300 = 3.36 ´ 105 Pa 24. r = p1 + p1¢ + p2 n RT n1¢ RT n2 RT = 1 + + V V V 1.4 1.4 0.4 ù RT = é0.7 ´ + 0.3 ´ + êë 28 14 4 úû V 8.314 ´ 1500 0.7 0.3 1ù =é + + ´ 10 10 ûú ëê 20 5 ´ 10 -3 3.3 = ´ 8.314 ´ 3 ´ 105 Pa 20 = 4.11 ´ 105 Pa 1 1 25. RKE = 2 ´ kT = Iw2 2 2 2 ´ 1.38 ´ 10 -23 ´ 300 2kT Þ w= = I 8.28 ´ 10 -38 ´ 10 -7 6 ´ 1.38 \ w = 1012 = 1012 rad/s 8.28 gp 26. v = r 2 rv2 1.3 ´ (330) Þ g= = = 1.398 p 1.013 ´ 105 f +2 2 ~ = Þf = -5 f 0.398 5 7 n1C p1 + n2 C p2 3 ´ 2 + 2 ´ 2 27. g = = 3 5 n1CV + n2 CV 3 ´ + 2´ 1 2 2 2 15 + 14 29 = = = 1.53 9 + 10 19 3 28. K = pV 2 3 p2 V2 K2 3 15 Þ =2 = × = 4.5 3 K1 2 5 p1 V1 2

K 2 = 4.5 K f +2 58 29. C p = R = 29 Þ f = -2 =5 2 R pV pT = p × nR Þ p2 V = constant 1 Þ pV 1/ 2 = constant Þ a = , 2 f +4 f R c= R+ = R = 29 1 2 2 12 58 f = - 4 =3 R 3 2 30. TKE = of total energy and RKE = of 5 5 total energy, so the gas is diatomic. 3 3 TKE = kT = ´ 1.38 ´ 10 -23 ´ 300 J 2 2 = 6.21 ´ 10 -21 J/molecule 5 DQ = nCV DT = 1 × ´ 8.314 ´ 1 = 20.8 J 2 n1C p1 + n2 C p2 31. C p = n1 + n2 2.5 R + 3.5 R = = 3R 1+1 n1`CV + n2 CV 1 2 CV = n1 + n2 1.5 R + 2.5 R = = 2R 1+1 C 3R g= P = = 1.5 CV 2R Þ

32. g =

n1C p1 + n2 C p2 n1CV + n2 CV 1

=

C p1 CV

2

=

( n1 + n2 )C p1 ( n1 + n2 )CV

1

=g

1

b

33. p = aV Þ pV - b = constant DQ C= = 0 for adiabatic process for nDT which pV g = constant comparing, we get, b = - g


34. p = kV Þ pV -1 = constant Þ pV = a constant Þ a = - 1 R R R C = CV + = CV + = CV + 1-a 1+1 2 35. v rms =

3 ´ 8.314 ´ 373.15 3 RT = M 2 ´ 10 -3

= 2.16 ´ 10 3 m /s = 2.16 km/s

36. v rms =

(500) 2 + (600) 2 + (700) 2 + (800) 2 + (900) 2

5 100 = 25 + 36 + 49 + 64 + 81 5

= 714 m/s 500 + 600 + 700 + 800 + 900 v av = 5 = 20 (5 + 6 + 7 + 8 + 9) = 700 m/s v rms ¹ v av 3 37. KE = pV 2 Þ N ´ 6 ´ 10 -26 = 1.5 ´ 2 ´ 105 ´ 100 ´ 10 -3 ´ 10 -3 3 ´ 10 \ N= = 5 ´ 1026 6 ´ 10 -26 5000 = ´ 6.023 ´ 1023 = 830.15 Na 6.023 \ n = 8300.15 moles v v 38. Frequency of collision, n = = 2 3 l 2 3 ×V 1 3 RT = M 2 3V p RT RT \n = = = nRT 4 VM 4 nM 4× ×M p =

2 ´ 10 +5 4 ´ 1 ´ 46 ´ 10 -3

= 41.04 ´ 10 3 /s 3 3 39. KE = pV = ´ 105 ´ 2 ´ 10 -6 = 0.3 J 2 2

| 43

-6

m 50 ´ 10 = = 6.25 ´ 1020 m1 8 ´ 10 -26 K 0.3 30 \K 1 = = = ´ 10 -22 J 20 N 6.25 ´ 10 6.25 N=

= 4.8 ´ 10 -22 J 3 RT0 40. v 0 = M0 (a)

v = v0

T 573 = = 1.4 Þ v = 1.4 v 0 T0 293

(b) v = v 0 as RMS speed changes with temperature and not with pressure. M0 M0 v 1 (c) = = = v0 M 3M 0 3 v0 Þ v= = 0.58 v 0 3 MH gRT gRT¢ 2 41. = ÞT = × T¢ MH MO MO 2

2

2

2 = ´ 320 = 20 K = - 253 ° C 32 1 GMm 42. mv2e = = g e Re m 2 Re 3 RT Þ v e = 2 g e Re = v H = 2 M 2 g e Re M Þ Te = 3R 2 ´ 9.8 ´ 6367 . ´ 10 6 ´ 2 ´ 10 -3 = 3 ´ 8.314 = 10007 K 2 g m Rm M and Tm = 3R 2 ´ 1.6 ´ 1.75 ´ 10 6 ´ 2 ´ 10 -3 = 3 ´ 8.314 = 449 K 3 3 43. (a) KE = kT = ´ 1.38 ´ 10 -23 ´ 300 J 2 2 -21 = 6.21 ´ 10 J 3 (b) KE¢ = kT × N a = 6.023 ´ 1023 2 ´ 6.21 ´ 10 -21 J = 3740 J


(c) v rms =

3 ´ 8.314 ´ 300 3 RT = M 32 ´ 10 -3

= 483.6 m /s ¢

Objective Questions (Level 1) 1. v av = 2. v rms = =

8 RT pM

T=

12 + 02 + 22 + 3 2 4 14 = 3.5 m /s 4

Dl = - aDq = - 12 ´ 10 -6 ´ 50 l = - 600 ´ 10 -6 = - 6 ´ 10 -4 V T 4. V µ T Þ 2 = 2 = 2; V1 T1 V V1 DV = 2 = 2 - 1 = 1 = 100% V V1 3.

5. KE µ T Þ

K 2 T2 2E = = =2 K 1 T1 E

Þ T2 = 2T1 = 2 ´ 283 K \ T2 = 566 K = 293 ° C f n 6. TE = kT = kT. 2 2 v T2 1200 7. 2 = = = 2 Þ v2 = 2 v1 v1 T1 300 8. (a) pav = m1v is different for different m1 3 (b) (KE) molecule = kT is same for any 2 gas. 3 pV 3 (c) (KE)/ V = = p is different as 2 V 2 p is different for different. 3 pV 3 p (d) (KE) m = is different as = 2 m 2 r p is different. r pV p V pV1 pV2 p( V1 + V2 ) 9. 1 1 + 2 2 = + = RT1 RT2 RT RT RT

10. a =

p( V1 + V2 ) T T p( V1 + V2 ) = 1 2 p1 V1 p V p1 V1 T2 + p2 V2 T1 + 2 2 T1 T2

0.08 ´ 10 -3 Dl = l Dq 10 ´ 10 -2 ´ 100

= 8 ´ 10 -6 / ° C DV = VgDq = 3 VaDq = 3 ´ 100 cc ´ 8 ´ 10 -6 ´ 100 = 0.24 cc Þ V¢ = 100 cc + 0.24 cc = 100.24 cc 11. T = T0 + tan 45° V = T0 + V pV = nRT = nR( T0 + V ) = nRT0 + nRV nRT0 b or p = nR + Þ p=a+ V V ie, p versus V graph will be hyperbola. 12. p2 V = constant 2

Þ

æ nRT ö ç ÷ V = constant. è V ø T2 µ V

Þ Þ

T2 = T1

V2 3 V0 = = 3 V1 V0

T2 = 3 T1 = 3 T0 1 2 1 m 2 13. p = rv rms = v rms 3 3 V Þ mT = constant m1 T2 310 = = = 1.1 m2 T1 280 Þ

14. As temperature of vessels A and B are some so is average velocity of O2 , i.e., u. pV 15. N = nN a = Na RT 10 -13 ´ 10 -6 = ´ 6.023 ´ 10 +23 8.314 ´ 300


= 16.

602.3 = 24 8.314 ´ 3

9 63 ´ 28 g = g 20 5 mp m m 21. r = = = RT ( n1 + n2 ) RT V ( n1 + n2 ) p 5 12 ´ 1.01 ´ 10 ´ 10 -3 = = 0.12 kg/m 3 (2 + 2) ´ 8.314 ´ 300 =

1 GMm GMm mv2 = = mgh 2 R R+h Þ

h~ -

\

h=

| 45

v2 3 RT = 2 g 2 gM 3 ´ 8.314 ´ 273 2 ´ 10 ´ 28 ´ 10 -3

m

kM V Þ pV = constant is for isothermal process, i. e., T = constant p2 23. = constant r 22. p = kr =

= 1216 . ´ 10 3 m » 12 km l a 19 17. 1 = 2 = and l1 - l2 = 30 cm l2 a 1 11 11 Þ l1 = l1 = 30 cm 19 19 Þ l1 = ´ 30 cm = 7125 . cm 8 11 and l2 = l1 = 41.25 cm 19 18. p µ V µ T ÞV = constant p

Þ p2 V constant Þ pT = constant r /2 p2 r2 1 = = = p1 r1 r 2 p Þ p2 = 1 2 T2 p1 = = 2 T1 p2 T2 = 2T1 = 2T 1 as pT = constant Þ p µ T i. e., p - T graph is hyperbola.

Þ V

19. V = V0 + tan q × T, pV = nRT = \ Þ or

m p( V0 + tan q ´ T) = RT M 1 ém tan q = RT - pV0 ù úû pT êë M mR V0 tan q = pM T

m RT M

\ tan q remains same when m ® 2 m and p ® 2 p pV p V 20. n1 = 1 and n2 = 2 RT1 RT2 n1 p1 T2 10 ´ 300 600 20 = × = = = n2 T1 p2 5 ´ 330 330 11 11 \ n2 = n1 20 11 \ Dm = m1 - m2 = m1 = m1 20

24. p2 V = constant Þ PT = constant and T 2 V -1 = constant. p2 V1 V 1 = = = p1 V2 4V 2 p p Þ p2 = 1 = 2 2 T2 V2 4V = = =2 T1 V1 V Þ T2 = 2T1 = 2T 1 as p µ Þ p-T graph is hyperbola. T C -0 F - 32 F - MP 25. = = 100 - 0 212 - 32 BP - MP Þ ice point = 32° F and steam point = 212° F


26. p µ V at ‘a’ p = p0 and V = V0 and at ‘b’ , m p = 2 p0 and V = 2V0 , r = V r b Va V0 1 Þ = = = ra Vb 2V0 2 1 Þ r b = Pa 2 Tb PV 2 p0 × 2V0 = b b = =4 Ta Pa Va P0 V0 Tb = 4 Ta U µ T Þ Ub = 4Ua as P µ V Þ T µ V2 Þ Parabola passing through origin 27. (a) TKE = 3 nRT is independent of type 2 of gas ® true. (b) In one degree of freedom for one 1 mole of gas, V = RT 2 Þ

(c) false (d) false 28. V µ T ÞV = tan q × T m pV = nRT = RT M mRT mR Þ p × T tan q = Þ tan q + M MP m1 m2 tan q1 > tan q2 Þ > p1 p2 Þ all a, b, c and d are possible. 29. pV = nRT N m/V n m Þ p = RT = a RT = RT V M M 3 kN a T 3 RT 3 kT v rms = = = M M M Þ (a) and (d) are correct.

JEE Corner ¢

Assertion & Reason 1. Assertion is false. p

isobaric h oc

ic or

isothermal

is

T

2. Assertion and reason are both true but reason is not correct explanation of assertion. As at low temperature atoms in molecules are tightly bound such that they cannot oscillate. 2 3. pV = nRT = × KE 3 2 KE 2 Þ p= Þ p = E. 3 V 3 Assertion and reason are both true but reason cannot explain assertion.

4. Internal energy remains same in train frame of reference, so temperature do not change, but KE of gas molecules in ground frame increases. 5. According to equipartition theory, energy is equally distributed for each degree of freedom, so assertion is false. 6. At high temperature and low pressure intermolecular distance is much larger than size of the molecules and intermolecular forces can be neglected. So, assertion and reason are both true but not correct explanation. 7. At 4°C, volume is minimum or density is maximum i. e., liquid will overflow on increasing or decreasing temperature This reason is false.


8. Temperature remains constant as pressure is double and volume is halfed, so internal energy remains constant. So reason partially explains assertion.

| 47

æ MR ö nR ÷÷ T Þ slope µ m; reason T = çç r è Mp ø is correct explanation.

10. V =

9. Assertion and reason are both true but not correct explanation. ¢

Match The Columns 1.

(a) TKE = 3 nRT = 3 × 2RT = 3 RT ® 2 2 2 2 (b) RKE = nRT = × 2RT = 2RT® 2 2 (c) PE = ® (d TKE = 5 nRT = 5RT ® 2 )

r

(d)

p s s

1 aTDq increases with 2 ® increasing temperature

p

DT =

5. (a) p = constant® q p

¾® q 2.

U T

p

1 Þ VT = constant V Þ pT 2 = constant and pV 2 = constant (a) U increases Þ T increases Þ P decreases ® r (b) p increase Þ V decreases ® r (c) U increases Þ T increases ® q T TV constant (d) as = = = increase V V2 V2 Vdecreases ® q 8 3. x1 = 3, x2 = , x 3 = 2 and x4 = r p (a) ® r, (b) ® s, (c) ® q, (d) ® s U = pÞ T µ

(b) V µ T 1 Þ µU r U

¾® R p

(c)

T

¾® p V

(d) V µ T Þ

4. (a) (b) (c)

density of water is ® maximum of 4°C depends of change in ® density of solid and liquid depends of change in ® density of solid and liquid

s

1 µT r T

s s

¾® r r

18. First Law of Thermodynamics Introductory Exercise 18.1 1. (a) DW = pDV = - 1.7 ´ 105 (1.2 - 0.8) J 4

= - 6.8 ´ 10 J (b) DV = 1.1 ´ 105 J Þ DQ = DU + DW = - 17.8 ´ 104 J i.e., 1.78 ´ 105 J of heat has flown out of the gas. (c) No, it is independent of the type of the gas. 2. (a) In p - V graph of cyclic process, clockwise rotation gives positive work and anticlockwise gives negative work. And as loop 1 has greater area than loop 2, that is why total work done by the system is positive. (b) As in cyclic process change in internal energy is zero, that’s why for positive work done by the system, heat flows into the system. (c) In loop ‘1’ work done is positive so, heat flows into the system and in loop ‘2’ work done is negative so heat flows out of the system.

3. As the box is insulated i.e., no heat exchange takes place with surrounding and as the gas expands against vacuum i.e., zero pressure that’s why no work has been done and there is no change in internal energy. Thus, temperature do not change, internal energy and gas does not do any work. f 3 4. U = nRT = nRT 2 2 2U 2 ´ 100 Þ n= = 3 RT 3 ´ 8.314 ´ 300 = 0.0267 mole. 5. DQ = ms Dq = 1 ´ 387 ´ 30 J = 11610 J m DV = Vg Dq = ´ 3 a Dq r 1 = ´ 3 ´ 7 ´ 10 -6 ´ 30 8.92 ´ 10 3 = 7.1 ´ 10 -8 m - 3 DW = p DV = 1.01 ´ 105 ´ 7.06 ´ 10 - 8 = 7.13 ´ 10 - 3 J DU = DQ - DW = 11609.99 J

Introductory Exercise 18.2 1. (a) At constant volume, DU = 0 Þ DW = 0 DQ = nCV DT DQ 200 Þ D= = = 16.04 K nCV 1 ´ 3 ´ 8.314 2 \ Tf = Ti + DT = 300 + 16.04 = 316.04 K

(b) At constant pressure, DQ 200 DT = = = 9.62 K nC p 1 ´ 5 ´ 8.314 2 Þ Tf = 300 + 9.62 K = 309.62 K 2. For adiabatic process, pVg = constant = c (say) V V V c \ ò f p dV = ò f g dV = cò f V - g dV Vi Vi V Vi

49 | First Law of Thermodynamics

=c

=

V -g +1 -g+1

-g + 1

Vf

=c

Vi -g + 1 pf Vf × Vf

Vf

-g + 1

- Vi

1-g - g +1

- pi Vig × V f

1-g p f V f - pi Vi pi Vi - p f V f (Proved) = = 1-g g -1 3. DW AB = + 500 J,DQ AB = + 250 J Þ DU AB = - 250 J DW AC = + 700 J, DQ AC = + 300 J Þ DU AC = - 400 J (a) Path BC is isochoric process, i.e., DWBC = 0 \ DQBC = DUBC = DU AC - DU AB = - 400 J - ( - 250 J) = - 150 J (b) DWCDA = DWCD + DWDA = - 800J + 0 = - 800 J (Work is negative as volume is decreasing) DUCDA = DU AC = - DU AC = 400 J Þ DQCDA = DWCDA + DUCDA = - 800 J + 400 J = - 400 J -2 5 pV 1 ´ 10 ´ 2 ´ 10 4. (a) T = = nR 1 ´ 8.314 = 240.6 K 5 -3 pDV 2 ´ 10 ´ 5 ´ 10 (b) DW = = 5 g -1 -1 3 3 10 J = 2/ 3 5. (a) p2 p2 p2 æç 1 1 ö÷ = 2m i 2m f 2 çè m i m f ÷ø (10 ´ 10 -3 ´ 200) 2 é 1 1 ù = ê ú -3 2 2.01 û ë10 ´ 10 1 = 2 é100 - ù = 199 J 2 ûú ëê DQ DQ (b) DQ = nCV DT Þ DT = = m nCV CV M DK =

=

M DQ m ´ 3R

=

200 ´ 199 2010 ´ 3 ´ 8.314

= 0.8 ° C 6. DW = DW AB + DWBC + DWCD + DWDA æ p ö = nRT1 mçç 1 ÷÷ + p2 ( VC - VBC ) è p2 ø æp ö + nRTm çç 2 ÷÷ + p1 ( V1 - V2 ) è p1 ø æp ö = nR( T2 - T1 ) ln çç 2 ÷÷ + p1 V2 - p1 V1 è p1 ø + p1 V1 - p1 V2 æ p2 ö = ( p2 V2 - p1 V1 ) ln çç ÷÷ è p1 ø 7. DW ABCA = (+)ve Þ DW AB = (+)ve, DWBC = 0, DWCA = ( -)ve For BC, DQ = ( -)ve Þ DUBC = ( -)ve and DWBC = 0 For CA, DU = ( -)ve Þ DQCA = ( -)ve as DWCA = ( -)ve. DU AB BC CA Total

+ 0

DW + 0 +

DQ + +

For AB, as DU ABCA = 0 and DUBC = ( -)ve, DUCA = ( -) ve Þ DU AB = ( -) ve As DQ ABCA = DW ABCA = ( +)ve and DQBC = ( -)ve DQCA = ( -)ve Þ DQ AB = ( +)ve In isobaric process, DW = pDV = nR DT = 0.2 ´ 8.314 ´ (300 - 200) = 166.3 J 1 9. DW = ò p dV = ò aV 2 dV = aV 3 3 1 5 = ´ 5 ´ 1.01 ´ 10 ´ (2 3 - 1 3 ) 3 = 1.18 ´ 10 6 J

First Law of Thermodynamics

Introductory Exercise 18.3 p

1.

Þ

p0

C

5V0

CV

=

29.39 = 1.4 21.07

1600 = 1 × C p × 72 C p = 22.22

Þ Þ

B V0

Cp

2. DQ = DU + DW ; DQ = nC p DT

A

p0/5

g=

CV = C p - R = 13.9 Þ g =

V

æV ö DWBB = nRT ln çç B ÷÷ = 3 R ´ 273 ln 5 è VA ø = 10959 J DWBC = 0 DQ = DU + DW DU = DQ - DW = 80000 - 10959 = 69041 Tf = 5Ti = 5 ´ 273 K = 1365 K DQ ABC = DQ AB + DQBC = DWBC + 0 + 0 + DUBC DQBC DQBC = nCV DT Þ CV = nDT 69041 = = 21.07 3 ´ 4 ´ 273

Cp CV

= 1.6

DW = DQ - DU = 1600 - nCV DT = 1600 - 1 ´ 13.9 ´ 72 = 1600 - 1000.8 J = 599.2 J and DU = nCV DT = 1 ´ 13.9 ´ 72 = 1001 = 1 kJ 1 3. DW = DpDV 2 1 = ´ 20 ´ 1.01 ´ 105 ´ 1 ´ 10 -3 2 = 10 ´ 101 = 1010 J nDW 100 ´ 1010 J \p= = Dt 60 s = 1.68 kW

C p = CV + R = 29.39

AIEEE Corner ¢

Subjectve Questions (Level 1) 1. DU = DQ - DW = 254 J + 73 J = 327 J 2 DQ 2 ´ 200 DQ 2. (a) DT = = = nCV 3 nR 2 ´ 1 ´ 8.314 = 16 K Þ Tf = Ti + DT = 316 K

(b) DT¢ =

Þ T¢f

2 DQ 2 ´ 200 DQ = = nC p 5 nR 5 ´ 1 ´ 8.314

= 9.6 K = Ti + DT¢ = 309.6 K

3. DU = nCV DT, in adiabatic process, DQ = 0 and DU = - DW

| 50


where, DW =

DU 900 = = 14.43 3 nCV 5 ´ ´ 8.314 2 Þ Tf = Ti - DT = 127 ° C - 14.43 ° C = 112.6° C DT =

1-g nRDT for all process. DU = g -1

Þ 4.

nR DT

DV = 0 Þ DW = 0

5 \ DQ = DU = nCV DT = n × R DT 2 5 5 = ( p f V f - pi Vi ) = ( p f - Vi ) V 2 2 5 5 5 = (5 ´ 10 - 10 ) ´ 10 ´ 10 -3 2 5 = ´ 4 ´ 105 ´ 10 -2 = 104 J 2 5 5. DQ1 = DU1 = nCV DT = n × R (3 Ti - Ti ) 2 = 5 nRTi 5 DQ2 = nC p DT = n × R (6Ti - 3 Ti ) 2 = 7.5 nRTi 12.5 nRTi 12.5 R DQ \c = = = = 2.5 R n DT n (6Ti - Ti ) 5 p 1 6. DW AB = 0, DWBC = 0 ´ V0 = p0 V0 2 2 1 = × nRT0 = 300 R 2 p A

p0

p0/5

B

2V0

æ 1 1 ö÷ 1 1 ö = rm ç = rm æç ÷ çr ÷ r 1000 999.9 è ø i ø è f 105 ´ 2 ´ 0.1 == - 0.02 J 1000 ´ 999.9 (work done is negative as volume decreases) DQ = ms Dq = 2 ´ 4200 ´ 4 = 33600 J DU = DQ - DW = 33600.02 J m 10. DW = pDV » pV f = p r 5 -3 10 ´ 10 ´ 10 = = 1666.67 J 0.6 DQ = msDq + mL = 10 -2 ´ 4200 ´ 100 + 10 -2 ´ 25 . ´ 10 6 = 4200 J + 25000 J = 29200 J DU = DQ - DW = 29200 J - 1666.67 J = 27533.33 J = 2.75 ´ 104 J = 1.013 ´ 105 ´ 1670 ´ 10 -6 = 1.013 ´ 167 J = 169.2 J DQ = mL = 10 -3 ´ 2.256 ´ 10 6 J = 2256 J \ DU = DQ - DW = (2256 - 169.2) J = 2086.8 J » 2087 J

V

DQ = ( DU + DW ) AB + ( DU + DW ) BC = DU AB + DUBC + DWBC = 0 + 300 R (As TA = TC ) = 2.49 ´ 10 3 J = 2.49 kJ 7. DU = DQ - DW = 1200 J - 2100 J = 900 J

9. DW = rDV = r( V f - Vi )

11. DW = p DV

C

V0

8. When gas expands it does positive work on the surrounding and for this purpose heat has to be supplied into the system.

12.

DW = pDV = - 2.3 ´ 105 ´ 0.5 = - 1.15 ´ 105 J DU = - 1.4 ´ 105 J Þ DQ = DU + DW


= - (1.4 + 1.15) ´ 105 J = - 2.55 ´ 105 J Thus, 2.55 ´ 105 J of heat flows out of the system and it is independent of the type of the gas. 13. In a cyclic process, U = 0 Þ DQ = DW (a) \Wh = (Q1 + Q2 + Q3 + Q4 ) - (W1 + W2 + W3 ) = (5960 - 5585 - 2980 + 3645) - (2200 - 825 - 1100) = 1040 - 275 = 765 J work done 1040 (b) h = = = 10.83% heat supplied 9605 1 ¾® ¾® 1 AB ´ AC = ´ 2 p0 ´ V0 2 2 = p0 V0 2 p0 V0 p V (b) TC = and TA = 0 0 nR nR p V Þ DQCA = nC p DT = - nC p × 0 0 nR p0 V0 5 5 = - R× = - p0 V0 2 R 2 3 p0 V0 , DQ AB = nCV DT TB = nR p V ö 3 æ3 p V = n ´ Rç 0 0 - 0 0 ÷ 2 è nR nR ø 3 = ´ 2 p0 V0 = 3 p0 V0 2 (c) DQ AB + DQBC + DQCA = DW 5 = 3 p0 V0 + DQBC - p0 V0 = p0 V0 2 p0 V0 Þ DQBC = 2 (d) Temperature is maximum at a point D lying somewhere between B and C where the product pV is maximum. 2 p0 p=+ 5 p0 V0 æ - 2 p0 ö Þ pV = çç V + 5 p0 ÷÷ V V è ø 0

14. (a) DW =

| 52

2 p0 2 V + 5 p0 V V0 d For pV = maximum ( pV ) = 0 dV 2p Þ - 2V × 0 + 5 p0 = 0 V0 5V0 Þ V = 4 2 p0 5V0 5 p=× + 5 p0 = p0 V0 4 2 ( pV ) max \ Tmax = nR 5 5 p0 × V0 25 p0 V0 4 \ p= 2 = 1× R 8R nRTA 2R ´ 300 15. V A = = = 3 ´ 10 -3 R pA 2 ´ 105 =-

p 2 atm

A

B

1 atm

C D V

VB = VC =

2R ´ 400 2 ´ 105 2R ´ 400 105 2R ´ 300

= 4 ´ 10 -3 R, = 8 ´ 10 -3 R

= 6 ´ 10 -3 R 10 DW = 2 ´ 105 ( 4 - 3) ´ 10 -3 R 8 + 2R ´ 400 ln æç ö÷ + 1 ´ 105 (6 - 8) ´ 10 -3 è4ø K 3ö æ + 2R ´ 300 ln ç ÷ è6ø V0 =

\ DW = 200R + 800R ln 2 - 200R - 600R ln 2 = 2000R ln 2 = 1153 J As DQ = DW = 1153 J and DU = 0 cyclic process.


V ö 1 æ 3V 16. DW = ç 0 - 0 ÷ ( pB - p0 ) 2è 2 2 ø V ö 1 æ 3V + ç 0 - 0 ÷ ( p0 - p0 ) 2è 2 2 ø 1 1 = V0 ( pB - p0 ) + V0 ( p0 - pD ) 2 2 1 = V0 ( pB - pD ) 2 p V 3 p0 where, pB = p0 + 0 × 0 = V0 2 2 p0 V0 p0 and pD = p0 × = V0 2 2 1 æ3 1 1 ö \DW = V0 ç p0 - p0 ÷ = p0 V0 2 è2 2 ø 2 V ö 1 æ 3V V ö æ 3V DW ABC = p0 ç 0 - 0 ÷ + ç 0 - 0 ÷ 2 ø 2è 2 2 ø è 2 1 æ3 ö ( pB - p0 ) = p0 V0 + V0 ç p0 - p0 ÷ 2 è2 ø 5 = p0 V0 4 DU ABC = nCV ( TC - TA ) 3 V0 V ù é p p0 0 ú 3 ê 0 2 2 =3 p V = n× R ê 0 0 ú 2 ê nR nR ú 2 ë û 5 11 \DQsupplied = p0 V0 = p0 V0 4 4 1 p0 V0 2 h= 2 = = 0.1818 = 18.18% 11 11 p0 V0 4 17. (a) As the cyclic process is clockwise i.e., work done is positive, so heat is absorbed by the system. (b) In cyclic process work done is equal to the net heat absorbed (as change in internal energy is zero) so, work done in one cycle is 7200 J. (c) In anticlockwise rotation, work done is negative and heat is liberated by the system, and its magnitude is 7200 J. 18. (a) As area under clockwise loop is more than that at anticlockwise loop, so network done is positive.

(b) In loop I work done is positive and in loop II work done is negative. (c) As network done in one cycle is positive so heat flows into the system. (d) In loop I heat flows into the system and in loop II heat flows out of the system. p V 1.01 ´ 105 ´ 22.4 19. TA = A A = nR 10 3 ´ 8.314 = 273 K p V 2 pA VA TB = B A = = 2TA nR nR = 546 K = Tc nRTc nRTc nRTB Vc = = = pc pA pA 2nRTA = = 2V A = 44.8 m 3 pA 20. (a) DW = AB ´ BC = ( 4 - 1.5) ´ 10 -6 ´ ( 4 - 2) ´ 105 = 2.5 ´ 0.2 = 0.5 J (b) DQ = DW as DU = 0 in a cycle Þ DQ = 0.5 J 1 21. As r µ V r 2

2r0 p

r00 2p

21

p0

3

3

2p0

p

1

p0 V0/2

V0

V


æ V / 2ö (a) DW12 = nRT0 ln çç 0 ÷÷ è V0 ø M = - p0 V0 ln 2 = - p0 ln 2 r0 V ö æ DW23 = 2 p0 ç V0 - 0 ÷ = p0 V0 2 ø è M ; DW31 = 0 = p0 r0 (b) DQ231 = DQ23 + DQ31 = nCV DT23 + DW23 + nCV DT31 V ö æ 2 p0 0 ÷ ç2p V 3 2 ÷+ p V = n ´ Rç 0 0 0 0 2 ç nR nR ÷ ç ÷ è ø V ö æ 2 p0 ´ 0 ÷ çp V 3 2 ÷+ + n´ Rç 0 0 2 ç nR nR ÷ ç ÷ è ø 3 5 p0 V = p0 V0 + p0 V0 = p0 V0 2 2 \ Heat rejected = DQ231 - DW 5 = p0 V0 - p0 V0 + pV ln 2 0 2 3 = p0 V0 + p0 V0 ln 2 2 p M 3 5 = p0 V0 æç + ln 2 ö÷ - 0 æç - ln 2 ö÷ p0 è 2 è2 ø ø work done DW (c) h = = heat supplied DQ231 p V - p0 V0 ln 2 2 = 0 0 = (1 - ln 2) 5 3 p0 V0 2 22. DW AB = p0 (3 V0 - 2V0 ) = p0 V0 ; p c

p0

a (200 K) 2V0

b (300 K) 3V0

V

| 54

DWBC = 0, DWCA = ? DQ = DW AB + DWBC + DWCA - 800 J = P0 V0 + 0 + DWCA Þ DWCA = - 800 J - p0 V0 1 = - 800 J - nRTA 2 \ DWCA = - 800 J - 200R = - 2463 J p V - pA VA 23. DW AB = B B 1-g p A

C B V0

2V0

V

3 ( p A V A - pB VB ) 2 æT ö 3 3 = nR( TA - TB ) = nRTB çç A - 1 ÷÷; TV g - 1 2 2 è TB ø =

5 æ ö -1 3 çæ 2 ö 3 ÷ 3 = nRTB ç ÷ - 1 = nRTB (22 / 3 - 1) ç ÷÷ 2 2 çè T ø è ø æ V ö DWBC = nRTB lnçç 0 ÷÷ è 2V0 ø

= -nRTB ln 2 and DWCA = 0 Heat Supplied 3 DQCA = DUCA = nR( TA - TC ) 2 3 = nR( TA - TB ) 2 æT ö 3 3 = nR TB çç A - 1 ÷÷ = nRTB (22 / 3 - 1) 2 è TD ø 2 DW \ h= DQCA 3 nRTB (22 / 3 - 1) - nRTB ln 2 + 0 2 = 3 nRTB (22 / 3 - 1) 2


h =1 -

ln 2 2 × 2/ 3 = 1 - 07867 . = 0 .213 3 2 -1

= 21.3% ¢

Objective Questions (Level 1) 1.

3 3 ´ 1 ´ RT = RT 2 2 2U Þ T= 3R 2V0 TD = = 300 K Þ U0 = 450 R, 3R 4 V0 TA = = 600 K 3R æ 2V ö DW = DW AB + WCD = nRTA ln çç 0 ÷÷ è V0 ø æ V0 ö + nRTD ln çç ÷÷ = nR( TA - TD ) ln 2 è 2V0 ø U = nCV T =

p 2

1

p 2

2. DW12 = pDV = nRDT = 2R ´ 300 = 600R DW23 = ?; DW31 = 0. As, DQ = DW12 + DW23 + DW31 = - 300 J = 600R + DW23 + 0 Þ DW23 = - 300 J - 600R = - 5288 J 3. nC p DT1 = nCV DT2

Þ 4. TV

4 T

=

k 2 V 2

1 1 R pV = nR ( T2 - T1 ) = ( T2 - T1 ) 2 2 2

7. p µ V 2 , W = ò p dV = ò kV 2 dV

= constant

Þ pV × V n - 1 = pV n = constant ln p + n ln V = ln c Dp Dp DV =-n Þ= np = B p V D V /V V 4

5. 2 ® 3 and 4 ® 1 are isobaric. 3 ® 4 is close to isothermal and 1 ® 2 is isochoric. 1 2

1

3

6. W = ò p dV = ò kVdV =

7 5 ´ 30 = DT2 2 2 DT2 = 42 K n-1

4 V

= 1 ´ R ´ (600 - 300) ln 2 = 300R ln 2 = DQ

Þ

3

3 T

1 1 kV 3 = pV 3 3 1 = nR ( Tf - Ti ) = ( +) ve 3 æ Vf ö ÷ 8. DW = - nRT ln ç çV ÷ è i ø 1 = - nRT ln æç ö÷ = nRT ln 2 è2ø =

9. DU = 600 J - 150 J = 450 J 3 = nCV DT = R × nDT 2


600 J 3 DQ 600 = = R´ nDT 450 J 2 450 3 R 2 3 4 = R ´ = 2R 2 3

C=

10. DW1 = ( +) ve, DW2 = 0, DW3 = ( -) ve and DU1 = DU2 = DU3 as DQ = DU + DW Þ Q1 > Q2 > Q3 11. U = 2 p0 2V0 - 2 p0 V0 = 2 p0 V0 and DW = p0 (2V0 - V0 ) = p0 V0 \ DQ = DU + DW = 3 p0 V0 12. In adiabatic compression, temperature of the gas increases and as pV µ T so, pV increases. 13. As DW1 < DW2 while DU1 = DU2 Þ Þ Þ

DQ1 < DQ2 C1 < C2 C1 <1 C2

14. DW = nR( 4 T - T) + + nR (3 T - 5T) +

nR (5T - 4 T) 1-g nR ( T - 3 T)

1-g nRT 2nRT = 3 nRT - 2nRT + 1-g 1-g nRT = nRT + g -1 g nRT = ( g - 1 + 1) = nRT g -1 g -1 5/ 3 = × 1 RT = 2.5RT 5/ 3 - 1 15. Up = constant 3 nM 3 2 T = nRT × = n MR 2 V 2 V Þ T µ V i.e., isobaric process. 3/2 DU DU 3 = = = DW DQ - DU 5 - 3 2 2 2

=

CV C p - CV

=

16. DW = 50 ´ (0.4 - 0.1) +

| 56

CV R 1 ´ 50 ´ (0.2 - 0.1) 2

= 15 + 2.5 = 27.5 J DU = 2.5 J Þ DQ = DU + DW = 20 J 17. W1 = ò

2 V0 V0

pdV = p(2V0 - V0 ) = pV0

1 kV 2 2 1 3 3 = k ( 4 V02 - V02 ) = kV02 = PV0 2 2 2 Þ W1 < W2 W2 = ò

2 V0

V0

kVdV =

18. DW = p r1 r2 = p ab r - r1 ( p2 - p1 ) =p 2 × 2 2 p = ( p2 - p1 ) ( V2 - V1 ) 4 nRT dx 19. W = ò PdV = ò dV = nRT ò V -b x x=V -b dx = dV 2V = nRT ln x = nRT ln ( V - b) V = nRT [ln (2V - b) - ln ( V - b)] ½ 2V - b½ ½ 2V - b½ = nRT ln ½ ½ = RT ln ½ ½ ½ V - b½ ½ V - b½ as n = 1 mole 20. AB is isochoric process, so, DW AB = 0 BC is isothermal process, so, æV ö æV DWBC = nRT2 ln çç 2 ÷÷ = RT2 ln çç 2 è V1 ø è V1

ö ÷÷ ø

CA is close to isobaric process, so, DWCA = nRT = nR ( T1 - T2 ) = R ( T1 - T2 ) 21. DQ = DU + DW = - DQ + DW Þ DW = 2 DQ DU = nCV DT = n

f n RD T = RDT; 2 g -1


g=

f +2

f 2 f = g -1

=1 +

2 f

positive. Looking at the graph, area can be assumed to be equal so, WDEF = - W ABC .

DW = ò p dV = 2 DQ 2 nRDT

2n RD T g -1

nRDT for polytropic 1-g 1-a process with pV a = constant 1-g 2 1 \ = Þ1 - a = 1-g 1-a 2 1 g Þ 1- + =a 2 2 1+ g 1 or a= +g= 2 2 \ pV a = constant =

1+ g 2

= constant

22. DW AB = 0, DU AB = 600 J p 8 atm

3atm

B

C

A

2×10–4

5×10–4

DWadiabatic

= 0.693 pV p f × 2V - pi V = 1-g r

æ V ö pi ç ÷ × 2V - pi V pV (21 - r - 1) 2V ø è = = 1-g 1-r æ 1 - 21 - r = pV ç ç r -1 è

æ 1 - 4 -1/ 3 ö ö ÷ ÷ = pV ç ÷ ç 2/3 ÷ ø è ø

= 0.55 pV So, work done is minimum in adiabatic process.

-1

g -1 2

DWisobaric = pDV = p(2V - V ) = pV æ 2V ö DWisothermal = nRT ln ç ÷ = pV ln 2 è V ø

=

= TV a - 1 = TV = TV

24.

V

DWBC = 8 ´ 105 (5 - 2) ´ 10 -4 = 240 J DUBC = QBC - WBC = 200 - 240 = - 40 J DU = DU AB + DUBC + DUCA = 0 in cyclic process. \ DUCA = - DU AB - DUBC = - 600 J + 40 J = - 560 J 23. Starting and ending points along x-axis in graph are not clear, so nothing can be said about the magnitude of work. It can only be said that work done in ABC is negative and that in DEF is

25. DQ = DU + DW 7 5 RT0 = 10 ´ RDT + 10RDT = 35RDT 2 2 T0 = 100T = 10 ( T - T0 ) Þ 11 T0 = 10 T Þ T = 1.1 T0 pV0 pV = RT0 R ´ 1.1 T0 11 Þ V = V0 = 1.1 V0 10 26. DW = (3 p0 - p0 )(2V0 - V0 ) = 2 p0 V0 p V ö 3 æ3 p V DQsupplied = n R ç 0 0 - 0 0 ÷ 2 è nR nR ø 5 æ 3 p0 × 2V0 3 p0 V0 ö + n× R ç ÷ 2 è nR nR ø 2p V 3p V 3 5 = nR × 0 0 + nR 0 0 2 nR 2 nR 15 21 = 3 p0 V0 + p0 V0 = p0 V0 2 2 2 p0 V0 DW 4 h= = = 21 Dr p0 V0 21 2


| 58

27. DW12 < DW13 can be seen from area under the curve, while DV1 = DV2 Þ DQ12 < DQ13 Þ Q2 < Q1 or Q1 > Q2 28. DWCA = p0 ( V0 - 2V0 ) = - p0 V0 3 and DUCA = - p0 V0 2 5 Þ DQCA = - p0 V0 2 29. DQ AB = 200 kJ = nCV DT; p C

B

A V

800 T ln VB = 9 ´ 104 J 225 Þ T ln VD = 2 DW ABCD = DW AB + DWBC + DWCD + DWDA æ VB ö pC pC - pB pB ÷+ = nRT ln çç ÷ 1-g è VA ø Þ

+ pC ( VD - VC ) + 0 5 10 nRT B = 9 ´ 104 + + 105 (2 - 1) 5 13 3 4 5 = 19 ´ 10 - (10 - 800 TB ) 2 4 = 4 ´ 10 + 1200 TB 2.4 ´ 105 ´ 1 = 4 ´ 104 + 1200 ´ 100 ´ 8

DUBC = - 100 kJ and DWBC = - 50 kJ DW AB = 0 Þ DU AB = 200 kJ, DQCA = 0 = 4 ´ 105 J DU ABC = DU AB + DUBC + DUCA = 0 p V - pB VB or 200 kJ - 100 kJ + DUCA = 0 31. DW = DW AB + C C + DWCD 1-g DUCA = - 100 kJ DQ AB + DQBC + DQCA 103N/m2 = 200 kJ + ( - 100 kJ - 50 kJ) + 0 A 2.4 = 50 kJ B DW AB + DWBC + DWCA = 0 + 200 kJ + DWCA C 1 D = DQ ABC = 50 kJ \ DWCA = - 150 kJ m3 1 2 20 ´ 10 -3 20 ´ 10 3 30. DQ = DW = p ab = p ´ ´ 2 2 2 ´ 105 - 9 ´ 104 p = 9 ´ 104 + - 1 ´ 105 1 - 5/ 3 2.4 = 102A p J 3 B = 9 ´ 104 + ´ 11 ´ 104 - 10 ´ 104 æ VB ö 2 4 31. DW AB = nRT ln ç ÷ = Q AB = 9 ´ 10 J 33 è 1 ø = æç - 1 ö÷ ´ 104 = 15.5 ´ 104 C 2 è ø 1 D 1 32. DW = ò p dV = ò kVdV = kV 2 2 V 1

2


1 1 1 pV = nRT0 = RT0 2 2 2 3 DU = nCV DT = 1 × RT0 2 3 1ö æ Þ DQ = ç + ÷RT0 = 2RT0 è 2 2ø

37. h = 1 -

pV p2 V = nR nR 2 p V = constant

33. pT = constant = p Þ p

A B V

p 2 \ p20 V0 = æç 0 ö÷ V Þ V = 4 V0 è 2 ø p0 × 4 V0 p V Þ T= 2 = 2 0 0 = 2T0 nR nR 3 \ DU = nCV DT = 2 ´ R (2T0 - T0 ) 2 p0 V0 3 = 3R × = p0 V0 2R 2 æV ö 35. DWBC = nRT0 ln çç C ÷÷ è VB ø æp ö æV ö = nRT0 ln çç B ÷÷ = 2 × nRT0 ln çç B ÷÷ p è Cø è VA ø æp ö = 2nRT0 ln çç A ÷÷ è pB ø 2

æp ö æ p ö \ ln çç B ÷÷ = ln çç 0 ÷÷ = ln 4 è p0 / 2 ø è pC ø Þ

38. As the volume is adiabatically decreased, temperature of the gas increases and as the time elapsed, temperature normalizes i.e., decreases and so pressure also decreases. 39. As the compression is quick , the process is adiabatic while leads to heating of the gas.

C

Þ

Tsink

300 =1 Tsource 600 1 = 1 - = 0.5 = 50% 2

=

pB = 4 pC p p pC = B = 0 4 8

36. As, DWa > DWb Þ DW1 > DW2 while, DU1 = D U2 Þ DQ1 > DQ2

40. pV g = constant nRT g = V = nRTV g - 1 V Þ TV g - 1 = constant T1 æ V2 ö =ç ÷ T2 çè V1 ÷ø

g -1

5

æL = çç 2 è L1

ö3 ÷÷ ø

æ ngT ö 41. pV g = constant = pç ÷ è p ø Þ Þ

2

-1

æL = çç 2 è L1

ö3 ÷÷ ø

g

p1 - g T g = constant pg - 1 µ T g g g -1

pµT 7/5 7 g As = = for diatom gases. 7 g -1 -1 2 5 3.5 \ p µ T Þ a = 3.5 nRDT 42. pV x = constant , DW = , 1-x 5 D U = n × RD T 2 nRDT 5 + nRDT 1-x 2 DQ C= = nDT nDT 5 R = R+ <0 2 1-x 5 R 2 R< Þx -1 < 2 x -1 5 Þ


7 Þ x < 1.4 but x > 1 as for x < 1, 5 C will become positive. \ 1 < x < 1.4 n1CV + n2 CV 13 1 2 43. CV = = R n1 + n2 6 5 5 2´ R + 4 ´ R 2 2 = 15 R (a) 2+ 4 6 5 3 2´ R + 4 ´ R 2 2 = 11 R (b) 2+ 4 6 3 5 2´ R + 4 ´ R 2 2 = 13 R and (c) 2+ 4 6 6 3 2´ R + 4 ´ R 2 2 = 12 R (d) 2+ 4 6

3 R ( TC - TB ) 2 p Vö 3 æp V = n ´ Rç C - B ÷ 2 è nR nR ø 3 1 2 = æç p A - p A ö÷ V 2 è3 3 ø

x<

=n´

1 1 3 3 p A V = - ´ pB V = - × nRTB 2 2 2 4 3 25 = - ´ 1´ ´ 850 = - 5312.5 J 4 3

=-

49. DW AB = ( +) ve, TA = TB p p0

46. pV

g

æ nRT ö = constant = pç ÷ è p ø

g

Tµ p

5/ 3 - 1

\

5/ 3

Tµ p

\ TB TA

2 /5

æ p ö = çç B ÷÷ è pA ø

2 /5

ÞT µ p

æ 2p = çç c è 3 pc

2 /5

ö ÷÷ ø

\ TB = 0.85TA = 850 K 25 1´ ´ 150 nRT 3 47. DW AB = = 5 1-g -1 3 = 75 ´ 25 J = 1875 J 48. DWBC = 0, DQBC = DUBC

V

p0 3 V + p0 2V0 2 p0 nRT 3 =V + p0 V 2V0 2 p0 3 p0 2 T =V + V0 2nRV0 2nR p=-

Þ

y = ax2 + bx is parabola . p nRT 3 Again, p = × + p0 2V0 p 2

g -1

Þ

2V0

Þ

p1 - g T g = constant

Þ

B

V0

or

g

A

p0/2

Passage 44 & 45 pV 1 44. DW ABCA = ´ p ´ V = = DQnet 2 2 45. CA ® isobaric and BC ® isochoric, Cp 5 \ =g= Cv 3

| 60

= 0.85

Þ is also equation of parabola. While going from A to B temperature first increases ad than decreases. 50. pV 2 = constant k 1 dV = kæç - ö÷ 2 è Vø V f = - pV i = pi Vi - p f V f = nR( Ti - Tf ) = - nR ( Tf - Ti ) = ( -) ve as Tf > Ti as Ti < Tf Þ Ui < U f Þ DU = ( +) ve DW = ò p dV = ò


3 RT0 + 4 RT0 ln 2 2 = 3 RT0 + 4 RT0 ln 2

DQ = nCV DT - nRDT = n(CV - R) DT = ( +) ve as CV > R i.e., heat is given to the system.

=2´

52. ab ® isochoric, bc ® isobaric and

51. In cyclic process, DU = 0

ca ® isothermal.

p

p 2T0

b

c b

a T0

d

a V0

æ 2V DW = 0 + nR2T0 ln çç 0 è V0

c

V

2V0

V

ö ÷÷ ø

æ V + 0 + nRT0 ln çç 0 è 2V0

ö ÷÷ ø

= 2nRT0 ln 2 - nRT0 ln 2 = nRT0 ln 2 = ( +) ve i.e., DW > 0 DQsupplied = DUab + DWbc = nCV (2T0

DWab = 0, DUca = 0 as in ca density is increasing, so volume is decreasing i.e., DWca = ( -) ve, i.e., DWca < 0 in isochoric process DQab is positive for increase in temperature. 53. In isochoric process DW = 0.

æ 2V - T0 ) + nR 2T0 ln çç 0 è V0

ö ÷÷ ø

and in adiabatic process DQ = 0 Þ Q3 to be minimum Þ Q2 > Q1 > Q3

JEE Corner ¢

Assertion & Reasons 1. In adiabatic expression, DW = ( +) ve

depends on the path through which the gas was taken from initial to find state.

while DQ = 0 and as according to first law of thermodynamics, DQ = DU + DW Þ DU = - DW

3. Assertion is false, as first law can be applied for both real and ideal gases.

i.e., DU = ( -) ve this implies decrease in temperature. So, Assertion and reason are both true but not correct explanation.

4. During melting of ice its volume decreases, so work done by it is negative and that by atmosphere is positive. So, reason is true explanation of assertion.

2. Assertion is false, as work done is a path function and not a state function i.e., it

5. As DQ = DU + DW Þ DU = DQ - DW , where DU is state function while DQ and DW are path function as for definite


initial and final state DU is constant and so is Q - W . Thus assertion and reason are both true but not correct explanation. 6. Carnot’s engine is ideal heat engine with maximum efficiency but it is not also 100%. So assertion and reason are both true but not correct explanation. pV p2 V 7. pT = constant = p × = R nR Þ p2 V = constant dV V 1/ 2 \ DW = ò p dV = ò k = k× 1/ 2 V 2

= 2 k V = 2 kV = 2 p / V = 2 pV = 2nR ( Tf - Ti ) = 2nRT DT \ DW = ( +) ve for DT = ( +) ve nRT and T = constant. V ¢

| 62

Þ T2 µ V or, V µ T2 Thus assertion is true but reason is false. 8. In adiabatic changes for free expansion, Q = 0, W = 0 and DU = 0 as in free expansion no work is done against any force. For ideal gases pV = constant as DU = 0 Þ T = constant So, assertion and reason are both true but not correct explanation. 9. Assertion and reason are both true and correct explanation. 10. Assertion and reason are both true and correct explanation.

Match the Columns 1. (a) DW = pò dV = pV = nR ( Tf - Ti ) = nRT = 2RT ¾® r (b) DU = nCV T = 2 ´

3 R (2T - t) 2

= 3RT ¾® p nR (2T - T) 3 (c) DW = = - ´ 2RT 1 - 5/ 3 2 = - 3RT ¾® s (d)DU = nCV DT = 3 RT ¾® p 2. (a) In ab slope is more so, pressure is less nR as V = × T, but is constant and in p isobaric process. DW = pDV = nRDT and as DT is same in both process so, DW is same for both ¾® r (b) As DU = - nCV DT is same for both process ¾® r (c) As DQ = DU + DW , it is also same for both process ¾® s

(d) Nothing can be said about molar heat capacity ® s 3. (a) DW = ò pdV =ò

k dV dV = k ò V V

= 2 kV = 2 pV = 2nRDT ¾® p 3 (b) DU = nCV DT = nRDT ¾® s 2 3 (c) DQ = 2nRDT + nRDT 2 7 = nRDT ¾® s 2 (d) ¾® s 4. (a) DW = pDV = nRDT and DU = nCV DT Þ DW < DU ¾® q (b) DW = 0 Þ DQ = DU ,DU = ( -) ve ® p, r (c) DW = ( +) ve, DU = ( -) ve, DQ = 0 ® p (d) DW = ( +) ve, DU = 0, DQ = ( +) ve ® p 1 5. (a) DW AB = p0 V0 + p0 V0 2


=

3 p0 V0 ® s 2 (b) DU AB = DQ - DW 3 9 = + 6 p0 V0 - p0 V0 = + p0 V0 ® s 2 2 (c) DQ = + 6 p0 V0 p V ö æ 4 p0 V0 = nC ç - 0 0÷ nR ø è nR 3 p0 V0 = C R

Þ C = 2R ® p (d) DU p V ö æ 4 p0 V0 = nCV ç - 0 0÷ nR nR ø è p V 9 = 3CV 0 0 = p0 V0 R 2 3 Þ CV = R ® s 2

19. Calorimetry and Heat Tansfer Introductory Exercise 19.1. 1.

140 g

140 g

ice Q1 –15°C

ice 0°C

Q2

mg

200 g

water 0°C

water 0°C Q3

200 g

\

water 40°C

or

As Heat gain = Heat loss Q1 + Q2 = Q3 Þ 140 ´ 0.53 ´ 15 + m ´ 80 = 200 ´ 1 ´ 40 8000 - 1113 Þm = = 86 g is the mass of 80 ice melt \ Mass of water = 200 g + 86 g = 286 g and mass of ice = 140 g - 86 g = 54 g while final temperature of mixture is 0°C. 2.

16°C

A 12°C

23°C

B 19°C

C 28°C

0°C

ms A (16 - 12) = ms B (19 - 16) Þ 4s A = 3sB

Þ

or or Þ

ms B (23 - 19) = ms C (28 - 23) 4sB = 5 sC 4 ms A ( q - 12) = ms (28 - q) 5 3 4 s B ( q - 12) = s B (28 - q) 4 5 15 ( q - 12) = 16 (28 - q) 31q = 448 + 180 q = 20.26° C

3. mL = msDq Þ 80 cal = 1 cal/ °C (q - 0° C) Þ q = 80° C 4. As Heat gain = Heat loss Þ (100 - m) ´ 529 = m ´ 80 \ 100 ´ 529 = 609 m 100 ´ 529 Þm = g = 86.86 g of ice will 609 be formed. dq d dm 5. P = = ( msDq) = sDq dt dt dt dm P Þ = dt sDq 500 ´ 10 6 J/s dm \ = dt 4200 J/ kg ° C ´ 10° C 5 = ´ 104 kg/s = 12 . ´ 104 kg/s 4.2

65 | Calorimetry and Heat Tansfer

Introductory Excersise 19.2 242 = - 8.07 ° C 30 dQ 0.01 ´ 1 ´ (19 + 8.1) = dt 3.5 ´ 10 -2

1. Rest of the liquid will be heated due to conduction and not convection. 2 dQ k × 4 pr ( - dq) 2. = dt dr

= 7.74 W/m 2 0.44 kg dQ dm 5. = ×L= ´ 2.256 ´ 10 6 J/kg dt dt 300 s

r+dr

r

or q = -

= 3308.8 J/s kA q 50.2 ´ 0.15 ´ ( q - 100) = = t 1.2 ´ 10 -2

a

b

dQ dr × = - 4pk dq dt r2 T dQ b dr or = - 4 pk ò 2 dq ò 2 T1 dt a r dQ æ 1 1 ö or ç - ÷ = - 4 pk ( T2 - T1 ) dt è a b ø T - T2 dQ 4 pk ( T1 - T2 ) Þ = = 4 pkab 1 1 1 dt b-a a b dQ kADq 3. = dt t dQ t Þ k= × dt ADq m \ Unit of k = watt 2 = W/m - K m -k K 1 A D q1 K 2 A D q2 4. = l1 l2 \

19°C

q

= 627.5 ( q - 100) 3308.8 Þ q - 100 = = 5.27 627.5 Þ q = 105.27 ° C dQ kA [0 - ( - q)] dm 6. = = ×L dt y dt dy

A

y

Þ

–10°C

Þ 0.01

0.08

3.5 cm

001 . (19 - q) 3.5

=

2 cm

0.08 ( q + 10) 2

or 2 (19 - q) = 28 ( q + 10) or 38 - 280 = 30q

dy dV × L = rA ×L dt dt dy kAq = rAL y dt dy kq (Proved) = dt Lry =r

dQ = esAT 4 dt = 4 ´ 5.67 ´ 10 -8 ´ 4 p ( 4 ´ 10 -2 ) 2 ´ (3000) 4 2 4 = 0. 4 ´ 4 p ´ 5.67 ´ 4 ´ 3 J/s = 3.7 ´ 104 watt dQ Dq Dq K 8. = Þ Rth = = = KW -1 dq W dt Rth dt 7.

Calorimetry and Heat Tansfer

| 66

AIEEE Corner ¢

Subjective Questions (Level-1) Q3

Q2

Q1

1. ice ¾® Water ¾® Water ¾® steam 0 °C

0 °C

100 °C

100 °C

Q = Q1 + Q2 + Q3 = mL f + msDq + mLv = 10 [80 + 1 ´ 100 + 540] = 10 ´ 720 cal = 7200 cal 2. 10 g of water at 40°C do not have sufficient heat energy to melt 15 g of ice at 0°C , so there will be a mixture of ice-water at 0°C. Let the mass of ice left is mg. \ (15 - m) ´ 80 = 10 ´ 1 ´ 40 15 - m = 5 Þ m = 10 g \Mass of ice = 10 g and mass of water = (10 + 5) g = 15 g 3. 4 ´ s P (60 - 55) = 1 ´ s R ´ (55 - 50) Þ 4s P = s R 1 ´ s P (60 - 55) = 1 ´ s Q (55 - 50) Þ sP = s Q 1 ´ s Q (60 - q) = 1 ´ s R ( q - 50) or s P (60 - q) = 4 s P ( q - 50) 260 260 = 5q Þ q = = 52° C 5 3 dQ m ´ 336 ´ 10 J/ kg 4. = dt 4 ´ 60 s T

Þ 1 5. Q = ´ 2 Þv= \ v= = =

6. h mg Dh = msDq hgDh 0.4 ´ 10 ´ 0.5 1 \ Dq = = = °C s 800 400 = 2.5 ´ 10 -3 ° C K 1 A ( q - 0) K 2 A(100 - q) 7. = l l Þ ( K 1 + K 2 ) q = 100 K 2 100 K 2 100 ´ 46 \ q= = = 1055 . °C K 1 + K 2 390 ´ 46 8. iCD = i AC - iCB KA( q - 25) KA(100 - q) KA( q - 0) = l l /2 l /2 or or

q - 25 = 2 (100 - q) - 2q 5q = 225 Þ q = 45° C 45 - 25 Dq iCD = = =4W Rth 5

\

9. i A = iC + iD KA ( T1 - q) l

q°C 0°C

1

5

7

= 1400 J/ kg = 1400 mW/ kg m × sDq m ´ 4200 ( q - 0) c = = t 2 ´ 60 s 1400 ´ 2 ´ 60 \ =q 4200

t

q = 40° C 1 mv2 = ms Dq + mL 2 4 ( sDq + L) 4 (125 ´ 300 + 25 . ´ 104 ) 4 ´ (3.75 + 2.5) ´ 104 4 ´ 6.25 ´ 104 = 500 m /s

=

KA( q - T3 )

+

KA( q - T2 )

3l /2 3l /2 2 2 Þ T1 - q = ( q - T3 ) + ( q - T2 ) 3 3 2 4 æ or T1 + ( T2 + T3 ) = q ç 1 + ö÷ 3 3ø è 2 T1 + ( T2 + T3 ) 3 Þ q= 7/3 3 T + 2 ( T2 + T3 ) = 1 7 KA(200 - q1 ) 2 KA( q1 - q2 ) 10. = l l


=

3 KA( q2 - 100)

(60 - 30)

6 t 45 æ 60 + 30 ö - 20 ÷ Þ t = 9 min ç 2 è ø

l \ 200 - q1 = 2 ( q1 - q2 ) = 3 ( q2 - 100) Þ 3 q1 - 2q2 = 200 q1 + 3 q2 = 500 Þ q1 = 11. 25 =

¢

-11q2 = - 1300 1300 q2 = = 118.2° C 11

1/2 +

Objective Questions (Level-1) 1.

1 [200 + 2q2 ] = 145.45° C 3 400 ´ 10 -4 ( q - 100) 400 ´ 10 -4 ( q - 0) 1/ 2

S 1/2 m

3 KA(35 - q) 10

=

KA( q - 0)

20 Þ 6 (35 - q) = q 6 ´ 35 Þ q= = 30° C 7 \ Dq A = 35 - 30 = 5° C TS l N 350 2. = = = 0.69 TN l S 510 According to Wien’s law 1 K × 4 A Dq

25 W ★ 100°C

=

0°C 1/2 m

25 = 8 ´ 10 -2 [ q - 100 + q] or 312.5 = 2q - 100 412.5 Þ q= = 206.25 2 \ Dq1 = 106.25 and Dq2 = 206.25 Dq1 106.25° C \ = = 212.5° C/m Dl 1/ 2 m Dq2 206.25 ° C and = = 412.5 °C/m Dl 1/ 2 m dQ 12. = esAT 4 = 0.6 ´ 5.67 ´ 10 -8 dt ´ 2 ´ (0.1) 2 ´ (1073) 4 = 0.6 ´ 5.67 ´ (10.73) 4 ´ 10 -2 ´ 2 = 902 W dQ ö = esAT 4 and æ dQ ö = sAT 4 13. æç ÷ ç ÷ è dt ø1 è dt ø2 ( dQ/ dt) 1 210 Þ e= = = 0.3 ( dQ/ dt) 2 700 (80 - 50) c æ 80 + 50 ö 14. =ç - 20 ÷c 5 2 è ø 6 Þ K = ; 45

dQ ö 3. æç ÷ è dt ø2

æ dQ ö = 4 ç ÷ è dt ø1

l /2 KD q l

=2

dm ö æ dm ö = 0.2 g/s Þ æç ÷ =2ç ÷ è dt ø2 è dt ø1 dQ 4 pK ( q - 0) 4 p K (100 - q) 4. = = 2a - a 3 a - 2a dt a × 2a 3 a × 2a 2q = 6 (100 - q) 6 Þ q = ´ 100 = 75° C 8 K 1 A( T2 - T1 ) K 2 A( T3 - T2 ) 5. = d 3d 1 Þ K 1 ( T2 - T1 ) = K 2 ( T3 - T2 ) 3 1 Þ K1 = K2 Þ K1 : K2 = 1 : 3 3 dQ ö æ dQ ö é2 K × 2 A × Dq ù é KA Dq ù = 2 6. æç ÷ ç ÷ =ê úû êë l úû 2l è dt ø2 è dt ø1 ë Þ

dQ ö æ dQ ö = 8 cal/s Þ æç ÷ =2ç ÷ è dt ø2 è dt ø1 q, q, + dq

7. 0°C

dx x


dQ K × Adq K 0 (1 + ax) A d q = = dt dx dx l K 0 A 100 dx \ò = dq 0 1 + ax P ò0 P=

102 ´ 10 -4 1 ln (1 + ax) |l0 = × q |100 =1 0 a 1 Þ ln (1 + al) - ln 1 = 1 Þ ln (1 + al) - ln 1 = 1 Þ ln (1 + al) = 1 or 1 + al = e1 1 or l = ( e - 1) = e - 1 = 1.7 m a l2 T1 2 2 8. = = Þ l2 = l m l1 T2 3 3 9. Heat required to boil 1 g of ice is 180 cal while 1 g of steam can release 540 cal during condenstion. So, temperture of the mixture will be 100°C with 2/3 g steam and 4.3 g water. 10. T1 < T2 < T3 as temperature of a body decreases in rate of cooling also decreases such that time increases for equal temperature difference. 11. Conduction is maximum for which thermal resistance is minimum, as l Rth µ 2 then for r (a) 50 (b) 25 (c) 100 (d) 33.33, So option ‘b’ has minimum resistance. 12. Slope of temperature versus heat graph gives increase of specific heat or heat capacity and the portion DE is the gaseous state. 13. dQ = m sdt = maT 3 dT Q a 42 a 15a Þ = T |1 = (16 - 1) = m 4 4 4 14. Resistance becomes 1/4th in parallel of that in series, so times taken will also become 1/4th ie, 12/4 = 3 min. 15. ms1 ´ 12 = ms2 ´ 8 Þ s1 : s2 = 2 : 3

16.

KA( T - Tc ) 2l Þ Þ Þ

=

| 68

KA( Tc - 2 T) l

T T + 2T = Tc + c 2 2 1+ 2 3 T= Tc 2 2 3 Tc = T 1+ 2

17. P = (1000 - 160) W = 840 W 2 ´ 4200 ´ 50 = t 42 ´ 104 \ t= = 500 s = 8 min 20 s 840 dQ KA( T2 - T) 2 KA( T - T1 ) 18. = = dt x 4x 1 1 Þ T2 - T = T - T1 2 2 1 3 Þ T2 + T1 = T 2 2 2æ 1 1 Þ T = ç T2 + T1 ö÷ = (2T2 + T1 ) 3è 2 ø 3 dQ KA é 1 \ = T2 - (2T2 + T1 ) ù dt x ëê 3 ûú KA 1 = [3 T2 - 2T2 - T1 ] ´ x 3 KA 1 = ( T2 - T1 ) ´ x 3 1 Þ f = 3 1 19. Dq µ K Dq A K 1 Þ = B = Dq B KA 2 1 Þ Dq A = Dq B = 18 ° C 2 ¢

More than One Correct Options

20. Amount of heat radiated or absorbed depends upon. Surface type, surface area, surface temperature and temperature of surrounding, so (a) and (b) are correct.


21.

KA( 40 - q) l

Þ Þ or So,

=

KA( q - 30)

+

KA( q - 20)

l 40 - q = 2q - 50 3 q = 90° q = 30° C (b) and (d) are correct.

l

22. m ´ s ´ (2q - q 0 ) = m ´ 2s ´ ( q 0 - q) 4 Þ 4q = 3q 0 Þ q 0 = q 3 c1 : c2 = m1 : s2 = s1 : s2 = 1 : 2

So, (b) and (c) are correct. 23. In series rate of R = R1 + R2 qq 1 1 1 Þ = + Þq= 1 2 q q1 q2 q1 + q2 1 1 1 In parallel = + R R1 R2 1 Þ q = q1 + q2 as q µ R So, (b) and (c) are correct. 24. (a), (c) and (d) are correct.

JEE Corner ¢

Assertion and Reason 1. Assertion is false. 2. According to Wien’s law assertion and reason are correct.

2 -2 (c) e = E = [ML T ] = [MT -3 ] At [L2 T] q (d) Rth = dq = dQ/ dt [ML2 T -2 T -1 ]

3. Assertion and reason are true but not correct explanation. 4. Assertion is true but reason is false as resistance becomes 1/4th. 5. Assertion and reason are both false. 6. Assertion is false as this statement was not given by Newton. 7. Assertion and reason are both true with correct explanation. 8. Both are true explanation.

but

not

correct

9. Assertion is false as temperature at different points become different. 10. As mass of follow sphere is less so cooling will be faster. So, both are true with correct explanation. ¢

Match the Columns

1. 2 -2 -1 (a) s = ( dQ/ dt) = ML T T AT4 L2 q4 -3 -4 = [MT q ] (b) b = lT = Lq

s

p

r s

= [M-1L-2 T 3q]

2. (a) (b) (c) (d)

3.

Slope of line ab Length of line bc µ m Solid + liquid ® bc Only liquid ® cd

KA (100 - q b ) l

= =

s r s q

KA ( q b - q d ) l KA ( q d + 80)

l \ 100 - q b = q b - q d and 100 - q b = q d + 80 q - 2q b = - 100ü \ d Þ -3 q b = - 120 q d + q b = 20 ýþ q b = 40° C Þ q d = - 20° C 40 - 20 qc = q f = = 10° C 2 \ (a) ® q, (b)® p, (c)® p, (d) ® r Þ

4. (a) ms ( q1 - q) = 2ms (2q - q1 ) 5 Þ 3 q1 = 5q Þ q1 = q ® q 3


(b) ms ( q2 - q) = 3 ms (3 q - q2 ) 5 Þ 4 q2 = 10 q Þ q2 = q ® p 2 (c) 2ms ( q 3 - 2q) = 3 ms (3 q - q 3 ) 13 Þ 5q 3 = 13 q Þ q 3 = q®s 2 (d) ms ( q4 - q) + 2ms ( q4 - 2q) = 3 ms(3 q - q4 ) 7 Þ 6q4 = 14 q Þ q4 = q ® r 3

5. (a) (b) (c) (d)

1 dQ J ¾® = m dq kg ° C dQ c = ms = m = J°/C ¾® mdq dQ i= = J/s ¾® dt E L= = J/kg ¾® m s=

q s r s

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Waves & Thermodynamics

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