26.

F = av2 = 103 × 10 × 10–4 × 20 × 20  75  105 N    1 

2
2 
 729     4    10  
F = 400  10
2
m    9  
27. Apply Bernoulli’s theorem between Piston
 4  10 4 
1 2  729 75  10 3  
and hole PA  gh  P0  v e  81 
2
= 9 [942 × 10–7 J]
Assuming there is no atmospheric pressure
Gain in surface energy
on piston
U = 9 × 942 × 10–7 – 942 × 10–7
5  105 1
 103  10  10  1.01  105   103  v e2 = 8 × 942 × 10–7 = 7536 × 10–7 J
 2
ve = 17.8 m/s = 7.5 × 10–4 J
28. P2A – P1A = 5.4 × 105 × g 30. Surface area of soap bubble = 2 × 4R2

5.4  106 Work done = change in surface energy × Ts

P2 – P1 = = 5.4 × 2 × 102 × 10 =
500 = Ts × 8 × R22  R12 
10.8 × 103
22 3
1 1 = 2 × 10–2 × 8 × × 49 × × 10–4
P2 + 0 + V22 = P1 + 0 + V12 7 4
2 2
= 18.48 × 10–4 J
1 1
P2 – P1 = (V12 – V22) = = (V1 –
2 2 31. E/ A
Y
V2)(V1 + V2) 

1
10.8 × 103 = × 1.2(V1 – V2) × 2 × 3 × 102
2 YA
 F 
10.8 × 10 = 3.6(V1 – V2) 
V1 – V2 = 30  A   A 
     
 V1  V2  30  1  2
 V   100  300  100  10%
   2 A1  2
  
29. Initial surface energy = TA  1 A 2  1
Where T is surface tension and A is  2 1 2
  
surface area 0.2 2.4  2.4 1

 75  10 5 N    2  6.9 10 2 mm
Ui     4(1 10 2 )2 
 10 2 m   32. Heat given by block to get 0ºC temperature
= 75 × 10–3 × 4 × 10–4 = 942 × 10–7 J Q1 = 5 × (0.39 × 103) × (500 – 0)
To get final radius of drops by volume = 975 × 103 J
conservation Heat absorbed by ice to melt m mass
4 3 4  Q2 = m × (335 × 103)J
R  729  r 3 
3 3  Q1 = Q2
R = Initial radius m × (335 × 103) = 975 × 103
r = final radius 975
m  2.910kg
R R 1 335
r   cm
(729)1/3 9 9
Final surface energy
Uf = 729 (TA)
33. R = R0 (1 + T) for (i) and (ii)
3 = R0 (1 +  (30 – 0)) 800
K
2 = R0 (1 +  (10 – 0)) 3

3 1  30 TH = 266.7 K

2 1  10 38. 1
WDE  (600  300)3 J
1 2
  0.033
30 = 1350 J

34.   = 6.241 – 6.230 = 0.011 cm WEF = – 300 × 3 = – 900 J

WDEF = 450 J
  =   
39. y = 2 sin (t – kx)
0.011 = 6.230 × 1.4 × 10–5 ( – 27)
Maximum particle velocity = A 
0.011  105
  27  
6.230  1.4 Wave velocity 
k
 = 153.11 nearest is 152.7°C.  1 2
 A k 
35. No of moles of H2 = 8 moles k A 
 = 2A
No of moles of O2 = 4 moles
= 4  cm
Total moles = 12 moles
40.  2 2 
y  0.5 sin  400t  x
At STP 1 mole occupy = 22.4 = 22.4 × 103 cm3    
12 moles will occupy = 12 × 22.4 × 103 cm3
2
 26.8 × 104 cm3  400

36. Constant entropy means process is 2
K
adiabatic 
PV = constant 
v [v  400m / s]
k
V1
V2  VP max = 4Vwave
8
  4
P1V1 = P2 V2 A  4    A 
k 2
5/3
V  2A 20
P1V1  P2  1     5
 8 4 4
P V 5/3 41. From the given equation k = 8 m–1 and  =
PV
1 1
5/3
 2 1 4 rad/s
32
P2 = 32P1 
Velocity of wave =
k
37. TL
  1 4
TH v= = 0.5m/s
8
1 T 1 k
 1 L ...(i) 42. f
2 TH 2 m
 T  40  YA
1  L   0.65 ...(ii) k
 TH  L

 1  YA
f  
 2  mL
43. 48. 41  5 C  0

95  5 100  0
36
 C  100  40C  313K
90
49. m  m

While approaching PV = const d p    const
v  d
 c 
  0  
 c  v cos   p d2
 const  32
While receding d d1

 c  p1  d1   1
 7/5
1
  0       
 c  v cos   p2  d2   32  128
44. Initially beat frequency = 5 Hz
T1 P1V1 1 1
so, A  340  5  345Hz, or 335 Hz   32 
T2 P2 V2 128 4
after filing frequency increases slightly n1Cv1  n2 Cv 2
50. Cv / mix 
so, new value of frequency of A > A n1  n2
Now, beat frequency = 2Hz 3R 5R
1.  3.
2  9R   R
2
 A  340  2  342Hz, or 338 Hz  2
1 3 4 4
hence, original frequency of A is A = 335
=3
Hz
45. fb = f1 – f2
v v 40
 
4.08 4.16 12
 v = 707.2
46. A1V1 = A2V2
750 × 10–4V1 = 500 × 10–6 × 0.3
500  3 10 3
V1  m / s = 2 × 10–3 m/s
750
dh
 –2 10 3 m / s
dt
47.

4T 4T
P2  P0  & P1  P2 
6 3
4T
 P1  P0  2
2