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50 Day's Plan For Physics : (Fluid 1 & 2)

ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A. 3 4 4 3 1 3 2 2 3 3 4 1 2 2 4 4 2 1 1 1 3 1 2 1 2 3 2 2 2
Q. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
A. 1 2 3 2 3 2 3 3 2 1 1 1 1 2 4 1 3 2 1 1 3 2 4 2 1 1 3 2 1 3

HINT – SHEET

1. Ans ( 3 ) 5. Ans ( 1 )
We know that, dv
F = ηA
dx
vterminal ∝ r2 in CGS units,
v1 r2 6
∴ = 12 103 = η × 103 ×
v2 r2 0.6
when 8 drops combine, we have ⇒ η = 0.1 poise

8× 4 3 4 3
πr = πr
3 1 3 2
6. Ans ( 3 )
F = 6 π η rv, F' = 6 π η (2r)(2v) = 4F
⇒ r2 = 2r1
16 4
7. Ans ( 2 )
∴ = ⇒ v2 = 64 cm/sec The velocity first of all increases with time and
v2 16
2. Ans ( 4 ) after that it acquires the terminal velocity given by
Q ∝ pr4
4 2 (ρ − σ)
Q2 p2 r v= g
= ( )( 2 ) 9 η
Q
2p
p1 r1
a/2
4 8. Ans ( 2 )
= ( )( )
p a ∵ v = const.
Q
Q2 = ∴ FGravity = FBoyant + FViscous
8
3. Ans ( 4 ) ⇒ W = W + FV
By stoke's law 2
⇒ FV = W
viscous force F = 6 π η r ν 2

For same liquid η = constant 10. Ans ( 3 )

and velocity ν is same Fv + Th = W
F1 r 1 Fv = W – Th = V ρ Bg – V ρ Lg
F∝r⇒ = =
F2 4r 4
ρL
= V ρ Bg ( 1− )

4. Ans ( 3 ) = mg ( 1−
y
)
ρB

By theory x
HS-1/5
11. Ans ( 4 ) 17. Ans ( 2 )
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Stress ΔV P
γ= ⇒ Stress = γ × Strain =
Strain V B
= 2 × 1011 × 10 – 3 = 2 × 108 N/m2 10 × 105
=
W eight 40 × 109
Now ⇒ Stress = = 0.25 × 10 – 4
Area
⇒ Weight = Stress × Area = 2.5 × 10 – 5
Weight = 2 × 108 × π (0.5 × 10 – 3)2
= 157 N
18. Ans ( 1 )
From graph
12. Ans ( 1 ) Slope = F
Fℓ
Y= , F = AY Δℓ Δℓ
AΔℓ ℓ
2
also Y = F L
F1 A1 ℓ 2 1 1 AΔℓ
= × =( ) ×2=
F2 A2 ℓ 1 2 2 Slope = F
=YA
Δℓ L
13. Ans ( 2 ) ⇒ more slope more Y
wℓ mgℓ
So YA > YB
∵ Δℓ =
AY
=
πr2 Y 19. Ans ( 1 )
2 In both case Y, A & F remain same so breaking
Δℓ1 m r ℓ Y
⇒ = 1 ×( 2 ) × 1 × 2 stress will be also same.
Δℓ2 m2 r1 ℓ 2 Y1

Δℓ1 3M 1 1 3a
20. Ans ( 1 )
= × ×a× =
Δℓ2 2M b 2 c 2b2 c Slope of stress vs strain curve is equal to Young's
14. Ans ( 2 ) modulus
P
B= ⇒ YA > YB
− ( ΔV )
V
345 × 106 Separation b/w elastic limit is breaking point is
⇒ 138 × 109 =
− ( ΔV )
V larger means that the material is more ductile.
ΔV 345 × 106
⇒ =− 9
= – 2.5 × 10 – 3 hence A is less ductile than B
V 138 × 10
ΔV
⇒%
V
= −2.5 × 10−3 × 100 = – 0.25 % 21. Ans ( 3 )
'Negative sign' indicates that volume decreases, if
pressure is increased and vice-versa.
22. Ans ( 1 )
15. Ans ( 4 )
F /A
Y = here Δ ℓ = ℓ
(Δℓ/ℓ)
F = AY = (0.1 × 10 – 4) × 2 × 1011
= 2 × 106 N 23. Ans ( 2 )
16. Ans ( 4 ) In stretching of a spring shape charges therefore
Shearing strain, ϕ = x = 0.02cm = 0.002 shear modulus is used.
L 0.1m Ycopper < Ysteel

HS-2/5
24. Ans ( ) 31. Ans ( 1 )
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F M(g + a) Equating pressure,

Maximum stress = =
A πr2min F 1200
= 2
or F = 3 kg f
πR 2
M(g + a)

 900(9.8 + 2.2) π[20R]
∴ rmin = √ =
32. Ans ( 2 )


π × Max. stress 3
⎷
π× π
× 108
= 6 × 10 – 3 m = 6 mm For solid to be in equilibrium its weight
25. Ans ( 1 ) = buoyant force due to both liquids
F x Let A = area of cross section of solid
η=
A Δx ρ s = density of solid
6 × 103 10
= × = 6 × 109 N/m2 ⇒ (AL) ρ sg = (A × L ) × 2ρg + (A × 3L ) × ρg
100 × 10 −6 0.1 4 4
26. Ans ( 2 ) ⇒ ρs =
ρ 3ρ
+
5
= ρ
2 4 4
Δ P = h ρ g = 200 × 103 × 10 N/m2 = 2 × 106 N/m2
ΔP 2 × 106
33. Ans ( 3 )
K= = FA P A ρA ghA AA
ΔV 0.1 = A A = =1
V 100 FB P B AB ρB ghB AB
= 2 × 109 N/m2
34. Ans ( 2 )
27. Ans ( 3 ) Pressure of liquid column = h ρ g
Decrease in tension = Buoyant Force P = 0.4 × 900 × 10Nm – 2
= ρ ω Vg Force on the base = P × area
= 103 × 3.6 × 10 – 4 × 10 = 3.6 N = P × 2 × 10 – 3m2
28. Ans ( 2 ) = 0.4 × 900 × 10 × 2 × 10 – 1 N
Using Pascal's law = 7.2 N
Patm + ρ oil g(10) = Patm + ρ wg(8) 35. Ans ( 3 )
ρ oil =
4
5
ρ w = 0.8 g/cm3 Case-I:
29. Ans ( 2 ) m = mass of boat
Wa = 210 gm mg = ρAℓ1 g
Ww = 180 gm m = ρAℓ1
W ℓ = 120 gm
ρℓ Wa − Wℓ Case-II:
=
ρw Wa − Ww m1 = mass of man
210 − 120
R.D. of liquid = =3 (m + m1 ) g = ρAℓ2 g
210 − 180
Wa ρ
R.D. of metal = = m m + m1 = ρAℓ2

= 210
=7
Wa − Ww ρw
Subtract
210 − 180 ρA (ℓ2 − ℓ1 ) = m1
0.3
1000 × 24 × = m1
30. Ans ( 2 ) m1 = 72 kg
100
5
P0 = 10 = Px + ρ g(h/5) = ρ gh
36. Ans ( 2 )
Also Px = 4 ρgh = 4
× 105 2ρ1 ρ2 2×1×2 4
5 5 ρ= = = .
Px = 0.8 × 105 Pa ρ1 + ρ2 1+2 3

HS-3/5
37. Ans ( 3 ) 43. Ans ( 1 )
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Pressure head Maximum height of water in tank obtained when

5
P 10 incoming volume of water is equal to outgoing
= = = 10 m
ρg 1000 × 10 volume of water
38. Ans ( 3 ) Q = A1v1 = A × 2gH √

A1V1 = A2V2 70 = 1 × 2 × 980 × H

√

(2 × 10 – 2) (2) = (0.01) V2 H = 2.5 cm

V2 = 4 m/s
44. Ans ( 2 )
4 × 10 + 1 (103)(2)2 = P + 1 (103)(4)2
4
1
2 2 From P1 − P2 = ρ (v22 − v21 )
P = 4 × 104 – 6 × 103 = 3.4 × 104 Pascal 2
1
F = (P1 − P2 ) A = ρA (v22 − v21 )
39. Ans ( 2 ) Force
1
2
From Bernoulli's theorem F = × 1.28 × 10 [(140)2 − (110)2 ]
2
1 F 48000 N
Patm + 0 = Pout + PV 2
2
1
Patm − Pout =
2
PV 2 45. Ans ( 4 )
1 2 Speed of efflux v = √ 2gh
= × 1.2 × (40)
2 ∴ Range = 2√h(H − h)
Δ P = 960 N/m2
H
Force acting on the roof F = ( Δ P)A for maximum range h =
2
= 960 × 250 Rmax. = H
= 2.4 × 105 N 46. Ans ( 1 )
Pressure inside the roof is greater than outside the According to Bernoulli's theorem.
roof, so force will act upward direction. 1 2
ρv + ρgh + P = const.
40. Ans ( 1 ) 2
So if A↓ v↑­and P↓
1 2
Pclose = Popen + ρv
2 and if A­v↓ and P­↑
5.5 × 105 = 5 × 105 + 1 × 103 v2
on solving v = 10 m/s
2 47. Ans ( 3 )
Water in capillary never overflows
41. Ans ( 1 ) 48. Ans ( 2 )
0.4 m below the tap speed is v then
In case of soap bubble,
use v2 = u2 + 2as
W = T × 2 × ΔA
⇒ v2 = 12 + 2 (10) (0.4)
= 0.03 × 2× 40 × 10 – 4 = 2.4 × 10 – 4J.
v = 3 m/s
By equation of continuity 49. Ans ( 1 )
A2 v = A1u h = ℓ sin θ
A × 3 = 0.2 × 1
0.2
A= = 0.066 m2
3
42. Ans ( 1 )
4(H – 4) = 6(H – 6) h 2
ℓ= = = 4cm
or 2H = 36 – 16 = 20 or H = 10 cm sin θ 1/2

HS-4/5
50. Ans ( 1 ) 55. Ans ( 1 )
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2T cos θ 4T
h= Δp =
rρg R
= 2 × 70 × 1 = 4 cm 4T 4 × 117.6 4 × 117.6
R= = = = 0.8cm
1
× 1 × 980 Δp (ρgh) 1 × 980 × (0.6)
28
51. Ans ( 3 ) R = 8 mm
Work done in forming a bubble
W = 8 π R2T 56. Ans ( 1 )
2T 2 × 0.072
W ∝ R2 Pex = = = 240 P a
r ( 0.6 × 10−3 )
As V ∝ R3
2V R
3 57. Ans ( 3 )
= ( 2 ) ⇒ R2 = 21/3 R1
V R1 2T ℓ = Mg ⇒ M = 2T ℓ
So W2 = W × 22/3 g
2 × 3 × 10−2 × (10 × 10−2 )
52. Ans ( 2 ) =
10
= 0.6 g
W = 4 π TR2 (n1/3 – 1) 58. Ans ( 2 )
W = 4 × 3.14 × 72 × 10 – 3 × 10 – 6 × 99 Excess pressure inside soap bubble of radius 'r' =
W = 8.95 × 10 – 5 J 4T
53. Ans ( 4 ) r
ΔP ∝
1
4 3 4 3 r
n× πr × πR
as 'r' increases with time, Δ P also decreased with
3 3
1000 r3 = R3 time.

r 1
59. Ans ( 1 )
R = 10r ⇒ = The force due to surface tension is always
R 10
tangential on the free surface which depresses to
Esmall = 4 π r × T Elarge = 4 π R2 × T
2
the more projected parts.
ESmall
=
4πr2 T r
2
1
2
=( ) =( ) =
1 60. Ans ( 3 )
EL arg e 4πR2 T R 10 100 4T
54. Ans ( 2 ) ∵ Pex =
1
r
Excess pressure in air bubble just below the water ∴ Pex ∝
r
2T The air pressure is greater inside the smaller
surface. P1 =
r bubble.
2T
Excess pressure inside a drop P2 = Hence air flows from the smaller to the larger
r
So P1 = P2 bubble.

HS-5/5