Class 11-science H C VERMA Solutions Physics Chapter 13 - Fluid Mechanics
Fluid Mechanics Exercise 273
Solution 1
Pressure due to water
⇒
P = h
g = 4 ×
1000 ×
10
= 40000 N/m²
Since pressure is proportional to height of water. If tap is open height of water decreases, so pressure decreases.
Solution 2
(a)
= 
=
(b) Pressure at bottom of mercury
Solution 3
Solution 4
(a) 
(b)
Cylindrical water column is balanced by bottom of glass and remaining by side wall.
Solution 5
Now atmospheric pressure will not act.
(a)
(b)
No change will occur in answer if height, area and volume is kept same irrespective of shape of container.
Solution 6
Fluid Mechanics Exercise 274
Solution 7
F= PA
= 107 N/m2
Force depends upon area not on orientation.
Solution 8
(a)
= 60000N
(b) Force of strip of width 'dx'
= x×1000×10×2dx
= 20,000 xdx
(c) Torque = force × distance
(d)Total force by water on side wall
(e) Total torque on side wall
Solution 9
Weight difference in air and water is due to buoyancy force
Solving, equation (1) and (2)
Solution 10
Now,
Solution 11
N + B = mg
⇒N = mg - B
= 160 ×
10-3 × 10 - Vwg
= 
= 
⇒N = 1.4
Solution 12
(a) In equilibrium,
Weight of boat = Buoyancy force
(b) Let Vfbe the volume of boat filled with water before water starts coming in from the sides. So,
Fraction
of boat's volume filled = 
=
Solution 13
Let side of ice cube be x.
In equilibrium,
g + mg =
g
(
x3)
+ 0.5 =x3w
(0.9x103)x3 + 0.5 = x3(103)
X=17 cm
Solution 14
Volume of ice inside kerosene oil and water be Vk and Vwrespectively
Vice = Vk + Vw ----- (1)
Since, Ice is in equilibrium,
Miceg = FB
g=
g+
g
(0.9)=
--------- (2)
Solving equation (1) and (2)
=1
Solution 15
Let external edge of iron cube be x
In equilibrium,
Weight of iron cube = Buyoncy force
g = g
6x[(0.1)x2(8000)] = x3(1000)
X=4.8cm
Solution 16
In equilibrium
(
+
)g
= (
+
)
g
(0.2+)=(
/
+ /
)
(0.2+mpb) = (0.2/0.8 + mpb/11.3)
mpb = 0.0548 kg
Solution 17
In equilibrium,
mpbg + mwg = Vwood
water
g
mpb +200 = 200/0.8
mpb = 50g
Solution 18
Initially,
mg = 
b
(12)3g=(12)2(12/5)(
)g
=13.6/5
gm/cc
Let x be the height of water column after the water is poured
mg=x(12)2
+(12-x)(12)2g
(13.6/5)(12)3 = x(12)2+(12-x)(12)2(13.6)
X=10.4cm
Solution 19
In equilibrium,
mg=
[
- 
(6)3]= 
g
=0.865
=865kg/m3
Solution 20
(
+
)g=V
(5)3
+ 0.1×1000 = (5)3
=
0.8gm/cc
Solution 21
Reading by
spring balance = mg-buoyancy
force
= mg - V
g
= mg - 
g
= [1 - (1/7800)(1.293)]g
= [1 - (1/800)(1.293)]g
= 1.0015
Solution 22
T=2
Water behaves as spring having spring
constant K=Ag
T=2
= 2
T= 0.5 sec
Solution 23
In equilibrium
Kx+Vwg = mg
(500 × 
)x + 
(5)2(
)(1)(980)
= (5)2(20)(8)(980)
X=23.5 cm
T=2
Water behaves as spring of spring
constant 
= Ag
Now, spring and water spring are in parallel combination
= + 
T = 2
T= 0.935 sec
Solution 24
(a)
In equilibrium,
Kx + Vg = mg
(50)x+(0.5/800)(1000)(10) = (0.5)(10)
X=-0.025 m
X= - 2.5 cm
So, spring will be in compression
(b)Since, system is completely inside water so unbalanced force will be due to spring only. Buyoncy force doesn't change
T=2
= 2
T = /5
Solution 25
Let the length of the edge of the ice block when it just leaves contact with the bottom of glass be x and height of water after melting be h.
Weight = Buoyancy force
(x)3g = (x2h)
g
0.9x = h ----- (1)
Volume of water formed by melting of ice
(4)3
- (x)3 = r2h
- x2h ------ (2)
Solving equation (1) and (2)
X=2.26cm
Solution 26
= 
+
g = +
g+
- 
= la/g
Solution 27
From continuity equation
A1V1+ A2V2 = A3V3
(12×d)(20) +
(8×d)(16)
= (16×d)(
)
= 23 km/hr
Solution 28
A) Discharge = Area × velocity
(1×10-6) = (4×10-6)
= 0.25 m/s
B) A1V1 = A2V2
(4)(0.25) = (2)
= 0.5 m/s
C) From Bernoulli's equation
PA + 
= PB + 
PA + 
= PB + 
PA - PB = 94 N/m2
Solution 29
(a) VA = 0.25 m/s
(b) VB = 0.5 m/s
(c) PA + +
ghA
= PB + +ghB
PA + = PB + +1000×10×(hB-hA)
[here, hB-hA=
cm]
PA - PB = 0 N/m2
Solution 30
(a) VA = 0.25 m/s
(b) VB = 0.5 m/s
(c) PA + +ghA
= PB + +ghB
PA + = PB + +1000×10×(hB-hA)
[here, hB-hA=
cm]
PA - PB = 188 N/m2
Solution 31
(a) AaVa = AbVb
(1)(10) = (0.5)(Vb)
Vb = 20 cm/s
(b) PA + +ghA
= PB + +ghB
PA + 
+1x1000xhA= PB +
+1x1000xhB
[here, hB-hA=5cm]
PB-PA = 4850 dyne/cm2
PB-PA = 485 N/m2
Solution 32
AaVa = AbVb
4Va = 2Vb
2Va = Vb ----- (1)
PA + +ghA
= PB + +ghB
PA + 
= PB +
[here, hA=hB=0cm]
PA-PB = 
2×1×1000
=
VA = 36.5 cm/sec
∴ Rate of discharge = VAAa
= (36.5)(4)
Q = 146 cm3/sec
Solution 33
Q = AaVa = AbVb
500 = 5(Va) = 2(Vb)
Va = 100 cm/sec
Vb = 250 cm/sec
PA + = PB +
PA - PB = 
(VB2
- VA2)
980 × 13.6 ×
h = 
[(250)2 - (100)2]
h = 1.97 cm
Solution 34
(a) V = 
= 
V = 4m/s
(b) V = 
= 
V = 
m/s
(c) Q = 
=
av
dV = (2mm2)(
)dt
(d) dV
= -Adh = a()dt
-= 
t= 6.5 hrs
Solution 35
Let h be the height of hole from bottom of tank
V = 
T = 
∴ R = vT
= 
To maximize Range,
= 0
H-2h = 0
h= 
