Chapter 13 : Fluid Mechanics - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

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JEE : Chapter 13 : Fluid Mechanics - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


13.1
SOLUTIONS TO CONCEPTS 
CHAPTER 13
1. p = h ? g
It is necessary to specify that the tap is closed. Otherwise pressure will gradually decrease, as h 
decrease, because, of the tap is open, the pressure at the tap is atmospheric.
2. a) Pressure at the bottom of the tube should be same when considered for both limbs.
From the figure are shown,
p
g
+ ?
Hg
× h
2
× g = p
a
+ ?
Hg
× h
1
× g
? p
g
= p
a
+ ?
Hg
× g(h
1
– h
2
)
b) Pressure of mercury at the bottom of u tube
       p = p
a
+ ?
Hg
h
1
× g
3. From the figure shown 
p
a
+ h ?g = p
a
+ mg/A
? h ?g = mg/A
? h = 
Ap
m
?
4. a) Force exerted at the bottom.
= Force due to cylindrical water colum + atm. Force
= A × h × ?
w
× g + p
a
× A
= A(h ?
w
g + p
a
)
b) To find out the resultant force exerted by the sides of the glass, from the freebody, diagram of water 
inside the glass
p
a
× A + mg = A × h × ?
w
× g + F
s
+ p
a
× A
? mg = A × h × ?
w
× g + F
s
This force is provided by the sides of the glass. ?
5. If the glass will be covered by a jar and the air is pumped out, the atmospheric pressure has no effect.
So, 
a) Force exerted on the bottom.
= (h ?
w
g) × A
b) mg = h × ?
w
× g × A × F
s
.
c) It glass of different shape is used provided the volume, height and area remain same, no change in 
answer will occur.
6. Standard atmospheric pressure is always pressure exerted by 76 cm Hg column 
= (76 × 13.6 × g) Dyne/cm
2
.
If water is used in the barometer.
Let h ? height of water column.
? h × ?
w
× g ?
7. a) F = P × A = (h ?
w
× g) A
b) The force does not depend on the orientation of the rock as long as the surface area remains same.
8. a) F = A h ? g.
b) The force exerted by water on the strip of width ?x as shown,
dF = p × A
= (x ?g) × A
c) Inside the liquid force act in every direction due to adhesion.
di = F × r
d) The total force by the water on that side is given by
F = 
?
1
0
20000 x ?x ? F = 20,000 
1
0
2
] 2 / x [
e) The torque by the water on that side will be,
Pa
Pa
Gas 
Pa
h 
pa 
45 kg 
A =900 cm
2
Page 2


13.1
SOLUTIONS TO CONCEPTS 
CHAPTER 13
1. p = h ? g
It is necessary to specify that the tap is closed. Otherwise pressure will gradually decrease, as h 
decrease, because, of the tap is open, the pressure at the tap is atmospheric.
2. a) Pressure at the bottom of the tube should be same when considered for both limbs.
From the figure are shown,
p
g
+ ?
Hg
× h
2
× g = p
a
+ ?
Hg
× h
1
× g
? p
g
= p
a
+ ?
Hg
× g(h
1
– h
2
)
b) Pressure of mercury at the bottom of u tube
       p = p
a
+ ?
Hg
h
1
× g
3. From the figure shown 
p
a
+ h ?g = p
a
+ mg/A
? h ?g = mg/A
? h = 
Ap
m
?
4. a) Force exerted at the bottom.
= Force due to cylindrical water colum + atm. Force
= A × h × ?
w
× g + p
a
× A
= A(h ?
w
g + p
a
)
b) To find out the resultant force exerted by the sides of the glass, from the freebody, diagram of water 
inside the glass
p
a
× A + mg = A × h × ?
w
× g + F
s
+ p
a
× A
? mg = A × h × ?
w
× g + F
s
This force is provided by the sides of the glass. ?
5. If the glass will be covered by a jar and the air is pumped out, the atmospheric pressure has no effect.
So, 
a) Force exerted on the bottom.
= (h ?
w
g) × A
b) mg = h × ?
w
× g × A × F
s
.
c) It glass of different shape is used provided the volume, height and area remain same, no change in 
answer will occur.
6. Standard atmospheric pressure is always pressure exerted by 76 cm Hg column 
= (76 × 13.6 × g) Dyne/cm
2
.
If water is used in the barometer.
Let h ? height of water column.
? h × ?
w
× g ?
7. a) F = P × A = (h ?
w
× g) A
b) The force does not depend on the orientation of the rock as long as the surface area remains same.
8. a) F = A h ? g.
b) The force exerted by water on the strip of width ?x as shown,
dF = p × A
= (x ?g) × A
c) Inside the liquid force act in every direction due to adhesion.
di = F × r
d) The total force by the water on that side is given by
F = 
?
1
0
20000 x ?x ? F = 20,000 
1
0
2
] 2 / x [
e) The torque by the water on that side will be,
Pa
Pa
Gas 
Pa
h 
pa 
45 kg 
A =900 cm
2
Chapter-13
13.2
i = 
?
1
0
20000 x ?x (1 – x) ? 20,000 
1
0
3 2
] 3 / x 2 / x [ ? ?
9. Here, m
0
= m
Au
+ m
cu
= 36 g …(1)
Let V be the volume of the ornament in cm
3
So, V × ?
w
× g = 2 × g
? (V
au
+ V
cu
) × ?
w
× g = 2 × g
?? g
m m
w
au au
? ?
?
?
?
?
?
?
?
?
?
?
?
?= 2 × g
? 1
9 . 8
m
3 . 19
m
Au Au
? ?
?
?
?
?
?
? = 2 
? 8.9 m
Au
+ 19.3 m
cu
= 2 × 19.3 × 8.9 = 343.54 …(2)
From equation (1) and (2), 8.9 m
Au
+ 19.3 m
cu
= 343.54
?
g 225 . 2 m
36 9 . 8 ) m m ( 9 . 8
cu
cu Au
?
? ? ?
So, the amount of copper in the ornament is 2.2 g. ?
10. g V
M
w c
Au
Au
? ?
?
?
?
?
?
?
?
?
?
?
= 2 × g (where V
c
= volume of cavity)
11. mg = U + R (where U = Upward thrust)
? mg – U = R
? R = mg – v ?
w
g (because, U = v ?
w
g)
= mg –
?
m
× ?
w
× g ?
12. a) Let V
i
? volume of boat inside water = volume of water displace in m
3
.
Since, weight of the boat is balanced by the buoyant force.
? mg = V
i
× ?
w
× g
b) Let, v
1
? volume of boat filled with water before water starts coming in from the sides.
mg + v
1
?
w
× g = V × ?
w
× g. ?
13. Let x ? minimum edge of the ice block in cm.
So, mg + W
ice
= U. (where U = Upward thrust)
? 0.5 × g + x
3
× ?
ice
× g = x
3
× ?
w
× g ?
14. V
ice
= V
k
+ V
w
V
ice 
× ?
ice
× g = V
k
× ?
k
× g + V
w
× ?
w
× g
? (V
k
+ V
w
) × ?
ice
= V
k
× ?
k
+ V
w
× ?
w
? 1
V
V
k
w
? . ?
15. V
ii
g = V ? w g
16. (m
w
+ m
pb
)g = (V
w
+ V
pb
) ? × g
? (m
w
+ m
pb
) = ?
?
?
?
?
?
?
?
?
?
?
?
pb
pb
w
w
m
m
17. Mg = w ? (m
w
+ m
pb
)g = V
w
× ? × g
18. Given, x = 12 cm
Length of the edge of the block ?
Hg 
= 13.6 gm/cc
Given that, initially 1/5 of block is inside mercuty.
Let ?
b
? density of block in gm/cc.
? (x)
3
× ?
b
× g = (x)
2
× (x/5) × ?
Hg
× g
? 12
3
× ?
b
= 12
2
× 12/5 × 13.6
? ?
b
= 
5
6 . 13
gm/cc
Page 3


13.1
SOLUTIONS TO CONCEPTS 
CHAPTER 13
1. p = h ? g
It is necessary to specify that the tap is closed. Otherwise pressure will gradually decrease, as h 
decrease, because, of the tap is open, the pressure at the tap is atmospheric.
2. a) Pressure at the bottom of the tube should be same when considered for both limbs.
From the figure are shown,
p
g
+ ?
Hg
× h
2
× g = p
a
+ ?
Hg
× h
1
× g
? p
g
= p
a
+ ?
Hg
× g(h
1
– h
2
)
b) Pressure of mercury at the bottom of u tube
       p = p
a
+ ?
Hg
h
1
× g
3. From the figure shown 
p
a
+ h ?g = p
a
+ mg/A
? h ?g = mg/A
? h = 
Ap
m
?
4. a) Force exerted at the bottom.
= Force due to cylindrical water colum + atm. Force
= A × h × ?
w
× g + p
a
× A
= A(h ?
w
g + p
a
)
b) To find out the resultant force exerted by the sides of the glass, from the freebody, diagram of water 
inside the glass
p
a
× A + mg = A × h × ?
w
× g + F
s
+ p
a
× A
? mg = A × h × ?
w
× g + F
s
This force is provided by the sides of the glass. ?
5. If the glass will be covered by a jar and the air is pumped out, the atmospheric pressure has no effect.
So, 
a) Force exerted on the bottom.
= (h ?
w
g) × A
b) mg = h × ?
w
× g × A × F
s
.
c) It glass of different shape is used provided the volume, height and area remain same, no change in 
answer will occur.
6. Standard atmospheric pressure is always pressure exerted by 76 cm Hg column 
= (76 × 13.6 × g) Dyne/cm
2
.
If water is used in the barometer.
Let h ? height of water column.
? h × ?
w
× g ?
7. a) F = P × A = (h ?
w
× g) A
b) The force does not depend on the orientation of the rock as long as the surface area remains same.
8. a) F = A h ? g.
b) The force exerted by water on the strip of width ?x as shown,
dF = p × A
= (x ?g) × A
c) Inside the liquid force act in every direction due to adhesion.
di = F × r
d) The total force by the water on that side is given by
F = 
?
1
0
20000 x ?x ? F = 20,000 
1
0
2
] 2 / x [
e) The torque by the water on that side will be,
Pa
Pa
Gas 
Pa
h 
pa 
45 kg 
A =900 cm
2
Chapter-13
13.2
i = 
?
1
0
20000 x ?x (1 – x) ? 20,000 
1
0
3 2
] 3 / x 2 / x [ ? ?
9. Here, m
0
= m
Au
+ m
cu
= 36 g …(1)
Let V be the volume of the ornament in cm
3
So, V × ?
w
× g = 2 × g
? (V
au
+ V
cu
) × ?
w
× g = 2 × g
?? g
m m
w
au au
? ?
?
?
?
?
?
?
?
?
?
?
?
?= 2 × g
? 1
9 . 8
m
3 . 19
m
Au Au
? ?
?
?
?
?
?
? = 2 
? 8.9 m
Au
+ 19.3 m
cu
= 2 × 19.3 × 8.9 = 343.54 …(2)
From equation (1) and (2), 8.9 m
Au
+ 19.3 m
cu
= 343.54
?
g 225 . 2 m
36 9 . 8 ) m m ( 9 . 8
cu
cu Au
?
? ? ?
So, the amount of copper in the ornament is 2.2 g. ?
10. g V
M
w c
Au
Au
? ?
?
?
?
?
?
?
?
?
?
?
= 2 × g (where V
c
= volume of cavity)
11. mg = U + R (where U = Upward thrust)
? mg – U = R
? R = mg – v ?
w
g (because, U = v ?
w
g)
= mg –
?
m
× ?
w
× g ?
12. a) Let V
i
? volume of boat inside water = volume of water displace in m
3
.
Since, weight of the boat is balanced by the buoyant force.
? mg = V
i
× ?
w
× g
b) Let, v
1
? volume of boat filled with water before water starts coming in from the sides.
mg + v
1
?
w
× g = V × ?
w
× g. ?
13. Let x ? minimum edge of the ice block in cm.
So, mg + W
ice
= U. (where U = Upward thrust)
? 0.5 × g + x
3
× ?
ice
× g = x
3
× ?
w
× g ?
14. V
ice
= V
k
+ V
w
V
ice 
× ?
ice
× g = V
k
× ?
k
× g + V
w
× ?
w
× g
? (V
k
+ V
w
) × ?
ice
= V
k
× ?
k
+ V
w
× ?
w
? 1
V
V
k
w
? . ?
15. V
ii
g = V ? w g
16. (m
w
+ m
pb
)g = (V
w
+ V
pb
) ? × g
? (m
w
+ m
pb
) = ?
?
?
?
?
?
?
?
?
?
?
?
pb
pb
w
w
m
m
17. Mg = w ? (m
w
+ m
pb
)g = V
w
× ? × g
18. Given, x = 12 cm
Length of the edge of the block ?
Hg 
= 13.6 gm/cc
Given that, initially 1/5 of block is inside mercuty.
Let ?
b
? density of block in gm/cc.
? (x)
3
× ?
b
× g = (x)
2
× (x/5) × ?
Hg
× g
? 12
3
× ?
b
= 12
2
× 12/5 × 13.6
? ?
b
= 
5
6 . 13
gm/cc
Chapter-13
13.3
After water poured, let x = height of water column.
V
b
= V
Hg
+ V
w
= 12
3
Where V
Hg
and V
w
are volume of block inside mercury and water respectively
?(V
b
× ?
b
× g) = (V
Hg
× ?
Hg
× g) + (V
w
× ?
w
× g)
? (V
Hg
+ V
w
) ?
b
= V
Hg
× ?
Hg
+ V
w
× ?
w
.
? (V
Hg
+ V
w
) × 
5
6 . 13
= V
Hg
× 13.6 + V
w
× 1
? (12)
3
× 
5
6 . 13
= (12 – x) × (12)
2
× 13.6 + (x) × (12)
2
× 1
? x = 10.4 cm
19. Here, Mg = Upward thrust
? V ?g = (V/2) ( ?
w
) × g (where ?
w
= density of water) ?
?
w
3
2
3
1
3
2
r
3
4
2
1
r
3
4
r
3
4
? ? ?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
? 1 r
2
1
) r r (
3
2
3
1
3
2
? ? ? ? ? = 865 kg/m
3
. ?
20. W
1
+ W
2
= U.
? mg + V × ?
s
× g = V × ?
w
× g (where ?
s
= density of sphere in gm/cc)
? 1 – ?
s
= 0.19
? ?
s
= 1 – (0.19) = 0.8 gm/cc
So, specific gravity of the material is 0.8. ?
21. W
i
= mg – V
i
?
air
× g = g
m
m
air
i
?
?
?
?
?
?
?
?
?
?
?
W
w
= mg – V
w
?
air
g = g
m
m
air
w
?
?
?
?
?
?
?
?
?
?
? ?
22. Driving force U = V ?
w
g
? a = ?r
2
(X) × ?
w
g  ? T = 
on Accelerati
nt displaceme
2 ?
23. a) F + U = mg (where F = kx)
? kx + V ?
w
g = mg
b) F = kX + V ?
w
× g
? ma = kX + ?r
2
× (X) × ?
w
× g = (k + ?r
2
× ?
w
× g)X
? ?
2
× (X) = ) X (
m
) g r k (
w
2
?
? ? ? ? ?
? T = 
g r K
m
2
w
2
? ? ? ? ?
?
24. a) mg = kX + V ?
w
g
b) a = kx/m
w
2
x = kx/m
T = k / m 2 ?
25. Let x ? edge of ice block
When it just leaves contact with the bottom of the glass.
h ? height of water melted from ice
W = U
? x
3
× ?
ice
× g = x
2
× h × ?
w
× g
Again, volume of water formed, from melting of ice is given by,
4
3
– x
3
= ? × r
2
× h – x
2
h ( because amount of water = ( ?r
2
– x
2
)h)
? 4
3
– x
3
= ? × 3
2
× h – x
2
h
Putting h = 0.9 x ? x = 2.26 cm. ?
Page 4


13.1
SOLUTIONS TO CONCEPTS 
CHAPTER 13
1. p = h ? g
It is necessary to specify that the tap is closed. Otherwise pressure will gradually decrease, as h 
decrease, because, of the tap is open, the pressure at the tap is atmospheric.
2. a) Pressure at the bottom of the tube should be same when considered for both limbs.
From the figure are shown,
p
g
+ ?
Hg
× h
2
× g = p
a
+ ?
Hg
× h
1
× g
? p
g
= p
a
+ ?
Hg
× g(h
1
– h
2
)
b) Pressure of mercury at the bottom of u tube
       p = p
a
+ ?
Hg
h
1
× g
3. From the figure shown 
p
a
+ h ?g = p
a
+ mg/A
? h ?g = mg/A
? h = 
Ap
m
?
4. a) Force exerted at the bottom.
= Force due to cylindrical water colum + atm. Force
= A × h × ?
w
× g + p
a
× A
= A(h ?
w
g + p
a
)
b) To find out the resultant force exerted by the sides of the glass, from the freebody, diagram of water 
inside the glass
p
a
× A + mg = A × h × ?
w
× g + F
s
+ p
a
× A
? mg = A × h × ?
w
× g + F
s
This force is provided by the sides of the glass. ?
5. If the glass will be covered by a jar and the air is pumped out, the atmospheric pressure has no effect.
So, 
a) Force exerted on the bottom.
= (h ?
w
g) × A
b) mg = h × ?
w
× g × A × F
s
.
c) It glass of different shape is used provided the volume, height and area remain same, no change in 
answer will occur.
6. Standard atmospheric pressure is always pressure exerted by 76 cm Hg column 
= (76 × 13.6 × g) Dyne/cm
2
.
If water is used in the barometer.
Let h ? height of water column.
? h × ?
w
× g ?
7. a) F = P × A = (h ?
w
× g) A
b) The force does not depend on the orientation of the rock as long as the surface area remains same.
8. a) F = A h ? g.
b) The force exerted by water on the strip of width ?x as shown,
dF = p × A
= (x ?g) × A
c) Inside the liquid force act in every direction due to adhesion.
di = F × r
d) The total force by the water on that side is given by
F = 
?
1
0
20000 x ?x ? F = 20,000 
1
0
2
] 2 / x [
e) The torque by the water on that side will be,
Pa
Pa
Gas 
Pa
h 
pa 
45 kg 
A =900 cm
2
Chapter-13
13.2
i = 
?
1
0
20000 x ?x (1 – x) ? 20,000 
1
0
3 2
] 3 / x 2 / x [ ? ?
9. Here, m
0
= m
Au
+ m
cu
= 36 g …(1)
Let V be the volume of the ornament in cm
3
So, V × ?
w
× g = 2 × g
? (V
au
+ V
cu
) × ?
w
× g = 2 × g
?? g
m m
w
au au
? ?
?
?
?
?
?
?
?
?
?
?
?
?= 2 × g
? 1
9 . 8
m
3 . 19
m
Au Au
? ?
?
?
?
?
?
? = 2 
? 8.9 m
Au
+ 19.3 m
cu
= 2 × 19.3 × 8.9 = 343.54 …(2)
From equation (1) and (2), 8.9 m
Au
+ 19.3 m
cu
= 343.54
?
g 225 . 2 m
36 9 . 8 ) m m ( 9 . 8
cu
cu Au
?
? ? ?
So, the amount of copper in the ornament is 2.2 g. ?
10. g V
M
w c
Au
Au
? ?
?
?
?
?
?
?
?
?
?
?
= 2 × g (where V
c
= volume of cavity)
11. mg = U + R (where U = Upward thrust)
? mg – U = R
? R = mg – v ?
w
g (because, U = v ?
w
g)
= mg –
?
m
× ?
w
× g ?
12. a) Let V
i
? volume of boat inside water = volume of water displace in m
3
.
Since, weight of the boat is balanced by the buoyant force.
? mg = V
i
× ?
w
× g
b) Let, v
1
? volume of boat filled with water before water starts coming in from the sides.
mg + v
1
?
w
× g = V × ?
w
× g. ?
13. Let x ? minimum edge of the ice block in cm.
So, mg + W
ice
= U. (where U = Upward thrust)
? 0.5 × g + x
3
× ?
ice
× g = x
3
× ?
w
× g ?
14. V
ice
= V
k
+ V
w
V
ice 
× ?
ice
× g = V
k
× ?
k
× g + V
w
× ?
w
× g
? (V
k
+ V
w
) × ?
ice
= V
k
× ?
k
+ V
w
× ?
w
? 1
V
V
k
w
? . ?
15. V
ii
g = V ? w g
16. (m
w
+ m
pb
)g = (V
w
+ V
pb
) ? × g
? (m
w
+ m
pb
) = ?
?
?
?
?
?
?
?
?
?
?
?
pb
pb
w
w
m
m
17. Mg = w ? (m
w
+ m
pb
)g = V
w
× ? × g
18. Given, x = 12 cm
Length of the edge of the block ?
Hg 
= 13.6 gm/cc
Given that, initially 1/5 of block is inside mercuty.
Let ?
b
? density of block in gm/cc.
? (x)
3
× ?
b
× g = (x)
2
× (x/5) × ?
Hg
× g
? 12
3
× ?
b
= 12
2
× 12/5 × 13.6
? ?
b
= 
5
6 . 13
gm/cc
Chapter-13
13.3
After water poured, let x = height of water column.
V
b
= V
Hg
+ V
w
= 12
3
Where V
Hg
and V
w
are volume of block inside mercury and water respectively
?(V
b
× ?
b
× g) = (V
Hg
× ?
Hg
× g) + (V
w
× ?
w
× g)
? (V
Hg
+ V
w
) ?
b
= V
Hg
× ?
Hg
+ V
w
× ?
w
.
? (V
Hg
+ V
w
) × 
5
6 . 13
= V
Hg
× 13.6 + V
w
× 1
? (12)
3
× 
5
6 . 13
= (12 – x) × (12)
2
× 13.6 + (x) × (12)
2
× 1
? x = 10.4 cm
19. Here, Mg = Upward thrust
? V ?g = (V/2) ( ?
w
) × g (where ?
w
= density of water) ?
?
w
3
2
3
1
3
2
r
3
4
2
1
r
3
4
r
3
4
? ? ?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
?
?
?
?
? ? ?
? 1 r
2
1
) r r (
3
2
3
1
3
2
? ? ? ? ? = 865 kg/m
3
. ?
20. W
1
+ W
2
= U.
? mg + V × ?
s
× g = V × ?
w
× g (where ?
s
= density of sphere in gm/cc)
? 1 – ?
s
= 0.19
? ?
s
= 1 – (0.19) = 0.8 gm/cc
So, specific gravity of the material is 0.8. ?
21. W
i
= mg – V
i
?
air
× g = g
m
m
air
i
?
?
?
?
?
?
?
?
?
?
?
W
w
= mg – V
w
?
air
g = g
m
m
air
w
?
?
?
?
?
?
?
?
?
?
? ?
22. Driving force U = V ?
w
g
? a = ?r
2
(X) × ?
w
g  ? T = 
on Accelerati
nt displaceme
2 ?
23. a) F + U = mg (where F = kx)
? kx + V ?
w
g = mg
b) F = kX + V ?
w
× g
? ma = kX + ?r
2
× (X) × ?
w
× g = (k + ?r
2
× ?
w
× g)X
? ?
2
× (X) = ) X (
m
) g r k (
w
2
?
? ? ? ? ?
? T = 
g r K
m
2
w
2
? ? ? ? ?
?
24. a) mg = kX + V ?
w
g
b) a = kx/m
w
2
x = kx/m
T = k / m 2 ?
25. Let x ? edge of ice block
When it just leaves contact with the bottom of the glass.
h ? height of water melted from ice
W = U
? x
3
× ?
ice
× g = x
2
× h × ?
w
× g
Again, volume of water formed, from melting of ice is given by,
4
3
– x
3
= ? × r
2
× h – x
2
h ( because amount of water = ( ?r
2
– x
2
)h)
? 4
3
– x
3
= ? × 3
2
× h – x
2
h
Putting h = 0.9 x ? x = 2.26 cm. ?
Chapter-13
13.4
26. If p
a
? atm. Pressure
A ? area of cross section
h ? increase in hright
p
a
A + A × L × ? × a
0
= pa
A
+ h ?g × A
? hg = a
0
L ? a
0
L/g ?
27. Volume of water, discharged from Alkananda + vol are of water discharged from Bhagirathi = Volume of 
water flow in Ganga.
28. a) a
A
× V
A
= Q
A
b) a
A
× V
A
= a
B
× V
B
c) 1/2 ?v
A
2
+ p
A
= 1/2 ?v
B
2
+ p
B
? (p
A
– p
B
) = 1/2 ? (v
B
2
– v
A
2
) ?
29. From Bernoulli’s equation, 1/2 ?v
A
2
+ ?gh
A
+ p
A
= 1/2 ?v
B
2
+ ?gh
B
+ p
B
.
? P
A
– P
B
= (1/2) ? (v
B
2
– v
A
2
) + ?g (h
B
– h
A
) ?
30. 1/2 ?v
B
2
+ ?gh
B
+ p
B
= 1/2 ?v
A
2
+ ?gh
A
+ p
A
  
31. 1/2 ?v
A
2
+ ?gh
A
+ p
A
=1/2 ?v
B
2
+ ?gh
B
+ p
B
? P
B
– P
A
= 1/2 ?(v
A
2
– v
B
2
) + ?g (h
A
– h
B
) ?
32.
B B A A
a v a v ? ?
? ?
? 1/2 ?v
A
2
+ ?gh
A
+ p
A
= 1/2 ?v
B
2
+ ?gh
B
+ p
B
? 1/2 ?v
A
2
+ p
A
= 1/2 ?v
B
2
+ p
B
? P
A
– P
B
= 1/2 ?(v
B
2
– v
B
2
)
Rate of flow = v
a
× a
A
33. V
A
a
A
= v
B
a
B
?
A
B A
a
a
B
v
?
5v
A
= 2v
B
? v
B
= (5/2)v
A
1/2 ?v
A
2
+ ?gh
A
+ p
A
= 1/2 ?v
B
2
+ ?gh
B
+ p
B
? P
A
– P
B
= 1/2 ? (v
B
2
– v
B
2
) (because P
A
– P
B
= h ?
m
g) ?
34. P
A
+ (1/2) ?v
A
2
= P
B
+ (1/2) ?v
B
2
? p
A
– p
B
= (1/2) ?v
B
2
{v
A
= 0}
? ?gh = (1/2) ?v
B
2
{p
A
= p
atm 
+ ?gh}
? v
B
= gh 2
a) v = gh 2
b) v = gh ) 2 / h ( g 2 ?
c) v = gh 2
v = av × dt 
AV = av
? A × 2gh a
dt
dh
? ? ? dh = 
A
dt gh 2 a ? ?
d) dh = 
A
dt gh 2 a ? ?
? T = ] H H [
g
2
a
A
2 1
?
35. v = ) h H ( g 2 ?
t = g / h 2
x = v × t = g / h 2 ) h H ( g 2 ? ? = ) h Hh ( 4
2
?
So, ? 0 ) h Hh (
dh
d
2
? ? ?
?
?
?
?
?
? 0 = H – 2h  ? h = H/2.
? ? ? ?
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