NEET  >  HC Verma Solutions: Chapter 13 - Fluid Mechanics

# HC Verma Solutions: Chapter 13 - Fluid Mechanics - Physics Class 11 - NEET

``` 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
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
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
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
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.
? ? ? ?
```

## FAQs on HC Verma Solutions: Chapter 13 - Fluid Mechanics - Physics Class 11 - NEET

 1. How do I solve numerical problems related to fluid mechanics?
Ans. To solve numerical problems related to fluid mechanics, you need to understand the basic principles and equations of fluid mechanics, such as Bernoulli's equation, Pascal's law, and Archimedes' principle. You should also be familiar with various properties of fluids, such as pressure, density, and viscosity. Practice solving different types of problems, including those involving fluid flow, buoyancy, and hydrostatics, to enhance your problem-solving skills in this subject.
 2. What are some common applications of fluid mechanics in daily life?
Ans. Fluid mechanics has several applications in our daily lives. Some common examples include the flow of water through pipes, the functioning of pumps and turbines, the operation of hydraulic systems in vehicles and machinery, the flight of airplanes and birds, the behavior of blood flow in our bodies, and the movement of oil and gas through pipelines. Understanding fluid mechanics helps us comprehend and design various systems and processes that involve the flow and behavior of fluids.
 3. How does viscosity affect fluid flow?
Ans. Viscosity is a measure of a fluid's resistance to flow. In fluid mechanics, viscosity plays a crucial role in determining the rate of flow and behavior of fluids. High viscosity fluids, such as honey or molasses, flow slowly due to their strong internal friction. On the other hand, low viscosity fluids, such as water or air, flow easily as they have less internal friction. Viscosity affects parameters like the Reynolds number, which determines whether a flow is laminar or turbulent. It also affects the pressure drop in pipes and the rate of diffusion of pollutants in the environment.
 4. What is the principle of buoyancy?
Ans. The principle of buoyancy, also known as Archimedes' principle, states that an object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. This principle explains why objects float or sink in fluids. If the weight of the object is less than the buoyant force, it floats, and if the weight is greater, it sinks. The buoyant force is dependent on the density of the fluid and the volume of the displaced fluid. This principle is widely used in designing ships, submarines, and hot air balloons.
 5. How does Bernoulli's principle explain the lift generated by an airplane wing?
Ans. Bernoulli's principle states that as the speed of a fluid increases, its pressure decreases. This principle explains the lift generated by an airplane wing, known as the Bernoulli lift. When an airplane moves through the air, the shape of its wing causes the air above it to move faster than the air below it. According to Bernoulli's principle, the faster-moving air above the wing exerts less pressure compared to the slower-moving air below the wing. This pressure difference creates an upward lift force, allowing the airplane to stay airborne.

## Physics Class 11

130 videos|483 docs|210 tests

## Physics Class 11

130 videos|483 docs|210 tests

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