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Mechanical Properties of Fluids

143213 Water from a hose pipe of radius $5 cm$ strikes a wall normally at a speed of $5 {ms}^{- 1}$ . The force exerted on the wall in Newton is

1

13.5 π

2

6.25 π

3

62.5 π

4

27 π

5

125 π

Explanation:

C Given,
Radius of hose pipe (r) $= 5 cm = 5 \times 10^{- 2} m$
Speed of water $(v) = 5 m / s$
Density of water $(ρ) = 1000 kg / m^{3}$
We know that,
$F = ρ A v^{2}$
$F = ρ π r^{2} v^{2}$
$F = 1000 \times π \times {(5 \times 10^{- 2})}^{2} \times 5^{2}$
$F = 1000 \times π \times 25 \times 10^{- 4} \times 25$
$F = π \times 625 \times 10^{- 1}$
$F = 62.5 π N$

Kerala CEE - 2016

Mechanical Properties of Fluids

143214 A liquid is filled up to a height of $20 cm$ in a cylindrical vessel. The speed of liquid coming out of a small hole at the bottom of the vessel is (take, $g = 10 {ms}^{- 2}$ )

1

1.2 {ms}^{- 1}

2

1 {ms}^{- 1}

3

2 {ms}^{- 1}

4

3.2 {ms}^{- 1}

5

1.4 {ms}^{- 1}

Explanation:

C Given,
Height of cylindrical vessel $(h) = 20 cm = 0.2 m$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know that,
Speed of liquid coming out of a small hole at the bottom of the vessel $(v) = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 0.2}$
$v = \sqrt{4}$
$v = 2 m / s$

Kerala CEE - 2015

Mechanical Properties of Fluids

143215 A large open tank has two holes in its wall. One is a square hole of side ' $a$ ' at a depth of $x$ from the top and the other is a circular hole of radius $r$ at a depth $4 x$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then $r$ is equal to

1

2 π a

2 a

3

\frac{a}{2 π}

4

\frac{a}{π}

5

\frac{a}{\sqrt{2 π}}

Explanation:

E We know that,
Velocity of efflux when the hole is at depth $h$ ,
$V = \sqrt{2 gh}$
Rate of flow of water from square hole,
$Q_{1} = A_{1} V_{1} = a^{2} \sqrt{2 gx}$
Rate of flow of water from circular hole,
$Q_{2} = A_{2} V_{2} = π r^{2} \sqrt{2 g (4 x)}$
It is given that the water flowing out per second from both holes is same,
So, $Q_{1} = Q_{2}$
$a^{2} \sqrt{2 g x} = π r^{2} \sqrt{2 g (4 x)}$
$r = \frac{a}{\sqrt{2 π}}$

Kerala CEE - 2011

Mechanical Properties of Fluids

143216 A tank is filled with water of density $1 {gcm}^{- 3}$ and oil of density $0.9 g {cm}^{- 3}$ . The height of water layer is $100 cm$ and of the oil layer is $400 cm$ . If $g = 980 cm s^{- 2}$ , then the velocity of efflux from an opening in the bottom of the tank is

1

\sqrt{920 \times 980} {cms}^{- 1}

2

\sqrt{900 \times 980} {cms}^{- 1}

3

\sqrt{1000 \times 980} {cms}^{- 1}

4

\sqrt{92 \times 980} {cms}^{- 1}

Explanation:

A Given $ρ_{w} 1 {gcm}^{- 3}, ρ_{0} = 0.9 {cm}^{- 3} h_{w} = 100 cm$ , $h_{0} = 400 cm$ ,
We know,
Pressure at the bottom of the tank $(P) = h_{w} ρ_{w} g + h_{o} ρ_{o} g$ We also know the pressure at the bottom of tank must be equal to pressure due to water of height $h$ . Then,
$h ρ_{w} g = h_{w} ρ_{w} g + h_{o} ρ_{o} g$
$h = h_{w} + \frac{h_{o} ρ_{o}}{ρ_{w}}$
$= 100 + \frac{400 \times 0.9}{1}$
$h = 100 + 360 = 460$
According to Torricelli's theorem,
$V = \sqrt{2 gh}$
$= \sqrt{2 \times 980 \times 460}$
$= \sqrt{920 \times 980} {cms}^{- 1}$

UPSEE - 2015

Mechanical Properties of Fluids

143218 Bottom of a cylindrical vessel has a hole of area A. If water is filled up to a height $h$ , it flows out in $t$ seconds. If water is filled to a height $4 h$ , it will flow out in time

1

t

2

4 t

3

2 t

4

\frac{t}{4}

Explanation:

A From continuity equation,
$AV = Constant$
For stream line flow, the product of cross-section area and velocity remains unchanged. In the narrowest part of the pipe, velocity will be maximum.
From Bernoulli's theorem,
$P + \frac{1}{2} ρ V^{2} + ρ gh = Constant$
It is given that water is flowing through a horizontal pipe. So, potential energy ( $ρ g h)$ will remain constant at all points.
From equation of the velocity is maximum in the narrowest part of the pipe. So, the pressure will be minimum.
Hence, the velocity is maximum at the narrowest part of the pipe and pressure is minimum.

}$

Mechanical Properties of Fluids

143213 Water from a hose pipe of radius $5 cm$ strikes a wall normally at a speed of $5 {ms}^{- 1}$ . The force exerted on the wall in Newton is

1

13.5 π

2

6.25 π

3

62.5 π

4

27 π

5

125 π

Explanation:

C Given,
Radius of hose pipe (r) $= 5 cm = 5 \times 10^{- 2} m$
Speed of water $(v) = 5 m / s$
Density of water $(ρ) = 1000 kg / m^{3}$
We know that,
$F = ρ A v^{2}$
$F = ρ π r^{2} v^{2}$
$F = 1000 \times π \times {(5 \times 10^{- 2})}^{2} \times 5^{2}$
$F = 1000 \times π \times 25 \times 10^{- 4} \times 25$
$F = π \times 625 \times 10^{- 1}$
$F = 62.5 π N$

Kerala CEE - 2016

Mechanical Properties of Fluids

143214 A liquid is filled up to a height of $20 cm$ in a cylindrical vessel. The speed of liquid coming out of a small hole at the bottom of the vessel is (take, $g = 10 {ms}^{- 2}$ )

1

1.2 {ms}^{- 1}

2

1 {ms}^{- 1}

3

2 {ms}^{- 1}

4

3.2 {ms}^{- 1}

5

1.4 {ms}^{- 1}

Explanation:

C Given,
Height of cylindrical vessel $(h) = 20 cm = 0.2 m$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know that,
Speed of liquid coming out of a small hole at the bottom of the vessel $(v) = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 0.2}$
$v = \sqrt{4}$
$v = 2 m / s$

Kerala CEE - 2015

Mechanical Properties of Fluids

143215 A large open tank has two holes in its wall. One is a square hole of side ' $a$ ' at a depth of $x$ from the top and the other is a circular hole of radius $r$ at a depth $4 x$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then $r$ is equal to

1

2 π a

2 a

3

\frac{a}{2 π}

4

\frac{a}{π}

5

\frac{a}{\sqrt{2 π}}

Explanation:

E We know that,
Velocity of efflux when the hole is at depth $h$ ,
$V = \sqrt{2 gh}$
Rate of flow of water from square hole,
$Q_{1} = A_{1} V_{1} = a^{2} \sqrt{2 gx}$
Rate of flow of water from circular hole,
$Q_{2} = A_{2} V_{2} = π r^{2} \sqrt{2 g (4 x)}$
It is given that the water flowing out per second from both holes is same,
So, $Q_{1} = Q_{2}$
$a^{2} \sqrt{2 g x} = π r^{2} \sqrt{2 g (4 x)}$
$r = \frac{a}{\sqrt{2 π}}$

Kerala CEE - 2011

Mechanical Properties of Fluids

143216 A tank is filled with water of density $1 {gcm}^{- 3}$ and oil of density $0.9 g {cm}^{- 3}$ . The height of water layer is $100 cm$ and of the oil layer is $400 cm$ . If $g = 980 cm s^{- 2}$ , then the velocity of efflux from an opening in the bottom of the tank is

1

\sqrt{920 \times 980} {cms}^{- 1}

2

\sqrt{900 \times 980} {cms}^{- 1}

3

\sqrt{1000 \times 980} {cms}^{- 1}

4

\sqrt{92 \times 980} {cms}^{- 1}

Explanation:

A Given $ρ_{w} 1 {gcm}^{- 3}, ρ_{0} = 0.9 {cm}^{- 3} h_{w} = 100 cm$ , $h_{0} = 400 cm$ ,
We know,
Pressure at the bottom of the tank $(P) = h_{w} ρ_{w} g + h_{o} ρ_{o} g$ We also know the pressure at the bottom of tank must be equal to pressure due to water of height $h$ . Then,
$h ρ_{w} g = h_{w} ρ_{w} g + h_{o} ρ_{o} g$
$h = h_{w} + \frac{h_{o} ρ_{o}}{ρ_{w}}$
$= 100 + \frac{400 \times 0.9}{1}$
$h = 100 + 360 = 460$
According to Torricelli's theorem,
$V = \sqrt{2 gh}$
$= \sqrt{2 \times 980 \times 460}$
$= \sqrt{920 \times 980} {cms}^{- 1}$

UPSEE - 2015

Mechanical Properties of Fluids

143218 Bottom of a cylindrical vessel has a hole of area A. If water is filled up to a height $h$ , it flows out in $t$ seconds. If water is filled to a height $4 h$ , it will flow out in time

1

t

2

4 t

3

2 t

4

\frac{t}{4}

Explanation:

A From continuity equation,
$AV = Constant$
For stream line flow, the product of cross-section area and velocity remains unchanged. In the narrowest part of the pipe, velocity will be maximum.
From Bernoulli's theorem,
$P + \frac{1}{2} ρ V^{2} + ρ gh = Constant$
It is given that water is flowing through a horizontal pipe. So, potential energy ( $ρ g h)$ will remain constant at all points.
From equation of the velocity is maximum in the narrowest part of the pipe. So, the pressure will be minimum.
Hence, the velocity is maximum at the narrowest part of the pipe and pressure is minimum.

}$

Mechanical Properties of Fluids

143213 Water from a hose pipe of radius $5 cm$ strikes a wall normally at a speed of $5 {ms}^{- 1}$ . The force exerted on the wall in Newton is

1

13.5 π

2

6.25 π

3

62.5 π

4

27 π

5

125 π

Explanation:

C Given,
Radius of hose pipe (r) $= 5 cm = 5 \times 10^{- 2} m$
Speed of water $(v) = 5 m / s$
Density of water $(ρ) = 1000 kg / m^{3}$
We know that,
$F = ρ A v^{2}$
$F = ρ π r^{2} v^{2}$
$F = 1000 \times π \times {(5 \times 10^{- 2})}^{2} \times 5^{2}$
$F = 1000 \times π \times 25 \times 10^{- 4} \times 25$
$F = π \times 625 \times 10^{- 1}$
$F = 62.5 π N$

Kerala CEE - 2016

Mechanical Properties of Fluids

143214 A liquid is filled up to a height of $20 cm$ in a cylindrical vessel. The speed of liquid coming out of a small hole at the bottom of the vessel is (take, $g = 10 {ms}^{- 2}$ )

1

1.2 {ms}^{- 1}

2

1 {ms}^{- 1}

3

2 {ms}^{- 1}

4

3.2 {ms}^{- 1}

5

1.4 {ms}^{- 1}

Explanation:

C Given,
Height of cylindrical vessel $(h) = 20 cm = 0.2 m$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know that,
Speed of liquid coming out of a small hole at the bottom of the vessel $(v) = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 0.2}$
$v = \sqrt{4}$
$v = 2 m / s$

Kerala CEE - 2015

Mechanical Properties of Fluids

143215 A large open tank has two holes in its wall. One is a square hole of side ' $a$ ' at a depth of $x$ from the top and the other is a circular hole of radius $r$ at a depth $4 x$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then $r$ is equal to

1

2 π a

2 a

3

\frac{a}{2 π}

4

\frac{a}{π}

5

\frac{a}{\sqrt{2 π}}

Explanation:

E We know that,
Velocity of efflux when the hole is at depth $h$ ,
$V = \sqrt{2 gh}$
Rate of flow of water from square hole,
$Q_{1} = A_{1} V_{1} = a^{2} \sqrt{2 gx}$
Rate of flow of water from circular hole,
$Q_{2} = A_{2} V_{2} = π r^{2} \sqrt{2 g (4 x)}$
It is given that the water flowing out per second from both holes is same,
So, $Q_{1} = Q_{2}$
$a^{2} \sqrt{2 g x} = π r^{2} \sqrt{2 g (4 x)}$
$r = \frac{a}{\sqrt{2 π}}$

Kerala CEE - 2011

Mechanical Properties of Fluids

143216 A tank is filled with water of density $1 {gcm}^{- 3}$ and oil of density $0.9 g {cm}^{- 3}$ . The height of water layer is $100 cm$ and of the oil layer is $400 cm$ . If $g = 980 cm s^{- 2}$ , then the velocity of efflux from an opening in the bottom of the tank is

1

\sqrt{920 \times 980} {cms}^{- 1}

2

\sqrt{900 \times 980} {cms}^{- 1}

3

\sqrt{1000 \times 980} {cms}^{- 1}

4

\sqrt{92 \times 980} {cms}^{- 1}

Explanation:

A Given $ρ_{w} 1 {gcm}^{- 3}, ρ_{0} = 0.9 {cm}^{- 3} h_{w} = 100 cm$ , $h_{0} = 400 cm$ ,
We know,
Pressure at the bottom of the tank $(P) = h_{w} ρ_{w} g + h_{o} ρ_{o} g$ We also know the pressure at the bottom of tank must be equal to pressure due to water of height $h$ . Then,
$h ρ_{w} g = h_{w} ρ_{w} g + h_{o} ρ_{o} g$
$h = h_{w} + \frac{h_{o} ρ_{o}}{ρ_{w}}$
$= 100 + \frac{400 \times 0.9}{1}$
$h = 100 + 360 = 460$
According to Torricelli's theorem,
$V = \sqrt{2 gh}$
$= \sqrt{2 \times 980 \times 460}$
$= \sqrt{920 \times 980} {cms}^{- 1}$

UPSEE - 2015

Mechanical Properties of Fluids

143218 Bottom of a cylindrical vessel has a hole of area A. If water is filled up to a height $h$ , it flows out in $t$ seconds. If water is filled to a height $4 h$ , it will flow out in time

1

t

2

4 t

3

2 t

4

\frac{t}{4}

Explanation:

A From continuity equation,
$AV = Constant$
For stream line flow, the product of cross-section area and velocity remains unchanged. In the narrowest part of the pipe, velocity will be maximum.
From Bernoulli's theorem,
$P + \frac{1}{2} ρ V^{2} + ρ gh = Constant$
It is given that water is flowing through a horizontal pipe. So, potential energy ( $ρ g h)$ will remain constant at all points.
From equation of the velocity is maximum in the narrowest part of the pipe. So, the pressure will be minimum.
Hence, the velocity is maximum at the narrowest part of the pipe and pressure is minimum.

}$

Mechanical Properties of Fluids

143213 Water from a hose pipe of radius $5 cm$ strikes a wall normally at a speed of $5 {ms}^{- 1}$ . The force exerted on the wall in Newton is

1

13.5 π

2

6.25 π

3

62.5 π

4

27 π

5

125 π

Explanation:

C Given,
Radius of hose pipe (r) $= 5 cm = 5 \times 10^{- 2} m$
Speed of water $(v) = 5 m / s$
Density of water $(ρ) = 1000 kg / m^{3}$
We know that,
$F = ρ A v^{2}$
$F = ρ π r^{2} v^{2}$
$F = 1000 \times π \times {(5 \times 10^{- 2})}^{2} \times 5^{2}$
$F = 1000 \times π \times 25 \times 10^{- 4} \times 25$
$F = π \times 625 \times 10^{- 1}$
$F = 62.5 π N$

Kerala CEE - 2016

Mechanical Properties of Fluids

143214 A liquid is filled up to a height of $20 cm$ in a cylindrical vessel. The speed of liquid coming out of a small hole at the bottom of the vessel is (take, $g = 10 {ms}^{- 2}$ )

1

1.2 {ms}^{- 1}

2

1 {ms}^{- 1}

3

2 {ms}^{- 1}

4

3.2 {ms}^{- 1}

5

1.4 {ms}^{- 1}

Explanation:

C Given,
Height of cylindrical vessel $(h) = 20 cm = 0.2 m$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know that,
Speed of liquid coming out of a small hole at the bottom of the vessel $(v) = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 0.2}$
$v = \sqrt{4}$
$v = 2 m / s$

Kerala CEE - 2015

Mechanical Properties of Fluids

143215 A large open tank has two holes in its wall. One is a square hole of side ' $a$ ' at a depth of $x$ from the top and the other is a circular hole of radius $r$ at a depth $4 x$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then $r$ is equal to

1

2 π a

2 a

3

\frac{a}{2 π}

4

\frac{a}{π}

5

\frac{a}{\sqrt{2 π}}

Explanation:

E We know that,
Velocity of efflux when the hole is at depth $h$ ,
$V = \sqrt{2 gh}$
Rate of flow of water from square hole,
$Q_{1} = A_{1} V_{1} = a^{2} \sqrt{2 gx}$
Rate of flow of water from circular hole,
$Q_{2} = A_{2} V_{2} = π r^{2} \sqrt{2 g (4 x)}$
It is given that the water flowing out per second from both holes is same,
So, $Q_{1} = Q_{2}$
$a^{2} \sqrt{2 g x} = π r^{2} \sqrt{2 g (4 x)}$
$r = \frac{a}{\sqrt{2 π}}$

Kerala CEE - 2011

Mechanical Properties of Fluids

143216 A tank is filled with water of density $1 {gcm}^{- 3}$ and oil of density $0.9 g {cm}^{- 3}$ . The height of water layer is $100 cm$ and of the oil layer is $400 cm$ . If $g = 980 cm s^{- 2}$ , then the velocity of efflux from an opening in the bottom of the tank is

1

\sqrt{920 \times 980} {cms}^{- 1}

2

\sqrt{900 \times 980} {cms}^{- 1}

3

\sqrt{1000 \times 980} {cms}^{- 1}

4

\sqrt{92 \times 980} {cms}^{- 1}

Explanation:

A Given $ρ_{w} 1 {gcm}^{- 3}, ρ_{0} = 0.9 {cm}^{- 3} h_{w} = 100 cm$ , $h_{0} = 400 cm$ ,
We know,
Pressure at the bottom of the tank $(P) = h_{w} ρ_{w} g + h_{o} ρ_{o} g$ We also know the pressure at the bottom of tank must be equal to pressure due to water of height $h$ . Then,
$h ρ_{w} g = h_{w} ρ_{w} g + h_{o} ρ_{o} g$
$h = h_{w} + \frac{h_{o} ρ_{o}}{ρ_{w}}$
$= 100 + \frac{400 \times 0.9}{1}$
$h = 100 + 360 = 460$
According to Torricelli's theorem,
$V = \sqrt{2 gh}$
$= \sqrt{2 \times 980 \times 460}$
$= \sqrt{920 \times 980} {cms}^{- 1}$

UPSEE - 2015

Mechanical Properties of Fluids

143218 Bottom of a cylindrical vessel has a hole of area A. If water is filled up to a height $h$ , it flows out in $t$ seconds. If water is filled to a height $4 h$ , it will flow out in time

1

t

2

4 t

3

2 t

4

\frac{t}{4}

Explanation:

A From continuity equation,
$AV = Constant$
For stream line flow, the product of cross-section area and velocity remains unchanged. In the narrowest part of the pipe, velocity will be maximum.
From Bernoulli's theorem,
$P + \frac{1}{2} ρ V^{2} + ρ gh = Constant$
It is given that water is flowing through a horizontal pipe. So, potential energy ( $ρ g h)$ will remain constant at all points.
From equation of the velocity is maximum in the narrowest part of the pipe. So, the pressure will be minimum.
Hence, the velocity is maximum at the narrowest part of the pipe and pressure is minimum.

}$

Mechanical Properties of Fluids

143213 Water from a hose pipe of radius $5 cm$ strikes a wall normally at a speed of $5 {ms}^{- 1}$ . The force exerted on the wall in Newton is

1

13.5 π

2

6.25 π

3

62.5 π

4

27 π

5

125 π

Explanation:

C Given,
Radius of hose pipe (r) $= 5 cm = 5 \times 10^{- 2} m$
Speed of water $(v) = 5 m / s$
Density of water $(ρ) = 1000 kg / m^{3}$
We know that,
$F = ρ A v^{2}$
$F = ρ π r^{2} v^{2}$
$F = 1000 \times π \times {(5 \times 10^{- 2})}^{2} \times 5^{2}$
$F = 1000 \times π \times 25 \times 10^{- 4} \times 25$
$F = π \times 625 \times 10^{- 1}$
$F = 62.5 π N$

Kerala CEE - 2016

Mechanical Properties of Fluids

143214 A liquid is filled up to a height of $20 cm$ in a cylindrical vessel. The speed of liquid coming out of a small hole at the bottom of the vessel is (take, $g = 10 {ms}^{- 2}$ )

1

1.2 {ms}^{- 1}

2

1 {ms}^{- 1}

3

2 {ms}^{- 1}

4

3.2 {ms}^{- 1}

5

1.4 {ms}^{- 1}

Explanation:

C Given,
Height of cylindrical vessel $(h) = 20 cm = 0.2 m$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know that,
Speed of liquid coming out of a small hole at the bottom of the vessel $(v) = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 0.2}$
$v = \sqrt{4}$
$v = 2 m / s$

Kerala CEE - 2015

Mechanical Properties of Fluids

143215 A large open tank has two holes in its wall. One is a square hole of side ' $a$ ' at a depth of $x$ from the top and the other is a circular hole of radius $r$ at a depth $4 x$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then $r$ is equal to

1

2 π a

2 a

3

\frac{a}{2 π}

4

\frac{a}{π}

5

\frac{a}{\sqrt{2 π}}

Explanation:

E We know that,
Velocity of efflux when the hole is at depth $h$ ,
$V = \sqrt{2 gh}$
Rate of flow of water from square hole,
$Q_{1} = A_{1} V_{1} = a^{2} \sqrt{2 gx}$
Rate of flow of water from circular hole,
$Q_{2} = A_{2} V_{2} = π r^{2} \sqrt{2 g (4 x)}$
It is given that the water flowing out per second from both holes is same,
So, $Q_{1} = Q_{2}$
$a^{2} \sqrt{2 g x} = π r^{2} \sqrt{2 g (4 x)}$
$r = \frac{a}{\sqrt{2 π}}$

Kerala CEE - 2011

Mechanical Properties of Fluids

143216 A tank is filled with water of density $1 {gcm}^{- 3}$ and oil of density $0.9 g {cm}^{- 3}$ . The height of water layer is $100 cm$ and of the oil layer is $400 cm$ . If $g = 980 cm s^{- 2}$ , then the velocity of efflux from an opening in the bottom of the tank is

1

\sqrt{920 \times 980} {cms}^{- 1}

2

\sqrt{900 \times 980} {cms}^{- 1}

3

\sqrt{1000 \times 980} {cms}^{- 1}

4

\sqrt{92 \times 980} {cms}^{- 1}

Explanation:

A Given $ρ_{w} 1 {gcm}^{- 3}, ρ_{0} = 0.9 {cm}^{- 3} h_{w} = 100 cm$ , $h_{0} = 400 cm$ ,
We know,
Pressure at the bottom of the tank $(P) = h_{w} ρ_{w} g + h_{o} ρ_{o} g$ We also know the pressure at the bottom of tank must be equal to pressure due to water of height $h$ . Then,
$h ρ_{w} g = h_{w} ρ_{w} g + h_{o} ρ_{o} g$
$h = h_{w} + \frac{h_{o} ρ_{o}}{ρ_{w}}$
$= 100 + \frac{400 \times 0.9}{1}$
$h = 100 + 360 = 460$
According to Torricelli's theorem,
$V = \sqrt{2 gh}$
$= \sqrt{2 \times 980 \times 460}$
$= \sqrt{920 \times 980} {cms}^{- 1}$

UPSEE - 2015

Mechanical Properties of Fluids

143218 Bottom of a cylindrical vessel has a hole of area A. If water is filled up to a height $h$ , it flows out in $t$ seconds. If water is filled to a height $4 h$ , it will flow out in time

1

t

2

4 t

3

2 t

4

\frac{t}{4}

Explanation:

A From continuity equation,
$AV = Constant$
For stream line flow, the product of cross-section area and velocity remains unchanged. In the narrowest part of the pipe, velocity will be maximum.
From Bernoulli's theorem,
$P + \frac{1}{2} ρ V^{2} + ρ gh = Constant$
It is given that water is flowing through a horizontal pipe. So, potential energy ( $ρ g h)$ will remain constant at all points.
From equation of the velocity is maximum in the narrowest part of the pipe. So, the pressure will be minimum.
Hence, the velocity is maximum at the narrowest part of the pipe and pressure is minimum.

}$