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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360886 A hole is made at the bottom of the tank filled with water
$(d e n s i t y = 1000 k g / m^{3})$ . If the total pressure at the bottom of the tank is three atomspheres ( 1 atmosphere $= 10^{5} N / m^{2}$ ), then the velocity of efflux is

1

\sqrt{400} m / s

2

\sqrt{200} m / s

3

\sqrt{600} m / s

4

\sqrt{500} m / s

Explanation:

Difference in pressure energy = difference in kinetic energy
$∴ (p_{0} - p_{0}) = \frac{1}{2} ρ v^{2}$
or $v = 2 \sqrt{\frac{p_{0}}{ρ}}$
$= 2 \sqrt{\frac{10^{5}}{10^{3}}}$
$= 20 m / s$ or $\sqrt{400} m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360887 A wide vessel with a small hole in the bottom is filled with water and kerosene. Neglecting viscosity, the velocity of water flow $v$ if the thickness of water layer is $h_{1}$ and that of kerosene layer is $h_{2}$ is (density of water is $ρ_{1} g$ / cc and that of kerosene is $ρ_{2} g / c c) (ρ_{1} > ρ_{2})$

1

v = \sqrt{2 g (h_{1} + h_{2})}

2

v = \sqrt{2 g (h_{1} + h_{2} \frac{ρ_{2}}{ρ_{1}})}

3

v = \sqrt{2 g (h_{1} ρ_{1} + h_{2} ρ_{2})}

4

v = \sqrt{2 g (h_{1} \frac{ρ_{1}}{ρ_{2}} + h_{2})}

Explanation:

supporting img

From Bernoulli's equation
$\begin{matrix} ρ_{2} g h_{2} + ρ_{1} g h_{1} = \frac{1}{2} ρ_{1} v^{2} \\ \Rightarrow v^{2} = \frac{2 g (ρ_{1} h_{1} + ρ_{2} h_{2})}{ρ_{1}} \Rightarrow v = \sqrt{2 g (h_{1} + \frac{ρ_{2}}{ρ_{1}} h_{2})} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360888 There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $h$ . As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to
supporting img

1

h^{1 / 2}

2

h^{3 / 2}

3

h^{2}

4

h

Explanation:

Here, $v_{1} = \sqrt{2 g (h + x)}; v_{2} = \sqrt{2 g x}$
Let, $a =$ area of cross-section of each hole
$ρ =$ density of the liquid
supporting img

The force exerted on the lower hole towards left
$= a ρ v_{1}^{2}$
Similarly, the force exerted on the upper hole towards right
$= a ρ v_{2}^{2}$
Net force on the tank, $F = a ρ (v_{1}^{2} - v_{2}^{2})$
$\begin{aligned} F = a ρ [2 g (h + x) - 2 g x] = 2 a ρ g h \\ \Rightarrow F \propto h \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360889 If a small orifice is made at a height of $0.25 m$ from the ground, the horizontal range of water stream will be
supporting img

1

46.5 c m

2

56.6 c m

3

76.6 c m

4

86.6 c m

Explanation:

Given, height of small orificc from ground is $h = 0.25 m$
supporting img
Total height of water tank, $H = 1 m$
$∴$ Range of water stream, is given by
$R = 2 \sqrt{h (H - h)}$
$= 2 \sqrt{(H - 0.25) (0.25)}$
$= 2 \sqrt{(1 - 0.25) 0.25} = 2 \sqrt{0.75 \times 0.25}$
$= 0.866 m = 86.6 c m$
So, correct option is (4)

AIIMS - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360890 The pressure of water in a pipe when tap is closed is $5.5 \times 10^{5} N / m^{2}$ . When tap gets open, pressure reduces to $5 \times 10^{5} N / m^{2}$ . The velocity with which water comes out on opening the tap is

1

10 m / s

2

5 m / s

3

20 m / s

4

15 m / s

Explanation:

Decrease in pressure energy $=$ increase in kinetic energy
$or Δ p = \frac{1}{2} ρ v^{2}$
$∴ v = \sqrt{\frac{2 (Δ p)}{ρ}}$
$= \sqrt{\frac{2 \times 0.5 \times 10^{5}}{10^{3}}}$
$= 10 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360886 A hole is made at the bottom of the tank filled with water
$(d e n s i t y = 1000 k g / m^{3})$ . If the total pressure at the bottom of the tank is three atomspheres ( 1 atmosphere $= 10^{5} N / m^{2}$ ), then the velocity of efflux is

1

\sqrt{400} m / s

2

\sqrt{200} m / s

3

\sqrt{600} m / s

4

\sqrt{500} m / s

Explanation:

Difference in pressure energy = difference in kinetic energy
$∴ (p_{0} - p_{0}) = \frac{1}{2} ρ v^{2}$
or $v = 2 \sqrt{\frac{p_{0}}{ρ}}$
$= 2 \sqrt{\frac{10^{5}}{10^{3}}}$
$= 20 m / s$ or $\sqrt{400} m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360887 A wide vessel with a small hole in the bottom is filled with water and kerosene. Neglecting viscosity, the velocity of water flow $v$ if the thickness of water layer is $h_{1}$ and that of kerosene layer is $h_{2}$ is (density of water is $ρ_{1} g$ / cc and that of kerosene is $ρ_{2} g / c c) (ρ_{1} > ρ_{2})$

1

v = \sqrt{2 g (h_{1} + h_{2})}

2

v = \sqrt{2 g (h_{1} + h_{2} \frac{ρ_{2}}{ρ_{1}})}

3

v = \sqrt{2 g (h_{1} ρ_{1} + h_{2} ρ_{2})}

4

v = \sqrt{2 g (h_{1} \frac{ρ_{1}}{ρ_{2}} + h_{2})}

Explanation:

supporting img

From Bernoulli's equation
$\begin{matrix} ρ_{2} g h_{2} + ρ_{1} g h_{1} = \frac{1}{2} ρ_{1} v^{2} \\ \Rightarrow v^{2} = \frac{2 g (ρ_{1} h_{1} + ρ_{2} h_{2})}{ρ_{1}} \Rightarrow v = \sqrt{2 g (h_{1} + \frac{ρ_{2}}{ρ_{1}} h_{2})} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360888 There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $h$ . As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to
supporting img

1

h^{1 / 2}

2

h^{3 / 2}

3

h^{2}

4

h

Explanation:

Here, $v_{1} = \sqrt{2 g (h + x)}; v_{2} = \sqrt{2 g x}$
Let, $a =$ area of cross-section of each hole
$ρ =$ density of the liquid
supporting img

The force exerted on the lower hole towards left
$= a ρ v_{1}^{2}$
Similarly, the force exerted on the upper hole towards right
$= a ρ v_{2}^{2}$
Net force on the tank, $F = a ρ (v_{1}^{2} - v_{2}^{2})$
$\begin{aligned} F = a ρ [2 g (h + x) - 2 g x] = 2 a ρ g h \\ \Rightarrow F \propto h \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360889 If a small orifice is made at a height of $0.25 m$ from the ground, the horizontal range of water stream will be
supporting img

1

46.5 c m

2

56.6 c m

3

76.6 c m

4

86.6 c m

Explanation:

Given, height of small orificc from ground is $h = 0.25 m$
supporting img
Total height of water tank, $H = 1 m$
$∴$ Range of water stream, is given by
$R = 2 \sqrt{h (H - h)}$
$= 2 \sqrt{(H - 0.25) (0.25)}$
$= 2 \sqrt{(1 - 0.25) 0.25} = 2 \sqrt{0.75 \times 0.25}$
$= 0.866 m = 86.6 c m$
So, correct option is (4)

AIIMS - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360890 The pressure of water in a pipe when tap is closed is $5.5 \times 10^{5} N / m^{2}$ . When tap gets open, pressure reduces to $5 \times 10^{5} N / m^{2}$ . The velocity with which water comes out on opening the tap is

1

10 m / s

2

5 m / s

3

20 m / s

4

15 m / s

Explanation:

Decrease in pressure energy $=$ increase in kinetic energy
$or Δ p = \frac{1}{2} ρ v^{2}$
$∴ v = \sqrt{\frac{2 (Δ p)}{ρ}}$
$= \sqrt{\frac{2 \times 0.5 \times 10^{5}}{10^{3}}}$
$= 10 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360886 A hole is made at the bottom of the tank filled with water
$(d e n s i t y = 1000 k g / m^{3})$ . If the total pressure at the bottom of the tank is three atomspheres ( 1 atmosphere $= 10^{5} N / m^{2}$ ), then the velocity of efflux is

1

\sqrt{400} m / s

2

\sqrt{200} m / s

3

\sqrt{600} m / s

4

\sqrt{500} m / s

Explanation:

Difference in pressure energy = difference in kinetic energy
$∴ (p_{0} - p_{0}) = \frac{1}{2} ρ v^{2}$
or $v = 2 \sqrt{\frac{p_{0}}{ρ}}$
$= 2 \sqrt{\frac{10^{5}}{10^{3}}}$
$= 20 m / s$ or $\sqrt{400} m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360887 A wide vessel with a small hole in the bottom is filled with water and kerosene. Neglecting viscosity, the velocity of water flow $v$ if the thickness of water layer is $h_{1}$ and that of kerosene layer is $h_{2}$ is (density of water is $ρ_{1} g$ / cc and that of kerosene is $ρ_{2} g / c c) (ρ_{1} > ρ_{2})$

1

v = \sqrt{2 g (h_{1} + h_{2})}

2

v = \sqrt{2 g (h_{1} + h_{2} \frac{ρ_{2}}{ρ_{1}})}

3

v = \sqrt{2 g (h_{1} ρ_{1} + h_{2} ρ_{2})}

4

v = \sqrt{2 g (h_{1} \frac{ρ_{1}}{ρ_{2}} + h_{2})}

Explanation:

supporting img

From Bernoulli's equation
$\begin{matrix} ρ_{2} g h_{2} + ρ_{1} g h_{1} = \frac{1}{2} ρ_{1} v^{2} \\ \Rightarrow v^{2} = \frac{2 g (ρ_{1} h_{1} + ρ_{2} h_{2})}{ρ_{1}} \Rightarrow v = \sqrt{2 g (h_{1} + \frac{ρ_{2}}{ρ_{1}} h_{2})} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360888 There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $h$ . As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to
supporting img

1

h^{1 / 2}

2

h^{3 / 2}

3

h^{2}

4

h

Explanation:

Here, $v_{1} = \sqrt{2 g (h + x)}; v_{2} = \sqrt{2 g x}$
Let, $a =$ area of cross-section of each hole
$ρ =$ density of the liquid
supporting img

The force exerted on the lower hole towards left
$= a ρ v_{1}^{2}$
Similarly, the force exerted on the upper hole towards right
$= a ρ v_{2}^{2}$
Net force on the tank, $F = a ρ (v_{1}^{2} - v_{2}^{2})$
$\begin{aligned} F = a ρ [2 g (h + x) - 2 g x] = 2 a ρ g h \\ \Rightarrow F \propto h \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360889 If a small orifice is made at a height of $0.25 m$ from the ground, the horizontal range of water stream will be
supporting img

1

46.5 c m

2

56.6 c m

3

76.6 c m

4

86.6 c m

Explanation:

Given, height of small orificc from ground is $h = 0.25 m$
supporting img
Total height of water tank, $H = 1 m$
$∴$ Range of water stream, is given by
$R = 2 \sqrt{h (H - h)}$
$= 2 \sqrt{(H - 0.25) (0.25)}$
$= 2 \sqrt{(1 - 0.25) 0.25} = 2 \sqrt{0.75 \times 0.25}$
$= 0.866 m = 86.6 c m$
So, correct option is (4)

AIIMS - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360890 The pressure of water in a pipe when tap is closed is $5.5 \times 10^{5} N / m^{2}$ . When tap gets open, pressure reduces to $5 \times 10^{5} N / m^{2}$ . The velocity with which water comes out on opening the tap is

1

10 m / s

2

5 m / s

3

20 m / s

4

15 m / s

Explanation:

Decrease in pressure energy $=$ increase in kinetic energy
$or Δ p = \frac{1}{2} ρ v^{2}$
$∴ v = \sqrt{\frac{2 (Δ p)}{ρ}}$
$= \sqrt{\frac{2 \times 0.5 \times 10^{5}}{10^{3}}}$
$= 10 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360886 A hole is made at the bottom of the tank filled with water
$(d e n s i t y = 1000 k g / m^{3})$ . If the total pressure at the bottom of the tank is three atomspheres ( 1 atmosphere $= 10^{5} N / m^{2}$ ), then the velocity of efflux is

1

\sqrt{400} m / s

2

\sqrt{200} m / s

3

\sqrt{600} m / s

4

\sqrt{500} m / s

Explanation:

Difference in pressure energy = difference in kinetic energy
$∴ (p_{0} - p_{0}) = \frac{1}{2} ρ v^{2}$
or $v = 2 \sqrt{\frac{p_{0}}{ρ}}$
$= 2 \sqrt{\frac{10^{5}}{10^{3}}}$
$= 20 m / s$ or $\sqrt{400} m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360887 A wide vessel with a small hole in the bottom is filled with water and kerosene. Neglecting viscosity, the velocity of water flow $v$ if the thickness of water layer is $h_{1}$ and that of kerosene layer is $h_{2}$ is (density of water is $ρ_{1} g$ / cc and that of kerosene is $ρ_{2} g / c c) (ρ_{1} > ρ_{2})$

1

v = \sqrt{2 g (h_{1} + h_{2})}

2

v = \sqrt{2 g (h_{1} + h_{2} \frac{ρ_{2}}{ρ_{1}})}

3

v = \sqrt{2 g (h_{1} ρ_{1} + h_{2} ρ_{2})}

4

v = \sqrt{2 g (h_{1} \frac{ρ_{1}}{ρ_{2}} + h_{2})}

Explanation:

supporting img

From Bernoulli's equation
$\begin{matrix} ρ_{2} g h_{2} + ρ_{1} g h_{1} = \frac{1}{2} ρ_{1} v^{2} \\ \Rightarrow v^{2} = \frac{2 g (ρ_{1} h_{1} + ρ_{2} h_{2})}{ρ_{1}} \Rightarrow v = \sqrt{2 g (h_{1} + \frac{ρ_{2}}{ρ_{1}} h_{2})} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360888 There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $h$ . As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to
supporting img

1

h^{1 / 2}

2

h^{3 / 2}

3

h^{2}

4

h

Explanation:

Here, $v_{1} = \sqrt{2 g (h + x)}; v_{2} = \sqrt{2 g x}$
Let, $a =$ area of cross-section of each hole
$ρ =$ density of the liquid
supporting img

The force exerted on the lower hole towards left
$= a ρ v_{1}^{2}$
Similarly, the force exerted on the upper hole towards right
$= a ρ v_{2}^{2}$
Net force on the tank, $F = a ρ (v_{1}^{2} - v_{2}^{2})$
$\begin{aligned} F = a ρ [2 g (h + x) - 2 g x] = 2 a ρ g h \\ \Rightarrow F \propto h \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360889 If a small orifice is made at a height of $0.25 m$ from the ground, the horizontal range of water stream will be
supporting img

1

46.5 c m

2

56.6 c m

3

76.6 c m

4

86.6 c m

Explanation:

Given, height of small orificc from ground is $h = 0.25 m$
supporting img
Total height of water tank, $H = 1 m$
$∴$ Range of water stream, is given by
$R = 2 \sqrt{h (H - h)}$
$= 2 \sqrt{(H - 0.25) (0.25)}$
$= 2 \sqrt{(1 - 0.25) 0.25} = 2 \sqrt{0.75 \times 0.25}$
$= 0.866 m = 86.6 c m$
So, correct option is (4)

AIIMS - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360890 The pressure of water in a pipe when tap is closed is $5.5 \times 10^{5} N / m^{2}$ . When tap gets open, pressure reduces to $5 \times 10^{5} N / m^{2}$ . The velocity with which water comes out on opening the tap is

1

10 m / s

2

5 m / s

3

20 m / s

4

15 m / s

Explanation:

Decrease in pressure energy $=$ increase in kinetic energy
$or Δ p = \frac{1}{2} ρ v^{2}$
$∴ v = \sqrt{\frac{2 (Δ p)}{ρ}}$
$= \sqrt{\frac{2 \times 0.5 \times 10^{5}}{10^{3}}}$
$= 10 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360886 A hole is made at the bottom of the tank filled with water
$(d e n s i t y = 1000 k g / m^{3})$ . If the total pressure at the bottom of the tank is three atomspheres ( 1 atmosphere $= 10^{5} N / m^{2}$ ), then the velocity of efflux is

1

\sqrt{400} m / s

2

\sqrt{200} m / s

3

\sqrt{600} m / s

4

\sqrt{500} m / s

Explanation:

Difference in pressure energy = difference in kinetic energy
$∴ (p_{0} - p_{0}) = \frac{1}{2} ρ v^{2}$
or $v = 2 \sqrt{\frac{p_{0}}{ρ}}$
$= 2 \sqrt{\frac{10^{5}}{10^{3}}}$
$= 20 m / s$ or $\sqrt{400} m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360887 A wide vessel with a small hole in the bottom is filled with water and kerosene. Neglecting viscosity, the velocity of water flow $v$ if the thickness of water layer is $h_{1}$ and that of kerosene layer is $h_{2}$ is (density of water is $ρ_{1} g$ / cc and that of kerosene is $ρ_{2} g / c c) (ρ_{1} > ρ_{2})$

1

v = \sqrt{2 g (h_{1} + h_{2})}

2

v = \sqrt{2 g (h_{1} + h_{2} \frac{ρ_{2}}{ρ_{1}})}

3

v = \sqrt{2 g (h_{1} ρ_{1} + h_{2} ρ_{2})}

4

v = \sqrt{2 g (h_{1} \frac{ρ_{1}}{ρ_{2}} + h_{2})}

Explanation:

supporting img

From Bernoulli's equation
$\begin{matrix} ρ_{2} g h_{2} + ρ_{1} g h_{1} = \frac{1}{2} ρ_{1} v^{2} \\ \Rightarrow v^{2} = \frac{2 g (ρ_{1} h_{1} + ρ_{2} h_{2})}{ρ_{1}} \Rightarrow v = \sqrt{2 g (h_{1} + \frac{ρ_{2}}{ρ_{1}} h_{2})} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360888 There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $h$ . As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to
supporting img

1

h^{1 / 2}

2

h^{3 / 2}

3

h^{2}

4

h

Explanation:

Here, $v_{1} = \sqrt{2 g (h + x)}; v_{2} = \sqrt{2 g x}$
Let, $a =$ area of cross-section of each hole
$ρ =$ density of the liquid
supporting img

The force exerted on the lower hole towards left
$= a ρ v_{1}^{2}$
Similarly, the force exerted on the upper hole towards right
$= a ρ v_{2}^{2}$
Net force on the tank, $F = a ρ (v_{1}^{2} - v_{2}^{2})$
$\begin{aligned} F = a ρ [2 g (h + x) - 2 g x] = 2 a ρ g h \\ \Rightarrow F \propto h \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360889 If a small orifice is made at a height of $0.25 m$ from the ground, the horizontal range of water stream will be
supporting img

1

46.5 c m

2

56.6 c m

3

76.6 c m

4

86.6 c m

Explanation:

Given, height of small orificc from ground is $h = 0.25 m$
supporting img
Total height of water tank, $H = 1 m$
$∴$ Range of water stream, is given by
$R = 2 \sqrt{h (H - h)}$
$= 2 \sqrt{(H - 0.25) (0.25)}$
$= 2 \sqrt{(1 - 0.25) 0.25} = 2 \sqrt{0.75 \times 0.25}$
$= 0.866 m = 86.6 c m$
So, correct option is (4)

AIIMS - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360890 The pressure of water in a pipe when tap is closed is $5.5 \times 10^{5} N / m^{2}$ . When tap gets open, pressure reduces to $5 \times 10^{5} N / m^{2}$ . The velocity with which water comes out on opening the tap is

1

10 m / s

2

5 m / s

3

20 m / s

4

15 m / s

Explanation:

Decrease in pressure energy $=$ increase in kinetic energy
$or Δ p = \frac{1}{2} ρ v^{2}$
$∴ v = \sqrt{\frac{2 (Δ p)}{ρ}}$
$= \sqrt{\frac{2 \times 0.5 \times 10^{5}}{10^{3}}}$
$= 10 m / s$

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