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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361390 A spherical ball of radius $1 \times 10^{- 4} m$ and density $10^{5} k g / m^{3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately (The coefficient of viscosity of water is $9.8 \times 10^{- 6} N s / m^{2})$

1

2249 m

2

2518 m

3

2396 m

4

2296 m

Explanation:

The terminal velocity is given by:
$v = \frac{2}{9} r^{2} \frac{(ρ - σ) g}{η}$
$v = \frac{2}{9} \times \frac{{(10^{- 4})}^{2} (10^{5} - 10^{3}) \times 10}{9.8 \times 10^{- 6}} = 224.5 m / s$
Velocity of ball just before entering the water tank is
$v = \sqrt{2 g h}$
$\Rightarrow h = \frac{v^{2}}{2 g}; h = \frac{{(224.5)}^{2}}{2 \times 10} = 2518 m$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361391 A small steel ball of mass $m$ and radius $r$ is falling under gravity through a viscous liquid of coefficient of viscocity $η$ . If $g$ is the value of acceleration due to gravity, then the terminal velocity of the ball is proportional to (ignore buoyancy)

1

\frac{m g}{r η}

2

\frac{m g η}{r}

3

\frac{m g r}{η}

4

m g η r

Explanation:

As the steel ball is small therefore upthrust can be neglected.
Now, $6 π η r v_{0} = m g$ or $v_{0} \propto \frac{m g}{η r}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361392 A metal sphere of radius ' $R$ ' and density ' $ρ_{1}$ ' is dropped in a liquid of density ' $σ$ ' moves with terminal velocity ' $v$ '. Another metal sphere of same radius and density ' $ρ_{2}$ ' is dropped in the same liquid, its terminal velocity will be

1

v [(ρ_{2} + σ) / (ρ_{1} + σ)]

2

v [(ρ_{1} + σ) / (ρ_{2} + σ)]

3

v [(ρ_{2} - σ) / (ρ_{1} - σ)]

4

v [(ρ_{1} - σ) / (ρ_{2} - σ)]

Explanation:

The terminal velocity of the sphere
$v = \frac{2}{9} R^{2} \frac{(ρ_{1} - σ) g}{η} (1)$
Here $R$ = radius of sphere,
$ρ_{1} =$ density of sphere
and $σ =$ density of liquid
Now, if density of metal sphere is changed to $ρ_{2}$ , then terminal velocity
$v_{2} = \frac{2}{9} \frac{R^{2} (ρ_{2} - σ) g}{η} (2)$
Divide eq.(1) and (2), we get
$\frac{v_{2}}{v} = \frac{(ρ_{2} - σ)}{(ρ_{1} - σ)} or v_{2} = \frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$
Hence the terminal velocity is $\frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361393 If the terminal speed of a sphere of gold (density $= 19.5 \times 10^{3} k g / m^{3})$ is $0.5 m / s$ in a viscous liquid $($ density $= 3 k g / m^{3})$ . Find the terminal velocity of a sphere of silver (density $= 10.5 \times 10^{3} kg / m^{3}$ ) of the same size in the same liquid.

1

0.30 m / s

2

0.20 m / s

3

0.51 m / s

4

0.22 m / s

Explanation:

$V_{T} = \frac{2}{9} \frac{r^{3} (ρ - σ) g}{η} ρ =$ density of sphere
$σ =$ density of liquid
$∴ V \propto (ρ - σ) \Rightarrow \frac{V_{g}}{V_{s}} = \frac{ρ_{g} - σ}{ρ_{s} - σ}$
$\frac{0.5}{V_{s}} = \frac{19.5 - 3}{10.5 - 3} = \frac{16.5}{7.5} \Rightarrow V_{s} = \frac{0.5 \times 7.5}{16.5}$
$V_{s} = 0.22 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361390 A spherical ball of radius $1 \times 10^{- 4} m$ and density $10^{5} k g / m^{3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately (The coefficient of viscosity of water is $9.8 \times 10^{- 6} N s / m^{2})$

1

2249 m

2

2518 m

3

2396 m

4

2296 m

Explanation:

The terminal velocity is given by:
$v = \frac{2}{9} r^{2} \frac{(ρ - σ) g}{η}$
$v = \frac{2}{9} \times \frac{{(10^{- 4})}^{2} (10^{5} - 10^{3}) \times 10}{9.8 \times 10^{- 6}} = 224.5 m / s$
Velocity of ball just before entering the water tank is
$v = \sqrt{2 g h}$
$\Rightarrow h = \frac{v^{2}}{2 g}; h = \frac{{(224.5)}^{2}}{2 \times 10} = 2518 m$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361391 A small steel ball of mass $m$ and radius $r$ is falling under gravity through a viscous liquid of coefficient of viscocity $η$ . If $g$ is the value of acceleration due to gravity, then the terminal velocity of the ball is proportional to (ignore buoyancy)

1

\frac{m g}{r η}

2

\frac{m g η}{r}

3

\frac{m g r}{η}

4

m g η r

Explanation:

As the steel ball is small therefore upthrust can be neglected.
Now, $6 π η r v_{0} = m g$ or $v_{0} \propto \frac{m g}{η r}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361392 A metal sphere of radius ' $R$ ' and density ' $ρ_{1}$ ' is dropped in a liquid of density ' $σ$ ' moves with terminal velocity ' $v$ '. Another metal sphere of same radius and density ' $ρ_{2}$ ' is dropped in the same liquid, its terminal velocity will be

1

v [(ρ_{2} + σ) / (ρ_{1} + σ)]

2

v [(ρ_{1} + σ) / (ρ_{2} + σ)]

3

v [(ρ_{2} - σ) / (ρ_{1} - σ)]

4

v [(ρ_{1} - σ) / (ρ_{2} - σ)]

Explanation:

The terminal velocity of the sphere
$v = \frac{2}{9} R^{2} \frac{(ρ_{1} - σ) g}{η} (1)$
Here $R$ = radius of sphere,
$ρ_{1} =$ density of sphere
and $σ =$ density of liquid
Now, if density of metal sphere is changed to $ρ_{2}$ , then terminal velocity
$v_{2} = \frac{2}{9} \frac{R^{2} (ρ_{2} - σ) g}{η} (2)$
Divide eq.(1) and (2), we get
$\frac{v_{2}}{v} = \frac{(ρ_{2} - σ)}{(ρ_{1} - σ)} or v_{2} = \frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$
Hence the terminal velocity is $\frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361393 If the terminal speed of a sphere of gold (density $= 19.5 \times 10^{3} k g / m^{3})$ is $0.5 m / s$ in a viscous liquid $($ density $= 3 k g / m^{3})$ . Find the terminal velocity of a sphere of silver (density $= 10.5 \times 10^{3} kg / m^{3}$ ) of the same size in the same liquid.

1

0.30 m / s

2

0.20 m / s

3

0.51 m / s

4

0.22 m / s

Explanation:

$V_{T} = \frac{2}{9} \frac{r^{3} (ρ - σ) g}{η} ρ =$ density of sphere
$σ =$ density of liquid
$∴ V \propto (ρ - σ) \Rightarrow \frac{V_{g}}{V_{s}} = \frac{ρ_{g} - σ}{ρ_{s} - σ}$
$\frac{0.5}{V_{s}} = \frac{19.5 - 3}{10.5 - 3} = \frac{16.5}{7.5} \Rightarrow V_{s} = \frac{0.5 \times 7.5}{16.5}$
$V_{s} = 0.22 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361390 A spherical ball of radius $1 \times 10^{- 4} m$ and density $10^{5} k g / m^{3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately (The coefficient of viscosity of water is $9.8 \times 10^{- 6} N s / m^{2})$

1

2249 m

2

2518 m

3

2396 m

4

2296 m

Explanation:

The terminal velocity is given by:
$v = \frac{2}{9} r^{2} \frac{(ρ - σ) g}{η}$
$v = \frac{2}{9} \times \frac{{(10^{- 4})}^{2} (10^{5} - 10^{3}) \times 10}{9.8 \times 10^{- 6}} = 224.5 m / s$
Velocity of ball just before entering the water tank is
$v = \sqrt{2 g h}$
$\Rightarrow h = \frac{v^{2}}{2 g}; h = \frac{{(224.5)}^{2}}{2 \times 10} = 2518 m$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361391 A small steel ball of mass $m$ and radius $r$ is falling under gravity through a viscous liquid of coefficient of viscocity $η$ . If $g$ is the value of acceleration due to gravity, then the terminal velocity of the ball is proportional to (ignore buoyancy)

1

\frac{m g}{r η}

2

\frac{m g η}{r}

3

\frac{m g r}{η}

4

m g η r

Explanation:

As the steel ball is small therefore upthrust can be neglected.
Now, $6 π η r v_{0} = m g$ or $v_{0} \propto \frac{m g}{η r}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361392 A metal sphere of radius ' $R$ ' and density ' $ρ_{1}$ ' is dropped in a liquid of density ' $σ$ ' moves with terminal velocity ' $v$ '. Another metal sphere of same radius and density ' $ρ_{2}$ ' is dropped in the same liquid, its terminal velocity will be

1

v [(ρ_{2} + σ) / (ρ_{1} + σ)]

2

v [(ρ_{1} + σ) / (ρ_{2} + σ)]

3

v [(ρ_{2} - σ) / (ρ_{1} - σ)]

4

v [(ρ_{1} - σ) / (ρ_{2} - σ)]

Explanation:

The terminal velocity of the sphere
$v = \frac{2}{9} R^{2} \frac{(ρ_{1} - σ) g}{η} (1)$
Here $R$ = radius of sphere,
$ρ_{1} =$ density of sphere
and $σ =$ density of liquid
Now, if density of metal sphere is changed to $ρ_{2}$ , then terminal velocity
$v_{2} = \frac{2}{9} \frac{R^{2} (ρ_{2} - σ) g}{η} (2)$
Divide eq.(1) and (2), we get
$\frac{v_{2}}{v} = \frac{(ρ_{2} - σ)}{(ρ_{1} - σ)} or v_{2} = \frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$
Hence the terminal velocity is $\frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361393 If the terminal speed of a sphere of gold (density $= 19.5 \times 10^{3} k g / m^{3})$ is $0.5 m / s$ in a viscous liquid $($ density $= 3 k g / m^{3})$ . Find the terminal velocity of a sphere of silver (density $= 10.5 \times 10^{3} kg / m^{3}$ ) of the same size in the same liquid.

1

0.30 m / s

2

0.20 m / s

3

0.51 m / s

4

0.22 m / s

Explanation:

$V_{T} = \frac{2}{9} \frac{r^{3} (ρ - σ) g}{η} ρ =$ density of sphere
$σ =$ density of liquid
$∴ V \propto (ρ - σ) \Rightarrow \frac{V_{g}}{V_{s}} = \frac{ρ_{g} - σ}{ρ_{s} - σ}$
$\frac{0.5}{V_{s}} = \frac{19.5 - 3}{10.5 - 3} = \frac{16.5}{7.5} \Rightarrow V_{s} = \frac{0.5 \times 7.5}{16.5}$
$V_{s} = 0.22 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361390 A spherical ball of radius $1 \times 10^{- 4} m$ and density $10^{5} k g / m^{3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately (The coefficient of viscosity of water is $9.8 \times 10^{- 6} N s / m^{2})$

1

2249 m

2

2518 m

3

2396 m

4

2296 m

Explanation:

The terminal velocity is given by:
$v = \frac{2}{9} r^{2} \frac{(ρ - σ) g}{η}$
$v = \frac{2}{9} \times \frac{{(10^{- 4})}^{2} (10^{5} - 10^{3}) \times 10}{9.8 \times 10^{- 6}} = 224.5 m / s$
Velocity of ball just before entering the water tank is
$v = \sqrt{2 g h}$
$\Rightarrow h = \frac{v^{2}}{2 g}; h = \frac{{(224.5)}^{2}}{2 \times 10} = 2518 m$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361391 A small steel ball of mass $m$ and radius $r$ is falling under gravity through a viscous liquid of coefficient of viscocity $η$ . If $g$ is the value of acceleration due to gravity, then the terminal velocity of the ball is proportional to (ignore buoyancy)

1

\frac{m g}{r η}

2

\frac{m g η}{r}

3

\frac{m g r}{η}

4

m g η r

Explanation:

As the steel ball is small therefore upthrust can be neglected.
Now, $6 π η r v_{0} = m g$ or $v_{0} \propto \frac{m g}{η r}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361392 A metal sphere of radius ' $R$ ' and density ' $ρ_{1}$ ' is dropped in a liquid of density ' $σ$ ' moves with terminal velocity ' $v$ '. Another metal sphere of same radius and density ' $ρ_{2}$ ' is dropped in the same liquid, its terminal velocity will be

1

v [(ρ_{2} + σ) / (ρ_{1} + σ)]

2

v [(ρ_{1} + σ) / (ρ_{2} + σ)]

3

v [(ρ_{2} - σ) / (ρ_{1} - σ)]

4

v [(ρ_{1} - σ) / (ρ_{2} - σ)]

Explanation:

The terminal velocity of the sphere
$v = \frac{2}{9} R^{2} \frac{(ρ_{1} - σ) g}{η} (1)$
Here $R$ = radius of sphere,
$ρ_{1} =$ density of sphere
and $σ =$ density of liquid
Now, if density of metal sphere is changed to $ρ_{2}$ , then terminal velocity
$v_{2} = \frac{2}{9} \frac{R^{2} (ρ_{2} - σ) g}{η} (2)$
Divide eq.(1) and (2), we get
$\frac{v_{2}}{v} = \frac{(ρ_{2} - σ)}{(ρ_{1} - σ)} or v_{2} = \frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$
Hence the terminal velocity is $\frac{v (ρ_{2} - σ)}{(ρ_{1} - σ)}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361393 If the terminal speed of a sphere of gold (density $= 19.5 \times 10^{3} k g / m^{3})$ is $0.5 m / s$ in a viscous liquid $($ density $= 3 k g / m^{3})$ . Find the terminal velocity of a sphere of silver (density $= 10.5 \times 10^{3} kg / m^{3}$ ) of the same size in the same liquid.

1

0.30 m / s

2

0.20 m / s

3

0.51 m / s

4

0.22 m / s

Explanation:

$V_{T} = \frac{2}{9} \frac{r^{3} (ρ - σ) g}{η} ρ =$ density of sphere
$σ =$ density of liquid
$∴ V \propto (ρ - σ) \Rightarrow \frac{V_{g}}{V_{s}} = \frac{ρ_{g} - σ}{ρ_{s} - σ}$
$\frac{0.5}{V_{s}} = \frac{19.5 - 3}{10.5 - 3} = \frac{16.5}{7.5} \Rightarrow V_{s} = \frac{0.5 \times 7.5}{16.5}$
$V_{s} = 0.22 m / s$