QBManager

Mechanical Properties of Fluids

143242 To what height water should be filled in a container of height $21 cm$ , so that it appears as half filled when viewed from the top $($ Take $_{a} μ_{w} = \frac{4}{3})$

1

12 cm

2

15 cm

3

10.5 cm

4

7 cm

Explanation:

A Given,
$_{a} μ_{w} = \frac{4}{3}$
Apparent depth $h^{'} = (21 - x)$
Real depth $h = x$
refractive index of water $= \frac{Real depth}{Apparent depth}$
$∴ μ = \frac{h}{h^{'}}$
$\frac{4}{3} = \frac{x}{21 - x}$
$x = 12 cm$

BCECE-2012

Mechanical Properties of Fluids

143243 A wooden ball of density $ρ$ is immersed in a liquid of density $σ$ to a depth $h$ below the surface of water and then released. The height to which the ball jumps out of water is:

1

(1 - \frac{ρ}{σ}) h

2

(1 + \frac{ρ}{σ}) h

3

(\frac{ρ}{σ} - 1) h

4

(\frac{σ}{ρ} - 1) h

Explanation:

D
Weight of the ball $= V ρ g$
upward thrust on ball $= V σ g - V ρ g$ Upward acceleration $(a) = \frac{V σ g - V ρ g}{V ρ} = (\frac{σ - ρ}{ρ}) g$
$∵ v^{2} = u^{2} + 2 as = 2$ as $[∵ u = 0]$
$v^{2} = 2 \times (\frac{σ - ρ}{ρ}) gh$
For the motion of the ball outside the water,
$K.E. = P.E. at height H$
$\frac{1}{2} {mv}^{2} = mgH$
$H = \frac{v^{2}}{2 g} = \frac{2 gh (σ - ρ) / ρ}{2 g}$
$H = (\frac{σ - ρ}{ρ}) h = (\frac{σ}{ρ} - 1) h$

SCRA-2012

Mechanical Properties of Fluids

143244 A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. What should be the ratio of radius of the cavity to that of outer radius of sphere?

1

\frac{\sqrt[3]{3}}{2}

2

\frac{\sqrt[3]{5}}{2}

3

\frac{\sqrt[3]{7}}{2}

4

\frac{\sqrt[3]{9}}{2}

Explanation:

C Weight of sphere $=$ Weight of water displaced $W_{S} = W_{w}$
$V_{s} \times ρ_{s} \times g = V_{w} \times ρ_{w} \times g$
Here, $ρ_{s} = 8 ρ_{w}$
$\frac{4}{3} π (R^{3} - r^{3}) ρ_{s} \times g = \frac{4}{3} π R^{3} \times ρ_{w} g$
$8 (R^{3} - r^{3}) = R^{3}$
$\frac{r^{3}}{R^{3}} = \frac{7}{8}$
$\frac{r}{R} = \frac{\sqrt[3]{7}}{2}$

SCRA-2010

Mechanical Properties of Fluids

143245 An ideal fluid flows through a pipe of circular cross-section with diameters $5 cm$ and $10 cm$ as shown in the figure. The ratio of velocities of fluid at $A$ and $B$ is :

1

4 : 1

2

1 : 4

3

2 : 1

4

1 : 2

Explanation:

A Given,
$d_{A} = 5 cm$
$d_{B} = 10 cm$
By continuity equation,
$a_{A} v_{A} = a_{B} v_{B}$
$\frac{v_{A}}{v_{B}} = \frac{a_{B}}{a_{A}}$
$= \frac{\frac{π}{4} \times d_{B}^{2}}{\frac{π}{4} \times d_{A}^{2}}$
$= {(\frac{d_{B}}{d_{A}})}^{2}$
$= {(\frac{10}{5})}^{2}$
$\frac{v_{A}}{v_{B}} = \frac{4}{1}$
$V_{A} : V_{B} = 4 : 1$

Karnataka CET-2016

Mechanical Properties of Fluids

143242 To what height water should be filled in a container of height $21 cm$ , so that it appears as half filled when viewed from the top $($ Take $_{a} μ_{w} = \frac{4}{3})$

1

12 cm

2

15 cm

3

10.5 cm

4

7 cm

Explanation:

A Given,
$_{a} μ_{w} = \frac{4}{3}$
Apparent depth $h^{'} = (21 - x)$
Real depth $h = x$
refractive index of water $= \frac{Real depth}{Apparent depth}$
$∴ μ = \frac{h}{h^{'}}$
$\frac{4}{3} = \frac{x}{21 - x}$
$x = 12 cm$

BCECE-2012

Mechanical Properties of Fluids

143243 A wooden ball of density $ρ$ is immersed in a liquid of density $σ$ to a depth $h$ below the surface of water and then released. The height to which the ball jumps out of water is:

1

(1 - \frac{ρ}{σ}) h

2

(1 + \frac{ρ}{σ}) h

3

(\frac{ρ}{σ} - 1) h

4

(\frac{σ}{ρ} - 1) h

Explanation:

D
Weight of the ball $= V ρ g$
upward thrust on ball $= V σ g - V ρ g$ Upward acceleration $(a) = \frac{V σ g - V ρ g}{V ρ} = (\frac{σ - ρ}{ρ}) g$
$∵ v^{2} = u^{2} + 2 as = 2$ as $[∵ u = 0]$
$v^{2} = 2 \times (\frac{σ - ρ}{ρ}) gh$
For the motion of the ball outside the water,
$K.E. = P.E. at height H$
$\frac{1}{2} {mv}^{2} = mgH$
$H = \frac{v^{2}}{2 g} = \frac{2 gh (σ - ρ) / ρ}{2 g}$
$H = (\frac{σ - ρ}{ρ}) h = (\frac{σ}{ρ} - 1) h$

SCRA-2012

Mechanical Properties of Fluids

143244 A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. What should be the ratio of radius of the cavity to that of outer radius of sphere?

1

\frac{\sqrt[3]{3}}{2}

2

\frac{\sqrt[3]{5}}{2}

3

\frac{\sqrt[3]{7}}{2}

4

\frac{\sqrt[3]{9}}{2}

Explanation:

C Weight of sphere $=$ Weight of water displaced $W_{S} = W_{w}$
$V_{s} \times ρ_{s} \times g = V_{w} \times ρ_{w} \times g$
Here, $ρ_{s} = 8 ρ_{w}$
$\frac{4}{3} π (R^{3} - r^{3}) ρ_{s} \times g = \frac{4}{3} π R^{3} \times ρ_{w} g$
$8 (R^{3} - r^{3}) = R^{3}$
$\frac{r^{3}}{R^{3}} = \frac{7}{8}$
$\frac{r}{R} = \frac{\sqrt[3]{7}}{2}$

SCRA-2010

Mechanical Properties of Fluids

143245 An ideal fluid flows through a pipe of circular cross-section with diameters $5 cm$ and $10 cm$ as shown in the figure. The ratio of velocities of fluid at $A$ and $B$ is :

1

4 : 1

2

1 : 4

3

2 : 1

4

1 : 2

Explanation:

A Given,
$d_{A} = 5 cm$
$d_{B} = 10 cm$
By continuity equation,
$a_{A} v_{A} = a_{B} v_{B}$
$\frac{v_{A}}{v_{B}} = \frac{a_{B}}{a_{A}}$
$= \frac{\frac{π}{4} \times d_{B}^{2}}{\frac{π}{4} \times d_{A}^{2}}$
$= {(\frac{d_{B}}{d_{A}})}^{2}$
$= {(\frac{10}{5})}^{2}$
$\frac{v_{A}}{v_{B}} = \frac{4}{1}$
$V_{A} : V_{B} = 4 : 1$

Karnataka CET-2016

Mechanical Properties of Fluids

143242 To what height water should be filled in a container of height $21 cm$ , so that it appears as half filled when viewed from the top $($ Take $_{a} μ_{w} = \frac{4}{3})$

1

12 cm

2

15 cm

3

10.5 cm

4

7 cm

Explanation:

A Given,
$_{a} μ_{w} = \frac{4}{3}$
Apparent depth $h^{'} = (21 - x)$
Real depth $h = x$
refractive index of water $= \frac{Real depth}{Apparent depth}$
$∴ μ = \frac{h}{h^{'}}$
$\frac{4}{3} = \frac{x}{21 - x}$
$x = 12 cm$

BCECE-2012

Mechanical Properties of Fluids

143243 A wooden ball of density $ρ$ is immersed in a liquid of density $σ$ to a depth $h$ below the surface of water and then released. The height to which the ball jumps out of water is:

1

(1 - \frac{ρ}{σ}) h

2

(1 + \frac{ρ}{σ}) h

3

(\frac{ρ}{σ} - 1) h

4

(\frac{σ}{ρ} - 1) h

Explanation:

D
Weight of the ball $= V ρ g$
upward thrust on ball $= V σ g - V ρ g$ Upward acceleration $(a) = \frac{V σ g - V ρ g}{V ρ} = (\frac{σ - ρ}{ρ}) g$
$∵ v^{2} = u^{2} + 2 as = 2$ as $[∵ u = 0]$
$v^{2} = 2 \times (\frac{σ - ρ}{ρ}) gh$
For the motion of the ball outside the water,
$K.E. = P.E. at height H$
$\frac{1}{2} {mv}^{2} = mgH$
$H = \frac{v^{2}}{2 g} = \frac{2 gh (σ - ρ) / ρ}{2 g}$
$H = (\frac{σ - ρ}{ρ}) h = (\frac{σ}{ρ} - 1) h$

SCRA-2012

Mechanical Properties of Fluids

143244 A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. What should be the ratio of radius of the cavity to that of outer radius of sphere?

1

\frac{\sqrt[3]{3}}{2}

2

\frac{\sqrt[3]{5}}{2}

3

\frac{\sqrt[3]{7}}{2}

4

\frac{\sqrt[3]{9}}{2}

Explanation:

C Weight of sphere $=$ Weight of water displaced $W_{S} = W_{w}$
$V_{s} \times ρ_{s} \times g = V_{w} \times ρ_{w} \times g$
Here, $ρ_{s} = 8 ρ_{w}$
$\frac{4}{3} π (R^{3} - r^{3}) ρ_{s} \times g = \frac{4}{3} π R^{3} \times ρ_{w} g$
$8 (R^{3} - r^{3}) = R^{3}$
$\frac{r^{3}}{R^{3}} = \frac{7}{8}$
$\frac{r}{R} = \frac{\sqrt[3]{7}}{2}$

SCRA-2010

Mechanical Properties of Fluids

143245 An ideal fluid flows through a pipe of circular cross-section with diameters $5 cm$ and $10 cm$ as shown in the figure. The ratio of velocities of fluid at $A$ and $B$ is :

1

4 : 1

2

1 : 4

3

2 : 1

4

1 : 2

Explanation:

A Given,
$d_{A} = 5 cm$
$d_{B} = 10 cm$
By continuity equation,
$a_{A} v_{A} = a_{B} v_{B}$
$\frac{v_{A}}{v_{B}} = \frac{a_{B}}{a_{A}}$
$= \frac{\frac{π}{4} \times d_{B}^{2}}{\frac{π}{4} \times d_{A}^{2}}$
$= {(\frac{d_{B}}{d_{A}})}^{2}$
$= {(\frac{10}{5})}^{2}$
$\frac{v_{A}}{v_{B}} = \frac{4}{1}$
$V_{A} : V_{B} = 4 : 1$

Karnataka CET-2016

Mechanical Properties of Fluids

143242 To what height water should be filled in a container of height $21 cm$ , so that it appears as half filled when viewed from the top $($ Take $_{a} μ_{w} = \frac{4}{3})$

1

12 cm

2

15 cm

3

10.5 cm

4

7 cm

Explanation:

A Given,
$_{a} μ_{w} = \frac{4}{3}$
Apparent depth $h^{'} = (21 - x)$
Real depth $h = x$
refractive index of water $= \frac{Real depth}{Apparent depth}$
$∴ μ = \frac{h}{h^{'}}$
$\frac{4}{3} = \frac{x}{21 - x}$
$x = 12 cm$

BCECE-2012

Mechanical Properties of Fluids

143243 A wooden ball of density $ρ$ is immersed in a liquid of density $σ$ to a depth $h$ below the surface of water and then released. The height to which the ball jumps out of water is:

1

(1 - \frac{ρ}{σ}) h

2

(1 + \frac{ρ}{σ}) h

3

(\frac{ρ}{σ} - 1) h

4

(\frac{σ}{ρ} - 1) h

Explanation:

D
Weight of the ball $= V ρ g$
upward thrust on ball $= V σ g - V ρ g$ Upward acceleration $(a) = \frac{V σ g - V ρ g}{V ρ} = (\frac{σ - ρ}{ρ}) g$
$∵ v^{2} = u^{2} + 2 as = 2$ as $[∵ u = 0]$
$v^{2} = 2 \times (\frac{σ - ρ}{ρ}) gh$
For the motion of the ball outside the water,
$K.E. = P.E. at height H$
$\frac{1}{2} {mv}^{2} = mgH$
$H = \frac{v^{2}}{2 g} = \frac{2 gh (σ - ρ) / ρ}{2 g}$
$H = (\frac{σ - ρ}{ρ}) h = (\frac{σ}{ρ} - 1) h$

SCRA-2012

Mechanical Properties of Fluids

143244 A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. What should be the ratio of radius of the cavity to that of outer radius of sphere?

1

\frac{\sqrt[3]{3}}{2}

2

\frac{\sqrt[3]{5}}{2}

3

\frac{\sqrt[3]{7}}{2}

4

\frac{\sqrt[3]{9}}{2}

Explanation:

C Weight of sphere $=$ Weight of water displaced $W_{S} = W_{w}$
$V_{s} \times ρ_{s} \times g = V_{w} \times ρ_{w} \times g$
Here, $ρ_{s} = 8 ρ_{w}$
$\frac{4}{3} π (R^{3} - r^{3}) ρ_{s} \times g = \frac{4}{3} π R^{3} \times ρ_{w} g$
$8 (R^{3} - r^{3}) = R^{3}$
$\frac{r^{3}}{R^{3}} = \frac{7}{8}$
$\frac{r}{R} = \frac{\sqrt[3]{7}}{2}$

SCRA-2010

Mechanical Properties of Fluids

143245 An ideal fluid flows through a pipe of circular cross-section with diameters $5 cm$ and $10 cm$ as shown in the figure. The ratio of velocities of fluid at $A$ and $B$ is :

1

4 : 1

2

1 : 4

3

2 : 1

4

1 : 2

Explanation:

A Given,
$d_{A} = 5 cm$
$d_{B} = 10 cm$
By continuity equation,
$a_{A} v_{A} = a_{B} v_{B}$
$\frac{v_{A}}{v_{B}} = \frac{a_{B}}{a_{A}}$
$= \frac{\frac{π}{4} \times d_{B}^{2}}{\frac{π}{4} \times d_{A}^{2}}$
$= {(\frac{d_{B}}{d_{A}})}^{2}$
$= {(\frac{10}{5})}^{2}$
$\frac{v_{A}}{v_{B}} = \frac{4}{1}$
$V_{A} : V_{B} = 4 : 1$

Karnataka CET-2016

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: