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Mechanical Properties of Fluids

143330 A body floats with one-third volume outside when put in water and three-fourth of its volume outside when put in another liquid. The density of the liquid is

1

9 / 4 g {cm}^{- 3}

2

8 / 3 g {cm}^{- 3}

3

3 / 8 g {cm}^{- 3}

4

4 / 9 g {cm}^{- 3}

Explanation:

B According to question -
In equilibrium condition -
Weight of body $=$ Buoyant force
$mg = ρ_{w} g_{d}$
$ρ_{b} v = ρ_{w} 2 v / 3$
$ρ_{b} = \frac{2 ρ_{w}}{3}$
Now, when body put in another liquid having density $ρ$ then in equilibrium condition
$mg = ρ (\frac{v}{4}) g$
$ρ_{b} vg = ρ (\frac{v}{4}) g$
$ρ = 4 ρ_{b}$
$ρ = 4 \times \frac{2 ρ_{w}}{3} = \frac{8}{3} ρ_{w}$
$ρ = \frac{8}{3} gm / {cm}^{3} [∵ ρ_{w} = 1 gm / {cm}^{3}]$

SCRA-2013

Mechanical Properties of Fluids

143332 A swimmer can swim in still water with speed $v$ and the river flowing with velocity $\frac{v}{2}$ . To cross the river in shortest time, he should swim making angel $θ$ with the upstream. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance?

1

\cos θ

2

\cot θ

3

\sin θ

4

\tan θ

Explanation:

C Given that speed of swimmer in still water $(v_{b}) = v$
Let width be $w$
speed of with respect to river $(v_{r}) = v / 2$ , angle of velocity swimmer with up stream $= θ$

Shortest path

Shortest time
The relative velocity of the swimmer with respect to water is $v_{b} \sin θ$ .
Time taken to cross the river in the shortest distance ( $t)$ $= \frac{w}{v_{b} \sin θ}$
Shortest time taken to cross the river is $t^{'} = \frac{w}{v_{b}}$
Ratio of times taken for shortest time to that of shortest path
$\frac{t^{'}}{t} = \frac{w / v_{b}}{w / \sin θ}$
$\frac{t^{'}}{t} = \sin θ$

AP EAMCET (18.09.2020) Shift-II

Mechanical Properties of Fluids

143333 A $10 {cm}^{3}$ cube floats in water with a height of 4 ${cm}^{3}$ remaining above the surface. The density of the material from which the cube is made is,

1

0.6 g {cm}^{- 3}

2

1.0 g {cm}^{- 3}

3

0.4 g {cm}^{- 3}

4

0.24 g {cm}^{- 3}

5 none of these

Explanation:

A According to the question,
Density of cube material $=$ Volume of cube submerged in water
Density of water
$\frac{d}{1} = \frac{10 - 4}{10}$
$d = \frac{6}{10}$
$d = 0.6 gm / {cm}^{3}$

KERALA CEE 2004

Mechanical Properties of Fluids

143334 A wooden block is floating on water kept in a beaker. $40 %$ of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $g / 2$ . The block will then

1

sink

2 float with

10 %

above the water surface

3 float with

40 %

above the water surface

4 float with

70 %

above the water surface

Explanation:

B
By the Archimedes principle weight of wooden block = world is place
Let $V$ be the Volume of the block
Density of the liquid
Net weight $= m g + \frac{m g}{2} = \frac{3 m g}{2}$
$\frac{3 mg}{2} = V_{i} ρ g$
$mg = \frac{60}{100} V \times ρ \times g$
$mg = \frac{3}{5} V \times ρ \times g$
From equation (i) and (ii), we get -
$\frac{3}{2} (\frac{3}{5} V ρ g) = V_{i} ρ g$
$V_{i} = \frac{90}{100} V$
Flout $= 10 %$

Mechanical Properties of Fluids

143330 A body floats with one-third volume outside when put in water and three-fourth of its volume outside when put in another liquid. The density of the liquid is

1

9 / 4 g {cm}^{- 3}

2

8 / 3 g {cm}^{- 3}

3

3 / 8 g {cm}^{- 3}

4

4 / 9 g {cm}^{- 3}

Explanation:

B According to question -
In equilibrium condition -
Weight of body $=$ Buoyant force
$mg = ρ_{w} g_{d}$
$ρ_{b} v = ρ_{w} 2 v / 3$
$ρ_{b} = \frac{2 ρ_{w}}{3}$
Now, when body put in another liquid having density $ρ$ then in equilibrium condition
$mg = ρ (\frac{v}{4}) g$
$ρ_{b} vg = ρ (\frac{v}{4}) g$
$ρ = 4 ρ_{b}$
$ρ = 4 \times \frac{2 ρ_{w}}{3} = \frac{8}{3} ρ_{w}$
$ρ = \frac{8}{3} gm / {cm}^{3} [∵ ρ_{w} = 1 gm / {cm}^{3}]$

SCRA-2013

Mechanical Properties of Fluids

143332 A swimmer can swim in still water with speed $v$ and the river flowing with velocity $\frac{v}{2}$ . To cross the river in shortest time, he should swim making angel $θ$ with the upstream. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance?

1

\cos θ

2

\cot θ

3

\sin θ

4

\tan θ

Explanation:

C Given that speed of swimmer in still water $(v_{b}) = v$
Let width be $w$
speed of with respect to river $(v_{r}) = v / 2$ , angle of velocity swimmer with up stream $= θ$

Shortest path

Shortest time
The relative velocity of the swimmer with respect to water is $v_{b} \sin θ$ .
Time taken to cross the river in the shortest distance ( $t)$ $= \frac{w}{v_{b} \sin θ}$
Shortest time taken to cross the river is $t^{'} = \frac{w}{v_{b}}$
Ratio of times taken for shortest time to that of shortest path
$\frac{t^{'}}{t} = \frac{w / v_{b}}{w / \sin θ}$
$\frac{t^{'}}{t} = \sin θ$

AP EAMCET (18.09.2020) Shift-II

Mechanical Properties of Fluids

143333 A $10 {cm}^{3}$ cube floats in water with a height of 4 ${cm}^{3}$ remaining above the surface. The density of the material from which the cube is made is,

1

0.6 g {cm}^{- 3}

2

1.0 g {cm}^{- 3}

3

0.4 g {cm}^{- 3}

4

0.24 g {cm}^{- 3}

5 none of these

Explanation:

A According to the question,
Density of cube material $=$ Volume of cube submerged in water
Density of water
$\frac{d}{1} = \frac{10 - 4}{10}$
$d = \frac{6}{10}$
$d = 0.6 gm / {cm}^{3}$

KERALA CEE 2004

Mechanical Properties of Fluids

143334 A wooden block is floating on water kept in a beaker. $40 %$ of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $g / 2$ . The block will then

1

sink

2 float with

10 %

above the water surface

3 float with

40 %

above the water surface

4 float with

70 %

above the water surface

Explanation:

B
By the Archimedes principle weight of wooden block = world is place
Let $V$ be the Volume of the block
Density of the liquid
Net weight $= m g + \frac{m g}{2} = \frac{3 m g}{2}$
$\frac{3 mg}{2} = V_{i} ρ g$
$mg = \frac{60}{100} V \times ρ \times g$
$mg = \frac{3}{5} V \times ρ \times g$
From equation (i) and (ii), we get -
$\frac{3}{2} (\frac{3}{5} V ρ g) = V_{i} ρ g$
$V_{i} = \frac{90}{100} V$
Flout $= 10 %$

Mechanical Properties of Fluids

143330 A body floats with one-third volume outside when put in water and three-fourth of its volume outside when put in another liquid. The density of the liquid is

1

9 / 4 g {cm}^{- 3}

2

8 / 3 g {cm}^{- 3}

3

3 / 8 g {cm}^{- 3}

4

4 / 9 g {cm}^{- 3}

Explanation:

B According to question -
In equilibrium condition -
Weight of body $=$ Buoyant force
$mg = ρ_{w} g_{d}$
$ρ_{b} v = ρ_{w} 2 v / 3$
$ρ_{b} = \frac{2 ρ_{w}}{3}$
Now, when body put in another liquid having density $ρ$ then in equilibrium condition
$mg = ρ (\frac{v}{4}) g$
$ρ_{b} vg = ρ (\frac{v}{4}) g$
$ρ = 4 ρ_{b}$
$ρ = 4 \times \frac{2 ρ_{w}}{3} = \frac{8}{3} ρ_{w}$
$ρ = \frac{8}{3} gm / {cm}^{3} [∵ ρ_{w} = 1 gm / {cm}^{3}]$

SCRA-2013

Mechanical Properties of Fluids

143332 A swimmer can swim in still water with speed $v$ and the river flowing with velocity $\frac{v}{2}$ . To cross the river in shortest time, he should swim making angel $θ$ with the upstream. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance?

1

\cos θ

2

\cot θ

3

\sin θ

4

\tan θ

Explanation:

C Given that speed of swimmer in still water $(v_{b}) = v$
Let width be $w$
speed of with respect to river $(v_{r}) = v / 2$ , angle of velocity swimmer with up stream $= θ$

Shortest path

Shortest time
The relative velocity of the swimmer with respect to water is $v_{b} \sin θ$ .
Time taken to cross the river in the shortest distance ( $t)$ $= \frac{w}{v_{b} \sin θ}$
Shortest time taken to cross the river is $t^{'} = \frac{w}{v_{b}}$
Ratio of times taken for shortest time to that of shortest path
$\frac{t^{'}}{t} = \frac{w / v_{b}}{w / \sin θ}$
$\frac{t^{'}}{t} = \sin θ$

AP EAMCET (18.09.2020) Shift-II

Mechanical Properties of Fluids

143333 A $10 {cm}^{3}$ cube floats in water with a height of 4 ${cm}^{3}$ remaining above the surface. The density of the material from which the cube is made is,

1

0.6 g {cm}^{- 3}

2

1.0 g {cm}^{- 3}

3

0.4 g {cm}^{- 3}

4

0.24 g {cm}^{- 3}

5 none of these

Explanation:

A According to the question,
Density of cube material $=$ Volume of cube submerged in water
Density of water
$\frac{d}{1} = \frac{10 - 4}{10}$
$d = \frac{6}{10}$
$d = 0.6 gm / {cm}^{3}$

KERALA CEE 2004

Mechanical Properties of Fluids

143334 A wooden block is floating on water kept in a beaker. $40 %$ of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $g / 2$ . The block will then

1

sink

2 float with

10 %

above the water surface

3 float with

40 %

above the water surface

4 float with

70 %

above the water surface

Explanation:

B
By the Archimedes principle weight of wooden block = world is place
Let $V$ be the Volume of the block
Density of the liquid
Net weight $= m g + \frac{m g}{2} = \frac{3 m g}{2}$
$\frac{3 mg}{2} = V_{i} ρ g$
$mg = \frac{60}{100} V \times ρ \times g$
$mg = \frac{3}{5} V \times ρ \times g$
From equation (i) and (ii), we get -
$\frac{3}{2} (\frac{3}{5} V ρ g) = V_{i} ρ g$
$V_{i} = \frac{90}{100} V$
Flout $= 10 %$

Mechanical Properties of Fluids

143330 A body floats with one-third volume outside when put in water and three-fourth of its volume outside when put in another liquid. The density of the liquid is

1

9 / 4 g {cm}^{- 3}

2

8 / 3 g {cm}^{- 3}

3

3 / 8 g {cm}^{- 3}

4

4 / 9 g {cm}^{- 3}

Explanation:

B According to question -
In equilibrium condition -
Weight of body $=$ Buoyant force
$mg = ρ_{w} g_{d}$
$ρ_{b} v = ρ_{w} 2 v / 3$
$ρ_{b} = \frac{2 ρ_{w}}{3}$
Now, when body put in another liquid having density $ρ$ then in equilibrium condition
$mg = ρ (\frac{v}{4}) g$
$ρ_{b} vg = ρ (\frac{v}{4}) g$
$ρ = 4 ρ_{b}$
$ρ = 4 \times \frac{2 ρ_{w}}{3} = \frac{8}{3} ρ_{w}$
$ρ = \frac{8}{3} gm / {cm}^{3} [∵ ρ_{w} = 1 gm / {cm}^{3}]$

SCRA-2013

Mechanical Properties of Fluids

143332 A swimmer can swim in still water with speed $v$ and the river flowing with velocity $\frac{v}{2}$ . To cross the river in shortest time, he should swim making angel $θ$ with the upstream. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance?

1

\cos θ

2

\cot θ

3

\sin θ

4

\tan θ

Explanation:

C Given that speed of swimmer in still water $(v_{b}) = v$
Let width be $w$
speed of with respect to river $(v_{r}) = v / 2$ , angle of velocity swimmer with up stream $= θ$

Shortest path

Shortest time
The relative velocity of the swimmer with respect to water is $v_{b} \sin θ$ .
Time taken to cross the river in the shortest distance ( $t)$ $= \frac{w}{v_{b} \sin θ}$
Shortest time taken to cross the river is $t^{'} = \frac{w}{v_{b}}$
Ratio of times taken for shortest time to that of shortest path
$\frac{t^{'}}{t} = \frac{w / v_{b}}{w / \sin θ}$
$\frac{t^{'}}{t} = \sin θ$

AP EAMCET (18.09.2020) Shift-II

Mechanical Properties of Fluids

143333 A $10 {cm}^{3}$ cube floats in water with a height of 4 ${cm}^{3}$ remaining above the surface. The density of the material from which the cube is made is,

1

0.6 g {cm}^{- 3}

2

1.0 g {cm}^{- 3}

3

0.4 g {cm}^{- 3}

4

0.24 g {cm}^{- 3}

5 none of these

Explanation:

A According to the question,
Density of cube material $=$ Volume of cube submerged in water
Density of water
$\frac{d}{1} = \frac{10 - 4}{10}$
$d = \frac{6}{10}$
$d = 0.6 gm / {cm}^{3}$

KERALA CEE 2004

Mechanical Properties of Fluids

143334 A wooden block is floating on water kept in a beaker. $40 %$ of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $g / 2$ . The block will then

1

sink

2 float with

10 %

above the water surface

3 float with

40 %

above the water surface

4 float with

70 %

above the water surface

Explanation:

B
By the Archimedes principle weight of wooden block = world is place
Let $V$ be the Volume of the block
Density of the liquid
Net weight $= m g + \frac{m g}{2} = \frac{3 m g}{2}$
$\frac{3 mg}{2} = V_{i} ρ g$
$mg = \frac{60}{100} V \times ρ \times g$
$mg = \frac{3}{5} V \times ρ \times g$
From equation (i) and (ii), we get -
$\frac{3}{2} (\frac{3}{5} V ρ g) = V_{i} ρ g$
$V_{i} = \frac{90}{100} V$
Flout $= 10 %$

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