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Mechanical Properties of Fluids

143209 An object of mass $10 kg$ is released from rest in a liquid. If the object moves a distance of $2 m$ while sinking in time duration of $1 s$ , then the mass of the liquid displaced by the submerged object is
(Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

5 kg

2

6 kg

3

3 kg

4

4 kg

Explanation:

B Let mass of liquid displaced by submerged object is $x kg$

Given mass of object $= 10 kg$
distance $s = 2 m$
time $t = 1 \sec$
From figure, net force $= 10 g - xg$
$10 \times a = 10 \times 10 - xg$
$a = \frac{100 - 10 x}{10}$
$∵ s = ut + \frac{1}{2} {at}^{2}$
$2 = 0 + \frac{1}{2} (\frac{100 - 10 x}{10}) \times 1 (u = 0 for rest)$
$40 = 100 - 10 x$
$x = 6 kg$

AP EAMCET-05.07.2022

Mechanical Properties of Fluids

143211 A vessel contains oil (density $0.8 g {cm}^{- 3}$ ) over mercury (density $13.6 g {cm}^{- 3}$ ). A homogenous sphere floats with half volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g {cm}^{- 3}$ is:

1 12.8

2 7.2

3 6.4

4 3.3

Explanation:

B Given,
Oil density $(ρ_{Oil}) = 0.8 g / {cm}^{3}$
Mercury density $(ρ_{Hg}) = 13.6 g / {cm}^{3}$

We know,
Weight of sphere $=$ Buoyant force
$v ρ g = \frac{v}{2} \times ρ_{Hg} \times g + \frac{v}{2} \times ρ_{oil} \times g$
$ρ = \frac{ρ_{Hg}}{2} + \frac{ρ_{oil}}{2}$
$ρ = \frac{13.6}{2} + \frac{0.8}{2}$
$ρ = \frac{14.4}{2}$
$ρ = 7.2 g / {cm}^{3}$

AMU-2009

Mechanical Properties of Fluids

143186 Find the apparent weight of a metallic block of density $5 g {cm}^{- 3}$ and dimensions $5 cm \times 5 cm \times 5 cm$ , in water.

1

5 \times 5 \times 5 \times 5 gf

2

4 \times 4 \times 4 \times 4 gf

3

5 \times 4 \times 4 \times 4 gf

4

4 \times 5 \times 5 \times 5 gf

Explanation:

D Apparent weight $=$ weight in air $[1 - \frac{1}{ρ}]$
$= v ρ g [1 - \frac{1}{ρ}]$
$= 5 \times 5 \times 5 \times 5 \times g [1 - \frac{1}{5}]$
weight $= 4 \times 5 \times 5 \times 5 gf$

AP EAMCET (22.09.2020) Shift-II

Mechanical Properties of Fluids

143198 A body of density d, is counterpoised by Mg of weight of density $d_{2}$ in air of density $d$ . Then, the true mass of the body is-

1

M

2

M (1 - \frac{d}{d_{2}})

3

M (1 - \frac{d}{d_{1}})

4

M (\frac{1 - d / d_{2}}{1 - d / d_{1}})

Explanation:

D Let the $M_{1}$ is the mass of the body.
Apparent weight of standard weight in air = Actual weight-Upthrust due to displaced air
$\Rightarrow M_{1} g - \frac{M_{1}}{d_{1}} d g = M g - \frac{M}{d_{2}} d g$
$\begin{aligned} M_{1} g (1 - \frac{d}{d_{1}}) = M g (1 - \frac{d}{d_{2}}) \\ M_{1} = \frac{M (1 - \frac{d}{d_{2}})}{(1 - \frac{d}{d_{1}})} \end{aligned}$

JCECE-2012

Mechanical Properties of Fluids

143209 An object of mass $10 kg$ is released from rest in a liquid. If the object moves a distance of $2 m$ while sinking in time duration of $1 s$ , then the mass of the liquid displaced by the submerged object is
(Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

5 kg

2

6 kg

3

3 kg

4

4 kg

Explanation:

B Let mass of liquid displaced by submerged object is $x kg$

Given mass of object $= 10 kg$
distance $s = 2 m$
time $t = 1 \sec$
From figure, net force $= 10 g - xg$
$10 \times a = 10 \times 10 - xg$
$a = \frac{100 - 10 x}{10}$
$∵ s = ut + \frac{1}{2} {at}^{2}$
$2 = 0 + \frac{1}{2} (\frac{100 - 10 x}{10}) \times 1 (u = 0 for rest)$
$40 = 100 - 10 x$
$x = 6 kg$

AP EAMCET-05.07.2022

Mechanical Properties of Fluids

143211 A vessel contains oil (density $0.8 g {cm}^{- 3}$ ) over mercury (density $13.6 g {cm}^{- 3}$ ). A homogenous sphere floats with half volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g {cm}^{- 3}$ is:

1 12.8

2 7.2

3 6.4

4 3.3

Explanation:

B Given,
Oil density $(ρ_{Oil}) = 0.8 g / {cm}^{3}$
Mercury density $(ρ_{Hg}) = 13.6 g / {cm}^{3}$

We know,
Weight of sphere $=$ Buoyant force
$v ρ g = \frac{v}{2} \times ρ_{Hg} \times g + \frac{v}{2} \times ρ_{oil} \times g$
$ρ = \frac{ρ_{Hg}}{2} + \frac{ρ_{oil}}{2}$
$ρ = \frac{13.6}{2} + \frac{0.8}{2}$
$ρ = \frac{14.4}{2}$
$ρ = 7.2 g / {cm}^{3}$

AMU-2009

Mechanical Properties of Fluids

143186 Find the apparent weight of a metallic block of density $5 g {cm}^{- 3}$ and dimensions $5 cm \times 5 cm \times 5 cm$ , in water.

1

5 \times 5 \times 5 \times 5 gf

2

4 \times 4 \times 4 \times 4 gf

3

5 \times 4 \times 4 \times 4 gf

4

4 \times 5 \times 5 \times 5 gf

Explanation:

D Apparent weight $=$ weight in air $[1 - \frac{1}{ρ}]$
$= v ρ g [1 - \frac{1}{ρ}]$
$= 5 \times 5 \times 5 \times 5 \times g [1 - \frac{1}{5}]$
weight $= 4 \times 5 \times 5 \times 5 gf$

AP EAMCET (22.09.2020) Shift-II

Mechanical Properties of Fluids

143198 A body of density d, is counterpoised by Mg of weight of density $d_{2}$ in air of density $d$ . Then, the true mass of the body is-

1

M

2

M (1 - \frac{d}{d_{2}})

3

M (1 - \frac{d}{d_{1}})

4

M (\frac{1 - d / d_{2}}{1 - d / d_{1}})

Explanation:

D Let the $M_{1}$ is the mass of the body.
Apparent weight of standard weight in air = Actual weight-Upthrust due to displaced air
$\Rightarrow M_{1} g - \frac{M_{1}}{d_{1}} d g = M g - \frac{M}{d_{2}} d g$
$\begin{aligned} M_{1} g (1 - \frac{d}{d_{1}}) = M g (1 - \frac{d}{d_{2}}) \\ M_{1} = \frac{M (1 - \frac{d}{d_{2}})}{(1 - \frac{d}{d_{1}})} \end{aligned}$

JCECE-2012

Mechanical Properties of Fluids

143209 An object of mass $10 kg$ is released from rest in a liquid. If the object moves a distance of $2 m$ while sinking in time duration of $1 s$ , then the mass of the liquid displaced by the submerged object is
(Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

5 kg

2

6 kg

3

3 kg

4

4 kg

Explanation:

B Let mass of liquid displaced by submerged object is $x kg$

Given mass of object $= 10 kg$
distance $s = 2 m$
time $t = 1 \sec$
From figure, net force $= 10 g - xg$
$10 \times a = 10 \times 10 - xg$
$a = \frac{100 - 10 x}{10}$
$∵ s = ut + \frac{1}{2} {at}^{2}$
$2 = 0 + \frac{1}{2} (\frac{100 - 10 x}{10}) \times 1 (u = 0 for rest)$
$40 = 100 - 10 x$
$x = 6 kg$

AP EAMCET-05.07.2022

Mechanical Properties of Fluids

143211 A vessel contains oil (density $0.8 g {cm}^{- 3}$ ) over mercury (density $13.6 g {cm}^{- 3}$ ). A homogenous sphere floats with half volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g {cm}^{- 3}$ is:

1 12.8

2 7.2

3 6.4

4 3.3

Explanation:

B Given,
Oil density $(ρ_{Oil}) = 0.8 g / {cm}^{3}$
Mercury density $(ρ_{Hg}) = 13.6 g / {cm}^{3}$

We know,
Weight of sphere $=$ Buoyant force
$v ρ g = \frac{v}{2} \times ρ_{Hg} \times g + \frac{v}{2} \times ρ_{oil} \times g$
$ρ = \frac{ρ_{Hg}}{2} + \frac{ρ_{oil}}{2}$
$ρ = \frac{13.6}{2} + \frac{0.8}{2}$
$ρ = \frac{14.4}{2}$
$ρ = 7.2 g / {cm}^{3}$

AMU-2009

Mechanical Properties of Fluids

143186 Find the apparent weight of a metallic block of density $5 g {cm}^{- 3}$ and dimensions $5 cm \times 5 cm \times 5 cm$ , in water.

1

5 \times 5 \times 5 \times 5 gf

2

4 \times 4 \times 4 \times 4 gf

3

5 \times 4 \times 4 \times 4 gf

4

4 \times 5 \times 5 \times 5 gf

Explanation:

D Apparent weight $=$ weight in air $[1 - \frac{1}{ρ}]$
$= v ρ g [1 - \frac{1}{ρ}]$
$= 5 \times 5 \times 5 \times 5 \times g [1 - \frac{1}{5}]$
weight $= 4 \times 5 \times 5 \times 5 gf$

AP EAMCET (22.09.2020) Shift-II

Mechanical Properties of Fluids

143198 A body of density d, is counterpoised by Mg of weight of density $d_{2}$ in air of density $d$ . Then, the true mass of the body is-

1

M

2

M (1 - \frac{d}{d_{2}})

3

M (1 - \frac{d}{d_{1}})

4

M (\frac{1 - d / d_{2}}{1 - d / d_{1}})

Explanation:

D Let the $M_{1}$ is the mass of the body.
Apparent weight of standard weight in air = Actual weight-Upthrust due to displaced air
$\Rightarrow M_{1} g - \frac{M_{1}}{d_{1}} d g = M g - \frac{M}{d_{2}} d g$
$\begin{aligned} M_{1} g (1 - \frac{d}{d_{1}}) = M g (1 - \frac{d}{d_{2}}) \\ M_{1} = \frac{M (1 - \frac{d}{d_{2}})}{(1 - \frac{d}{d_{1}})} \end{aligned}$

JCECE-2012

Mechanical Properties of Fluids

143209 An object of mass $10 kg$ is released from rest in a liquid. If the object moves a distance of $2 m$ while sinking in time duration of $1 s$ , then the mass of the liquid displaced by the submerged object is
(Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

5 kg

2

6 kg

3

3 kg

4

4 kg

Explanation:

B Let mass of liquid displaced by submerged object is $x kg$

Given mass of object $= 10 kg$
distance $s = 2 m$
time $t = 1 \sec$
From figure, net force $= 10 g - xg$
$10 \times a = 10 \times 10 - xg$
$a = \frac{100 - 10 x}{10}$
$∵ s = ut + \frac{1}{2} {at}^{2}$
$2 = 0 + \frac{1}{2} (\frac{100 - 10 x}{10}) \times 1 (u = 0 for rest)$
$40 = 100 - 10 x$
$x = 6 kg$

AP EAMCET-05.07.2022

Mechanical Properties of Fluids

143211 A vessel contains oil (density $0.8 g {cm}^{- 3}$ ) over mercury (density $13.6 g {cm}^{- 3}$ ). A homogenous sphere floats with half volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g {cm}^{- 3}$ is:

1 12.8

2 7.2

3 6.4

4 3.3

Explanation:

B Given,
Oil density $(ρ_{Oil}) = 0.8 g / {cm}^{3}$
Mercury density $(ρ_{Hg}) = 13.6 g / {cm}^{3}$

We know,
Weight of sphere $=$ Buoyant force
$v ρ g = \frac{v}{2} \times ρ_{Hg} \times g + \frac{v}{2} \times ρ_{oil} \times g$
$ρ = \frac{ρ_{Hg}}{2} + \frac{ρ_{oil}}{2}$
$ρ = \frac{13.6}{2} + \frac{0.8}{2}$
$ρ = \frac{14.4}{2}$
$ρ = 7.2 g / {cm}^{3}$

AMU-2009

Mechanical Properties of Fluids

143186 Find the apparent weight of a metallic block of density $5 g {cm}^{- 3}$ and dimensions $5 cm \times 5 cm \times 5 cm$ , in water.

1

5 \times 5 \times 5 \times 5 gf

2

4 \times 4 \times 4 \times 4 gf

3

5 \times 4 \times 4 \times 4 gf

4

4 \times 5 \times 5 \times 5 gf

Explanation:

D Apparent weight $=$ weight in air $[1 - \frac{1}{ρ}]$
$= v ρ g [1 - \frac{1}{ρ}]$
$= 5 \times 5 \times 5 \times 5 \times g [1 - \frac{1}{5}]$
weight $= 4 \times 5 \times 5 \times 5 gf$

AP EAMCET (22.09.2020) Shift-II

Mechanical Properties of Fluids

143198 A body of density d, is counterpoised by Mg of weight of density $d_{2}$ in air of density $d$ . Then, the true mass of the body is-

1

M

2

M (1 - \frac{d}{d_{2}})

3

M (1 - \frac{d}{d_{1}})

4

M (\frac{1 - d / d_{2}}{1 - d / d_{1}})

Explanation:

D Let the $M_{1}$ is the mass of the body.
Apparent weight of standard weight in air = Actual weight-Upthrust due to displaced air
$\Rightarrow M_{1} g - \frac{M_{1}}{d_{1}} d g = M g - \frac{M}{d_{2}} d g$
$\begin{aligned} M_{1} g (1 - \frac{d}{d_{1}}) = M g (1 - \frac{d}{d_{2}}) \\ M_{1} = \frac{M (1 - \frac{d}{d_{2}})}{(1 - \frac{d}{d_{1}})} \end{aligned}$

JCECE-2012