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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361006 Assertion :
For a floating body to be in stable equilibrium, its centre of buoyancy must be located above the centre of gravity.
Reason :
The torque produced by the weight of the body and the upthrust will restore body back to its normal position, after the body is disturbed.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361007 A $10 c m^{3}$ cube floats in water with a volume of $4 c m^{3}$ remaining above the surface. The density of the material from which the cube is made, is

1

0.6 g c m^{- 3}

2

1 g c m^{- 3}

3

0.4 g c m^{- 3}

4

0.24 g c m^{- 3}

Explanation:

According to Archimedes' principle
$ρ_{C} g V_{C} = ρ_{n} g V_{C S}$
where, $ρ_{C}$ and $ρ_{w}$ are densities of cube and water respectively; and $V_{C}$ and $V_{C S}$ are volumes of whole cube and submerged cube,respectively.
$\begin{aligned} \Rightarrow \frac{Volume of cube submerged}{Total volume of cube} \\ = \frac{Density of cube material}{Density of water} \end{aligned}$
$\Rightarrow \frac{10 - 4}{10} = \frac{d}{1}$
$\Rightarrow d = \frac{6}{10} = 0.6 g c m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361008 A solid floats with $1 / 4$ th of its volume above the surface of water, the density of the solid is

1

750 k g m^{- 3}

2

650 k g m^{- 3}

3

560 k g m^{- 3}

4

450 k g m^{- 3}

Explanation:

Let $V$ and $ρ$ be volume and density of solid respectively and $ρ^{'}$ be the density of water,
i.e. $ρ^{'} = 10^{3} k g m^{- 3}$
Weight of body $= V ρ g$
Volume of solid body outside water $= V / 4$
$∴$ Volume of solid body inside water $= V - V / 4 = 3 V / 4$
Weight of water displaced by solid
$= \frac{3 V}{4} \times 10^{3} \times g$
As solid body is floating, then
Weight of body $=$ Weight of water displaced by it
$V ρ g = \frac{3 V}{4} \times 10^{3} g$
$\Rightarrow ρ = \frac{3}{4} \times 1000 = 750 k g m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361009 Density of ice is $ρ$ and that of water is $σ$ . What will be the decrease in volume when a mass $M$ of ice melts?

1

\frac{M}{σ - ρ}

2

\frac{σ - ρ}{M}

3

\frac{σ + ρ}{M}

4

M [\frac{1}{ρ} - \frac{1}{σ}]

Explanation:

Volume of ice $= \frac{M}{ρ}$ , volume of water
$= \frac{M}{σ}$
$∴$ Change in volume $= \frac{M}{ρ} - \frac{M}{σ} = M (\frac{1}{ρ} - \frac{1}{σ})$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361006 Assertion :
For a floating body to be in stable equilibrium, its centre of buoyancy must be located above the centre of gravity.
Reason :
The torque produced by the weight of the body and the upthrust will restore body back to its normal position, after the body is disturbed.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361007 A $10 c m^{3}$ cube floats in water with a volume of $4 c m^{3}$ remaining above the surface. The density of the material from which the cube is made, is

1

0.6 g c m^{- 3}

2

1 g c m^{- 3}

3

0.4 g c m^{- 3}

4

0.24 g c m^{- 3}

Explanation:

According to Archimedes' principle
$ρ_{C} g V_{C} = ρ_{n} g V_{C S}$
where, $ρ_{C}$ and $ρ_{w}$ are densities of cube and water respectively; and $V_{C}$ and $V_{C S}$ are volumes of whole cube and submerged cube,respectively.
$\begin{aligned} \Rightarrow \frac{Volume of cube submerged}{Total volume of cube} \\ = \frac{Density of cube material}{Density of water} \end{aligned}$
$\Rightarrow \frac{10 - 4}{10} = \frac{d}{1}$
$\Rightarrow d = \frac{6}{10} = 0.6 g c m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361008 A solid floats with $1 / 4$ th of its volume above the surface of water, the density of the solid is

1

750 k g m^{- 3}

2

650 k g m^{- 3}

3

560 k g m^{- 3}

4

450 k g m^{- 3}

Explanation:

Let $V$ and $ρ$ be volume and density of solid respectively and $ρ^{'}$ be the density of water,
i.e. $ρ^{'} = 10^{3} k g m^{- 3}$
Weight of body $= V ρ g$
Volume of solid body outside water $= V / 4$
$∴$ Volume of solid body inside water $= V - V / 4 = 3 V / 4$
Weight of water displaced by solid
$= \frac{3 V}{4} \times 10^{3} \times g$
As solid body is floating, then
Weight of body $=$ Weight of water displaced by it
$V ρ g = \frac{3 V}{4} \times 10^{3} g$
$\Rightarrow ρ = \frac{3}{4} \times 1000 = 750 k g m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361009 Density of ice is $ρ$ and that of water is $σ$ . What will be the decrease in volume when a mass $M$ of ice melts?

1

\frac{M}{σ - ρ}

2

\frac{σ - ρ}{M}

3

\frac{σ + ρ}{M}

4

M [\frac{1}{ρ} - \frac{1}{σ}]

Explanation:

Volume of ice $= \frac{M}{ρ}$ , volume of water
$= \frac{M}{σ}$
$∴$ Change in volume $= \frac{M}{ρ} - \frac{M}{σ} = M (\frac{1}{ρ} - \frac{1}{σ})$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361006 Assertion :
For a floating body to be in stable equilibrium, its centre of buoyancy must be located above the centre of gravity.
Reason :
The torque produced by the weight of the body and the upthrust will restore body back to its normal position, after the body is disturbed.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361007 A $10 c m^{3}$ cube floats in water with a volume of $4 c m^{3}$ remaining above the surface. The density of the material from which the cube is made, is

1

0.6 g c m^{- 3}

2

1 g c m^{- 3}

3

0.4 g c m^{- 3}

4

0.24 g c m^{- 3}

Explanation:

According to Archimedes' principle
$ρ_{C} g V_{C} = ρ_{n} g V_{C S}$
where, $ρ_{C}$ and $ρ_{w}$ are densities of cube and water respectively; and $V_{C}$ and $V_{C S}$ are volumes of whole cube and submerged cube,respectively.
$\begin{aligned} \Rightarrow \frac{Volume of cube submerged}{Total volume of cube} \\ = \frac{Density of cube material}{Density of water} \end{aligned}$
$\Rightarrow \frac{10 - 4}{10} = \frac{d}{1}$
$\Rightarrow d = \frac{6}{10} = 0.6 g c m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361008 A solid floats with $1 / 4$ th of its volume above the surface of water, the density of the solid is

1

750 k g m^{- 3}

2

650 k g m^{- 3}

3

560 k g m^{- 3}

4

450 k g m^{- 3}

Explanation:

Let $V$ and $ρ$ be volume and density of solid respectively and $ρ^{'}$ be the density of water,
i.e. $ρ^{'} = 10^{3} k g m^{- 3}$
Weight of body $= V ρ g$
Volume of solid body outside water $= V / 4$
$∴$ Volume of solid body inside water $= V - V / 4 = 3 V / 4$
Weight of water displaced by solid
$= \frac{3 V}{4} \times 10^{3} \times g$
As solid body is floating, then
Weight of body $=$ Weight of water displaced by it
$V ρ g = \frac{3 V}{4} \times 10^{3} g$
$\Rightarrow ρ = \frac{3}{4} \times 1000 = 750 k g m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361009 Density of ice is $ρ$ and that of water is $σ$ . What will be the decrease in volume when a mass $M$ of ice melts?

1

\frac{M}{σ - ρ}

2

\frac{σ - ρ}{M}

3

\frac{σ + ρ}{M}

4

M [\frac{1}{ρ} - \frac{1}{σ}]

Explanation:

Volume of ice $= \frac{M}{ρ}$ , volume of water
$= \frac{M}{σ}$
$∴$ Change in volume $= \frac{M}{ρ} - \frac{M}{σ} = M (\frac{1}{ρ} - \frac{1}{σ})$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361006 Assertion :
For a floating body to be in stable equilibrium, its centre of buoyancy must be located above the centre of gravity.
Reason :
The torque produced by the weight of the body and the upthrust will restore body back to its normal position, after the body is disturbed.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361007 A $10 c m^{3}$ cube floats in water with a volume of $4 c m^{3}$ remaining above the surface. The density of the material from which the cube is made, is

1

0.6 g c m^{- 3}

2

1 g c m^{- 3}

3

0.4 g c m^{- 3}

4

0.24 g c m^{- 3}

Explanation:

According to Archimedes' principle
$ρ_{C} g V_{C} = ρ_{n} g V_{C S}$
where, $ρ_{C}$ and $ρ_{w}$ are densities of cube and water respectively; and $V_{C}$ and $V_{C S}$ are volumes of whole cube and submerged cube,respectively.
$\begin{aligned} \Rightarrow \frac{Volume of cube submerged}{Total volume of cube} \\ = \frac{Density of cube material}{Density of water} \end{aligned}$
$\Rightarrow \frac{10 - 4}{10} = \frac{d}{1}$
$\Rightarrow d = \frac{6}{10} = 0.6 g c m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361008 A solid floats with $1 / 4$ th of its volume above the surface of water, the density of the solid is

1

750 k g m^{- 3}

2

650 k g m^{- 3}

3

560 k g m^{- 3}

4

450 k g m^{- 3}

Explanation:

Let $V$ and $ρ$ be volume and density of solid respectively and $ρ^{'}$ be the density of water,
i.e. $ρ^{'} = 10^{3} k g m^{- 3}$
Weight of body $= V ρ g$
Volume of solid body outside water $= V / 4$
$∴$ Volume of solid body inside water $= V - V / 4 = 3 V / 4$
Weight of water displaced by solid
$= \frac{3 V}{4} \times 10^{3} \times g$
As solid body is floating, then
Weight of body $=$ Weight of water displaced by it
$V ρ g = \frac{3 V}{4} \times 10^{3} g$
$\Rightarrow ρ = \frac{3}{4} \times 1000 = 750 k g m^{- 3}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361009 Density of ice is $ρ$ and that of water is $σ$ . What will be the decrease in volume when a mass $M$ of ice melts?

1

\frac{M}{σ - ρ}

2

\frac{σ - ρ}{M}

3

\frac{σ + ρ}{M}

4

M [\frac{1}{ρ} - \frac{1}{σ}]

Explanation:

Volume of ice $= \frac{M}{ρ}$ , volume of water
$= \frac{M}{σ}$
$∴$ Change in volume $= \frac{M}{ρ} - \frac{M}{σ} = M (\frac{1}{ρ} - \frac{1}{σ})$

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