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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360954 A boat full of scrap iron is floating on water in a lake. If all the iron is dropped into the water,the level of water will

1 Go up

2 Fall down

3 Remain the same

4 Cannot be decided

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360955 A cube of ice floats partly in water partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil $= 0.8$ , specific gravity of ice $= 0.9$ )
supporting img

1

1 : 1

2

5 : 4

3

9 : 10

4

8 : 9

Explanation:

Weight of ice is balanced by buoyant force due to water and kerosene oil.
$∴ m g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$ρ_{ice} V_{ice} g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$\frac{ρ_{ice}}{ρ_{w}} V_{ice} = \frac{ρ_{k}}{ρ_{w}} V_{k} + V_{w}$
$\Rightarrow 0.9 V_{ice} = 0.8 V_{k} + V_{w}$
$(V o l u m e)_{i c e} = (V o l u m e)_{w a t e r} + (V o l u m e)_{k e r o s e n e}$
$∴ 0.9 (V_{w} + V_{k}) = 0.8 V_{k} + V_{w}$
$0.1 V_{w} = 0.1 V_{k} \Rightarrow \frac{V_{w}}{V_{k}} = \frac{1}{1}$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360956 A cubical block of density $10^{2} K g / m^{3}$ is floating in water. Find the fraction of the volume of the cube present outside water.

1 0.2

2 0.1

3 0.5

4 0.9

Explanation:

As the cube is in equilibrium
$F_{b} = F_{g}$
$ρ_{l} V^{'} g = ρ_{s} V g$
$ρ_{l} = 10^{3} K g / m^{3}, ρ_{s} = 10^{2} K g / m^{3}$
$V^{'}$ - volume of the cube present inside water
$V$ - volume of the cube
$\begin{aligned} 10^{3} V^{'} = 10^{2} V \\ \frac{V^{'}}{V} = \frac{1}{10} = 0.1 \end{aligned}$
The fractional volume present outside is 0.9

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360957 A cube of wood of mass $0.5 k g$ and density $800 k g m^{- 3}$ is fastened to the free end of a vertical spring of spring constant $k = 50 {N m}^{- 1}$ , fixed at the bottom. Now, the entire system is completely submerged in water. The elongation or compression of the spring in equilibrium is $\frac{β}{2} c m$ . The value of $β$ is
(Given, $g = 10 {m s}^{- 2}$ )
supporting img

1 7

2 5

3 3

4 2

Explanation:

$ρ_{s} < ρ_{w}$ . So, block tends to move up and thus the spring has elongation.
$B_{F} = m g \times k x$
$V ρ_{w} g = V ρ_{s} g + k x$
$\frac{0.8}{800} \times 1000 \times 10 = (0.5) (10) + 50 x$
supporting img
$x = 0.025 m = 2.5 c m = \frac{5}{2} c m$
$\Rightarrow β = 5$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360958 An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

1 Go down

2 Rise

3 First rise then go down

4 Remain same

Explanation:

According to law of flotation
$m_{i c e} g = V_{1} ρ_{ℓ} g (ρ_{ℓ} =$ density of liquid $)$
Volume of liquid displaced $V_{1} = \frac{m_{i c e}}{ρ_{ℓ}} (1)$
When ice melts volume of water formed
$V_{2} = \frac{m_{i c e}}{ρ_{W a t e r}} (2)$
As $ρ_{ℓ} < ρ_{Water}$ in the given situation $\Rightarrow V_{2} < V_{1}$ i.e., liquid level will fall.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360954 A boat full of scrap iron is floating on water in a lake. If all the iron is dropped into the water,the level of water will

1 Go up

2 Fall down

3 Remain the same

4 Cannot be decided

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360955 A cube of ice floats partly in water partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil $= 0.8$ , specific gravity of ice $= 0.9$ )
supporting img

1

1 : 1

2

5 : 4

3

9 : 10

4

8 : 9

Explanation:

Weight of ice is balanced by buoyant force due to water and kerosene oil.
$∴ m g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$ρ_{ice} V_{ice} g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$\frac{ρ_{ice}}{ρ_{w}} V_{ice} = \frac{ρ_{k}}{ρ_{w}} V_{k} + V_{w}$
$\Rightarrow 0.9 V_{ice} = 0.8 V_{k} + V_{w}$
$(V o l u m e)_{i c e} = (V o l u m e)_{w a t e r} + (V o l u m e)_{k e r o s e n e}$
$∴ 0.9 (V_{w} + V_{k}) = 0.8 V_{k} + V_{w}$
$0.1 V_{w} = 0.1 V_{k} \Rightarrow \frac{V_{w}}{V_{k}} = \frac{1}{1}$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360956 A cubical block of density $10^{2} K g / m^{3}$ is floating in water. Find the fraction of the volume of the cube present outside water.

1 0.2

2 0.1

3 0.5

4 0.9

Explanation:

As the cube is in equilibrium
$F_{b} = F_{g}$
$ρ_{l} V^{'} g = ρ_{s} V g$
$ρ_{l} = 10^{3} K g / m^{3}, ρ_{s} = 10^{2} K g / m^{3}$
$V^{'}$ - volume of the cube present inside water
$V$ - volume of the cube
$\begin{aligned} 10^{3} V^{'} = 10^{2} V \\ \frac{V^{'}}{V} = \frac{1}{10} = 0.1 \end{aligned}$
The fractional volume present outside is 0.9

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360957 A cube of wood of mass $0.5 k g$ and density $800 k g m^{- 3}$ is fastened to the free end of a vertical spring of spring constant $k = 50 {N m}^{- 1}$ , fixed at the bottom. Now, the entire system is completely submerged in water. The elongation or compression of the spring in equilibrium is $\frac{β}{2} c m$ . The value of $β$ is
(Given, $g = 10 {m s}^{- 2}$ )
supporting img

1 7

2 5

3 3

4 2

Explanation:

$ρ_{s} < ρ_{w}$ . So, block tends to move up and thus the spring has elongation.
$B_{F} = m g \times k x$
$V ρ_{w} g = V ρ_{s} g + k x$
$\frac{0.8}{800} \times 1000 \times 10 = (0.5) (10) + 50 x$
supporting img
$x = 0.025 m = 2.5 c m = \frac{5}{2} c m$
$\Rightarrow β = 5$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360958 An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

1 Go down

2 Rise

3 First rise then go down

4 Remain same

Explanation:

According to law of flotation
$m_{i c e} g = V_{1} ρ_{ℓ} g (ρ_{ℓ} =$ density of liquid $)$
Volume of liquid displaced $V_{1} = \frac{m_{i c e}}{ρ_{ℓ}} (1)$
When ice melts volume of water formed
$V_{2} = \frac{m_{i c e}}{ρ_{W a t e r}} (2)$
As $ρ_{ℓ} < ρ_{Water}$ in the given situation $\Rightarrow V_{2} < V_{1}$ i.e., liquid level will fall.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360954 A boat full of scrap iron is floating on water in a lake. If all the iron is dropped into the water,the level of water will

1 Go up

2 Fall down

3 Remain the same

4 Cannot be decided

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360955 A cube of ice floats partly in water partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil $= 0.8$ , specific gravity of ice $= 0.9$ )
supporting img

1

1 : 1

2

5 : 4

3

9 : 10

4

8 : 9

Explanation:

Weight of ice is balanced by buoyant force due to water and kerosene oil.
$∴ m g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$ρ_{ice} V_{ice} g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$\frac{ρ_{ice}}{ρ_{w}} V_{ice} = \frac{ρ_{k}}{ρ_{w}} V_{k} + V_{w}$
$\Rightarrow 0.9 V_{ice} = 0.8 V_{k} + V_{w}$
$(V o l u m e)_{i c e} = (V o l u m e)_{w a t e r} + (V o l u m e)_{k e r o s e n e}$
$∴ 0.9 (V_{w} + V_{k}) = 0.8 V_{k} + V_{w}$
$0.1 V_{w} = 0.1 V_{k} \Rightarrow \frac{V_{w}}{V_{k}} = \frac{1}{1}$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360956 A cubical block of density $10^{2} K g / m^{3}$ is floating in water. Find the fraction of the volume of the cube present outside water.

1 0.2

2 0.1

3 0.5

4 0.9

Explanation:

As the cube is in equilibrium
$F_{b} = F_{g}$
$ρ_{l} V^{'} g = ρ_{s} V g$
$ρ_{l} = 10^{3} K g / m^{3}, ρ_{s} = 10^{2} K g / m^{3}$
$V^{'}$ - volume of the cube present inside water
$V$ - volume of the cube
$\begin{aligned} 10^{3} V^{'} = 10^{2} V \\ \frac{V^{'}}{V} = \frac{1}{10} = 0.1 \end{aligned}$
The fractional volume present outside is 0.9

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360957 A cube of wood of mass $0.5 k g$ and density $800 k g m^{- 3}$ is fastened to the free end of a vertical spring of spring constant $k = 50 {N m}^{- 1}$ , fixed at the bottom. Now, the entire system is completely submerged in water. The elongation or compression of the spring in equilibrium is $\frac{β}{2} c m$ . The value of $β$ is
(Given, $g = 10 {m s}^{- 2}$ )
supporting img

1 7

2 5

3 3

4 2

Explanation:

$ρ_{s} < ρ_{w}$ . So, block tends to move up and thus the spring has elongation.
$B_{F} = m g \times k x$
$V ρ_{w} g = V ρ_{s} g + k x$
$\frac{0.8}{800} \times 1000 \times 10 = (0.5) (10) + 50 x$
supporting img
$x = 0.025 m = 2.5 c m = \frac{5}{2} c m$
$\Rightarrow β = 5$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360958 An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

1 Go down

2 Rise

3 First rise then go down

4 Remain same

Explanation:

According to law of flotation
$m_{i c e} g = V_{1} ρ_{ℓ} g (ρ_{ℓ} =$ density of liquid $)$
Volume of liquid displaced $V_{1} = \frac{m_{i c e}}{ρ_{ℓ}} (1)$
When ice melts volume of water formed
$V_{2} = \frac{m_{i c e}}{ρ_{W a t e r}} (2)$
As $ρ_{ℓ} < ρ_{Water}$ in the given situation $\Rightarrow V_{2} < V_{1}$ i.e., liquid level will fall.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360954 A boat full of scrap iron is floating on water in a lake. If all the iron is dropped into the water,the level of water will

1 Go up

2 Fall down

3 Remain the same

4 Cannot be decided

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360955 A cube of ice floats partly in water partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil $= 0.8$ , specific gravity of ice $= 0.9$ )
supporting img

1

1 : 1

2

5 : 4

3

9 : 10

4

8 : 9

Explanation:

Weight of ice is balanced by buoyant force due to water and kerosene oil.
$∴ m g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$ρ_{ice} V_{ice} g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$\frac{ρ_{ice}}{ρ_{w}} V_{ice} = \frac{ρ_{k}}{ρ_{w}} V_{k} + V_{w}$
$\Rightarrow 0.9 V_{ice} = 0.8 V_{k} + V_{w}$
$(V o l u m e)_{i c e} = (V o l u m e)_{w a t e r} + (V o l u m e)_{k e r o s e n e}$
$∴ 0.9 (V_{w} + V_{k}) = 0.8 V_{k} + V_{w}$
$0.1 V_{w} = 0.1 V_{k} \Rightarrow \frac{V_{w}}{V_{k}} = \frac{1}{1}$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360956 A cubical block of density $10^{2} K g / m^{3}$ is floating in water. Find the fraction of the volume of the cube present outside water.

1 0.2

2 0.1

3 0.5

4 0.9

Explanation:

As the cube is in equilibrium
$F_{b} = F_{g}$
$ρ_{l} V^{'} g = ρ_{s} V g$
$ρ_{l} = 10^{3} K g / m^{3}, ρ_{s} = 10^{2} K g / m^{3}$
$V^{'}$ - volume of the cube present inside water
$V$ - volume of the cube
$\begin{aligned} 10^{3} V^{'} = 10^{2} V \\ \frac{V^{'}}{V} = \frac{1}{10} = 0.1 \end{aligned}$
The fractional volume present outside is 0.9

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360957 A cube of wood of mass $0.5 k g$ and density $800 k g m^{- 3}$ is fastened to the free end of a vertical spring of spring constant $k = 50 {N m}^{- 1}$ , fixed at the bottom. Now, the entire system is completely submerged in water. The elongation or compression of the spring in equilibrium is $\frac{β}{2} c m$ . The value of $β$ is
(Given, $g = 10 {m s}^{- 2}$ )
supporting img

1 7

2 5

3 3

4 2

Explanation:

$ρ_{s} < ρ_{w}$ . So, block tends to move up and thus the spring has elongation.
$B_{F} = m g \times k x$
$V ρ_{w} g = V ρ_{s} g + k x$
$\frac{0.8}{800} \times 1000 \times 10 = (0.5) (10) + 50 x$
supporting img
$x = 0.025 m = 2.5 c m = \frac{5}{2} c m$
$\Rightarrow β = 5$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360958 An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

1 Go down

2 Rise

3 First rise then go down

4 Remain same

Explanation:

According to law of flotation
$m_{i c e} g = V_{1} ρ_{ℓ} g (ρ_{ℓ} =$ density of liquid $)$
Volume of liquid displaced $V_{1} = \frac{m_{i c e}}{ρ_{ℓ}} (1)$
When ice melts volume of water formed
$V_{2} = \frac{m_{i c e}}{ρ_{W a t e r}} (2)$
As $ρ_{ℓ} < ρ_{Water}$ in the given situation $\Rightarrow V_{2} < V_{1}$ i.e., liquid level will fall.

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360954 A boat full of scrap iron is floating on water in a lake. If all the iron is dropped into the water,the level of water will

1 Go up

2 Fall down

3 Remain the same

4 Cannot be decided

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360955 A cube of ice floats partly in water partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil $= 0.8$ , specific gravity of ice $= 0.9$ )
supporting img

1

1 : 1

2

5 : 4

3

9 : 10

4

8 : 9

Explanation:

Weight of ice is balanced by buoyant force due to water and kerosene oil.
$∴ m g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$ρ_{ice} V_{ice} g = ρ_{k} V_{k} g + ρ_{w} V_{w} g$
$\frac{ρ_{ice}}{ρ_{w}} V_{ice} = \frac{ρ_{k}}{ρ_{w}} V_{k} + V_{w}$
$\Rightarrow 0.9 V_{ice} = 0.8 V_{k} + V_{w}$
$(V o l u m e)_{i c e} = (V o l u m e)_{w a t e r} + (V o l u m e)_{k e r o s e n e}$
$∴ 0.9 (V_{w} + V_{k}) = 0.8 V_{k} + V_{w}$
$0.1 V_{w} = 0.1 V_{k} \Rightarrow \frac{V_{w}}{V_{k}} = \frac{1}{1}$

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360956 A cubical block of density $10^{2} K g / m^{3}$ is floating in water. Find the fraction of the volume of the cube present outside water.

1 0.2

2 0.1

3 0.5

4 0.9

Explanation:

As the cube is in equilibrium
$F_{b} = F_{g}$
$ρ_{l} V^{'} g = ρ_{s} V g$
$ρ_{l} = 10^{3} K g / m^{3}, ρ_{s} = 10^{2} K g / m^{3}$
$V^{'}$ - volume of the cube present inside water
$V$ - volume of the cube
$\begin{aligned} 10^{3} V^{'} = 10^{2} V \\ \frac{V^{'}}{V} = \frac{1}{10} = 0.1 \end{aligned}$
The fractional volume present outside is 0.9

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360957 A cube of wood of mass $0.5 k g$ and density $800 k g m^{- 3}$ is fastened to the free end of a vertical spring of spring constant $k = 50 {N m}^{- 1}$ , fixed at the bottom. Now, the entire system is completely submerged in water. The elongation or compression of the spring in equilibrium is $\frac{β}{2} c m$ . The value of $β$ is
(Given, $g = 10 {m s}^{- 2}$ )
supporting img

1 7

2 5

3 3

4 2

Explanation:

$ρ_{s} < ρ_{w}$ . So, block tends to move up and thus the spring has elongation.
$B_{F} = m g \times k x$
$V ρ_{w} g = V ρ_{s} g + k x$
$\frac{0.8}{800} \times 1000 \times 10 = (0.5) (10) + 50 x$
supporting img
$x = 0.025 m = 2.5 c m = \frac{5}{2} c m$
$\Rightarrow β = 5$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360958 An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

1 Go down

2 Rise

3 First rise then go down

4 Remain same

Explanation:

According to law of flotation
$m_{i c e} g = V_{1} ρ_{ℓ} g (ρ_{ℓ} =$ density of liquid $)$
Volume of liquid displaced $V_{1} = \frac{m_{i c e}}{ρ_{ℓ}} (1)$
When ice melts volume of water formed
$V_{2} = \frac{m_{i c e}}{ρ_{W a t e r}} (2)$
As $ρ_{ℓ} < ρ_{Water}$ in the given situation $\Rightarrow V_{2} < V_{1}$ i.e., liquid level will fall.