143341
A block of material with density $3 \mathrm{gm} / \mathrm{cc}$ is placed on a fluid of density $7 \mathrm{gm} / \mathrm{cc}$. The fraction of volume of the piece of material outside the fluid is
1 0.43
2 0.57
3 0.63
4 0.15
Explanation:
B Given that, Density of material of block $=3 \mathrm{gm} / \mathrm{cc}$ Density of fluid $=7 \mathrm{gm} / \mathrm{cc}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{\mathrm{A} l_{1}}{\mathrm{~A} l}=\frac{l_{1}}{l}$ $\mathrm{~F}_{\mathrm{B}}=\mathrm{mg}$ $\rho_{\mathrm{f}} \mathrm{V}_{1} \mathrm{~g}=\rho_{\mathrm{m}} \mathrm{Vg}$ $\rho_{\mathrm{f}} \times \mathrm{A} \times\left(l-l_{1}\right) \times \mathrm{g}=\rho_{\mathrm{m}} \times \mathrm{A} \times l \times \mathrm{g}$ $\frac{l-l_{1}}{l}=\frac{\rho_{\mathrm{m}}}{\rho_{\mathrm{f}}} \Rightarrow 1-\frac{l_{1}}{l}=\frac{3}{7}$ $\frac{l_{1}}{l}=\frac{4}{7}=0.57$
TS EAMCET 06.08.2021
Mechanical Properties of Fluids
143342
A raft of wood of mass $120 \mathrm{~kg}$ floats in water. The weight that can be put on the raft to make it just sink, should be $\left(d_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}\right)$
1 $80 \mathrm{~kg}$
2 $50 \mathrm{~kg}$
3 $60 \mathrm{~kg}$
4 $30 \mathrm{~kg}$
Explanation:
A Given that, Mass $(\mathrm{m})=120 \mathrm{~kg}, \quad \mathrm{~d}_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}$ $(120+\mathrm{m}) \mathrm{g}$ We know, $\rho=\frac{\mathrm{m}}{\mathrm{v}}$ $\mathrm{v}=\frac{\mathrm{m}}{\rho}=\frac{120}{600}=0.2 \mathrm{~m}^{3}$ $(120+\mathrm{m}) \mathrm{g}=\rho \mathrm{vg} \quad\left[\because \rho=10^{3} \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right]$ $120+\mathrm{m}=10^{3} \times 0.2$ $\mathrm{~m}=200-120$ $\mathrm{~m}=80 \mathrm{~kg}$
JIPMER-2005
Mechanical Properties of Fluids
143343
An ice-cube of density $900 \mathrm{~kg} / \mathrm{m}^{3}$ is floating in water of density $1000 \mathrm{~kg} / \mathrm{m}^{3}$. The percentage to volume of ice-cube outside the water is
1 $20 \%$
2 $35 \%$
3 $10 \%$
4 $25 \%$
Explanation:
C Let the total volume of the iceberg be $\mathrm{V}$ $\mathrm{V}_{\text {sub }}=$ Volume of an iceberg submerged $\rho_{\mathrm{b}}=$ Density of iceberg $=900 \mathrm{~kg} / \mathrm{m}^{3}$ $\rho_{\mathrm{w}}=$ Density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ For the flotation of the body, the weight of the body = Weight of water displaced $\mathrm{V} \rho g=\mathrm{V}_{\text {sub }} \rho g \Rightarrow \mathrm{V}_{\text {sub }}=\left(\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}}\right) \mathrm{V}$ $\mathrm{V}_{\text {out }} =\mathrm{V}-\mathrm{V}_{\text {sub }}$ $=\mathrm{V}-\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \mathrm{V}$ $\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} =\frac{\rho_{\mathrm{w}}-\rho_{\mathrm{h}}}{\rho_{\mathrm{w}}}$ Now by converting it in percentage, we get $=\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} \times 100=\frac{\rho_{\mathrm{W}}-\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \times 100$ $=\frac{1000-900}{1000} \times 100$ $=\frac{100}{1000} \times 100$ $=10 \%$ Therefore, the percentage volume outside the water is $10 \%$
JIPMER-2017
Mechanical Properties of Fluids
143339
Two identical blocks of ice $A$ and $B$ float in water as shown in figure. Then
1 Block A displaced a greater volume of water since the pressure acts on a smaller bottom area
2 Block B displaced a greater volume of water since the pressure is less on its bottom
3 The two blocks displace equal volumes of water since they have the same weight
4 Black A displaces a greater volume of water since its submerge end is lower in the water
Explanation:
C Two identical blocks of ice float in water as shown in figure, then block A displaces equal volume of water as block B displace. Since they have same weight according to Archimedes law.
TS EAMCET 31.07.2022
Mechanical Properties of Fluids
143344
A body of volume $V$ floats on water with $\frac{1}{3}$ of its volume above the surface. The volume of the object above the surface when floating on a liquid of specific gravity 1.5 is-
1 $\frac{3 \mathrm{~V}}{8}$
2 $\frac{4 \mathrm{~V}}{9}$
3 $\frac{5 \mathrm{~V}}{9}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
C Given that, Volume of object $=\mathrm{V}$ Volume above water surface $=\frac{\mathrm{V}}{3}$ Volume of water displaced $=\mathrm{V}-\frac{\mathrm{V}}{3}$ $=\frac{2 \mathrm{~V}}{3}$ Let density of water be $\rho_{\mathrm{w}}$ and density of object be $\rho$. We known that, Weight of object $=$ Weight of water displaced $\rho V g=\rho_{w} \times \frac{2 V}{3} \times g$ $\therefore \quad \rho=\frac{2}{3} \rho_{\mathrm{w}} \quad\left[\because \rho_{\mathrm{w}}=1 \mathrm{gm} / \mathrm{cm}^{3}\right]$ Density of new liquid $=1.5 \rho_{\mathrm{w}}$ Volume above surface in new liquid $=\mathrm{V}^{\prime}$ $\rho \mathrm{gV}=1.5 \rho_{\mathrm{w}} \mathrm{g}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \rho_{\mathrm{w}} \mathrm{V}=1.5 \rho_{\mathrm{w}}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \mathrm{~V}=1.5 \mathrm{~V}-1.5 \mathrm{~V}$ $2 \mathrm{~V}=4.5 \mathrm{~V}-4.5 \mathrm{~V}^{\prime}$ $4.5 \mathrm{~V}^{\prime}=2.5 \mathrm{~V}$ $\mathrm{V}^{\prime}=\frac{2.5}{4.5} \mathrm{~V}$ $\therefore \mathrm{V}^{\prime}=\frac{5}{9} \mathrm{~V}$
143341
A block of material with density $3 \mathrm{gm} / \mathrm{cc}$ is placed on a fluid of density $7 \mathrm{gm} / \mathrm{cc}$. The fraction of volume of the piece of material outside the fluid is
1 0.43
2 0.57
3 0.63
4 0.15
Explanation:
B Given that, Density of material of block $=3 \mathrm{gm} / \mathrm{cc}$ Density of fluid $=7 \mathrm{gm} / \mathrm{cc}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{\mathrm{A} l_{1}}{\mathrm{~A} l}=\frac{l_{1}}{l}$ $\mathrm{~F}_{\mathrm{B}}=\mathrm{mg}$ $\rho_{\mathrm{f}} \mathrm{V}_{1} \mathrm{~g}=\rho_{\mathrm{m}} \mathrm{Vg}$ $\rho_{\mathrm{f}} \times \mathrm{A} \times\left(l-l_{1}\right) \times \mathrm{g}=\rho_{\mathrm{m}} \times \mathrm{A} \times l \times \mathrm{g}$ $\frac{l-l_{1}}{l}=\frac{\rho_{\mathrm{m}}}{\rho_{\mathrm{f}}} \Rightarrow 1-\frac{l_{1}}{l}=\frac{3}{7}$ $\frac{l_{1}}{l}=\frac{4}{7}=0.57$
TS EAMCET 06.08.2021
Mechanical Properties of Fluids
143342
A raft of wood of mass $120 \mathrm{~kg}$ floats in water. The weight that can be put on the raft to make it just sink, should be $\left(d_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}\right)$
1 $80 \mathrm{~kg}$
2 $50 \mathrm{~kg}$
3 $60 \mathrm{~kg}$
4 $30 \mathrm{~kg}$
Explanation:
A Given that, Mass $(\mathrm{m})=120 \mathrm{~kg}, \quad \mathrm{~d}_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}$ $(120+\mathrm{m}) \mathrm{g}$ We know, $\rho=\frac{\mathrm{m}}{\mathrm{v}}$ $\mathrm{v}=\frac{\mathrm{m}}{\rho}=\frac{120}{600}=0.2 \mathrm{~m}^{3}$ $(120+\mathrm{m}) \mathrm{g}=\rho \mathrm{vg} \quad\left[\because \rho=10^{3} \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right]$ $120+\mathrm{m}=10^{3} \times 0.2$ $\mathrm{~m}=200-120$ $\mathrm{~m}=80 \mathrm{~kg}$
JIPMER-2005
Mechanical Properties of Fluids
143343
An ice-cube of density $900 \mathrm{~kg} / \mathrm{m}^{3}$ is floating in water of density $1000 \mathrm{~kg} / \mathrm{m}^{3}$. The percentage to volume of ice-cube outside the water is
1 $20 \%$
2 $35 \%$
3 $10 \%$
4 $25 \%$
Explanation:
C Let the total volume of the iceberg be $\mathrm{V}$ $\mathrm{V}_{\text {sub }}=$ Volume of an iceberg submerged $\rho_{\mathrm{b}}=$ Density of iceberg $=900 \mathrm{~kg} / \mathrm{m}^{3}$ $\rho_{\mathrm{w}}=$ Density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ For the flotation of the body, the weight of the body = Weight of water displaced $\mathrm{V} \rho g=\mathrm{V}_{\text {sub }} \rho g \Rightarrow \mathrm{V}_{\text {sub }}=\left(\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}}\right) \mathrm{V}$ $\mathrm{V}_{\text {out }} =\mathrm{V}-\mathrm{V}_{\text {sub }}$ $=\mathrm{V}-\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \mathrm{V}$ $\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} =\frac{\rho_{\mathrm{w}}-\rho_{\mathrm{h}}}{\rho_{\mathrm{w}}}$ Now by converting it in percentage, we get $=\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} \times 100=\frac{\rho_{\mathrm{W}}-\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \times 100$ $=\frac{1000-900}{1000} \times 100$ $=\frac{100}{1000} \times 100$ $=10 \%$ Therefore, the percentage volume outside the water is $10 \%$
JIPMER-2017
Mechanical Properties of Fluids
143339
Two identical blocks of ice $A$ and $B$ float in water as shown in figure. Then
1 Block A displaced a greater volume of water since the pressure acts on a smaller bottom area
2 Block B displaced a greater volume of water since the pressure is less on its bottom
3 The two blocks displace equal volumes of water since they have the same weight
4 Black A displaces a greater volume of water since its submerge end is lower in the water
Explanation:
C Two identical blocks of ice float in water as shown in figure, then block A displaces equal volume of water as block B displace. Since they have same weight according to Archimedes law.
TS EAMCET 31.07.2022
Mechanical Properties of Fluids
143344
A body of volume $V$ floats on water with $\frac{1}{3}$ of its volume above the surface. The volume of the object above the surface when floating on a liquid of specific gravity 1.5 is-
1 $\frac{3 \mathrm{~V}}{8}$
2 $\frac{4 \mathrm{~V}}{9}$
3 $\frac{5 \mathrm{~V}}{9}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
C Given that, Volume of object $=\mathrm{V}$ Volume above water surface $=\frac{\mathrm{V}}{3}$ Volume of water displaced $=\mathrm{V}-\frac{\mathrm{V}}{3}$ $=\frac{2 \mathrm{~V}}{3}$ Let density of water be $\rho_{\mathrm{w}}$ and density of object be $\rho$. We known that, Weight of object $=$ Weight of water displaced $\rho V g=\rho_{w} \times \frac{2 V}{3} \times g$ $\therefore \quad \rho=\frac{2}{3} \rho_{\mathrm{w}} \quad\left[\because \rho_{\mathrm{w}}=1 \mathrm{gm} / \mathrm{cm}^{3}\right]$ Density of new liquid $=1.5 \rho_{\mathrm{w}}$ Volume above surface in new liquid $=\mathrm{V}^{\prime}$ $\rho \mathrm{gV}=1.5 \rho_{\mathrm{w}} \mathrm{g}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \rho_{\mathrm{w}} \mathrm{V}=1.5 \rho_{\mathrm{w}}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \mathrm{~V}=1.5 \mathrm{~V}-1.5 \mathrm{~V}$ $2 \mathrm{~V}=4.5 \mathrm{~V}-4.5 \mathrm{~V}^{\prime}$ $4.5 \mathrm{~V}^{\prime}=2.5 \mathrm{~V}$ $\mathrm{V}^{\prime}=\frac{2.5}{4.5} \mathrm{~V}$ $\therefore \mathrm{V}^{\prime}=\frac{5}{9} \mathrm{~V}$
143341
A block of material with density $3 \mathrm{gm} / \mathrm{cc}$ is placed on a fluid of density $7 \mathrm{gm} / \mathrm{cc}$. The fraction of volume of the piece of material outside the fluid is
1 0.43
2 0.57
3 0.63
4 0.15
Explanation:
B Given that, Density of material of block $=3 \mathrm{gm} / \mathrm{cc}$ Density of fluid $=7 \mathrm{gm} / \mathrm{cc}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{\mathrm{A} l_{1}}{\mathrm{~A} l}=\frac{l_{1}}{l}$ $\mathrm{~F}_{\mathrm{B}}=\mathrm{mg}$ $\rho_{\mathrm{f}} \mathrm{V}_{1} \mathrm{~g}=\rho_{\mathrm{m}} \mathrm{Vg}$ $\rho_{\mathrm{f}} \times \mathrm{A} \times\left(l-l_{1}\right) \times \mathrm{g}=\rho_{\mathrm{m}} \times \mathrm{A} \times l \times \mathrm{g}$ $\frac{l-l_{1}}{l}=\frac{\rho_{\mathrm{m}}}{\rho_{\mathrm{f}}} \Rightarrow 1-\frac{l_{1}}{l}=\frac{3}{7}$ $\frac{l_{1}}{l}=\frac{4}{7}=0.57$
TS EAMCET 06.08.2021
Mechanical Properties of Fluids
143342
A raft of wood of mass $120 \mathrm{~kg}$ floats in water. The weight that can be put on the raft to make it just sink, should be $\left(d_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}\right)$
1 $80 \mathrm{~kg}$
2 $50 \mathrm{~kg}$
3 $60 \mathrm{~kg}$
4 $30 \mathrm{~kg}$
Explanation:
A Given that, Mass $(\mathrm{m})=120 \mathrm{~kg}, \quad \mathrm{~d}_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}$ $(120+\mathrm{m}) \mathrm{g}$ We know, $\rho=\frac{\mathrm{m}}{\mathrm{v}}$ $\mathrm{v}=\frac{\mathrm{m}}{\rho}=\frac{120}{600}=0.2 \mathrm{~m}^{3}$ $(120+\mathrm{m}) \mathrm{g}=\rho \mathrm{vg} \quad\left[\because \rho=10^{3} \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right]$ $120+\mathrm{m}=10^{3} \times 0.2$ $\mathrm{~m}=200-120$ $\mathrm{~m}=80 \mathrm{~kg}$
JIPMER-2005
Mechanical Properties of Fluids
143343
An ice-cube of density $900 \mathrm{~kg} / \mathrm{m}^{3}$ is floating in water of density $1000 \mathrm{~kg} / \mathrm{m}^{3}$. The percentage to volume of ice-cube outside the water is
1 $20 \%$
2 $35 \%$
3 $10 \%$
4 $25 \%$
Explanation:
C Let the total volume of the iceberg be $\mathrm{V}$ $\mathrm{V}_{\text {sub }}=$ Volume of an iceberg submerged $\rho_{\mathrm{b}}=$ Density of iceberg $=900 \mathrm{~kg} / \mathrm{m}^{3}$ $\rho_{\mathrm{w}}=$ Density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ For the flotation of the body, the weight of the body = Weight of water displaced $\mathrm{V} \rho g=\mathrm{V}_{\text {sub }} \rho g \Rightarrow \mathrm{V}_{\text {sub }}=\left(\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}}\right) \mathrm{V}$ $\mathrm{V}_{\text {out }} =\mathrm{V}-\mathrm{V}_{\text {sub }}$ $=\mathrm{V}-\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \mathrm{V}$ $\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} =\frac{\rho_{\mathrm{w}}-\rho_{\mathrm{h}}}{\rho_{\mathrm{w}}}$ Now by converting it in percentage, we get $=\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} \times 100=\frac{\rho_{\mathrm{W}}-\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \times 100$ $=\frac{1000-900}{1000} \times 100$ $=\frac{100}{1000} \times 100$ $=10 \%$ Therefore, the percentage volume outside the water is $10 \%$
JIPMER-2017
Mechanical Properties of Fluids
143339
Two identical blocks of ice $A$ and $B$ float in water as shown in figure. Then
1 Block A displaced a greater volume of water since the pressure acts on a smaller bottom area
2 Block B displaced a greater volume of water since the pressure is less on its bottom
3 The two blocks displace equal volumes of water since they have the same weight
4 Black A displaces a greater volume of water since its submerge end is lower in the water
Explanation:
C Two identical blocks of ice float in water as shown in figure, then block A displaces equal volume of water as block B displace. Since they have same weight according to Archimedes law.
TS EAMCET 31.07.2022
Mechanical Properties of Fluids
143344
A body of volume $V$ floats on water with $\frac{1}{3}$ of its volume above the surface. The volume of the object above the surface when floating on a liquid of specific gravity 1.5 is-
1 $\frac{3 \mathrm{~V}}{8}$
2 $\frac{4 \mathrm{~V}}{9}$
3 $\frac{5 \mathrm{~V}}{9}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
C Given that, Volume of object $=\mathrm{V}$ Volume above water surface $=\frac{\mathrm{V}}{3}$ Volume of water displaced $=\mathrm{V}-\frac{\mathrm{V}}{3}$ $=\frac{2 \mathrm{~V}}{3}$ Let density of water be $\rho_{\mathrm{w}}$ and density of object be $\rho$. We known that, Weight of object $=$ Weight of water displaced $\rho V g=\rho_{w} \times \frac{2 V}{3} \times g$ $\therefore \quad \rho=\frac{2}{3} \rho_{\mathrm{w}} \quad\left[\because \rho_{\mathrm{w}}=1 \mathrm{gm} / \mathrm{cm}^{3}\right]$ Density of new liquid $=1.5 \rho_{\mathrm{w}}$ Volume above surface in new liquid $=\mathrm{V}^{\prime}$ $\rho \mathrm{gV}=1.5 \rho_{\mathrm{w}} \mathrm{g}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \rho_{\mathrm{w}} \mathrm{V}=1.5 \rho_{\mathrm{w}}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \mathrm{~V}=1.5 \mathrm{~V}-1.5 \mathrm{~V}$ $2 \mathrm{~V}=4.5 \mathrm{~V}-4.5 \mathrm{~V}^{\prime}$ $4.5 \mathrm{~V}^{\prime}=2.5 \mathrm{~V}$ $\mathrm{V}^{\prime}=\frac{2.5}{4.5} \mathrm{~V}$ $\therefore \mathrm{V}^{\prime}=\frac{5}{9} \mathrm{~V}$
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Mechanical Properties of Fluids
143341
A block of material with density $3 \mathrm{gm} / \mathrm{cc}$ is placed on a fluid of density $7 \mathrm{gm} / \mathrm{cc}$. The fraction of volume of the piece of material outside the fluid is
1 0.43
2 0.57
3 0.63
4 0.15
Explanation:
B Given that, Density of material of block $=3 \mathrm{gm} / \mathrm{cc}$ Density of fluid $=7 \mathrm{gm} / \mathrm{cc}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{\mathrm{A} l_{1}}{\mathrm{~A} l}=\frac{l_{1}}{l}$ $\mathrm{~F}_{\mathrm{B}}=\mathrm{mg}$ $\rho_{\mathrm{f}} \mathrm{V}_{1} \mathrm{~g}=\rho_{\mathrm{m}} \mathrm{Vg}$ $\rho_{\mathrm{f}} \times \mathrm{A} \times\left(l-l_{1}\right) \times \mathrm{g}=\rho_{\mathrm{m}} \times \mathrm{A} \times l \times \mathrm{g}$ $\frac{l-l_{1}}{l}=\frac{\rho_{\mathrm{m}}}{\rho_{\mathrm{f}}} \Rightarrow 1-\frac{l_{1}}{l}=\frac{3}{7}$ $\frac{l_{1}}{l}=\frac{4}{7}=0.57$
TS EAMCET 06.08.2021
Mechanical Properties of Fluids
143342
A raft of wood of mass $120 \mathrm{~kg}$ floats in water. The weight that can be put on the raft to make it just sink, should be $\left(d_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}\right)$
1 $80 \mathrm{~kg}$
2 $50 \mathrm{~kg}$
3 $60 \mathrm{~kg}$
4 $30 \mathrm{~kg}$
Explanation:
A Given that, Mass $(\mathrm{m})=120 \mathrm{~kg}, \quad \mathrm{~d}_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}$ $(120+\mathrm{m}) \mathrm{g}$ We know, $\rho=\frac{\mathrm{m}}{\mathrm{v}}$ $\mathrm{v}=\frac{\mathrm{m}}{\rho}=\frac{120}{600}=0.2 \mathrm{~m}^{3}$ $(120+\mathrm{m}) \mathrm{g}=\rho \mathrm{vg} \quad\left[\because \rho=10^{3} \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right]$ $120+\mathrm{m}=10^{3} \times 0.2$ $\mathrm{~m}=200-120$ $\mathrm{~m}=80 \mathrm{~kg}$
JIPMER-2005
Mechanical Properties of Fluids
143343
An ice-cube of density $900 \mathrm{~kg} / \mathrm{m}^{3}$ is floating in water of density $1000 \mathrm{~kg} / \mathrm{m}^{3}$. The percentage to volume of ice-cube outside the water is
1 $20 \%$
2 $35 \%$
3 $10 \%$
4 $25 \%$
Explanation:
C Let the total volume of the iceberg be $\mathrm{V}$ $\mathrm{V}_{\text {sub }}=$ Volume of an iceberg submerged $\rho_{\mathrm{b}}=$ Density of iceberg $=900 \mathrm{~kg} / \mathrm{m}^{3}$ $\rho_{\mathrm{w}}=$ Density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ For the flotation of the body, the weight of the body = Weight of water displaced $\mathrm{V} \rho g=\mathrm{V}_{\text {sub }} \rho g \Rightarrow \mathrm{V}_{\text {sub }}=\left(\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}}\right) \mathrm{V}$ $\mathrm{V}_{\text {out }} =\mathrm{V}-\mathrm{V}_{\text {sub }}$ $=\mathrm{V}-\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \mathrm{V}$ $\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} =\frac{\rho_{\mathrm{w}}-\rho_{\mathrm{h}}}{\rho_{\mathrm{w}}}$ Now by converting it in percentage, we get $=\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} \times 100=\frac{\rho_{\mathrm{W}}-\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \times 100$ $=\frac{1000-900}{1000} \times 100$ $=\frac{100}{1000} \times 100$ $=10 \%$ Therefore, the percentage volume outside the water is $10 \%$
JIPMER-2017
Mechanical Properties of Fluids
143339
Two identical blocks of ice $A$ and $B$ float in water as shown in figure. Then
1 Block A displaced a greater volume of water since the pressure acts on a smaller bottom area
2 Block B displaced a greater volume of water since the pressure is less on its bottom
3 The two blocks displace equal volumes of water since they have the same weight
4 Black A displaces a greater volume of water since its submerge end is lower in the water
Explanation:
C Two identical blocks of ice float in water as shown in figure, then block A displaces equal volume of water as block B displace. Since they have same weight according to Archimedes law.
TS EAMCET 31.07.2022
Mechanical Properties of Fluids
143344
A body of volume $V$ floats on water with $\frac{1}{3}$ of its volume above the surface. The volume of the object above the surface when floating on a liquid of specific gravity 1.5 is-
1 $\frac{3 \mathrm{~V}}{8}$
2 $\frac{4 \mathrm{~V}}{9}$
3 $\frac{5 \mathrm{~V}}{9}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
C Given that, Volume of object $=\mathrm{V}$ Volume above water surface $=\frac{\mathrm{V}}{3}$ Volume of water displaced $=\mathrm{V}-\frac{\mathrm{V}}{3}$ $=\frac{2 \mathrm{~V}}{3}$ Let density of water be $\rho_{\mathrm{w}}$ and density of object be $\rho$. We known that, Weight of object $=$ Weight of water displaced $\rho V g=\rho_{w} \times \frac{2 V}{3} \times g$ $\therefore \quad \rho=\frac{2}{3} \rho_{\mathrm{w}} \quad\left[\because \rho_{\mathrm{w}}=1 \mathrm{gm} / \mathrm{cm}^{3}\right]$ Density of new liquid $=1.5 \rho_{\mathrm{w}}$ Volume above surface in new liquid $=\mathrm{V}^{\prime}$ $\rho \mathrm{gV}=1.5 \rho_{\mathrm{w}} \mathrm{g}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \rho_{\mathrm{w}} \mathrm{V}=1.5 \rho_{\mathrm{w}}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \mathrm{~V}=1.5 \mathrm{~V}-1.5 \mathrm{~V}$ $2 \mathrm{~V}=4.5 \mathrm{~V}-4.5 \mathrm{~V}^{\prime}$ $4.5 \mathrm{~V}^{\prime}=2.5 \mathrm{~V}$ $\mathrm{V}^{\prime}=\frac{2.5}{4.5} \mathrm{~V}$ $\therefore \mathrm{V}^{\prime}=\frac{5}{9} \mathrm{~V}$
143341
A block of material with density $3 \mathrm{gm} / \mathrm{cc}$ is placed on a fluid of density $7 \mathrm{gm} / \mathrm{cc}$. The fraction of volume of the piece of material outside the fluid is
1 0.43
2 0.57
3 0.63
4 0.15
Explanation:
B Given that, Density of material of block $=3 \mathrm{gm} / \mathrm{cc}$ Density of fluid $=7 \mathrm{gm} / \mathrm{cc}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{\mathrm{A} l_{1}}{\mathrm{~A} l}=\frac{l_{1}}{l}$ $\mathrm{~F}_{\mathrm{B}}=\mathrm{mg}$ $\rho_{\mathrm{f}} \mathrm{V}_{1} \mathrm{~g}=\rho_{\mathrm{m}} \mathrm{Vg}$ $\rho_{\mathrm{f}} \times \mathrm{A} \times\left(l-l_{1}\right) \times \mathrm{g}=\rho_{\mathrm{m}} \times \mathrm{A} \times l \times \mathrm{g}$ $\frac{l-l_{1}}{l}=\frac{\rho_{\mathrm{m}}}{\rho_{\mathrm{f}}} \Rightarrow 1-\frac{l_{1}}{l}=\frac{3}{7}$ $\frac{l_{1}}{l}=\frac{4}{7}=0.57$
TS EAMCET 06.08.2021
Mechanical Properties of Fluids
143342
A raft of wood of mass $120 \mathrm{~kg}$ floats in water. The weight that can be put on the raft to make it just sink, should be $\left(d_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}\right)$
1 $80 \mathrm{~kg}$
2 $50 \mathrm{~kg}$
3 $60 \mathrm{~kg}$
4 $30 \mathrm{~kg}$
Explanation:
A Given that, Mass $(\mathrm{m})=120 \mathrm{~kg}, \quad \mathrm{~d}_{\text {raft }}=600 \mathrm{~kg} / \mathrm{m}^{3}$ $(120+\mathrm{m}) \mathrm{g}$ We know, $\rho=\frac{\mathrm{m}}{\mathrm{v}}$ $\mathrm{v}=\frac{\mathrm{m}}{\rho}=\frac{120}{600}=0.2 \mathrm{~m}^{3}$ $(120+\mathrm{m}) \mathrm{g}=\rho \mathrm{vg} \quad\left[\because \rho=10^{3} \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right]$ $120+\mathrm{m}=10^{3} \times 0.2$ $\mathrm{~m}=200-120$ $\mathrm{~m}=80 \mathrm{~kg}$
JIPMER-2005
Mechanical Properties of Fluids
143343
An ice-cube of density $900 \mathrm{~kg} / \mathrm{m}^{3}$ is floating in water of density $1000 \mathrm{~kg} / \mathrm{m}^{3}$. The percentage to volume of ice-cube outside the water is
1 $20 \%$
2 $35 \%$
3 $10 \%$
4 $25 \%$
Explanation:
C Let the total volume of the iceberg be $\mathrm{V}$ $\mathrm{V}_{\text {sub }}=$ Volume of an iceberg submerged $\rho_{\mathrm{b}}=$ Density of iceberg $=900 \mathrm{~kg} / \mathrm{m}^{3}$ $\rho_{\mathrm{w}}=$ Density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ For the flotation of the body, the weight of the body = Weight of water displaced $\mathrm{V} \rho g=\mathrm{V}_{\text {sub }} \rho g \Rightarrow \mathrm{V}_{\text {sub }}=\left(\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}}\right) \mathrm{V}$ $\mathrm{V}_{\text {out }} =\mathrm{V}-\mathrm{V}_{\text {sub }}$ $=\mathrm{V}-\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \mathrm{V}$ $\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} =\frac{\rho_{\mathrm{w}}-\rho_{\mathrm{h}}}{\rho_{\mathrm{w}}}$ Now by converting it in percentage, we get $=\frac{\mathrm{V}_{\text {out }}}{\mathrm{V}} \times 100=\frac{\rho_{\mathrm{W}}-\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}} \times 100$ $=\frac{1000-900}{1000} \times 100$ $=\frac{100}{1000} \times 100$ $=10 \%$ Therefore, the percentage volume outside the water is $10 \%$
JIPMER-2017
Mechanical Properties of Fluids
143339
Two identical blocks of ice $A$ and $B$ float in water as shown in figure. Then
1 Block A displaced a greater volume of water since the pressure acts on a smaller bottom area
2 Block B displaced a greater volume of water since the pressure is less on its bottom
3 The two blocks displace equal volumes of water since they have the same weight
4 Black A displaces a greater volume of water since its submerge end is lower in the water
Explanation:
C Two identical blocks of ice float in water as shown in figure, then block A displaces equal volume of water as block B displace. Since they have same weight according to Archimedes law.
TS EAMCET 31.07.2022
Mechanical Properties of Fluids
143344
A body of volume $V$ floats on water with $\frac{1}{3}$ of its volume above the surface. The volume of the object above the surface when floating on a liquid of specific gravity 1.5 is-
1 $\frac{3 \mathrm{~V}}{8}$
2 $\frac{4 \mathrm{~V}}{9}$
3 $\frac{5 \mathrm{~V}}{9}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
C Given that, Volume of object $=\mathrm{V}$ Volume above water surface $=\frac{\mathrm{V}}{3}$ Volume of water displaced $=\mathrm{V}-\frac{\mathrm{V}}{3}$ $=\frac{2 \mathrm{~V}}{3}$ Let density of water be $\rho_{\mathrm{w}}$ and density of object be $\rho$. We known that, Weight of object $=$ Weight of water displaced $\rho V g=\rho_{w} \times \frac{2 V}{3} \times g$ $\therefore \quad \rho=\frac{2}{3} \rho_{\mathrm{w}} \quad\left[\because \rho_{\mathrm{w}}=1 \mathrm{gm} / \mathrm{cm}^{3}\right]$ Density of new liquid $=1.5 \rho_{\mathrm{w}}$ Volume above surface in new liquid $=\mathrm{V}^{\prime}$ $\rho \mathrm{gV}=1.5 \rho_{\mathrm{w}} \mathrm{g}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \rho_{\mathrm{w}} \mathrm{V}=1.5 \rho_{\mathrm{w}}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)$ $\frac{2}{3} \mathrm{~V}=1.5 \mathrm{~V}-1.5 \mathrm{~V}$ $2 \mathrm{~V}=4.5 \mathrm{~V}-4.5 \mathrm{~V}^{\prime}$ $4.5 \mathrm{~V}^{\prime}=2.5 \mathrm{~V}$ $\mathrm{V}^{\prime}=\frac{2.5}{4.5} \mathrm{~V}$ $\therefore \mathrm{V}^{\prime}=\frac{5}{9} \mathrm{~V}$
