142677
A piece of solid weights $120 \mathrm{~g}$ in air, $80 \mathrm{~g}$ in water and $60 \mathrm{~g}$ in a liquid. The relative density of the solid and that of the liquid are respectively,
1 3,2
2 $2, \frac{3}{4}$
3 $\frac{3}{2}, 2$
4 4,3
5 $3, \frac{3}{2}$
Explanation:
E Given that, Relative density of solid $=\frac{\text { weight in air }}{\text { weight in air }- \text { weight in water }}$ Relative density of solid $=\frac{120}{120-80}=\frac{120}{40}=3$ Relative density of liquid $=\frac{w t . \text { in air }- \text { wt. in liquid }}{w t . \text { in air }- \text { wt.in water }}$ Relative density of liquid $=\frac{120-60}{120-80}$ $=\frac{60}{40}=\frac{3}{2}$
Kerala CEE 2007
Mechanical Properties of Fluids
142678
A sphere of radius $R$ and density $\rho_{1}$ is dropped in a liquid of density $\sigma$. Its terminal velocity is $v_{1}$. If another sphere of radius $R$ and density $\rho_{2}$ is dropped in the same liquid, its terminal velocity will be
A Given, $\mathrm{R}=$ Radius of sphere, $\rho_{1}=$ Density of first sphere, $\sigma=$ Density of liquid, $\rho_{2}=$ Density of second sphere We know that, terminal velocity $\left(\mathrm{v}_{\mathrm{T}}\right)=\frac{2(\rho-\sigma) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Terminal velocity of first sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{1}=\frac{2\left(\rho_{1}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ And terminal velocity of second sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}==\frac{2\left(\rho_{2}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Dividing equation (ii) by equation (i), we get - $\frac{\left(\mathrm{v}_{\mathrm{T}}\right)_{2}}{\left(\mathrm{v}_{\mathrm{T}}\right)_{1}}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}\left(\mathrm{v}_{\mathrm{T}}\right)_{1}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)} \mathrm{v}_{1}$
UPSEE - 2013
Mechanical Properties of Fluids
142679
The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density $d$ will be
C According to the question - Where, $\mathrm{V}_{0}=$ Total volume of object $\mathrm{V}_{1}=$ Volume of object above liquid $\mathrm{V}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right)=$ Volume of object inside liquid From equilibrium, $\mathrm{mg}=\mathrm{F}_{\text {buoyancy }}$ $\mathrm{V}_{0} \times \mathrm{d}_{0} \times \mathrm{g}=\mathrm{V} \times \mathrm{d} \times \mathrm{g}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right) \times \mathrm{d}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{1} \times \mathrm{d}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{0} \times \mathrm{d}_{0}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\mathrm{V}_{0} \times\left(\mathrm{d}-\mathrm{d}_{0}\right)$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{0}}=\left(\frac{\mathrm{d}-\mathrm{d}_{0}}{\mathrm{~d}}\right)$
UPSEE - 2012
Mechanical Properties of Fluids
142686
A glass flask having mass $390 \mathrm{~g}$ and an interior volume of $500 \mathrm{~cm}^{3}$ floats on water. When it is less than half filled with water. The density of material of the flask is
1 $0.8 \mathrm{~g} / \mathrm{cc}$
2 $2.8 \mathrm{~g} / \mathrm{cc}$
3 $1.8 \mathrm{~g} / \mathrm{cc}$
4 $0.28 \mathrm{~g} / \mathrm{cc}$
Explanation:
B Let, volume of water displaced $=\mathrm{V}$ Mass of water displaced $=$ mass of flask + mass of water $\mathrm{V} \times 1 =390+250 \times 1$ $\mathrm{~V} =390+250$ $\mathrm{~V} =640 \mathrm{~cm}^{3}$ $\because$ Volume of flask material $=640-500=140 \mathrm{~cm}^{3}$ $\therefore \quad \text { Density } =\frac{\text { mass of flask }}{\text { volume of flask material }}$ $=\frac{390}{140}=2.78 \simeq 2.8 \mathrm{~g} / \mathrm{cc}$
Manipal UGET-2010
Mechanical Properties of Fluids
142687
The speed of the water in a river is $v$ near the surface. If the coefficient of viscosity of water is $\eta$ and the depth of the river is $H$, then the shearing stress between the horizontal layers of water is
1 $\eta \frac{\mathrm{H}}{\mathrm{v}}$
2 $\eta \frac{\mathrm{v}}{\mathrm{H}}$
3 $\frac{\mathrm{v}}{\eta \mathrm{H}}$
4 $\eta \mathrm{vH}$
Explanation:
B The given situation is shown below- Force of viscous drag $(F)=\eta A\left(\frac{d v}{d x}\right)$ Where, $\frac{\mathrm{dv}}{\mathrm{dx}}=$ Velocity gradient. In given case, $\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}}{\mathrm{H}}$ So, shearing stress $=\frac{F}{A}=\eta \frac{d v}{d x}=\eta \frac{v}{H}$
142677
A piece of solid weights $120 \mathrm{~g}$ in air, $80 \mathrm{~g}$ in water and $60 \mathrm{~g}$ in a liquid. The relative density of the solid and that of the liquid are respectively,
1 3,2
2 $2, \frac{3}{4}$
3 $\frac{3}{2}, 2$
4 4,3
5 $3, \frac{3}{2}$
Explanation:
E Given that, Relative density of solid $=\frac{\text { weight in air }}{\text { weight in air }- \text { weight in water }}$ Relative density of solid $=\frac{120}{120-80}=\frac{120}{40}=3$ Relative density of liquid $=\frac{w t . \text { in air }- \text { wt. in liquid }}{w t . \text { in air }- \text { wt.in water }}$ Relative density of liquid $=\frac{120-60}{120-80}$ $=\frac{60}{40}=\frac{3}{2}$
Kerala CEE 2007
Mechanical Properties of Fluids
142678
A sphere of radius $R$ and density $\rho_{1}$ is dropped in a liquid of density $\sigma$. Its terminal velocity is $v_{1}$. If another sphere of radius $R$ and density $\rho_{2}$ is dropped in the same liquid, its terminal velocity will be
A Given, $\mathrm{R}=$ Radius of sphere, $\rho_{1}=$ Density of first sphere, $\sigma=$ Density of liquid, $\rho_{2}=$ Density of second sphere We know that, terminal velocity $\left(\mathrm{v}_{\mathrm{T}}\right)=\frac{2(\rho-\sigma) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Terminal velocity of first sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{1}=\frac{2\left(\rho_{1}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ And terminal velocity of second sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}==\frac{2\left(\rho_{2}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Dividing equation (ii) by equation (i), we get - $\frac{\left(\mathrm{v}_{\mathrm{T}}\right)_{2}}{\left(\mathrm{v}_{\mathrm{T}}\right)_{1}}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}\left(\mathrm{v}_{\mathrm{T}}\right)_{1}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)} \mathrm{v}_{1}$
UPSEE - 2013
Mechanical Properties of Fluids
142679
The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density $d$ will be
C According to the question - Where, $\mathrm{V}_{0}=$ Total volume of object $\mathrm{V}_{1}=$ Volume of object above liquid $\mathrm{V}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right)=$ Volume of object inside liquid From equilibrium, $\mathrm{mg}=\mathrm{F}_{\text {buoyancy }}$ $\mathrm{V}_{0} \times \mathrm{d}_{0} \times \mathrm{g}=\mathrm{V} \times \mathrm{d} \times \mathrm{g}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right) \times \mathrm{d}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{1} \times \mathrm{d}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{0} \times \mathrm{d}_{0}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\mathrm{V}_{0} \times\left(\mathrm{d}-\mathrm{d}_{0}\right)$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{0}}=\left(\frac{\mathrm{d}-\mathrm{d}_{0}}{\mathrm{~d}}\right)$
UPSEE - 2012
Mechanical Properties of Fluids
142686
A glass flask having mass $390 \mathrm{~g}$ and an interior volume of $500 \mathrm{~cm}^{3}$ floats on water. When it is less than half filled with water. The density of material of the flask is
1 $0.8 \mathrm{~g} / \mathrm{cc}$
2 $2.8 \mathrm{~g} / \mathrm{cc}$
3 $1.8 \mathrm{~g} / \mathrm{cc}$
4 $0.28 \mathrm{~g} / \mathrm{cc}$
Explanation:
B Let, volume of water displaced $=\mathrm{V}$ Mass of water displaced $=$ mass of flask + mass of water $\mathrm{V} \times 1 =390+250 \times 1$ $\mathrm{~V} =390+250$ $\mathrm{~V} =640 \mathrm{~cm}^{3}$ $\because$ Volume of flask material $=640-500=140 \mathrm{~cm}^{3}$ $\therefore \quad \text { Density } =\frac{\text { mass of flask }}{\text { volume of flask material }}$ $=\frac{390}{140}=2.78 \simeq 2.8 \mathrm{~g} / \mathrm{cc}$
Manipal UGET-2010
Mechanical Properties of Fluids
142687
The speed of the water in a river is $v$ near the surface. If the coefficient of viscosity of water is $\eta$ and the depth of the river is $H$, then the shearing stress between the horizontal layers of water is
1 $\eta \frac{\mathrm{H}}{\mathrm{v}}$
2 $\eta \frac{\mathrm{v}}{\mathrm{H}}$
3 $\frac{\mathrm{v}}{\eta \mathrm{H}}$
4 $\eta \mathrm{vH}$
Explanation:
B The given situation is shown below- Force of viscous drag $(F)=\eta A\left(\frac{d v}{d x}\right)$ Where, $\frac{\mathrm{dv}}{\mathrm{dx}}=$ Velocity gradient. In given case, $\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}}{\mathrm{H}}$ So, shearing stress $=\frac{F}{A}=\eta \frac{d v}{d x}=\eta \frac{v}{H}$
142677
A piece of solid weights $120 \mathrm{~g}$ in air, $80 \mathrm{~g}$ in water and $60 \mathrm{~g}$ in a liquid. The relative density of the solid and that of the liquid are respectively,
1 3,2
2 $2, \frac{3}{4}$
3 $\frac{3}{2}, 2$
4 4,3
5 $3, \frac{3}{2}$
Explanation:
E Given that, Relative density of solid $=\frac{\text { weight in air }}{\text { weight in air }- \text { weight in water }}$ Relative density of solid $=\frac{120}{120-80}=\frac{120}{40}=3$ Relative density of liquid $=\frac{w t . \text { in air }- \text { wt. in liquid }}{w t . \text { in air }- \text { wt.in water }}$ Relative density of liquid $=\frac{120-60}{120-80}$ $=\frac{60}{40}=\frac{3}{2}$
Kerala CEE 2007
Mechanical Properties of Fluids
142678
A sphere of radius $R$ and density $\rho_{1}$ is dropped in a liquid of density $\sigma$. Its terminal velocity is $v_{1}$. If another sphere of radius $R$ and density $\rho_{2}$ is dropped in the same liquid, its terminal velocity will be
A Given, $\mathrm{R}=$ Radius of sphere, $\rho_{1}=$ Density of first sphere, $\sigma=$ Density of liquid, $\rho_{2}=$ Density of second sphere We know that, terminal velocity $\left(\mathrm{v}_{\mathrm{T}}\right)=\frac{2(\rho-\sigma) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Terminal velocity of first sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{1}=\frac{2\left(\rho_{1}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ And terminal velocity of second sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}==\frac{2\left(\rho_{2}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Dividing equation (ii) by equation (i), we get - $\frac{\left(\mathrm{v}_{\mathrm{T}}\right)_{2}}{\left(\mathrm{v}_{\mathrm{T}}\right)_{1}}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}\left(\mathrm{v}_{\mathrm{T}}\right)_{1}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)} \mathrm{v}_{1}$
UPSEE - 2013
Mechanical Properties of Fluids
142679
The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density $d$ will be
C According to the question - Where, $\mathrm{V}_{0}=$ Total volume of object $\mathrm{V}_{1}=$ Volume of object above liquid $\mathrm{V}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right)=$ Volume of object inside liquid From equilibrium, $\mathrm{mg}=\mathrm{F}_{\text {buoyancy }}$ $\mathrm{V}_{0} \times \mathrm{d}_{0} \times \mathrm{g}=\mathrm{V} \times \mathrm{d} \times \mathrm{g}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right) \times \mathrm{d}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{1} \times \mathrm{d}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{0} \times \mathrm{d}_{0}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\mathrm{V}_{0} \times\left(\mathrm{d}-\mathrm{d}_{0}\right)$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{0}}=\left(\frac{\mathrm{d}-\mathrm{d}_{0}}{\mathrm{~d}}\right)$
UPSEE - 2012
Mechanical Properties of Fluids
142686
A glass flask having mass $390 \mathrm{~g}$ and an interior volume of $500 \mathrm{~cm}^{3}$ floats on water. When it is less than half filled with water. The density of material of the flask is
1 $0.8 \mathrm{~g} / \mathrm{cc}$
2 $2.8 \mathrm{~g} / \mathrm{cc}$
3 $1.8 \mathrm{~g} / \mathrm{cc}$
4 $0.28 \mathrm{~g} / \mathrm{cc}$
Explanation:
B Let, volume of water displaced $=\mathrm{V}$ Mass of water displaced $=$ mass of flask + mass of water $\mathrm{V} \times 1 =390+250 \times 1$ $\mathrm{~V} =390+250$ $\mathrm{~V} =640 \mathrm{~cm}^{3}$ $\because$ Volume of flask material $=640-500=140 \mathrm{~cm}^{3}$ $\therefore \quad \text { Density } =\frac{\text { mass of flask }}{\text { volume of flask material }}$ $=\frac{390}{140}=2.78 \simeq 2.8 \mathrm{~g} / \mathrm{cc}$
Manipal UGET-2010
Mechanical Properties of Fluids
142687
The speed of the water in a river is $v$ near the surface. If the coefficient of viscosity of water is $\eta$ and the depth of the river is $H$, then the shearing stress between the horizontal layers of water is
1 $\eta \frac{\mathrm{H}}{\mathrm{v}}$
2 $\eta \frac{\mathrm{v}}{\mathrm{H}}$
3 $\frac{\mathrm{v}}{\eta \mathrm{H}}$
4 $\eta \mathrm{vH}$
Explanation:
B The given situation is shown below- Force of viscous drag $(F)=\eta A\left(\frac{d v}{d x}\right)$ Where, $\frac{\mathrm{dv}}{\mathrm{dx}}=$ Velocity gradient. In given case, $\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}}{\mathrm{H}}$ So, shearing stress $=\frac{F}{A}=\eta \frac{d v}{d x}=\eta \frac{v}{H}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Fluids
142677
A piece of solid weights $120 \mathrm{~g}$ in air, $80 \mathrm{~g}$ in water and $60 \mathrm{~g}$ in a liquid. The relative density of the solid and that of the liquid are respectively,
1 3,2
2 $2, \frac{3}{4}$
3 $\frac{3}{2}, 2$
4 4,3
5 $3, \frac{3}{2}$
Explanation:
E Given that, Relative density of solid $=\frac{\text { weight in air }}{\text { weight in air }- \text { weight in water }}$ Relative density of solid $=\frac{120}{120-80}=\frac{120}{40}=3$ Relative density of liquid $=\frac{w t . \text { in air }- \text { wt. in liquid }}{w t . \text { in air }- \text { wt.in water }}$ Relative density of liquid $=\frac{120-60}{120-80}$ $=\frac{60}{40}=\frac{3}{2}$
Kerala CEE 2007
Mechanical Properties of Fluids
142678
A sphere of radius $R$ and density $\rho_{1}$ is dropped in a liquid of density $\sigma$. Its terminal velocity is $v_{1}$. If another sphere of radius $R$ and density $\rho_{2}$ is dropped in the same liquid, its terminal velocity will be
A Given, $\mathrm{R}=$ Radius of sphere, $\rho_{1}=$ Density of first sphere, $\sigma=$ Density of liquid, $\rho_{2}=$ Density of second sphere We know that, terminal velocity $\left(\mathrm{v}_{\mathrm{T}}\right)=\frac{2(\rho-\sigma) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Terminal velocity of first sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{1}=\frac{2\left(\rho_{1}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ And terminal velocity of second sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}==\frac{2\left(\rho_{2}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Dividing equation (ii) by equation (i), we get - $\frac{\left(\mathrm{v}_{\mathrm{T}}\right)_{2}}{\left(\mathrm{v}_{\mathrm{T}}\right)_{1}}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}\left(\mathrm{v}_{\mathrm{T}}\right)_{1}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)} \mathrm{v}_{1}$
UPSEE - 2013
Mechanical Properties of Fluids
142679
The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density $d$ will be
C According to the question - Where, $\mathrm{V}_{0}=$ Total volume of object $\mathrm{V}_{1}=$ Volume of object above liquid $\mathrm{V}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right)=$ Volume of object inside liquid From equilibrium, $\mathrm{mg}=\mathrm{F}_{\text {buoyancy }}$ $\mathrm{V}_{0} \times \mathrm{d}_{0} \times \mathrm{g}=\mathrm{V} \times \mathrm{d} \times \mathrm{g}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right) \times \mathrm{d}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{1} \times \mathrm{d}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{0} \times \mathrm{d}_{0}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\mathrm{V}_{0} \times\left(\mathrm{d}-\mathrm{d}_{0}\right)$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{0}}=\left(\frac{\mathrm{d}-\mathrm{d}_{0}}{\mathrm{~d}}\right)$
UPSEE - 2012
Mechanical Properties of Fluids
142686
A glass flask having mass $390 \mathrm{~g}$ and an interior volume of $500 \mathrm{~cm}^{3}$ floats on water. When it is less than half filled with water. The density of material of the flask is
1 $0.8 \mathrm{~g} / \mathrm{cc}$
2 $2.8 \mathrm{~g} / \mathrm{cc}$
3 $1.8 \mathrm{~g} / \mathrm{cc}$
4 $0.28 \mathrm{~g} / \mathrm{cc}$
Explanation:
B Let, volume of water displaced $=\mathrm{V}$ Mass of water displaced $=$ mass of flask + mass of water $\mathrm{V} \times 1 =390+250 \times 1$ $\mathrm{~V} =390+250$ $\mathrm{~V} =640 \mathrm{~cm}^{3}$ $\because$ Volume of flask material $=640-500=140 \mathrm{~cm}^{3}$ $\therefore \quad \text { Density } =\frac{\text { mass of flask }}{\text { volume of flask material }}$ $=\frac{390}{140}=2.78 \simeq 2.8 \mathrm{~g} / \mathrm{cc}$
Manipal UGET-2010
Mechanical Properties of Fluids
142687
The speed of the water in a river is $v$ near the surface. If the coefficient of viscosity of water is $\eta$ and the depth of the river is $H$, then the shearing stress between the horizontal layers of water is
1 $\eta \frac{\mathrm{H}}{\mathrm{v}}$
2 $\eta \frac{\mathrm{v}}{\mathrm{H}}$
3 $\frac{\mathrm{v}}{\eta \mathrm{H}}$
4 $\eta \mathrm{vH}$
Explanation:
B The given situation is shown below- Force of viscous drag $(F)=\eta A\left(\frac{d v}{d x}\right)$ Where, $\frac{\mathrm{dv}}{\mathrm{dx}}=$ Velocity gradient. In given case, $\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}}{\mathrm{H}}$ So, shearing stress $=\frac{F}{A}=\eta \frac{d v}{d x}=\eta \frac{v}{H}$
142677
A piece of solid weights $120 \mathrm{~g}$ in air, $80 \mathrm{~g}$ in water and $60 \mathrm{~g}$ in a liquid. The relative density of the solid and that of the liquid are respectively,
1 3,2
2 $2, \frac{3}{4}$
3 $\frac{3}{2}, 2$
4 4,3
5 $3, \frac{3}{2}$
Explanation:
E Given that, Relative density of solid $=\frac{\text { weight in air }}{\text { weight in air }- \text { weight in water }}$ Relative density of solid $=\frac{120}{120-80}=\frac{120}{40}=3$ Relative density of liquid $=\frac{w t . \text { in air }- \text { wt. in liquid }}{w t . \text { in air }- \text { wt.in water }}$ Relative density of liquid $=\frac{120-60}{120-80}$ $=\frac{60}{40}=\frac{3}{2}$
Kerala CEE 2007
Mechanical Properties of Fluids
142678
A sphere of radius $R$ and density $\rho_{1}$ is dropped in a liquid of density $\sigma$. Its terminal velocity is $v_{1}$. If another sphere of radius $R$ and density $\rho_{2}$ is dropped in the same liquid, its terminal velocity will be
A Given, $\mathrm{R}=$ Radius of sphere, $\rho_{1}=$ Density of first sphere, $\sigma=$ Density of liquid, $\rho_{2}=$ Density of second sphere We know that, terminal velocity $\left(\mathrm{v}_{\mathrm{T}}\right)=\frac{2(\rho-\sigma) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Terminal velocity of first sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{1}=\frac{2\left(\rho_{1}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ And terminal velocity of second sphere - $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}==\frac{2\left(\rho_{2}-\sigma\right) \mathrm{R}^{2} \times \mathrm{g}}{9 \eta}$ Dividing equation (ii) by equation (i), we get - $\frac{\left(\mathrm{v}_{\mathrm{T}}\right)_{2}}{\left(\mathrm{v}_{\mathrm{T}}\right)_{1}}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)}\left(\mathrm{v}_{\mathrm{T}}\right)_{1}$ $\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\rho_{2}-\sigma\right)}{\left(\rho_{1}-\sigma\right)} \mathrm{v}_{1}$
UPSEE - 2013
Mechanical Properties of Fluids
142679
The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density $d$ will be
C According to the question - Where, $\mathrm{V}_{0}=$ Total volume of object $\mathrm{V}_{1}=$ Volume of object above liquid $\mathrm{V}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right)=$ Volume of object inside liquid From equilibrium, $\mathrm{mg}=\mathrm{F}_{\text {buoyancy }}$ $\mathrm{V}_{0} \times \mathrm{d}_{0} \times \mathrm{g}=\mathrm{V} \times \mathrm{d} \times \mathrm{g}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0}-\mathrm{V}_{1}\right) \times \mathrm{d}$ $\mathrm{V}_{0} \times \mathrm{d}_{0}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{1} \times \mathrm{d}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\left(\mathrm{V}_{0} \times \mathrm{d}\right)-\left(\mathrm{V}_{0} \times \mathrm{d}_{0}\right)$ $\mathrm{V}_{1} \mathrm{~d}=\mathrm{V}_{0} \times\left(\mathrm{d}-\mathrm{d}_{0}\right)$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{0}}=\left(\frac{\mathrm{d}-\mathrm{d}_{0}}{\mathrm{~d}}\right)$
UPSEE - 2012
Mechanical Properties of Fluids
142686
A glass flask having mass $390 \mathrm{~g}$ and an interior volume of $500 \mathrm{~cm}^{3}$ floats on water. When it is less than half filled with water. The density of material of the flask is
1 $0.8 \mathrm{~g} / \mathrm{cc}$
2 $2.8 \mathrm{~g} / \mathrm{cc}$
3 $1.8 \mathrm{~g} / \mathrm{cc}$
4 $0.28 \mathrm{~g} / \mathrm{cc}$
Explanation:
B Let, volume of water displaced $=\mathrm{V}$ Mass of water displaced $=$ mass of flask + mass of water $\mathrm{V} \times 1 =390+250 \times 1$ $\mathrm{~V} =390+250$ $\mathrm{~V} =640 \mathrm{~cm}^{3}$ $\because$ Volume of flask material $=640-500=140 \mathrm{~cm}^{3}$ $\therefore \quad \text { Density } =\frac{\text { mass of flask }}{\text { volume of flask material }}$ $=\frac{390}{140}=2.78 \simeq 2.8 \mathrm{~g} / \mathrm{cc}$
Manipal UGET-2010
Mechanical Properties of Fluids
142687
The speed of the water in a river is $v$ near the surface. If the coefficient of viscosity of water is $\eta$ and the depth of the river is $H$, then the shearing stress between the horizontal layers of water is
1 $\eta \frac{\mathrm{H}}{\mathrm{v}}$
2 $\eta \frac{\mathrm{v}}{\mathrm{H}}$
3 $\frac{\mathrm{v}}{\eta \mathrm{H}}$
4 $\eta \mathrm{vH}$
Explanation:
B The given situation is shown below- Force of viscous drag $(F)=\eta A\left(\frac{d v}{d x}\right)$ Where, $\frac{\mathrm{dv}}{\mathrm{dx}}=$ Velocity gradient. In given case, $\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}}{\mathrm{H}}$ So, shearing stress $=\frac{F}{A}=\eta \frac{d v}{d x}=\eta \frac{v}{H}$
