361385
The ratio of radii of two spheres are \({1: 2}\) when they are dropped in a viscous liquid. The ratio of their terminal speed is
1 \({1: 1}\)
2 \({2: 1}\)
3 \({1: 2}\)
4 \({1: 4}\)
Explanation:
Terminal velocity is directly proportional to the square of radius of the body. \({\therefore v \propto r^{2}}\) \(\frac{{{v_1}}}{{{v_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\;\;\;\left[ {\because \frac{{{r_2}}}{{{r_1}}} = \frac{1}{2}{\text{ Given}}} \right]\). So correct option is (4)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361386
The viscous force on a sphere moving in a liquid depends on
1 Coefficient of viscosity
2 Volume of the sphere
3 Velocity of the sphere
4 All of the above
Explanation:
Viscosity force on a moving sphere in a liquid is \(f=6 \pi \eta r v\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361387
A small ball of mass \(m\) and density \(\rho\) is dropped in a viscous liquid of density \(\rho_{0}\). After sometime, the ball falls with constant velocity. The viscous force on the ball is
1 \(m g\left(1-\rho \rho_{0}\right)\)
2 \(m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
3 \(m g\left(1+\dfrac{\rho}{\rho_{0}}\right)\)
4 \(m g\left(\dfrac{\rho_{0}}{\rho}-1\right)\)
Explanation:
The small ball is having mass \(m\) and density \(\rho\). Let the mass of liquid displaced be \(m^{\prime}\). Volume of ball is \(V=\dfrac{m}{\rho}\) \(\quad \quad (1)\) Volume of liquid displaced by ball,
\(V^{\prime}=\dfrac{m^{\prime}}{\rho_{0}}\) \(\quad \quad (2)\) Volume of ball = Volume of liquid displaced \(\dfrac{m}{\rho}=\dfrac{m^{\prime}}{\rho_{0}}\) \(m^{\prime}=\dfrac{m}{\rho} \rho_{0}\) \(\quad \quad (3)\) Buoyant force, \(F_{B}=\rho_{0} V^{\prime} g=m^{\prime} g\) Viscous force on the ball is, \(F_{V}=m g-m^{\prime} g\). \(F_{V}=m g-\left(\dfrac{m}{\rho} \rho_{0}\right) g=m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
JEE - 2024
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361388
Two spheres of the same material, but of radii \(R\) and \(3 R\) are allowed to fall vertically downwards through a liquid of density \(\sigma\). The ratio of their terminal velocities is
1 \(1: 3\)
2 \(1: 6\)
3 \(1: 9\)
4 \(1: 1\)
Explanation:
The terminal velocity is \(v \alpha r^{2} \Rightarrow \dfrac{v_{R}}{v_{3 R}}=\dfrac{1}{9}\).
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361389
The terminal speed attained by an aluminium sphere of radius \(1\;mm\) falling through water at \(20^\circ {\text{ }}C\) will be close to (Assume laminar flow, specific gravity of \(A l=2.7\) and \(\eta_{\text {water }}=8 \times 10^{-4}\) )
1 \(4.6\;\,m/s\)
2 \(6.9\,\;m/s\)
3 \(9.2\,\;m/s\)
4 \(2.3\,\;m/s\)
Explanation:
Terminal speed, \(v=\dfrac{2}{9} \dfrac{r^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) and \(\sigma\) are the densities of spherical body and medium respectively, \(r\) is the radius of the spherical body and \(\eta\) is the coefficient of viscosity of the medium. Here, \(r = 1\;mm = {10^{ - 3}}\) Specific gravity of \(A l=2.7\) Density of \(Al,\rho = 2.7 \times {10^3}\;kg/{m^3}\) Density of water, \(\sigma = {10^3}\;kg/{m^3}\) \(\eta = 8 \times {10^{ - 4}},g = 9.8\;m/{s^2}\) \(\therefore \quad v = \frac{{2{{\left( {{{10}^{ - 3}}} \right)}^2} \times \left( {2.7 \times {{10}^3} - {{10}^3}} \right) \times 9.8}}{{9 \times 8 \times {{10}^{ - 4}}}} = 4.6\;m/s\)
361385
The ratio of radii of two spheres are \({1: 2}\) when they are dropped in a viscous liquid. The ratio of their terminal speed is
1 \({1: 1}\)
2 \({2: 1}\)
3 \({1: 2}\)
4 \({1: 4}\)
Explanation:
Terminal velocity is directly proportional to the square of radius of the body. \({\therefore v \propto r^{2}}\) \(\frac{{{v_1}}}{{{v_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\;\;\;\left[ {\because \frac{{{r_2}}}{{{r_1}}} = \frac{1}{2}{\text{ Given}}} \right]\). So correct option is (4)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361386
The viscous force on a sphere moving in a liquid depends on
1 Coefficient of viscosity
2 Volume of the sphere
3 Velocity of the sphere
4 All of the above
Explanation:
Viscosity force on a moving sphere in a liquid is \(f=6 \pi \eta r v\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361387
A small ball of mass \(m\) and density \(\rho\) is dropped in a viscous liquid of density \(\rho_{0}\). After sometime, the ball falls with constant velocity. The viscous force on the ball is
1 \(m g\left(1-\rho \rho_{0}\right)\)
2 \(m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
3 \(m g\left(1+\dfrac{\rho}{\rho_{0}}\right)\)
4 \(m g\left(\dfrac{\rho_{0}}{\rho}-1\right)\)
Explanation:
The small ball is having mass \(m\) and density \(\rho\). Let the mass of liquid displaced be \(m^{\prime}\). Volume of ball is \(V=\dfrac{m}{\rho}\) \(\quad \quad (1)\) Volume of liquid displaced by ball,
\(V^{\prime}=\dfrac{m^{\prime}}{\rho_{0}}\) \(\quad \quad (2)\) Volume of ball = Volume of liquid displaced \(\dfrac{m}{\rho}=\dfrac{m^{\prime}}{\rho_{0}}\) \(m^{\prime}=\dfrac{m}{\rho} \rho_{0}\) \(\quad \quad (3)\) Buoyant force, \(F_{B}=\rho_{0} V^{\prime} g=m^{\prime} g\) Viscous force on the ball is, \(F_{V}=m g-m^{\prime} g\). \(F_{V}=m g-\left(\dfrac{m}{\rho} \rho_{0}\right) g=m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
JEE - 2024
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361388
Two spheres of the same material, but of radii \(R\) and \(3 R\) are allowed to fall vertically downwards through a liquid of density \(\sigma\). The ratio of their terminal velocities is
1 \(1: 3\)
2 \(1: 6\)
3 \(1: 9\)
4 \(1: 1\)
Explanation:
The terminal velocity is \(v \alpha r^{2} \Rightarrow \dfrac{v_{R}}{v_{3 R}}=\dfrac{1}{9}\).
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361389
The terminal speed attained by an aluminium sphere of radius \(1\;mm\) falling through water at \(20^\circ {\text{ }}C\) will be close to (Assume laminar flow, specific gravity of \(A l=2.7\) and \(\eta_{\text {water }}=8 \times 10^{-4}\) )
1 \(4.6\;\,m/s\)
2 \(6.9\,\;m/s\)
3 \(9.2\,\;m/s\)
4 \(2.3\,\;m/s\)
Explanation:
Terminal speed, \(v=\dfrac{2}{9} \dfrac{r^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) and \(\sigma\) are the densities of spherical body and medium respectively, \(r\) is the radius of the spherical body and \(\eta\) is the coefficient of viscosity of the medium. Here, \(r = 1\;mm = {10^{ - 3}}\) Specific gravity of \(A l=2.7\) Density of \(Al,\rho = 2.7 \times {10^3}\;kg/{m^3}\) Density of water, \(\sigma = {10^3}\;kg/{m^3}\) \(\eta = 8 \times {10^{ - 4}},g = 9.8\;m/{s^2}\) \(\therefore \quad v = \frac{{2{{\left( {{{10}^{ - 3}}} \right)}^2} \times \left( {2.7 \times {{10}^3} - {{10}^3}} \right) \times 9.8}}{{9 \times 8 \times {{10}^{ - 4}}}} = 4.6\;m/s\)
361385
The ratio of radii of two spheres are \({1: 2}\) when they are dropped in a viscous liquid. The ratio of their terminal speed is
1 \({1: 1}\)
2 \({2: 1}\)
3 \({1: 2}\)
4 \({1: 4}\)
Explanation:
Terminal velocity is directly proportional to the square of radius of the body. \({\therefore v \propto r^{2}}\) \(\frac{{{v_1}}}{{{v_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\;\;\;\left[ {\because \frac{{{r_2}}}{{{r_1}}} = \frac{1}{2}{\text{ Given}}} \right]\). So correct option is (4)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361386
The viscous force on a sphere moving in a liquid depends on
1 Coefficient of viscosity
2 Volume of the sphere
3 Velocity of the sphere
4 All of the above
Explanation:
Viscosity force on a moving sphere in a liquid is \(f=6 \pi \eta r v\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361387
A small ball of mass \(m\) and density \(\rho\) is dropped in a viscous liquid of density \(\rho_{0}\). After sometime, the ball falls with constant velocity. The viscous force on the ball is
1 \(m g\left(1-\rho \rho_{0}\right)\)
2 \(m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
3 \(m g\left(1+\dfrac{\rho}{\rho_{0}}\right)\)
4 \(m g\left(\dfrac{\rho_{0}}{\rho}-1\right)\)
Explanation:
The small ball is having mass \(m\) and density \(\rho\). Let the mass of liquid displaced be \(m^{\prime}\). Volume of ball is \(V=\dfrac{m}{\rho}\) \(\quad \quad (1)\) Volume of liquid displaced by ball,
\(V^{\prime}=\dfrac{m^{\prime}}{\rho_{0}}\) \(\quad \quad (2)\) Volume of ball = Volume of liquid displaced \(\dfrac{m}{\rho}=\dfrac{m^{\prime}}{\rho_{0}}\) \(m^{\prime}=\dfrac{m}{\rho} \rho_{0}\) \(\quad \quad (3)\) Buoyant force, \(F_{B}=\rho_{0} V^{\prime} g=m^{\prime} g\) Viscous force on the ball is, \(F_{V}=m g-m^{\prime} g\). \(F_{V}=m g-\left(\dfrac{m}{\rho} \rho_{0}\right) g=m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
JEE - 2024
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361388
Two spheres of the same material, but of radii \(R\) and \(3 R\) are allowed to fall vertically downwards through a liquid of density \(\sigma\). The ratio of their terminal velocities is
1 \(1: 3\)
2 \(1: 6\)
3 \(1: 9\)
4 \(1: 1\)
Explanation:
The terminal velocity is \(v \alpha r^{2} \Rightarrow \dfrac{v_{R}}{v_{3 R}}=\dfrac{1}{9}\).
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361389
The terminal speed attained by an aluminium sphere of radius \(1\;mm\) falling through water at \(20^\circ {\text{ }}C\) will be close to (Assume laminar flow, specific gravity of \(A l=2.7\) and \(\eta_{\text {water }}=8 \times 10^{-4}\) )
1 \(4.6\;\,m/s\)
2 \(6.9\,\;m/s\)
3 \(9.2\,\;m/s\)
4 \(2.3\,\;m/s\)
Explanation:
Terminal speed, \(v=\dfrac{2}{9} \dfrac{r^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) and \(\sigma\) are the densities of spherical body and medium respectively, \(r\) is the radius of the spherical body and \(\eta\) is the coefficient of viscosity of the medium. Here, \(r = 1\;mm = {10^{ - 3}}\) Specific gravity of \(A l=2.7\) Density of \(Al,\rho = 2.7 \times {10^3}\;kg/{m^3}\) Density of water, \(\sigma = {10^3}\;kg/{m^3}\) \(\eta = 8 \times {10^{ - 4}},g = 9.8\;m/{s^2}\) \(\therefore \quad v = \frac{{2{{\left( {{{10}^{ - 3}}} \right)}^2} \times \left( {2.7 \times {{10}^3} - {{10}^3}} \right) \times 9.8}}{{9 \times 8 \times {{10}^{ - 4}}}} = 4.6\;m/s\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361385
The ratio of radii of two spheres are \({1: 2}\) when they are dropped in a viscous liquid. The ratio of their terminal speed is
1 \({1: 1}\)
2 \({2: 1}\)
3 \({1: 2}\)
4 \({1: 4}\)
Explanation:
Terminal velocity is directly proportional to the square of radius of the body. \({\therefore v \propto r^{2}}\) \(\frac{{{v_1}}}{{{v_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\;\;\;\left[ {\because \frac{{{r_2}}}{{{r_1}}} = \frac{1}{2}{\text{ Given}}} \right]\). So correct option is (4)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361386
The viscous force on a sphere moving in a liquid depends on
1 Coefficient of viscosity
2 Volume of the sphere
3 Velocity of the sphere
4 All of the above
Explanation:
Viscosity force on a moving sphere in a liquid is \(f=6 \pi \eta r v\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361387
A small ball of mass \(m\) and density \(\rho\) is dropped in a viscous liquid of density \(\rho_{0}\). After sometime, the ball falls with constant velocity. The viscous force on the ball is
1 \(m g\left(1-\rho \rho_{0}\right)\)
2 \(m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
3 \(m g\left(1+\dfrac{\rho}{\rho_{0}}\right)\)
4 \(m g\left(\dfrac{\rho_{0}}{\rho}-1\right)\)
Explanation:
The small ball is having mass \(m\) and density \(\rho\). Let the mass of liquid displaced be \(m^{\prime}\). Volume of ball is \(V=\dfrac{m}{\rho}\) \(\quad \quad (1)\) Volume of liquid displaced by ball,
\(V^{\prime}=\dfrac{m^{\prime}}{\rho_{0}}\) \(\quad \quad (2)\) Volume of ball = Volume of liquid displaced \(\dfrac{m}{\rho}=\dfrac{m^{\prime}}{\rho_{0}}\) \(m^{\prime}=\dfrac{m}{\rho} \rho_{0}\) \(\quad \quad (3)\) Buoyant force, \(F_{B}=\rho_{0} V^{\prime} g=m^{\prime} g\) Viscous force on the ball is, \(F_{V}=m g-m^{\prime} g\). \(F_{V}=m g-\left(\dfrac{m}{\rho} \rho_{0}\right) g=m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
JEE - 2024
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361388
Two spheres of the same material, but of radii \(R\) and \(3 R\) are allowed to fall vertically downwards through a liquid of density \(\sigma\). The ratio of their terminal velocities is
1 \(1: 3\)
2 \(1: 6\)
3 \(1: 9\)
4 \(1: 1\)
Explanation:
The terminal velocity is \(v \alpha r^{2} \Rightarrow \dfrac{v_{R}}{v_{3 R}}=\dfrac{1}{9}\).
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361389
The terminal speed attained by an aluminium sphere of radius \(1\;mm\) falling through water at \(20^\circ {\text{ }}C\) will be close to (Assume laminar flow, specific gravity of \(A l=2.7\) and \(\eta_{\text {water }}=8 \times 10^{-4}\) )
1 \(4.6\;\,m/s\)
2 \(6.9\,\;m/s\)
3 \(9.2\,\;m/s\)
4 \(2.3\,\;m/s\)
Explanation:
Terminal speed, \(v=\dfrac{2}{9} \dfrac{r^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) and \(\sigma\) are the densities of spherical body and medium respectively, \(r\) is the radius of the spherical body and \(\eta\) is the coefficient of viscosity of the medium. Here, \(r = 1\;mm = {10^{ - 3}}\) Specific gravity of \(A l=2.7\) Density of \(Al,\rho = 2.7 \times {10^3}\;kg/{m^3}\) Density of water, \(\sigma = {10^3}\;kg/{m^3}\) \(\eta = 8 \times {10^{ - 4}},g = 9.8\;m/{s^2}\) \(\therefore \quad v = \frac{{2{{\left( {{{10}^{ - 3}}} \right)}^2} \times \left( {2.7 \times {{10}^3} - {{10}^3}} \right) \times 9.8}}{{9 \times 8 \times {{10}^{ - 4}}}} = 4.6\;m/s\)
361385
The ratio of radii of two spheres are \({1: 2}\) when they are dropped in a viscous liquid. The ratio of their terminal speed is
1 \({1: 1}\)
2 \({2: 1}\)
3 \({1: 2}\)
4 \({1: 4}\)
Explanation:
Terminal velocity is directly proportional to the square of radius of the body. \({\therefore v \propto r^{2}}\) \(\frac{{{v_1}}}{{{v_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\;\;\;\left[ {\because \frac{{{r_2}}}{{{r_1}}} = \frac{1}{2}{\text{ Given}}} \right]\). So correct option is (4)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361386
The viscous force on a sphere moving in a liquid depends on
1 Coefficient of viscosity
2 Volume of the sphere
3 Velocity of the sphere
4 All of the above
Explanation:
Viscosity force on a moving sphere in a liquid is \(f=6 \pi \eta r v\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361387
A small ball of mass \(m\) and density \(\rho\) is dropped in a viscous liquid of density \(\rho_{0}\). After sometime, the ball falls with constant velocity. The viscous force on the ball is
1 \(m g\left(1-\rho \rho_{0}\right)\)
2 \(m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
3 \(m g\left(1+\dfrac{\rho}{\rho_{0}}\right)\)
4 \(m g\left(\dfrac{\rho_{0}}{\rho}-1\right)\)
Explanation:
The small ball is having mass \(m\) and density \(\rho\). Let the mass of liquid displaced be \(m^{\prime}\). Volume of ball is \(V=\dfrac{m}{\rho}\) \(\quad \quad (1)\) Volume of liquid displaced by ball,
\(V^{\prime}=\dfrac{m^{\prime}}{\rho_{0}}\) \(\quad \quad (2)\) Volume of ball = Volume of liquid displaced \(\dfrac{m}{\rho}=\dfrac{m^{\prime}}{\rho_{0}}\) \(m^{\prime}=\dfrac{m}{\rho} \rho_{0}\) \(\quad \quad (3)\) Buoyant force, \(F_{B}=\rho_{0} V^{\prime} g=m^{\prime} g\) Viscous force on the ball is, \(F_{V}=m g-m^{\prime} g\). \(F_{V}=m g-\left(\dfrac{m}{\rho} \rho_{0}\right) g=m g\left(1-\dfrac{\rho_{0}}{\rho}\right)\)
JEE - 2024
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361388
Two spheres of the same material, but of radii \(R\) and \(3 R\) are allowed to fall vertically downwards through a liquid of density \(\sigma\). The ratio of their terminal velocities is
1 \(1: 3\)
2 \(1: 6\)
3 \(1: 9\)
4 \(1: 1\)
Explanation:
The terminal velocity is \(v \alpha r^{2} \Rightarrow \dfrac{v_{R}}{v_{3 R}}=\dfrac{1}{9}\).
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361389
The terminal speed attained by an aluminium sphere of radius \(1\;mm\) falling through water at \(20^\circ {\text{ }}C\) will be close to (Assume laminar flow, specific gravity of \(A l=2.7\) and \(\eta_{\text {water }}=8 \times 10^{-4}\) )
1 \(4.6\;\,m/s\)
2 \(6.9\,\;m/s\)
3 \(9.2\,\;m/s\)
4 \(2.3\,\;m/s\)
Explanation:
Terminal speed, \(v=\dfrac{2}{9} \dfrac{r^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) and \(\sigma\) are the densities of spherical body and medium respectively, \(r\) is the radius of the spherical body and \(\eta\) is the coefficient of viscosity of the medium. Here, \(r = 1\;mm = {10^{ - 3}}\) Specific gravity of \(A l=2.7\) Density of \(Al,\rho = 2.7 \times {10^3}\;kg/{m^3}\) Density of water, \(\sigma = {10^3}\;kg/{m^3}\) \(\eta = 8 \times {10^{ - 4}},g = 9.8\;m/{s^2}\) \(\therefore \quad v = \frac{{2{{\left( {{{10}^{ - 3}}} \right)}^2} \times \left( {2.7 \times {{10}^3} - {{10}^3}} \right) \times 9.8}}{{9 \times 8 \times {{10}^{ - 4}}}} = 4.6\;m/s\)
