Q. No.If an air bubble of diameter \(2~\text{mm}\) rises steadily through a liquid of density \(200~\text{kg/m}^{3}\) at a rate of \(0.5 ~\text{cm/s},\) then the coefficient of viscosity of liquid is: (in Poise) (Take \(g = 10~\text{m/s}^{2}\))
1. \(0.88\)
2. \(8.8\)
3. \(88.8\)
4. \(0.088\)
1. \(0.88\)
2. \(8.8\)
3. \(88.8\)
4. \(0.088\)
Subtopic: Stokes' Law |
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A spherical liquid drop of radius \(R\) acquires the terminal velocity \(v_1\) when falls through a gas of viscosity \(\eta .\) Now the drop is broken into \(64\) indentical droplets and each droplets acquires terminal velocity \(v_2\) falling through the same gas. The ratio of terminal velocities \(v_1/v_2 \) is:
1. \(4\)
2. \(0.25\)
3. \(32\)
4. \(16\)
1. \(4\)
2. \(0.25\)
3. \(32\)
4. \(16\)
Subtopic: Stokes' Law |
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A spherical body of radius \(r\) and density \(\sigma\) falls freely through a viscous liquid having density \(\rho\) and viscosity \(\eta\) and attains a terminal velocity \(v_0\). Estimated maximum error in the quantity \(\eta\) is:
(Ignore errors associated with \(\sigma\), \(\rho\) and \(g\), gravitational acceleration)
1. \(2 \dfrac{\Delta r}{r}-\dfrac{\Delta {v}_0}{{v}_0}\)
2. \(\dfrac{2 \Delta r}{r}+\dfrac{\Delta v_0}{v_0}\)
3. \(2\left[\dfrac{\Delta r}{r}+\dfrac{\Delta {v}_0}{{v}_0}\right]\)
4. \(2\left[\dfrac{\Delta r}{r}-\dfrac{\Delta {v}_0}{{v}_0}\right]\)
(Ignore errors associated with \(\sigma\), \(\rho\) and \(g\), gravitational acceleration)
1. \(2 \dfrac{\Delta r}{r}-\dfrac{\Delta {v}_0}{{v}_0}\)
2. \(\dfrac{2 \Delta r}{r}+\dfrac{\Delta v_0}{v_0}\)
3. \(2\left[\dfrac{\Delta r}{r}+\dfrac{\Delta {v}_0}{{v}_0}\right]\)
4. \(2\left[\dfrac{\Delta r}{r}-\dfrac{\Delta {v}_0}{{v}_0}\right]\)
Subtopic: Stokes' Law |
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The terminal velocity of a metallic ball of radius \(6~\text{mm}\) in a viscous fluid is \(20~\text{cm/s}\). The terminal velocity of another ball of same material and having radius \(3~\text{mm}\) in the same fluid will be: (in \(\text{cm/s}\))
1. \(3\)
2. \(4\)
3. \(5\)
4. \(6\)
1. \(3\)
2. \(4\)
3. \(5\)
4. \(6\)
Subtopic: Stokes' Law |
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A small metallic sphere of diameter \(2\) mm and density \(10.5\) g/cm3 is dropped in glycerine having viscosity \(10\) Poise and density \(1.5\) g/cm3 respectively. The terminal velocity attained by the sphere is: (in cm/s)
\(\left(\pi=\dfrac{22}{7} \text { and } g=10 ~\text{m/s}^2\right) \)
1. \(2.0\)
2. \(1.0\)
3. \(3.0\)
4. \(1.5\)
\(\left(\pi=\dfrac{22}{7} \text { and } g=10 ~\text{m/s}^2\right) \)
1. \(2.0\)
2. \(1.0\)
3. \(3.0\)
4. \(1.5\)
Subtopic: Stokes' Law |
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A solid steel ball of diameter \(3.6 ~\text{mm} \) acquired terminal velocity \(2.45 \times 10^{-2}~\text{ m/s} \) while falling under gravity through an oil of density \(925 ~\text{kg m}^{-3} .\) Take density of steel as \(7825 \text{ kg m}^{-3 } \) and \(g\) as \(9.8\text{ m/s}^2 . \) The viscosity of the oil in SI unit is:
1. \(2.38 \)
2. \(1.99 \)
3. \(1.68 \)
4. \(2.18 \)
1. \(2.38 \)
2. \(1.99 \)
3. \(1.68 \)
4. \(2.18 \)
Subtopic: Stokes' Law |
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A spherical water droplet of radius \(1~\mu \text{m}\) falls through air, where the buoyant force can be ignored. The coefficient of viscosity of air is \(1.8 \times 10^{-5} ~\text N ~\text{s} ~\text m^{-2} ,\) and the density of air is negligible compared to that of water (\(10^6~\text{g}~\text{m}^{-3}\)). If \(g=10~\text{m}~\text{s}^{-2},\) what is the terminal velocity of the droplet?
1. \(145.4 \times 10^{-6}~ \text{m s}^{-1} \)
2. \( 118.0 \times 10^{-6} ~ \text{m s}^{-1} \)
3. \( 132.6 \times 10^{-6} ~ \text{m s}^{-1} \)
4. \( 123.4 \times 10^{-6}~ \text{m s}^{-1} \)
1. \(145.4 \times 10^{-6}~ \text{m s}^{-1} \)
2. \( 118.0 \times 10^{-6} ~ \text{m s}^{-1} \)
3. \( 132.6 \times 10^{-6} ~ \text{m s}^{-1} \)
4. \( 123.4 \times 10^{-6}~ \text{m s}^{-1} \)
Subtopic: Stokes' Law |
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The velocity of a small ball of mass \(m\) and density \(d_1,\) when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is \(d_2,\) then the viscous force acting on the ball will be:
| 1. | \( m g\left(1-\dfrac{d_1}{d_2}\right) \) | 2. | \(m g\left(1-\dfrac{d_2}{d_1}\right) \) |
| 3. | \(m g\left(\dfrac{d_1}{d_2}-1\right) \) | 4. | \(m g\left(\dfrac{d_2}{d_1}-1\right)\) |
Subtopic: Stokes' Law |
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The terminal velocity \((v_{T})\) of a spherical raindrop depends on the radius \((r)\) of the raindrop as follows:
| 1. | \(r^{1/2}\) | 2. | \(r\) |
| 3. | \(r^{2}\) | 4. | \(r^{3}\) |
Subtopic: Stokes' Law |
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A small spherical ball of radius \(0.1~\text{mm}\) and density \(10^{4}~\text{kg-m}^{-3}\) falls freely under gravity through a distance of \(h\) before entering a tank of water. If after entering the water, the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water, then the value of \(h\) will be:
(given \(g=10~\text{m/s}^2,\) the viscosity of water \(=1.0\times10^{-5}~\text{N-sm}^{-2}\) )
1. \(15~\text m\)
2. \(25~\text m\)
3. \(20~\text m\)
4. \(10~\text m\)
(given \(g=10~\text{m/s}^2,\) the viscosity of water \(=1.0\times10^{-5}~\text{N-sm}^{-2}\) )
1. \(15~\text m\)
2. \(25~\text m\)
3. \(20~\text m\)
4. \(10~\text m\)
Subtopic: Stokes' Law |
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