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Q. No.In an experiment to verify Stokes's law, a small spherical ball of radius \(r\) and density \(\rho\) falls under gravity through a distance \(h\) in air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water surface, then the value of \(h\) is proportional to: (Ignore viscosity of air)
1. \(r\)
2. \(r^4\)
3. \(r^3\)
4. \(r^2\)
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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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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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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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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The diameter of an air bubble which was initially \(2\) mm, rises steadily through a solution of density \(1750\) kgm–3 at the rate of \(0.35\) cms–1. The coefficient of viscosity of the solution is (in the nearest integer):
(The density of air is negligible).
1. \(15\) poise
2. \(9\) poise
3. \(18\) poise
4. \(11\) poise
(The density of air is negligible).
1. \(15\) poise
2. \(9\) poise
3. \(18\) poise
4. \(11\) poise
Subtopic: Stokes' Law |
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A raindrop with radius \(R=0.2 ~\text {mm} \) falls from a cloud at a height \(h=2000 ~\text m \) above the ground. Assume that the drop is spherical through its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is:
(density of water \(f_w = 1000~ \text{kg m} ^{-3}\) and density of air \(f_a=1.2~\text{kg m}^{-3},\) \(g =10 ~\text {m/s}^2,\) coefficient of viscosity of air \(=1.8\times10^{-5} ~\text{Nsm} ^{-2}\))
1. \(4.94 ~\text {m s}^{-1}\)
2. \(14.4 ~\text {m s}^{-1}\)
3. \(43.56 ~\text {m s}^{-1}\)
4. \(250.6~\text {m s}^{-1}\)
(density of water \(f_w = 1000~ \text{kg m} ^{-3}\) and density of air \(f_a=1.2~\text{kg m}^{-3},\) \(g =10 ~\text {m/s}^2,\) coefficient of viscosity of air \(=1.8\times10^{-5} ~\text{Nsm} ^{-2}\))
1. \(4.94 ~\text {m s}^{-1}\)
2. \(14.4 ~\text {m s}^{-1}\)
3. \(43.56 ~\text {m s}^{-1}\)
4. \(250.6~\text {m s}^{-1}\)
Subtopic: Stokes' Law |
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In Millikan's oil drop experiment, the viscous force acting on an uncharged drop of radius \(2.0 \times 10^{-5} ~\text {m} \text { and density } 1.2 \times 10^3~ \text {kg m}^{-3} \) is:
\(\text{(Take viscosity of liquid }=1.8 \times 10^{-5}~\text{N s m}^{-2} \text {. }\)Neglect buoyancy due to air)
1. \(3.8 \times 10^{-11}~\text N\)
2. \(3.9 \times 10^{-10}~\text N\)
3. \(1.8 \times 10^{-10}~\text N\)
4. \(5.8 \times 10^{-10}~\text N\)
\(\text{(Take viscosity of liquid }=1.8 \times 10^{-5}~\text{N s m}^{-2} \text {. }\)Neglect buoyancy due to air)
1. \(3.8 \times 10^{-11}~\text N\)
2. \(3.9 \times 10^{-10}~\text N\)
3. \(1.8 \times 10^{-10}~\text N\)
4. \(5.8 \times 10^{-10}~\text N\)
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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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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