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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 water drop of radius 1 \(\mu m\) falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is \(1.8 \times 10^{-5} \text N \text s \text m^{-2} \) and its density is negligible as compared to that of water \(10^6 \text g \text m^{-3}\). Terminal velocity of the water drop is:
(Take acceleration due to gravity = 10 ms–2 )
1. \(145.4 \times 10^{-6} \mathrm{~ms}^{-1} \)
2. \( 118.0 \times 10^{-6} \mathrm{~ms}^{-1} \)
3. \( 132.6 \times 10^{-6} \mathrm{~ms}^{-1} \)
4. \( 123.4 \times 10^{-6} \mathrm{~ms}^{-1}\)
(Take acceleration due to gravity = 10 ms–2 )
1. \(145.4 \times 10^{-6} \mathrm{~ms}^{-1} \)
2. \( 118.0 \times 10^{-6} \mathrm{~ms}^{-1} \)
3. \( 132.6 \times 10^{-6} \mathrm{~ms}^{-1} \)
4. \( 123.4 \times 10^{-6} \mathrm{~ms}^{-1}\)
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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