Q. No.The dimensions of the quantity \(\Large\frac{\eta}{\rho}\) where \(\eta\) is the viscosity and \(\rho\) is the density are:
| 1. | \(\left [ \dfrac{L}{T} \right ]\) | 2. | \(\left [ \dfrac{L^2}{T} \right ]\) |
| 3. | \(\left [ \dfrac{L}{T^2} \right ]\) | 4. | \(\left [ \dfrac{L^3}{T^2} \right ]\) |
Subtopic: Stokes' Law |
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The viscous force on a ball of radius \(1 \mathrm{~mm}\) moving through a liquid of viscosity \(0.2 ~\mathrm{Ns} / \mathrm{m}^2\) at a speed of \(0.07 \mathrm{~m} / \mathrm{s}\) is:
1. \(2.63 \times 10^{-4} \mathrm{~N}\)
2. \(0.263 \mathrm{~N}\)
3. \(3.48 \times 10^{-4} \mathrm{~N}\)
4. \(0.348 \mathrm{~N}\)
1. \(2.63 \times 10^{-4} \mathrm{~N}\)
2. \(0.263 \mathrm{~N}\)
3. \(3.48 \times 10^{-4} \mathrm{~N}\)
4. \(0.348 \mathrm{~N}\)
Subtopic: Stokes' Law |
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Two balls of the same material whose radii are in the ratio of \(1: 2\) fall in a liquid. The ratio of their terminal velocities is given by:
1. \(2: 3\)
2. \(1: 4\)
3. \(3: 2\)
4. \(5: 9\)
1. \(2: 3\)
2. \(1: 4\)
3. \(3: 2\)
4. \(5: 9\)
Subtopic: Stokes' Law |
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Choose the correct statement among the following:
| 1. | The terminal velocity is directly proportional to the square of the radius of the body. |
| 2. | The terminal velocity is inversely proportional to viscosity of the medium. |
| 3. | The terminal velocity is proportional to the difference of densities of body and fluid. |
| 4. | All of these |
Subtopic: Stokes' Law |
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A graph is plotted between the square of radius \(\text{a}\) of a body and the terminal velocity \(v_T\) acquired by the body while falling through a viscous medium. The slope of the graph is given by where terminal velocity is plotted along y-axis and the square of the radius of body along \(x\)-axis: (where symbols have their usual meaning)
1. \(\dfrac{2}{9 \eta}(\sigma-\rho) g\)
2, \(\dfrac{3}{8}(\sigma-\rho) g\)
3. \(\dfrac{9}{2 \eta}(\sigma-\rho) g\)
4. \(\dfrac{2}{9 \eta}(\rho+\sigma) g\)
1. \(\dfrac{2}{9 \eta}(\sigma-\rho) g\)
2, \(\dfrac{3}{8}(\sigma-\rho) g\)
3. \(\dfrac{9}{2 \eta}(\sigma-\rho) g\)
4. \(\dfrac{2}{9 \eta}(\rho+\sigma) g\)
Subtopic: Stokes' Law |
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A liquid having a coefficient of viscosity \(\eta\) is filled in a container of cross-sectional area \(A. \) If viscous drag between two adjacent layers is \(F_0, \) then the velocity gradient is:
1. \(\dfrac{A \eta}{F_0}\)
2. \(\dfrac{A^2 \eta}{F_0}\)
3. \(\dfrac{F_0}{\eta A}\)
4. \(\dfrac{F_0^2}{\eta A^2}\)
1. \(\dfrac{A \eta}{F_0}\)
2. \(\dfrac{A^2 \eta}{F_0}\)
3. \(\dfrac{F_0}{\eta A}\)
4. \(\dfrac{F_0^2}{\eta A^2}\)
Subtopic: Stokes' Law |
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An air bubble of radius, \(r\) rises steadily through a liquid of density, \(\mathit{\rho}\) at the rate of \(v\). Neglecting the density of air, the coefficient of viscosity of the liquid is:
| 1. | \(\dfrac{2}{9}\dfrac{{r}^{2}{\rho}{g}}{v}\) | 2. | \(\dfrac{1}{3}\dfrac{{r}^{2}{\rho}{g}}{v}\) |
| 3. | \(\dfrac{1}{9}\dfrac{{r}^{2}{\rho}{g}}{v}\) | 4. | \(\dfrac{1}{4}\dfrac{{r}^{2}{\rho}{g}}{v}\) |
Subtopic: Stokes' Law |
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Spherical balls of radius \(R\) are falling in a viscous fluid of viscosity with a velocity \(v.\) The retarding viscous force acting on the spherical ball is:
| 1. | directly proportional to \(R\) but inversely proportional to \(v.\) |
| 2. | directly proportional to both radius \(R\) and velocity \(v.\) |
| 3. | inversely proportional to both radius \(R\) and velocity \(v.\) |
| 4. | inversely proportional to \(R\) but directly proportional to velocity \(v.\) |
Subtopic: Stokes' Law |
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The terminal speed of a raindrop falling through air depends on:
| 1. | the size of the raindrop |
| 2. | the viscosity of air |
| 3. | the acceleration due to gravity |
| 4. | all of the above |
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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