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Q. No.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 the upper layer of water in a river is \(36~\text{kmh}^{-1}.\) Shearing stress between horizontal layers of water is \(10^{-3}~\text{Nm}^{-2}.\) The depth of the river is:
(coefficient of viscosity of water is \(10^{-2}~\text {Pa-s}\) )
(coefficient of viscosity of water is \(10^{-2}~\text {Pa-s}\) )
| 1. | \(100~\text m\) | 2. | \(200~\text m\) |
| 3. | \(300~\text m\) | 4. | \(400~\text m\) |
Subtopic: Viscosity |
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Level 1: 80%+
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An ideal fluid with a density of \(800\) \(\text{kg}/\text{m}^3\) flows smoothly through a bent pipe that gradually narrows. The pipe's cross-sectional area decreases from \(a\) at the entrance to \(\dfrac{a}{2}\) at the exit. The pressure at the wider section exceeds the pressure at the narrower section by \(4100\) Pa. Given that the fluid velocity at the wider section is \(\dfrac{\sqrt x}{6}~\text{m}/\text{s}\) and the vertical height difference between the two sections is \(1~\text{m},\) then the value of \(x\) is: (use \(g=10~\text{m}/\text{s}^2\))
1. \(124\)
2. \(236\)
3. \(363\)
4. \(432\)
1. \(124\)
2. \(236\)
3. \(363\)
4. \(432\)
Subtopic: Bernoulli's Theorem |
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If \(\rho\) is the density and \(\eta\) is the coefficient of viscosity of fluid which flows with a speed \(v\) in the pipe of diameter \(d,\) the correct formula for the Reynolds number \(R_e\) is:
1. \( \mathrm{R}_e=\frac{\eta \mathrm{d}}{\rho \mathrm{v}} \)
2. \(\mathrm{R}_{\mathrm{e}}=\frac{\rho \mathrm{v}}{\eta \mathrm{d}} \)
3. \( \mathrm{R}_e=\frac{\rho v d}{\eta} \)
4. \(R_e=\frac{\eta}{\rho v d}\)
1. \( \mathrm{R}_e=\frac{\eta \mathrm{d}}{\rho \mathrm{v}} \)
2. \(\mathrm{R}_{\mathrm{e}}=\frac{\rho \mathrm{v}}{\eta \mathrm{d}} \)
3. \( \mathrm{R}_e=\frac{\rho v d}{\eta} \)
4. \(R_e=\frac{\eta}{\rho v d}\)
Subtopic: Properties of Fluids |
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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 area of the cross-section of a large tank is \(0.5~\text{m}^2\). It has a narrow opening near the bottom having an area of cross-section \(1~\text{cm}^2\). A load of \(25~\text{kg}\) is applied on the water at the top of the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of the water level in the tank is \(40~\text{cm}\) above the bottom, will be: [Take \(g = 10~\text{ms}^{-2}\)]
1. \(1~\text{ms}^{-1}\)
2. \(2~\text{ms}^{-1}\)
3. \(3~\text{ms}^{-1}\)
4. \(4~\text{ms}^{-1}\)
1. \(1~\text{ms}^{-1}\)
2. \(2~\text{ms}^{-1}\)
3. \(3~\text{ms}^{-1}\)
4. \(4~\text{ms}^{-1}\)
Subtopic: Bernoulli's Theorem |
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A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be:
1. \(2.8 \times 10^{-4} \text J\)
2. \(1.5 \times 10^{-3} \text J\)
3. \(1.9 \times 10^{-4} \text J\)
4. \(9.4 \times 10^{-5} \text J\)
1. \(2.8 \times 10^{-4} \text J\)
2. \(1.5 \times 10^{-3} \text J\)
3. \(1.9 \times 10^{-4} \text J\)
4. \(9.4 \times 10^{-5} \text J\)
Subtopic: Surface Tension |
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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 liquid of density \(750~\text{kgm}^{-3}\) flows smoothly through a horizontal pipe that tapers in cross-sectional area from \(A_1 = 1.2 \times 10^{-2}~\text{m}^2\) to \(A_2 = \dfrac{A_1}{2}.\) The pressure difference between the wide and narrow sections of the pipe is \(4500~\text{Pa}.\) The rate of flow of liquid is:
1. \(20\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
2. \(30\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
3. \(28\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
4. \(24\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
1. \(20\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
2. \(30\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
3. \(28\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
4. \(24\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
Subtopic: Bernoulli's Theorem |
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A tube of length 50 cm is filled completely with an incompressible liquid of mass 250 g and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity \(x \sqrt F\) rad s–1. If F is the force exerted by the liquid at the other end then the value of x will be:
1. 2
2. 3
3. 4
4. 5
1. 2
2. 3
3. 4
4. 5
Subtopic: Pressure |
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