Q: Two identical raindrops falling with terminal velocity v_t coalesce into a single larger drop. The terminal velocity of the combined drop is:

2^{2/3} v_t. Volume is conserved: \dfrac{4}{3}\pi R^3 = 2 \times \dfrac{4}{3}\pi r^3 \Rightarrow R = 2^{1/3}r. Since v_t \propto r^2, the new terminal velocity is (2^{1/3})^2 v_t = 2^{2/3} v_t.

Q: The SI unit of the coefficient of viscosity is:

\text{Pa}\cdot\text{s}. From F = \eta A \dfrac{dv}{dx}, \eta has units \dfrac{\text{N}/\text{m}^2}{\text{s}^{-1}} = \text{N}\cdot\text{s}/\text{m}^2 = \text{Pa}\cdot\text{s}. The CGS unit is the poise.

Question 1

According to Stokes' law, the viscous drag on a small sphere of radius r moving with velocity v through a fluid of viscosity \eta is:

Accepted Answer

6\pi\eta r v. Stokes' law gives F = 6\pi\eta r v for slow, streamline motion of a sphere through a viscous medium.

Question 2

The terminal velocity of a sphere of radius r and density \rho falling through a fluid of density \sigma and viscosity \eta is:

Accepted Answer

\dfrac{2}{9}\dfrac{r^2(\rho - \sigma)g}{\eta}. At terminal velocity the weight is balanced by upthrust plus viscous drag: \dfrac{4}{3}\pi r^3(\rho - \sigma)g = 6\pi\eta r v_t, giving v_t = \dfrac{2}{9}\dfrac{r^2(\rho - \sigma)g}{\eta}.

Question 3

The terminal velocity of a small sphere falling through a viscous liquid is proportional to:

Accepted Answer

r^2. From v_t = \dfrac{2}{9}\dfrac{r^2(\rho - \sigma)g}{\eta}, the terminal velocity varies as the square of the radius.

Question 4

If the radius of a sphere falling through a viscous fluid is doubled, its terminal velocity becomes:

Accepted Answer

4 times. Since v_t \propto r^2, doubling the radius multiplies the terminal velocity by 2^2 = 4.

Question 5

Two identical raindrops falling with terminal velocity v_t coalesce into a single larger drop. The terminal velocity of the combined drop is:

Accepted Answer

2^{2/3} v_t. Volume is conserved: \dfrac{4}{3}\pi R^3 = 2 	imes \dfrac{4}{3}\pi r^3 \Rightarrow R = 2^{1/3}r. Since v_t \propto r^2, the new terminal velocity is (2^{1/3})^2 v_t = 2^{2/3} v_t.

Question 6

The SI unit of the coefficient of viscosity is:

Accepted Answer

ext{Pa}\cdot	ext{s}. From F = \eta A \dfrac{dv}{dx}, \eta has units \dfrac{	ext{N}/	ext{m}^2}{	ext{s}^{-1}} = 	ext{N}\cdot	ext{s}/	ext{m}^2 = 	ext{Pa}\cdot	ext{s}. The CGS unit is the poise.

Question 7

The dimensional formula of the coefficient of viscosity is:

Accepted Answer

[M^1L^{-1}T^{-1}]. \eta = \dfrac{F}{A\,(dv/dx)} gives \dfrac{[M^1L^1T^{-2}]}{[L^2][T^{-1}]} = [M^1L^{-1}T^{-1}].

Question 8

On raising the temperature, the viscosity of a liquid and that of a gas respectively:

Accepted Answer

Decreases and increases. In liquids viscosity arises from intermolecular cohesion, which heating weakens. In gases it arises from momentum transfer by molecular collisions, which heating makes more vigorous.

Question 9

Newton's law of viscous flow expresses the tangential force between two layers of fluid as:

Accepted Answer

F = \eta A \dfrac{dv}{dx}. The viscous force is proportional to the area of contact A and to the velocity gradient \dfrac{dv}{dx} perpendicular to the flow, the constant of proportionality being \eta.

Question 10

The quantity \dfrac{dv}{dx} appearing in the expression for viscous force is called the:

Accepted Answer

Velocity gradient. It measures how rapidly the flow speed changes from layer to layer across the direction of flow, and has units 	ext{s}^{-1}.

Question 11

Poiseuille's formula for the volume of liquid flowing per second through a capillary tube of radius r and length L under a pressure difference \Delta P is:

Accepted Answer

\dfrac{\pi \Delta P r^4}{8\eta L}. Poiseuille's equation is Q = \dfrac{\pi \Delta P r^4}{8\eta L}. The fourth-power dependence on radius makes flow extremely sensitive to the bore of the tube.

Question 12

If the radius of a capillary tube is halved while the pressure difference is unchanged, the rate of laminar flow through it becomes:

Accepted Answer

\dfrac{1}{16} of the original. By Poiseuille's law Q \propto r^4, so halving the radius reduces the flow by a factor of 2^4 = 16.

Question 13

Two capillary tubes of equal length with radii in the ratio 1 : 2 are joined in series. The ratio of the pressure drops across them in steady laminar flow is:

Accepted Answer

16 : 1. In series the flow rate Q is common. From Q = \dfrac{\pi \Delta P r^4}{8\eta L}, \Delta P \propto \dfrac{1}{r^4}, so the ratio is \left(\dfrac{2}{1}\right)^4 = 16 : 1.

Question 14

Viscosity may be described as:

Accepted Answer

The internal friction that opposes relative motion between fluid layers. Adjacent layers moving at different speeds exert tangential dragging forces on one another; this internal friction is what we call viscosity.

Question 15

A body falling through a viscous fluid attains terminal velocity when:

Accepted Answer

Its weight is balanced by the sum of the upthrust and the viscous drag. The drag grows with speed until the net force vanishes: mg = F_B + F_{viscous}. With zero net force the body then falls at a constant terminal velocity.

Question 16

A steel ball dropped into a tall jar of glycerine eventually falls with constant velocity because:

Accepted Answer

The viscous drag increases with speed until it balances the net downward force. Drag is proportional to speed, so as the ball accelerates the drag rises. Once 6\pi\eta r v plus upthrust equals the weight, the acceleration is zero and the speed stays fixed.

Question 17

A sphere falls with terminal velocity through a liquid of density equal to the density of the sphere. The terminal velocity is then:

Accepted Answer

Zero. From v_t = \dfrac{2}{9}\dfrac{r^2(\rho - \sigma)g}{\eta}, setting \rho = \sigma makes the numerator vanish. The body simply floats in equilibrium and does not sink.

Question 18

A sphere of radius 1	ext{ mm} moves at 0.02	ext{ m/s} through a liquid of viscosity 1.0	ext{ Pa}\cdot	ext{s}. The viscous drag on it is:

Accepted Answer

3.77 	imes 10^{-4}	ext{ N}. F = 6\pi\eta r v = 6\pi(1.0)(10^{-3})(0.02) = 3.77 	imes 10^{-4}	ext{ N}.

Question 19

If the radius of a sphere falling through a viscous liquid is tripled, its terminal velocity becomes:

Accepted Answer

9 times. v_t \propto r^2, so tripling the radius multiplies the terminal velocity by 3^2 = 9.

Question 20

Eight identical raindrops, each falling at terminal velocity v_t, coalesce into one large drop. The terminal velocity of the big drop is:

Accepted Answer

4v_t. Volume conservation gives R = 8^{1/3}r = 2r. Since v_t \propto r^2, the new terminal velocity is 2^2 v_t = 4v_t.

Viscosity & Stokes' Law — MH-CET Physics MCQs with Solutions

Questions with solutions