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A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, distance between the plates in m is

0.1

0.05

0.005

0.0005

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Explanation

(d)

$A = (0.1)^{2} = 0.01 m^{2} η = 0.01 Poise = 0.001 decapoise (M . K . S . unit) dv = 0.1 m / s and F = 0.002 N F = ηA \frac{dv}{dx} ∴ dx = \frac{ηAdv}{F} = \frac{0.001 \times 0.01 \times 0.1}{0.002} = 0.0005 m$

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Spherical balls of radius 'r' are falling in a viscous fluid of viscosity ' $η$ ' with a velocity 'v'. The retarding viscous force acting on the spherical ball is

Inversely proportional to 'r' but directly proportional to velocity 'v'

Directly proportional to both radius 'r' and velocity 'v'

Inversely proportional to both radius 'r' and velocity 'v'

Directly proportional to 'r' but inversely proportional to 'v'

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Explanation

(b) $F = 6 πηrv$

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A small sphere of mass m is dropped from a great height. After it has fallen 100 m, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 m of fall is

Greater than the work done by air friction in the second 100 m

Less than the work done by air friction in the second 100 m

Equal to 100 mg

Greater than 100 mg

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Explanation

(b) In the first 100 m body starts from rest and its velocity goes on increasing and after 100 m it acquire maximum velocity (terminal velocity). Further, air friction i.e. viscous force which is proportional to velocity is low in the beginning and maximum at $v = v_{T}$ .
Hence work done against air friction in the first 100 m is less than the work done in next 100 m.

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Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be

10 cm per sec

2.5 cm per sec

$5 \times (4)^{\frac{1}{3}}$ cm per sec

$5 \times \sqrt{2}$ cm per sec

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Explanation

$\frac{4}{3} {πR}^{3} = \frac{4}{3} {πr}^{3} + \frac{4}{3} {πr}^{3} \Rightarrow R^{3} = 2 r^{3} \Rightarrow R = 2^{1 / 3} r$

If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $v \propto r^{2}$

$\frac{v_{2}}{v_{1}} = {(\frac{R}{r})}^{2} = {(\frac{2^{1 / 3}}{r})}^{2} \Rightarrow v_{2} = 2^{2 / 3} \times v_{1} = 2^{2 / 3} \times 5 = 5 \times 4^{1 / 3} cm / s$

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The rate of steady volume flow of water through a capillary tube of length 'l' and radius 'r' under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P)

$\frac{V}{16}$

$\frac{V}{17}$

$\frac{16 V}{17}$

$\frac{17 V}{16}$

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Explanation

(b) Rate of flow of liquid $V = \frac{P}{R}$

where liquid resistance $R = \frac{8 ηl}{{πr}^{4}}$

For another liquid resistance

$R' = \frac{8 ηl}{π {(\frac{r}{2})}^{4}} = \frac{8 ηl}{{πr}^{4}} . 16 = 16 R$

For the series combination

$V_{New} = \frac{P}{R + R'} = \frac{P}{R + 16 R} = \frac{P}{17 R} = \frac{V}{17}$

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A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is

$\frac{3 P}{4}$

$\frac{P}{2}$

$\frac{P}{4}$

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Explanation

(d) From

$V = \frac{{Pπr}^{4}}{8 ηl} \Rightarrow P = \frac{V 8 ηl}{{πr}^{4}} \Rightarrow \frac{P_{2}}{P_{1}} = \frac{V_{2}}{V_{1}} \times \frac{l_{2}}{l_{1}} \times {(\frac{r_{1}}{r_{2}})}^{4} = 2 \times 2 \times {(\frac{1}{2})}^{4} = \frac{1}{4} \Rightarrow P_{2} = \frac{P_{1}}{4} = \frac{P}{4}$

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We have two (narrow) capillary tubes T₁ and T₂. Their lengths are l₁ and l₂ and radii of cross-section are r₁ and r₂ respectively. The rate of flow of water under a pressure difference P through tube T₁ is 8cm³/sec. If l₁ = 2l₂ and r₁ =r₂, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)

4 cm³/sec

(16/3) cm³/sec

(8/17) cm³/sec

None of these

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Explanation

(b) $V = \frac{{Pπr}^{4}}{8 ηl} \Rightarrow P = \frac{V 8 ηl}{{πr}^{4}}$

For composite tube

$V_{1} = \frac{{Pπr}^{4}}{8 η (l + \frac{l}{2})} = \frac{2}{3} \frac{{πPr}^{4}}{8 ηl} = \frac{2}{3} \times 8 = \frac{16}{3} \frac{{cm}^{3}}{\sec} [∵ l_{1} = l = 2 l_{2} or l_{2} = \frac{l}{2}]$

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The Reynolds number of a flow is the ratio of

Gravity to viscous force

Gravity force to pressure force

Inertia forces to viscous force

Viscous forces to pressure forces

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Explanation

(c) $R_{e} = \frac{e v d}{η} o r R_{e} = \frac{e v^{2} d \times A}{A \times v \times η} \Rightarrow R_{e} = \frac{e v^{2} A}{η A v / d}$

Here, $η A v / d$ =viscous force and $e v^{2} A =$ Inertial force

Hence, $R_{e} = \frac{I n e r t i a l f o r c e}{v i s c o u s f o r c e}$

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Water is flowing through a tube of non-uniform cross-section ratio of the radius at entry and exit end of the pipe is 3 : 2. Then the ratio of velocities at entry and exit of liquid is -

4 : 9

9 : 4

8 : 27

1 : 1

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Explanation

(a) If velocities of water at entry and exit points are v₁ and v₂, then according to equation of continuity,

$A_{1} V_{1} = A_{2} V_{2} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}} = {(\frac{r_{2}}{r_{1}})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$

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An application of Bernoulli's equation for fluid flow is found in

Dynamic lift of an aeroplane

Viscosity meter

Capillary rise

Hydraulic press

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Explanation

Bernoulli's equation relates the pressure, velocity, and height of a flowing fluid. It is used to explain the dynamic lift of an aeroplane, where the faster air flow over the curved upper surface creates a lower pressure, resulting in an upward force on the wing.

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