First slide

Viscosity

Question

A rectangular metal plate has dimensions of 10 cm x 20 cm. A thin film of oil separates the plate from a fixed horizontal surface. The separation between the rectangular
plate and the horizontal surface is 0.2 mm. An ideal string is attached to the plate and passes over an ideal pulley to a mass m. When m = 125 gm, the metal plate moves at constant speed of 5 cm/s across the horizontal surface. Then the coefficient of viscosity of oil in dyne-s/ ${cm}^{2}$ is (Use g = 1000 cm/ $s^{2}$ )

Moderate

A

5
B

25
C

2.5
D

50

Solution

The coefficient of viscosity is the ratio of tangential stress on top surface of film (exerted by block) to that of velocity gradient( vertically downwards) of film. Since mass m moves with constant velocity, the string exerts a force equal to mg on plate towards right. Hence oil shall exert tangential force mg on plate towards left.
$η = \frac{\frac{F}{A}}{\frac{(v - 0)}{∆ x}} = \frac{\frac{125 \times 10000}{10 \times 20}}{\frac{(5 - 0)}{. 02}}$
= $2.5 dyne - s / {cm}^{2}$

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