JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 7)
A water drop of radius 1 $$\mu$$m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 $$\times$$ 10$$-$$5 Nsm$$-$$2 and its density is negligible as compared to that of water 106 gm$$-$$3. Terminal velocity of the water drop is :
(Take acceleration due to gravity = 10 ms$$-$$2)
145.4 $$\times$$ 10$$-$$6 ms$$-$$1
118.0 $$\times$$ 10$$-$$6 ms$$-$$1
132.6 $$\times$$ 10$$-$$6 ms$$-$$1
123.4 $$\times$$ 10$$-$$6 ms$$-$$1
Explanation
$$6\pi \eta rv = mg$$
$$6\pi \eta rv = {4 \over 3}\pi {r^3}\rho g$$
or $$v = {2 \over 9}{{\rho {r^2}g} \over \eta } = {2 \over 9} \times {{{{10}^3} \times {{({{10}^{ - 6}})}^2} \times 10} \over {1.8 \times {{10}^{ - 5}}}}$$
$$ = 123.4 \times {10^{ - 6}}$$ m/s
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