JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 20)
A small spherical ball of radius 0.1 mm and density 104 kg m$$-$$3 falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h will be ___________ m.
(Given g = 10 ms$$-$$2, viscosity of water = 1.0 $$\times$$ 10$$-$$5 N-sm$$-$$2).
Answer
20
Explanation
$$\sqrt {2gh} $$ = terminal speed
$$ \Rightarrow \sqrt {2gh} = {2 \over 9}{{{r^2}g(\rho - \rho ')} \over \eta }$$
$$ = {2 \over 9} \times {{{{10}^{ - 8}} \times 10 \times 9000} \over {{{10}^{ - 5}}}}$$
$$ \Rightarrow h = {{400} \over {2g}}$$
$$ \Rightarrow h = 20$$ m
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