JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 3)
A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is :
[Density of water fw = 1000 kg m$$-$$3 and Density of air fa = 1.2 kg m$$-$$3, g = 10 m/s2, Coefficient of viscosity of air = 1.8 $$\times$$ 10$$-$$5 Nsm$$-$$2]
[Density of water fw = 1000 kg m$$-$$3 and Density of air fa = 1.2 kg m$$-$$3, g = 10 m/s2, Coefficient of viscosity of air = 1.8 $$\times$$ 10$$-$$5 Nsm$$-$$2]
250.6 ms$$-$$1
43.56 ms$$-$$1
4.94 ms$$-$$1
14.4 ms$$-$$1
Explanation
At terminal speed
a = 0
Fnet = 0
mg = Fv = 6$$\pi$$ $$\eta $$Rv
$$v = {{mg} \over {6\pi \eta Rv}}$$
$$v = {{{\rho _w}{{4\pi } \over 3}{R^3}g} \over {6\pi \eta R}}$$
$$ = {{2{\rho _w}{R^2}g} \over {9\eta }}$$
$$ = {{400} \over {81}}$$ m/s
= 4.94 m/s
a = 0
Fnet = 0
mg = Fv = 6$$\pi$$ $$\eta $$Rv
$$v = {{mg} \over {6\pi \eta Rv}}$$
$$v = {{{\rho _w}{{4\pi } \over 3}{R^3}g} \over {6\pi \eta R}}$$
$$ = {{2{\rho _w}{R^2}g} \over {9\eta }}$$
$$ = {{400} \over {81}}$$ m/s
= 4.94 m/s
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