Question

Rain drop is falling under gravity on the earth surface. Its speed remain constant near the
earth surface, this constant speed is known as terminal velocity. This velocity depends on
viscosity of air n radius of drop (r.), difference in density of water and air (d) and acceleration
due to gravity (g). Then terminal velocity is [Given that: F = 6xrnV.):-​

Answer 1

a is the correct option.

Answer 2

Answer:

we have V(T) =$\frac{2}{9} \frac{{r}^{2}g }{n} d0$thus option d

Explanation:

we have given that F=

$6\pi \: n \: r \: v (t)$

in terminal velocity we have acceleration as zero,

so net force zero,

$m \gamma o = f$

=equation (1)

$m = do \times v$

$m = \frac{4}{3} \times do \times \: \pi \: \gamma o \: \times 3$

=equation (2)

substitute (2) in (1) we have ,

$\frac{4}{3} \pi {\gamma o}^{3} {do}^{3} g = 6\pi \: n \gamma ov(t)$

therefore we have V(T)=

$\frac{4}{18} \frac{ {\gamma o}^{2} g \: do}{n} \:$

V(T)=

$\frac{2}{g} \frac{ {\gamma o}^{2} g }{n} \: \: \: do$

= option(d).

terminal velocity :- when force due to gravity is vbalances by drag force , object no longer speeds up or slows down but will continue falling at a constant velocity. this velocity is called as terminal velocity

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