A raindrop (spherical drop of density ρ) falling from a large height initially at rest has a radius ro. As it falls its mass increases due to vapour (same density ρ) condensing around it. Rate of mass of vapour condensing around it is proportional to surface area of the drop (dmdt=k4πr2) and assume the spherical shape is retained (Neglect Buoyant force and air resistance). The terminal velocity is the final velocity attained by the drop. If this velocity is v0, then the time taken by it to reach this is

Newton’s II law for the falling drop at any instant mg−dmdtV=ma ….1 at terminal velocity V=Vo, a=0 ….2also dmdt=k4πr2⇒ddt(ρ43πr3)=k4πr2 ⇒drdt=kρ⇒r=ro+kρt ….3From 1, 2, 3 we get t=3Vog−ρroK

A raindrop (spherical drop of density ρ) falling from a large height initially at rest has a radius ro. As it falls its mass increases due to vapour (same density ρ) condensing around it. Rate of mass of vapour condensing around it is proportional to surface area of the drop (dmdt=k4πr2) and assume the spherical shape is retained (Neglect Buoyant force and air resistance). The terminal velocity is the final velocity attained by the drop. If this velocity is v0, then the time taken by it to reach this is

Newton’s II law for the falling drop at any instant mg−dmdtV=ma ….1 at terminal velocity V=Vo, a=0 ….2also dmdt=k4πr2⇒ddt(ρ43πr3)=k4πr2 ⇒drdt=kρ⇒r=ro+kρt ….3From 1, 2, 3 we get t=3Vog−ρroK

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A raindrop (spherical drop of density $ρ$ ) falling from a large height initially at rest has a radius $r_{o}$ . As it falls its mass increases due to vapour (same density $ρ$ ) condensing around it. Rate of mass of vapour condensing around it is proportional to surface area of the drop $(\frac{d m}{d t} = k 4 π r^{2})$ and assume the spherical shape is retained (Neglect Buoyant force and air resistance). The terminal velocity is the final velocity attained by the drop. If this velocity is $v_{0}$ , then the time taken by it to reach this is

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$\frac{v_{0}}{g} + \frac{ρ r_{0}}{k}$

$\frac{4 v_{0}}{3 g} - \frac{ρ r_{0}}{k}$

$\frac{3 v_{0}}{g} - \frac{ρ r_{0}}{k}$

infinity

answer is C.

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Detailed Solution

Newton’s II law for the falling drop at any instant
$m g - \frac{d m}{d t} V = m a$ ….1
at terminal velocity $V = V_{o}, a = 0$ ….2
also $\frac{d m}{d t} = k 4 π r^{2} \Rightarrow \frac{d}{d t} (ρ \frac{4}{3} π r^{3}) = k 4 π r^{2}$
$\Rightarrow \frac{d r}{d t} = \frac{k}{ρ} \Rightarrow r = r_{o} + \frac{k}{ρ} t$ ….3
From 1, 2, 3 we get $t = \frac{3 V_{o}}{g} - \frac{ρ r_{o}}{K}$