A rain drop (assumed as sphere) evaporates at a rate proportional to its surface area at any instant t. The differential equation governing rate of change of radius r of the drop is

Question

A rain drop (assumed as sphere) evaporates at a rate proportional to its surface area at any instant t. The differential equation governing rate of change of radius r of the drop is (A) (d2r/dt2)+2r=0 (B) (d2r/dt2)=3r (C) (d2r/dt2)=0 (D) none of these

Tardigrade Admin · Accepted Answer

Since V = (4/3) π r3 ∴ (dV/dt) = 4 π r2 (dr/dt) Given (dV/dt) propto 4 π r2 ∴ (dV/dt) = 4 π r2 K (say) ∴ 4 π r2 (dr/dt) = 4π r2 K ⇒ (dr/dt) = K ⇒ (d2r/dt2)=0.

Q. A rain drop (assumed as sphere) evaporates at a rate proportional to its surface area at any instant t. The differential equation governing rate of change of radius r of the drop is

$\frac{d ^{2} r}{d t ^{2}} + 2 r = 0$

$\frac{d ^{2} r}{d t ^{2}} = 3 r$

$\frac{d ^{2} r}{d t ^{2}} = 0$

none of these

Since $V = \frac{4}{3} π r^{3}$
$∴ \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
Given $\frac{d V}{d t} \propto 4 π r^{2}$
$∴ \frac{d V}{d t} = 4 π r^{2} K$ (say)
$∴ 4 π r^{2} \frac{d r}{d t} = 4 π r^{2} K$
$\Rightarrow \frac{d r}{d t} = K$
$\Rightarrow \frac{d ^{2} r}{d t ^{2}} = 0$ .

Q. A rain drop (assumed as sphere) evaporates at a rate proportional to its surface area at any instant t. The differential equation governing rate of change of radius r of the drop is

dt2d2r​+2r=0

dt2d2r​=3r

dt2d2r​=0

none of these

Since V=34​πr3 ∴dtdV​=4πr2dtdr​ Given dtdV​∝4πr2 ∴dtdV​=4πr2K (say) ∴4πr2dtdr​=4πr2K ⇒dtdr​=K ⇒dt2d2r​=0.

$\frac{d ^{2} r}{d t ^{2}} + 2 r = 0$

$\frac{d ^{2} r}{d t ^{2}} = 3 r$

$\frac{d ^{2} r}{d t ^{2}} = 0$

Since $V = \frac{4}{3} π r^{3}$
$∴ \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
Given $\frac{d V}{d t} \propto 4 π r^{2}$
$∴ \frac{d V}{d t} = 4 π r^{2} K$ (say)
$∴ 4 π r^{2} \frac{d r}{d t} = 4 π r^{2} K$
$\Rightarrow \frac{d r}{d t} = K$
$\Rightarrow \frac{d ^{2} r}{d t ^{2}} = 0$ .