Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is k 0, is

Question

Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is k >0, is (A) (d r/d t)+k=0 (B) ( dr / dt )- k =0 (C) ( dr / dt )= kr (D) ( dr / dt )=2 kr

Tardigrade Admin · Accepted Answer

(d v/d t)=-k(4 π r2) Put v=(4/3) π r3 or ( dv / dt )=4 π r 2 ( dr / dt ) Hence ( dr / dt )=- k

Q. Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is $k > 0$ , is

$\frac{d r}{d t} + k = 0$

$\frac{d r}{d t} - k = 0$

$\frac{d r}{d t} = k r$

$\frac{d r}{d t} = 2 k r$

$\frac{d v}{d t} = - k (4 π r^{2})$
Put $v = \frac{4}{3} π r^{3}$ or $\frac{d v}{d t} = 4 π r^{2} \frac{d r}{d t}$
Hence $\frac{d r}{d t} = - k$

Q. Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is k>0, is

dtdr​+k=0

dtdr​−k=0

dtdr​=kr

dtdr​=2kr