JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Kinematics

Consider an expanding sphere of instantaneous radius

R

whose total mass remains constant. The expansion is such that the instantaneous density

ρ

remains uniform throughout the volume. The rate of fractional change in density

(\frac{1}{ρ} \frac{d ρ}{d t})

is constant. The velocity

v

of any point on the surface of the expanding sphere is proportional to

Answer: (A)

Solution

Since mass is constant,

m = ρ \frac{4}{3} π R^{3}

. Differentiating:

0 = ρ \cdot 4 π R^{2} \frac{d R}{d t} + \frac{4}{3} π R^{3} \frac{d ρ}{d t}

. Rearranging:

- \frac{1}{ρ} \frac{d ρ}{d t} = \frac{3}{R} \frac{d R}{d t}

. Since

\frac{1}{ρ} \frac{d ρ}{d t}

is constant,

\frac{d R}{d t} \propto R

. The velocity

v = \frac{d R}{d t} \propto R

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option