Discussion Forum

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 $(\frac{1}{\rho}\frac{d\rho}{dt})$ is constant. The velocity  v of any point of the surface of the expanding sphere is proportional to

Answer:

Explanation:

$m=\frac{4\pi R^{3}}{3}\times\rho$

On taking log both sides, we have

ln(m)= ln($(\frac{4\pi ^{}}{3})$) + ln(ρ)+3ln(R)

On differentiating  with respect to time

$0=0+\frac{1}{\rho}\frac{d \rho}{dt}+\frac{3}{R}\frac{dR}{dt}$

   $\Rightarrow$              $(\frac{dR}{dt})= v \propto -R \times \frac{1}{\rho}(\frac{d \rho}{dt})$

    $v  \propto R$