A spherically symmetric gravitational system of particles has a mass density $\rho=\left\{\begin{array}{ll}\rho_{0} & \text { for } \mathrm{r} \leq \mathrm{R} \\ 0 & \text { for } \mathrm{r}>\mathrm{R}\end{array}\right.$ where $\mathrm{r}_{0}$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $\mathrm{V}$ as a function of distance $r(0<r<\infty)$ from the centre of the system is represented by

• A

  
• B

  
• Correct answer
  C

  
• D

  

Solution

Given mass density \(P=\left\{P_{0}\right.\) for \(r \leq R O\) for \(\left.r>R\right\}\) Since gravitational depends on mass, which is zero for \(r>R\), hence it only acts for \(r \leq R\). \(\begin{aligned} &\overrightarrow{\mathrm{F}}_{\text {graw }}=\overrightarrow{\mathrm{F}}_{\text {circular }} \\ &\Rightarrow \frac{\mathrm{GM}_{\text {sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}} \\ &\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text {sphere }}}{\mathrm{R}} \text { since } \mathrm{r} \leq \mathrm{R} \\ &\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text { straight line }+\mathrm{ve} \text { slope } \end{aligned}\) Beyond \(\mathrm{r}=\mathrm{R}\), mass density \(=0 \Rightarrow\) acceleration due to gravity \(=0 \Rightarrow\) velocity stays constant.

Hence, the correct answer is option: 'C'.