Q119 / 180 · AIIMS 2017 · Physics

A large solid sphere with uniformly distributed positive charge has a smooth narrow tunnel through its centre. A small particle with negative charge, initially at rest far from the sphere, approaches it along the line of the tunnel, reaches its surface with a speed $v$, and passes through the tunnel. Its speed at the centre of the sphere will be

  • A
    $0$
  • B
    V
  • C
    $\sqrt{2} v$
  • Correct answer
    D
    $\sqrt{1.5} v$

Solution

Potential at infinity $=V_{\infty}=0$ Potential at the surface of the sphere, $V_s=k \frac{Q}{R}$ Potential at the centre of the sphere, $V_c=\frac{3}{2} k \frac{Q}{R}$ Let $m$ and $-q$ be the mass and the charge of the particle respectively. Let $v_0=$ speed of the particle at the centre of the sphere. $\begin{aligned} & \frac{1}{2} m v^2=-q\left[V_{\infty}-V_s\right]=q k \frac{Q}{R} \\ & \frac{1}{2} m v_0^2=-q\left[V_{\infty}-V_c\right]=q \cdot \frac{3}{2} k \frac{Q}{R}\end{aligned}$ Dividing eqn. (ii) by eqn. (i), $\frac{v_0^2}{v^2}=\frac{3}{2}=1.5$ or $\quad v_0=\sqrt{1.5} v$

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

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