### JEE Main 12 Jan 2019 Morning Question 5

##### Question: There is a uniform spherically symmetric surface charge density at a distance $ R _0 $ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

#### Options:

A)

B)

C)

D)

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

At any instant ’t’, Total energy of charge distribution is constant i.e.,

$ \begin{aligned} \frac{1}{2}mV^2 + \frac{KQ^2}{2R} &= 0 + \frac{KQ^2}{2R_0} \\ \therefore \frac{1}{2}mV^2 &= \frac{KQ^2}{2R_0} - \frac{KQ^2}{2R} \\ \therefore V &= \sqrt{\frac{KQ^2}{m} \left(\frac{1}{R_0} - \frac{1}{R}\right)} = C\sqrt{\frac{1}{R_0} - \frac{1}{R}} \ \end{aligned} $

```
Also the slope of v-s curve will go on decreasing.
```

${\therefore} $ Graph is correctly shown by option (1)