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)
