• question_answer The bob of a simple pendulum is hanging vertically down from a fixed identical bob by means of a string of length$l.$ If both bobs are charged with a charge q each, time period of the pendulum is: (ignore the radii of the bobs) A)  $2\pi \sqrt{\frac{\ell }{g+\sqrt{\frac{{{q}^{2}}}{{{\ell }^{2}}m}}}}$           B)  $2\pi \sqrt{\frac{\ell }{g-\sqrt{\frac{{{q}^{2}}}{{{\ell }^{2}}m}}}}$C) $2\pi \sqrt{\frac{\ell }{g}}$               D)  $2\pi \sqrt{\frac{\ell }{g-\left( \frac{{{q}^{2}}}{\ell } \right)}}$

Charge on each bob = q $\therefore$ force of repulsion between them $=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{\ell }^{2}}}.$ As this force is mutual, value of g is not affected. Time period remains the same. $\therefore$$t=2\pi \sqrt{\frac{\ell }{g}}.$ Hence, the correction option is [c].