question_answer

A ball of mass m with a charge \[+q\] can rotate in a vertical plane at the end of a string of length L in a uniform electrostatic field whose lines of forces are directed vertically upwards. The horizontal velocity that must be imparted to the ball at top position so that tension in the string at the bottom position of the ball is 15 times the weight of ball, is

A) \[\sqrt{\frac{5L}{m}(qE+mg)}\] 

B) \[\sqrt{\frac{5L}{m}(2qE+mg)}\]

C) \[\sqrt{\frac{5L}{m}(qE+2mg)}\]

D)  \[\sqrt{\frac{5L}{m}(2qE+3mg)}\]

Correct Answer: C

Solution :

[c] \[{{T}_{P}}-mg+qE=\frac{mv_{P}^{2}}{L}\]  \[{{T}_{P}}=15mg\] \[mv_{P}^{2}=L\,\,(14mg+qE)\]           ...(i) From work-energy theorem \[(qE-mg)2L=\frac{1}{2}mv_{Q}^{2}-\frac{1}{2}mv_{P}^{2}\] \[{{V}_{Q}}=\sqrt{(5qE+10mg)\frac{L}{m}}=\sqrt{\frac{5L}{m}(qE+2mg)}\]