Question

A bob of a simple pendulum of mass 40gm with a positive charge $4\times {10}^{-6}$ is oscillating with time period ‘$T_1$’. An electric field of intensity $3.6\times {10}^4$ N/c is applied vertically upwards now time period is $T_2$. The value of $\frac{{}^{\prime }{T}_2^{\prime }}{T_1}$ is (g = 10 $m/s^2$)

• A. 0.16
• B. 0.64
• C. 1.25
• D. 0.8

Answer 1

The correct option is C 1.25

$T_1\alpha \frac{1}{\sqrt{g}};T_2\alpha \frac{1}{\sqrt{g-a}}\alpha \frac{1}{\sqrt{g-\frac{Eq}{m}}}$
$\frac{T_2}{T_1}=\sqrt{\frac{g}{g-\frac{Eq}{m}}}=\sqrt{\frac{1}{1-\frac{Eq}{m}}}=\frac{1}{\sqrt{1-\frac{3.6\times {10}^4\times 4\times {10}^{-6}}{40\times {10}^{-3}\times 10}}}$
$\frac{T_2}{T_1}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=\frac{10}{8}=1.25$