Question

A bob of a simple pendulum of mass $40\text{g}$ with a positive charge of $4\times {10}^{-6}\text{C}$ is oscillating with time period $T_1$. An electric field of intensity $3.6\times {10}^4\text{N/C}$ is applied vertically upwards, now the time period is $T_2$. The value of $\frac{T_2}{T_1}$ is

$(g=10\text{m}/{\text{s}}^2)$

• A. $0.85$
• B. $0.16$
• C. $0.64$
• D. $1.25$

Answer 1

The correct option is D $1.25$

Time period of the bob,

$T_1=2\pi \sqrt{\frac{l}{g}}...\left(1\right)$

In a vertically upward electric field, the time period is,

$T_2=2\pi \sqrt{\frac{l}{g_{eff}}}$


Here, $g_{eff}=\frac{W-F_e}{m}=\frac{mg-qE}{m}=g-\frac{qE}{m}$

$\therefore T_2=2\pi \sqrt{\frac{l}{g-\frac{qE}{m}}}...\left(2\right)$

On dividing $\left(2\right)$ by $\left(1\right)$,

$\frac{T_2}{T_1}=\sqrt{\frac{g}{g-\frac{qE}{m}}}=\sqrt{\frac{1}{1-\frac{qE}{mg}}}$

$=\sqrt{\frac{1}{1-\frac{4\times {10}^{-6}\times 3.6\times {10}^4}{40\times {10}^{-3}\times 10}}}$

$=1.25$