Q124 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
A bob of a simple pendulum has a mass of 4 g and a charge of $20 \mu \mathrm{C}$. If it is at rest in a uniform horizontal electric field of intensity $1000 \mathrm{Vm}^{-1}$, the angle that the pendulum makes with the vertical at equilibrium is:
- A$60^{\circ}$
- B$\tan ^{-1} 2$
- Correct answerC$\tan ^{-1} \dfrac{1}{2}$
- D$30^{\circ}$
Solution
The forces acting on the bob at equilibrium are the gravitational force $mg$ acting vertically downwards, the electric force $qE$ acting horizontally, and the tension $T$ in the string.
Let $\theta$ be the angle the string makes with the vertical. At equilibrium, the forces balance as follows:
$T \sin \theta = qE$
$T \cos \theta = mg$
Dividing the two equations, we get:
$\tan \theta = \dfrac{qE}{mg}$
Given values are $m = 4 \text{ g} = 4 \times 10^{-3} \text{ kg}$, $q = 20 \mu \text{C} = 20 \times 10^{-6} \text{ C}$, $E = 1000 \text{ Vm}^{-1}$, and taking $g = 10 \text{ ms}^{-2}$:
$\tan \theta = \dfrac{(20 \times 10^{-6}) \times 1000}{4 \times 10^{-3} \times 10}$
$\tan \theta = \dfrac{20 \times 10^{-3}}{40 \times 10^{-3}} = \dfrac{20}{40} = \dfrac{1}{2}$
Therefore, $\theta = \tan^{-1} \left( \dfrac{1}{2} \right)$.
Answer: $\tan ^{-1} \dfrac{1}{2}$
Let $\theta$ be the angle the string makes with the vertical. At equilibrium, the forces balance as follows:
$T \sin \theta = qE$
$T \cos \theta = mg$
Dividing the two equations, we get:
$\tan \theta = \dfrac{qE}{mg}$
Given values are $m = 4 \text{ g} = 4 \times 10^{-3} \text{ kg}$, $q = 20 \mu \text{C} = 20 \times 10^{-6} \text{ C}$, $E = 1000 \text{ Vm}^{-1}$, and taking $g = 10 \text{ ms}^{-2}$:
$\tan \theta = \dfrac{(20 \times 10^{-6}) \times 1000}{4 \times 10^{-3} \times 10}$
$\tan \theta = \dfrac{20 \times 10^{-3}}{40 \times 10^{-3}} = \dfrac{20}{40} = \dfrac{1}{2}$
Therefore, $\theta = \tan^{-1} \left( \dfrac{1}{2} \right)$.
Answer: $\tan ^{-1} \dfrac{1}{2}$
Hence, the correct answer is option: 'C'.