JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2019
- Hard
- MCQ
The bob of a simple pendulum has mass \(2\,g\) and a charge of \(5.0\,\mu C\). It is at rest in a uniform horizontal electric field of intensity \(2000\,\frac{V}{m}\). At equilibrium, the angle that the pendulum makes with the vertical is (take \(g = 10\,\frac{m}{{{s^2}}}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \({\tan ^{ - 1}}\left( {0.5} \right)\)
Step-by-step Solution
Detailed explanation
\(\tan \theta=\frac{\mathrm{qE}}{\mathrm{mg}}=\frac{5 \times 10^{-6} \times 2000}{2 \times 10^{-3} \times 10}\)
\(\tan \theta=\frac{1}{2} \Rightarrow \theta=\tan ^{-1}(0.5)\)
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