JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2019
- Hard
- MCQ
A simple pendulum of length \(L\) is placed between the plates of a parallel plate capacitor having electric field \(E,\) as shown in figure. Its bob has mass \(m\) and charge \(q.\) the time period of the pendulum is given by

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Answer & Solution
Correct Answer
(D) \(2\pi \sqrt {\frac{L}{{\sqrt {{g^2}\, + \,{{\left( {\frac{{qE}}{m}} \right)}^2}} }}} \)
Step-by-step Solution
Detailed explanation
\(g_{e f f}=\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}\)
\(T=2 \pi \sqrt{\frac{\ell}{g_{e f f}}}\)
\(2 \pi \sqrt{\frac{\ell}{\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}}}\)
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