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A simple pendulum has a length l and mass of the bob is m. The bob is given a chang q coulomb. The pendulum is suspended between the vertical plates of a charged parallel plate capacitor. If E is the electric field strength between the plates, the time period of the pendulum is given by -

Option: 1

2 \pi \sqrt{\frac{l}{\sqrt{g-\frac{q E}{m}}}}


Option: 2

2 \pi \sqrt{\frac{e}{\sqrt{g^2+\left(\frac{q E}{m}\right)^2}}}


Option: 3

2 \pi \sqrt{\frac{l}{g}}


Option: 4

2 \pi \sqrt{\frac{l}{\sqrt{q+\frac{q E}{m}}}}


Answers (1)

best_answer


Time period of the pendulum.
\mathrm{ \begin{aligned} & T=2 \pi \sqrt{\frac{l}{g}} \& \\ & T^{\prime}=2 \pi \sqrt{\frac{l}{\sqrt{g^2+\left(\frac{2 E}{m}\right)^2}}} \end{aligned} }

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Gautam harsolia

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