Simple pendulum length between |SIMPLEHARMOMOTIONJEEMain2019

JEE MAIN 2019

10-04-2019 S2

Question

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

Select the correct option:

$2 \pi \sqrt{\frac{L}{\left(g-\frac{q E}{m}\right)}}$

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

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

$2 \pi \sqrt{\frac{L}{\sqrt{g^2-\frac{q^2 E^2}{m^2}}}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Hard

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