JEE PYQ: Oscillations Question 29
Question 29 - 2019 (10 Apr 2019 Shift 2)
A simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having electric field $E$. Its bob has mass $m$ and charge $q$. The time period of the pendulum is given by:
(1) $2\pi\sqrt{\frac{L}{g + \frac{qE}{m}}}$
(2) $2\pi\sqrt{\frac{L}{g^2 - \frac{q^2E^2}{m^2}}}$
(3) $2\pi\sqrt{\frac{L}{g - \frac{qE}{m}}}$
(4) $2\pi\sqrt{\frac{L}{\sqrt{g^2 + \left(\frac{qE}{m}\right)^2}}}$
Show Answer
Answer: (4)
Solution
$g_{\text{eff}} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$. $T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}}$.
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