JEE Main 2019 — Physics solution

JEE Main2019Physics

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:

A
$2 π \frac{L}{g - \frac{q E}{m}}$
B
$2 π \frac{L}{g + \frac{q E}{m}}$
C
$2 π \frac{L}{g ^{2} + ( \frac{q E}{m} ) ^{2}}$
D
$2 π \frac{L}{g ^{2} - \frac{q ^{2} E ^{2}}{m ^{2}}}$

Solution

$t = 2 π \frac{L}{g _{e f f}} \Rightarrow g_{e f f} = g^{2} + (\frac{g E}{m})^{2} \Rightarrow t = 2 π \frac{L}{g ^{2} + ( \frac{q E}{m} ) ^{2}}$

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