JEE PYQ: Electrostatics Question 37
Question 37 - 2019 (08 Apr Shift 1)
The bob of a simple pendulum has mass 2 g and a charge of $5.0;\mu$C. It is at rest in a uniform horizontal electric field of intensity $2000$ V/m. At equilibrium, the angle that the pendulum makes with the vertical is: (take $g = 10$ m/s$^2$)
(1) $\tan^{-1}(2.0)$
(2) $\tan^{-1}(0.2)$
(3) $\tan^{-1}(5.0)$
(4) $\tan^{-1}(0.5)$
Show Answer
Answer: (4)
Solution
$\tan\theta = \frac{qE}{mg} = \frac{5 \times 10^{-6} \times 2000}{2 \times 10^{-3} \times 10} = \frac{10^{-2}}{2 \times 10^{-2}} = 0.5$. So $\theta = \tan^{-1}(0.5)$.
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