The bob of a simple pendulum has mass 2 g and a charge of $5.0 \mu \mathrm{C}$. It is at rest in a uniform horizontal electric field of intensity $2000 \frac{\mathrm{~V}}{\mathrm{~m}}$. At equilibrium, the angle that the pendulum makes with the vertical is: (take $g=10 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}$ )

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Laws of Motion

The bob of a simple pendulum has mass 2 g and a charge of

5.0 μ C

. It is at rest in a uniform horizontal electric field of intensity

2000 \frac{V}{m}

. At equilibrium, the angle that the pendulum makes with the vertical is: (take

g = 10 \frac{m}{s ^{2}}

)

(A) $tan^{- 1} (2.0)$
(B) $tan^{- 1} (0.2)$
(C) $tan^{- 1} (5.0)$
(D) $tan^{- 1} (0.5)$ ✓ Correct

Answer: (D)

Solution

At equilibrium the string tilts so that tension balances the downward weight and the horizontal electric force together, which means

tan θ = \frac{q E}{m g}

. The electric force is

q E = 5 \times 1 0^{- 6} \times 2000 = 0.01

N and the weight is

m g = 2 \times 1 0^{- 3} \times 10 = 0.02

N, so

tan θ = \frac{0.01}{0.02} = 0.5

and the pendulum makes an angle

tan^{- 1} (0.5)

with the vertical.

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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