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Q.

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

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a

$\tan^{- 1} (0.5)$

b

$\tan^{- 1} (0.2)$

c

$\tan^{1} (5.0)$

d

$\tan^{- 1} (2.0)$

answer is D.

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Detailed Solution

At equilibrium resultant on bob must be zero, so
$T = \cos θ = mg$ ....(i)
$Tsin θ = qE$ .........(ii)
Solving (i) and (ii) we get
$\begin{array}{l} \tan θ = \frac{qE}{mg} \\ \tan θ = \frac{5 \times 10^{- 6} \times 2000}{2 \times 10^{- 3} \times 10} = \frac{1}{2} \end{array}$

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$\Rightarrow \tan^{- 1} (\frac{1}{2})$
$[\begin{array}{l} Here, q = 5 \times 10^{- 6} C \\ E = 2000 v / m = 2 \times 10^{- 3} kg \end{array}]$

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