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A block of mass 8 kg is suspended by a rope of length 3m from the ceiling. A force of 40N is applied horizontally to the block, then the angle that the rope makes with the vertical in equilibrium is ( $g = 10 m / s^{2}$ ;neglect mass of the rope)

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$\sin^{- 1} (\frac{1}{2})$

$\sin^{- 1} (\frac{1}{3})$

$\tan^{- 1} (\frac{1}{2})$

$\tan^{- 1} (\frac{1}{3})$

answer is B.

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Detailed Solution

Use F.B.D and equilibrium
$\tan θ = \frac{F}{m g} = \frac{40}{80}$
$θ = \tan^{- 1} (\frac{1}{2})$
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