A block of mass $8 \mathrm{~kg}$ is suspended by a rope of length $3 \mathrm{~m}$ from the ceiling. A force of $40 \mathrm{~N}$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is (acceleration due to gravity $=10 \mathrm{~m} \mathrm{~s}^{-2}$, neglect the mass of the rope)

• A

  $\sin ^{-1}\left(\frac{1}{2}\right)$
• Correct answer
  B

  $\tan ^{-1}\left(\frac{1}{2}\right)$
• C

  $\sin ^{-1}\left(\frac{1}{3}\right)$
• D

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

Solution

Let ' $\theta$ ' be the angle that string makes in equilibrium. So, $\mathrm{F}=\mathrm{T} \sin \theta$...(i) $\mathrm{mg}=\mathrm{T} \cos \theta$...(ii) Dividing (ii) by (i), we get $\begin{aligned} & \frac{\mathrm{F}}{\mathrm{mg}}=\tan \theta \Rightarrow \tan \theta=\frac{40}{80}=\frac{1}{2} \\ & \theta=\tan ^{-1}\left(\frac{1}{2}\right)\end{aligned}$

Hence, the correct answer is option: 'B'.