AP EAMCET · PHYSICS · Laws of Motion
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 \(\sin ^{-1}\left(\frac{1}{2}\right)\)
- B \(\tan ^{-1}\left(\frac{1}{2}\right)\)
- C \(\sin ^{-1}\left(\frac{1}{3}\right)\)
- D \(\tan ^{-1}\left(\frac{1}{3}\right)\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}\left(\frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
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…
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