JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A mass of \(10\; kg\) is suspended by a rope of length \(4 \;\mathrm{m},\) from the ceiling. A force \(\mathrm{F}\) is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of \(45^{\circ}\) with the vertical. Then \(\mathrm{F}\) equals ........... \(N\) (Take \(\mathrm{g}=10 \;\mathrm{ms}^{-2}\) and the rope to be massless)
- A \(100\)
- B \(90\)
- C \(75\)
- D \(70\)
Answer & Solution
Correct Answer
(A) \(100\)
Step-by-step Solution
Detailed explanation
For equilibrium. \(\mathrm{T} \sin 45^{\circ}=\mathrm{F}\) and \(\mathrm{T} \cos 45^{\circ}=10 \mathrm{g}\) equation \((1) /(2)\) we get \(\mathrm{F}=10 \mathrm{g}\) \(=100 \mathrm{N}\)
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