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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2020
  • Medium
  • Numerical

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)

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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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