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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2019
  • Medium
  • Numerical

A mass of \(10\,kg\) is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of \(45^o\) at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is ..........  \(N\) \((g = 10\,ms^{-2})\) 

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Answer & Solution

Correct Answer

(D) \(100\)

Step-by-step Solution

Detailed explanation

\(T\,\,cos\,45^o = mg\)
\(T\,\,sin\,45^o = F\)
\(\Rightarrow F = mg = 100\,N.\)
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