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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

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 ..........  \(N\) \((g = 10\,ms^{-2})\) 

  1. A \(200\)
  2. B \(140\)
  3. C \(70\)
  4. D \(100\)
Verified Solution

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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