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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2023
  • Medium
  • MCQ

As per given figure, a weightless pulley \(P\) is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if \(g=\) \(10\,m / s ^2\) )

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Answer & Solution

Correct Answer

(B) \(4(\sqrt{3}+1) N\)

Step-by-step Solution

Detailed explanation

\(4 g \sin 60^{\circ}-T=4 a.....(1)\)

\(T - g \sin 30^{\circ}= a.......(2)\)

Solving \((1)\) and \((2)\) we get.

\(20 \sqrt{3}-T=4 T-20\)

\(T=4(\sqrt{3}+1)\,N\)

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