JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(g=\) \(10\,m / s ^2\) )
- A \((-4 \sqrt{3}+1) N\)
- B \(4(\sqrt{3}+1) N\)
- C \(-4(\sqrt{3}-1) N\)
- D \((4 \sqrt{3}-1) N\)
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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