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QJEEM23Phf74b: JEE Main 2023 Physics - Class 11 JEEM23Phf74b Solution: A stone of mass is tied to end of a massless string of length . …

A stone of mass 1 kg1 \mathrm{~kg} is tied to end of a massless string of length 1 m1 \mathrm{~m} . If the breaking tension of the string is 400 N400 \mathrm{~N} , then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :

Multiple Choice Options

  • A 20 ms−120 \mathrm{~ms}^{-1}
  • B 40 ms−140 \mathrm{~ms}^{-1}
  • C 400 ms−1400 \mathrm{~ms}^{-1}
  • D 10 ms−110 \mathrm{~ms}^{-1}

Step-by-Step Verified Solution

JEE Main 2023 (Online) 31st January Evening Shift Physics - Circular Motion Question 31 English Explanation

amp;Tsin⁡θ=mv2lsin⁡θamp;cos⁡θ=mgT........(1)amp;sin⁡2θ=mv2Tl........(2)\begin{aligned}& T \sin \theta=\frac{m v^{2}}{l \sin \theta} \\\\& \cos \theta=\frac{m g}{T} ........(1) \\\\& \sin^2 \theta=\frac{m v^{2}}{T l} ........(2)\end{aligned}

From (1) and (2),

1=(mgT)2+mv2Tl1=\left(\frac{m g}{T}\right)^{2}+\frac{m v^{2}}{T l}

⇒\Rightarrow 1=(10400)2+v24001=\left(\frac{10}{400}\right)^{2}+\frac{v^{2}}{400}

⇒\Rightarrow v2=399.78v^{2}=399.78

⇒\Rightarrow v=20 m/sv=20 \mathrm{~m} / \mathrm{s}

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