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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2025 · 24 January · Evening Shift · Physics · Laws of Motion

A string of length L is fixed at one end and carries a mass of M at...

A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes ((3)/(π)) rotations
A string of length LL is fixed at one end and carries a mass of MM at the other end. The mass makes (3π)\left(\frac{3}{\pi}\right) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ……\ldots\ldots ML.

Correct answer

36

Step-by-step solution

ω=3π×2π=6rad/sR=Lsin⁡θ\begin{aligned} & \omega=\frac{3}{\pi} \times 2 \pi=6 \mathrm{rad} / \mathrm{s} \\ & R=L \sin \theta\end{aligned} A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes ((3)/(π)) rotations
and T=Mg2+ω4R2T=M \sqrt{g^2+\omega^4 R^2} Also, Tsin⁡θ=Mω2⋅Lsin⁡θT \sin \theta=M \omega^2 \cdot L \sin \theta⇒T=M(36)L⇒T=36ML\begin{aligned} & \Rightarrow \quad T=M(36) \mathrm{L} \\ & \Rightarrow \quad T=36 \mathrm{ML} \end{aligned}$$
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