JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 23)
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A string of length $L$ is fixed at one end and carries a mass of $M$ at the other end. The mass makes $\left(\frac{3}{\pi}\right)$ rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is __________ ML.
Answer
36
Explanation
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$$\begin{aligned} & \mathrm{T} \cos \theta=\mathrm{mg} \\ & \mathrm{~T} \sin \theta=\mathrm{M} \omega^2 \mathrm{R} \\ & \text { Using equation (2) } \\ & \mathrm{T} \sin \theta=\mathrm{M} \omega^2(\mathrm{~L} \sin \theta) \\ & \mathrm{T}=\mathrm{M} \omega^2 \mathrm{~L}=\mathrm{M}\left(\frac{3}{\pi} \times 2 \pi\right)^2 \mathrm{~L} \\ & \mathrm{~T}=36 \mathrm{ML} \end{aligned}$$
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