Q128 / 150 · MHT CET 2022 (10 Aug Shift 1) · Physics

A string of $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $\frac{\pi}{2}$ revolutions per second around the vertical axis through the fixed end as shown in the figure, then the tension in the string is Figure

  • A
    $2 M L$
  • B
    $M L$
  • Correct answer
    C
    $16 M L$
  • D
    $4 M L$

Solution

Given: $\omega=\frac{2}{\pi}$ rev.per sec $\Rightarrow \omega=2 \pi \frac{2}{\pi} \mathrm{rad} / \mathrm{s}=4 \mathrm{rad} / \mathrm{s}$ The horizontal force balance reads: $T \sin \theta=M R \omega^2$ From figure $R=L \sin \theta$ $\Rightarrow T=M L \omega^2$ On introducing, $\omega=4 \mathrm{rad} / \mathrm{s}$ $\Rightarrow T=M L(4)^2=16 \mathrm{ML}$

Hence, the correct answer is option: 'C'.

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