A string of length ' $L$ ' fixed at one end carries a mass ' $m$ ' at the other end. The string makes…

A string of length ' $L$ ' fixed at one end carries a mass ' $m$ ' at the other end. The string makes $\frac{3}{\pi}$ r.p.s. around the vertical axis through fixed end. The tension in the string is
  1. 72 mL
  2. 18 mL
  3. 9 mL
  4. 36 mL

Solution

Consider the free body diagram On considering horizontal force balance: $\mathrm{T} \sin \theta=\mathrm{m} \omega^2 \mathrm{R}$ Using the geometry, $\mathrm{R}=\mathrm{L} \sin \theta$ $\therefore \mathrm{T}=\left(\mathrm{m} \omega^2 \mathrm{~L}\right)$ Given, $\omega=\left(\frac{3}{\pi}\right)$ r.p.s. $=\left(\frac{3}{\pi}\right)(2 \pi) \mathrm{rad} / \mathrm{s}=6 \mathrm{rad} / \mathrm{s}$ $\therefore \mathrm{T}=36 \mathrm{~mL}$ *

Asked in: MHT CET 2022 (08 Aug Shift 2)

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