Q119 / 150 · MHT CET 2022 (08 Aug Shift 2) · Physics
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
- A72 mL
- B18 mL
- C9 mL
- Correct answerD36 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. $=\left(\frac{3}{\pi}\right)(2 \pi) \mathrm{rad} / \mathrm{s}=6 \mathrm{rad} / \mathrm{s}$ $\therefore \mathrm{T}=36 \mathrm{~mL}$
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. $=\left(\frac{3}{\pi}\right)(2 \pi) \mathrm{rad} / \mathrm{s}=6 \mathrm{rad} / \mathrm{s}$ $\therefore \mathrm{T}=36 \mathrm{~mL}$
Hence, the correct answer is option: 'D'.