MHT CET · Physics · Work Power Energy
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
- A 72 mL
- B 18 mL
- C 9 mL
- D 36 mL
Answer & Solution
Correct Answer
(D) 36 mL
Step-by-step Solution
Detailed explanation
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. \(=\left(\frac{3}{\pi}\right)(2 \pi) \mathrm{rad} / \mathrm{s}=6 \mathrm{rad} / \mathrm{s}\)
\(\therefore \mathrm{T}=36 \mathrm{~mL}\)
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