MHT CET · Physics · Laws of Motion
A ball of mass 'm' is attached to the free end of an inextensible string of length ' \(\ell\) '. Let 'T' be the tension in the string. The ball is moving in horizontal circular path about the vertical axis. The angular velocity of the ball at any particular instant will be
- A \(\sqrt{\frac{\mathrm{T}}{\mathrm{m} \ell}}\)
- B \(\sqrt{\frac{\mathrm{T} \ell}{\mathrm{m}}}\)
- C \(\sqrt{\frac{\mathrm{m} \ell}{\mathrm{T}}}\)
- D \(\sqrt{\frac{\mathrm{Tm}}{\ell}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{\mathrm{T}}{\mathrm{m} \ell}}\)
Step-by-step Solution
Detailed explanation
A ball of mass ' \(m\) ' is attached to the free end of an inextensible string of length 'I'. Let 'T' be the tension in the string. The ball is moving in horizontal circular path about the vertical axis. The angular velocity of the ball at any particular instant will be \(\sqrt{\frac{T}{m l}}\).
Explanation:
\(\begin{array}{l}\mathrm{T}=\mathrm{ml} \omega^{2} \\\therefore \omega=\sqrt{\frac{T}{m l}}\end{array}\)
Explanation:
\(\begin{array}{l}\mathrm{T}=\mathrm{ml} \omega^{2} \\\therefore \omega=\sqrt{\frac{T}{m l}}\end{array}\)
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