JEE Advanced · Physics · 4. Motion in 2D
A ball of mass \((m) 0.5 \mathrm{~kg}\) is attached to the end of a string having length \((L) 0.5 \mathrm{~m}\). The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is \(324 \mathrm{~N}\). The maximum possible value of angular velocity of ball (in rad/s) is

- A 9
- B 18
- C 27
- D 36
Answer & Solution
Correct Answer
(D) 36
Step-by-step Solution
Detailed explanation

\(T \cos \theta\) component will cancel \(\mathrm{mg}\).
\(T \sin \theta\) component will provide necessary centripetal force to the ball towards centre \(C\).
\(
\begin{array}{rlrl}
& \therefore & T \sin \theta & =m r \omega^2=m(l \sin \theta) \omega^2 \\
& \text { or } & T & =m l \omega^2 \\
& \therefore & \omega & =\sqrt{\frac{T}{m l}} \\
& \text { or } & \omega_{\max } & =\sqrt{\frac{T_{\max }}{m l}}=\sqrt{\frac{324}{0.5 \times 0.5}} \\
& = & 36 \mathrm{rad} / \mathrm{s}
\end{array}
\)
\(\therefore\) Correct option is (d).
Analysis of Question
(i) Question is simple.
(ii) This is called the conical pendulum.
(iii) The interesting fact in this problem is that \(\omega\) or \(T\) is independent of \(\theta\).
\(\omega \propto \sqrt{T}\)
If \(\omega\) is increased, \(T\) will also increase.
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