JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
In case of vertical circular motion of a particle by a thread of length \(r\) if the tension in the thread is zero at an angle \(30^{\circ}\) shown in figure, the velocity at the bottom point (A) of the circular path is ( \(g =\) gravitational acceleration)
- A \(\sqrt{5gr}\)
- B \(\sqrt{\frac{7}{2}gr}\)
- C \(\sqrt{4gr}\)
- D \(\sqrt{\frac{5}{2}gr}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{7}{2}gr}\)
Step-by-step Solution
Detailed explanation
\(T + mg \cos 60^{\circ}=\frac{ mV ^2}{\ell}\) \(T =0\) \(V ^2=\frac{ g \ell}{2}\) here V is the speed at point A M.E.C. \(\frac{1}{2} mu ^2= mg \left(\ell+\ell \cos 60^{\circ}\right)+\frac{1}{2} mV ^2\) \(u ^2=3 g \ell+\frac{ g \ell}{2}\) \(u =\sqrt{\frac{7 g \ell}{2}}\)
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