Kinematics Question 488
Question: A stone of mass 1 kg tied to a light inextensible string of length $ L=\frac{10}{3}m $ it’s whirling in a circular path of radius $ L $ in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if $ g $ it’s taken to be $ 10m/{{\sec }^{2}} $ , the speed of the stone at the highest point of the circle it’s [CBSE PMT 1990]
Options:
A) $ 20m/\sec $
B) $ 10\sqrt{3}m/\sec $
C) $ 5\sqrt{2}m/\sec $
D) $ 10m/\sec $
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Answer:
Correct Answer: D
Solution:
Since the maximum tension $ T _{B} $ in the string moving in the vertical circle is at the bottom and minimum tension $ T _{T} $ is at the top.
$ T _{B}=\frac{mv _{B}^{2}}{L}+mg $ and $ T _{T}=\frac{mv _{T}^{2}}{L}-mg $
$ \frac{T _{B}}{T _{T}}=\frac{\frac{mv _{B}^{2}}{L}+mg}{\frac{mv _{T}^{2}}{L}-mg}=\frac{4}{1} $ or $ \frac{v _{B}^{2}+gL}{v _{T}^{2}-gL}=\frac{4}{1} $ or $ v _{B}^{2}+gL=4v _{T}^{2}-4gL $
but $ v _{B}^{2}=v _{T}^{2}+4gL $
$ v _{T}^{2}+4gL+gL=4v _{T}^{2}-4gL $
therefore $ 3v _{T}^{2}=9gL $
$ v _{T}^{2}=3\times g\times L=3\times 10\times \frac{10}{3} $
or $ v _{T}=10m/\sec $
BETA
