Kinematics Question 566

Question: The maximum and minimum tension in the string whirling in a circle of radius 2.5 m with constant velocity are in the ratio 5 : 3 then it’s velocity is [Pb. PET 2003]

Options:

A) $ \sqrt{98}m/s $

B)$ 7m/s $

C) $ \sqrt{490}m/s $

D) $ \sqrt{4.9} $

Show Answer

Answer:

Correct Answer: A

Solution:

In this problem it’s assumed that particle although moving in a vertical loop but it’s speed remain constant. Tension at lowest point $ {T _{\max }}=\frac{mv^{2}}{r}+mg $ Tension at highest point $ {T _{\min }}=\frac{mv^{2}}{r}-mg $

$ \frac{{T _{\max }}}{{T _{\min }}}=\frac{\frac{mv^{2}}{r}+mg}{\frac{mv^{2}}{r}-mg}=\frac{5}{3} $ by solving we get,$ v=\sqrt{4gr} $

$ =\sqrt{4\times 9.8\times 2.5} $

$ =\sqrt{98}m/s $



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