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JEE Main 2016 · Speed of Waves

Speed of Waves question from JEE Main 2016

Q36JEE Main 2016Medium

Speed of Waves

A uniform string of length 20 m20\text{ m} is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the supports is: (take g=10 ms−2g = 10\text{ ms}^{-2})

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Answer: (a)

  1. At distance yy from the lower end, v=gyv = \sqrt{gy}, so dt=dygydt = \dfrac{dy}{\sqrt{gy}}.
  2. t=∫0Ldygy=2Lg=22010t = \displaystyle\int_0^L \dfrac{dy}{\sqrt{gy}} = 2\sqrt{\dfrac{L}{g}} = 2\sqrt{\dfrac{20}{10}}
  3. So the answer is (a) 222\sqrt{2} s.
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