JEE Main 2016 · 3 Apr · Shift 1 · Physics Q11

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PhysicsSingle correct+4 / -1Question 11 of 90

A uniform string of length 20 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 support is :

(take g = 10 ms $^{-2}$ )

2\pi\sqrt2

2 s

2\sqrt2

\sqrt2

Answer and worked solution

Official answer: C

At height $y$ above the lower end, the tension is the weight of the string below, so $v=\sqrt{gy}$ . Then
$t=\int_0^L\frac{dy}{\sqrt{gy}}=2\sqrt{\frac Lg}=2\sqrt2\text{ s}.$

Topics:Speed of waves

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 11