JEE MAIN - Physics (2016 (Offline) - No. 16)

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 supports is :
(take $${\,\,g = 10m{s^{ - 2}}}$$ )
$$2\sqrt 2 s$$
$$2\pi \sqrt 2 s$$
$$2\pi \sqrt 2 s$$
$$2$$ $$s$$

Explanation

JEE Main 2016 (Offline) Physics - Waves Question 113 English Explanation
We know that velocity in string is given by

$$v = \sqrt {{T \over \mu }} \,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$

where $$\mu = {m \over {\rm I}} = {{mass\,\,\,of\,\,\,string} \over {length\,\,\,of\,\,\,string}}$$

The tension $$T = {m \over \ell } \times x \times g\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$(a)$$ and $$(b)$$ $${{dx} \over {dt}} = \sqrt {gx} $$

$${x^{ - 1/2}}\,dx = \sqrt g \,dt$$

$$\therefore$$ $$\int\limits_0^\ell {{x^{ - 1/2}}} dx - \sqrt g \int\limits_0^\ell {dt} $$

$$2\sqrt \ell = \sqrt g \times t$$

$$\therefore$$ $$t = 2\sqrt {{\ell \over g}} = 2\sqrt {{{20} \over {10}}} = 2\sqrt 2 $$

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