MCQ
A uniform string oflength $20\ m$ 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= 10 $ $ms^{-2}$ )
  • $2$$\sqrt 2 s$
  • B
    $\sqrt 2 s$
  • C
    $\;2\pi \sqrt 2 s$
  • D
    $2s$

Answer

Correct option: A.
$2$$\sqrt 2 s$
a
We know that velocity in string is given by

$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}$              $...(I)$

where $\mu=\frac{\mathrm{m}}{1}=\frac{\text { mass of string }}{\text { length of string }}$

The tension $\mathrm{T}=\frac{\mathrm{m}}{\ell} \times \mathrm{x} \times \mathrm{g}$                   $...(II)$

From $(a)$ and $(b)$

$\frac{d x}{d t}=\sqrt{g x}$

$\mathrm{x}^{-1 / 2} \mathrm{dx}=\sqrt{\mathrm{g}} \mathrm{dt} \quad \therefore \int_{0}^{\ell} \mathrm{x}^{-1 / 2} \mathrm{dx}-\sqrt{\mathrm{g}} \int_{0}^{\ell} \mathrm{dt}$

$2 \sqrt{l}=\sqrt{g} \times t \quad \therefore t=2 \sqrt{\frac{\ell}{g}}=2 \sqrt{\frac{20}{10}}=2 \sqrt{2}$

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