Q52 / 75 · JEE Main 2020 (09 Jan Shift 2) · Physics

A wire of length $L$ and mass per unit length $6.0 \times 10^{- 3} k g m^{- 1}$ is put under tension of $540 N .$ Two consecutive frequencies that it resonates at are: $420 H z$ and $490 H z .$ Then $L$ in meters is :

Correct answer
A
$2.1 m$
B
$1.1 m$
C
$8.1 m$
D
$5.1 m$

Solution

Fundamental frequency

= 490 - 420 = 70 H z

70 = \frac{1}{2 L} \sqrt{\frac{T}{μ}}

\Rightarrow 70 = \frac{1}{2 L} \sqrt{\frac{540}{6 \times 10^{- 3}}}

\Rightarrow L = \frac{1}{2 \times 70} \sqrt{90 \times 10^{3}} = \frac{300}{140}

\Rightarrow L \approx 2.14 m

Hence, the correct answer is option: 'A'.

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