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JEE Main 20209 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Evening Shift
Subject: Physics
Chapter: Waves and Sound

JEE Main 2020 · 9 January · Evening Shift · Physics · Waves and Sound

A wire of length L and mass per unit length 6.0 × 10 ^ - 3 kg m ^ -...

A wire of length

L

and mass per unit length

6.0 \times {10}^{- 3}kg{m}^{- 1}

is put under tension of

540N .

Two consecutive frequencies that it resonates at are:

420Hz

and

490Hz .

Then

L

in meters is :

Options

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

Correct answer

A. $2.1m$

Step-by-step solution

Fundamental frequency

= 490 - 420 = 70Hz

70 = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

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

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

⇒ L \approx 2.14m

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