Waves

JEE Main level, single correct, medium

70 of 142

Question 70 of 142

A wire of length LL and mass per unit length 6.0×10−3kgm−16.0\times {10}^{-3}kg{m}^{-1} is put under tension of 540N.540N. Two consecutive frequencies that it resonates at are: 420Hz420Hz and 490Hz.490Hz. Then LL in meters is :

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Answer: (A) 2.1m2.1m

Fundamental frequency =490−420=70Hz=490-420=70Hz
70=12LTμ70=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}
⇒70=12L5406×10−3\Rightarrow 70=\dfrac{1}{2L}\sqrt{\dfrac{540}{6\times {10}^{-3}}}
⇒L=12×7090×103=300140\Rightarrow L=\dfrac{1}{2\times 70}\sqrt{90\times {10}^{3}}=\dfrac{300}{140}
⇒L≈2.14m\Rightarrow L\approx 2.14m

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