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14.Waves and Sound
hard
A wire of length $L$ and mass per unit length $6.0\times 10^{-3}\; \mathrm{kgm}^{-1}$ is put under tension of $540\; \mathrm{N}$. Two consecutive frequencies that it resonates at are : $420\; \mathrm{Hz}$ and $490 \;\mathrm{Hz}$. Then $\mathrm{L}$ in meters is
A
$8.1 \;m$
B
$5.1 \;m$
C
$1.1 \;m$
D
$2.1 \;m$
(JEE MAIN-2020)
Solution
$\frac{\mathrm{nv}}{2 \ell}=420$
$\frac{(\mathrm{n}+1) \mathrm{v}}{2 \ell}=490$
$\frac{v}{2 \ell}=70$
$\ell=\frac{\mathrm{v}}{140}=\frac{1}{140} \sqrt{\frac{540}{6 \times 10^{-3}}}=\frac{1}{140} \sqrt{90 \times 10^{3}}$
$\ell=\frac{300}{140}=2.142$
Standard 11
Physics
