Q88 / 90 · JEE Main 2021 (31 Aug Shift 1) · Physics

A wire having a linear mass density $9.0 \times 10^{- 4} kg m^{- 1}$ is stretched between two rigid supports with a tension of $900 N .$ The wire resonates at a frequency of $500 Hz .$ The next higher frequency at which the same wire resonates is $550 Hz .$ The length of the wire is ___________ $m .$

Correct answer: 10

Solution

f_{n} = \frac{n v}{2 ℓ} = 500

f_{n + 1} = \frac{(n + 1) v}{2 ℓ} = 550

\frac{n + 1}{n} = \frac{11}{10}

Thus,

n = 10

Thus,

ℓ = \frac{nv}{2 f_{n}} (v = \sqrt{\frac{T}{μ}})

= \frac{10}{2 \times 500} \times \sqrt{\frac{900}{9 \times 10^{- 4}}} = 10 m

Hence, the correct answer is 10.

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