JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 23)
The speed of a transverse wave passing through a string of length $$50 \mathrm{~cm}$$ and mass $$10 \mathrm{~g}$$ is $$60 \mathrm{~ms}^{-1}$$. The area of cross-section of the wire is $$2.0 \mathrm{~mm}^{2}$$ and its Young's modulus is $$1.2 \times 10^{11} \mathrm{Nm}^{-2}$$. The extension of the wire over its natural length due to its tension will be $$x \times 10^{-5} \mathrm{~m}$$. The value of $$x$$ is __________.
Answer
15
Explanation
$\mathrm{V}_{\mathrm{w}}=\sqrt{\frac{\mathrm{T}}{\mu}}$
$60=\sqrt{\frac{\mathrm{T}}{10 \times 10^{-3}} \times 0.5}$
$\mathrm{~T}=\frac{(60)^2 \times 10^{-2}}{0.5}=72 \mathrm{~N}$
$\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}}=\frac{72 \times 0.5}{2 \times 10^{-6} \times 1.2 \times 10^{11}}$
$=\frac{72 \times 5}{24} \times 10^{-5}=15 \times 10^{-5}$
$60=\sqrt{\frac{\mathrm{T}}{10 \times 10^{-3}} \times 0.5}$
$\mathrm{~T}=\frac{(60)^2 \times 10^{-2}}{0.5}=72 \mathrm{~N}$
$\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}}=\frac{72 \times 0.5}{2 \times 10^{-6} \times 1.2 \times 10^{11}}$
$=\frac{72 \times 5}{24} \times 10^{-5}=15 \times 10^{-5}$
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