JEE PYQ: Thermal Properties Of Matter Question 6
Question 6 - 2020 (02 Sep 2020 Shift 2)
A wire of density $9 \times 10^{-3}$ kg cm$^{-3}$ is stretched between two clamps 1 m apart. The resulting strain in the wire is $4.9 \times 10^{-4}$. The lowest frequency of the transverse vibrations in the wire is (Young’s modulus of wire $Y = 9 \times 10^{10}$ Nm$^{-2}$), (to the nearest integer), ____.
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Answer: 35
Solution
$f = \frac{1}{2l}\sqrt{\frac{T}{\mu}} = \frac{1}{2l}\sqrt{\frac{Y \times \text{Strain}}{\sigma}} = \frac{1}{2}\sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{9 \times 10^3}} = \frac{1}{2} \times 70 = 35$ Hz
BETA
