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A $0.1 \mathrm{~kg}$ mass is suspended from a wire of negligible mass. The length of the wire is $1 \mathrm{~m}$ and its cross-sectional area is $4.9 \times 10^{-7} \mathrm{~m}^2$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 \mathrm{rad} \mathrm{s}^{-1}$. If the Young's modulus of the material of the wire is $n \times 10^9 \mathrm{Nm}^{-2}$, the value of $n$ is
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Verified Answer
The correct answer is:
4
$\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^9\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}$
Putting, $\omega=140 \operatorname{rad} \mathrm{s}^{-1}$ in above equation we get,
$$
n=4
$$
$\therefore$ Answer is 4 .
Putting, $\omega=140 \operatorname{rad} \mathrm{s}^{-1}$ in above equation we get,
$$
n=4
$$
$\therefore$ Answer is 4 .
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