A 0.1kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross sectional area is $4.9 \times 10^{- 7} m^{2}$ . If the mass is pulled slightly in the vertically downward direction and then released, it performs simple harmonic motion of angular frequency $140 rad / s$ . Then the Young’s modulus of material of the wire is(Assume that the Hooke’s law is valid throughout the motion )

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$5 \times 10^{8} N / m^{2}$

$4 \times 10^{9} N / m^{2}$

$4 \times 10^{8} N / m^{2}$

$14 \times 10^{9} N / m^{2}$

answer is A.

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Detailed Solution

We know that $ω = \sqrt{\frac{k}{m}}$ ……….. (i)
Here $Y = \frac{F L}{A l} \Rightarrow F = (\frac{Y A}{L}) l$
Comparing the above equation with $F = k l$ we get
$k = (\frac{Y A}{L})$ …….. (ii)
From equations (i) and (ii) we get
$∴ 140 = \sqrt{\frac{Y \times 4.9 \times 10^{- 7}}{0.1 \times 1}}$
$∴ Y = 4 \times 10^{9} N / m^{2}$
Therefore, the correct answer is (A).