A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross sectional area is $4.9 \times 10^{- 7} m^{2}$ . If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad $s^{- 1}$ . If the Young's modulus of the material of the wire is $n \times 10^{9} N m^{- 2}$ , the value of $n$ is

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answer is 4.

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Detailed Solution

As we know, $y = \frac{F L}{A l} \Rightarrow F = (\frac{Y A}{L}) l$

Also, $F = K l$

or, $K l = (\frac{Y A}{L}) l \Rightarrow K = \frac{Y A}{L}$

Angular frequency $ω = \sqrt{\frac{K}{m}} o r, ω = \sqrt{\frac{Y A}{m l}}$

or, $140 = \sqrt{\frac{n \times 10^{9} \times 4.9 \times 10^{- 7}}{0.1 \times 1}} (∵ Y = 9 \times 10^{9} g i v e n)$

$∴ n = 4$

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