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 4.9×10−7 m2 . 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 )

# 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 $4.9×{10}^{-7}{\text{\hspace{0.17em}m}}^{2}$ . If the mass is pulled  slightly in the vertically downward direction and then released, it performs simple harmonic motion of angular frequency  $140\text{\hspace{0.17em}rad}/\text{s}$. Then the Young’s modulus of material of the wire is(Assume that the Hooke’s law is valid throughout the motion )

1. A

$4×{10}^{9}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

2. B

$14×{10}^{9}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

3. C

$4×{10}^{8}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

4. D

$5×{10}^{8}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

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### Solution:

We know that  $\omega =\sqrt{\frac{k}{m}}$……….. (i)
Here  $Y=\frac{FL}{Al}⇒F=\left(\frac{YA}{L}\right)l$
Comparing the above equation with $F=kl$ we get
$k=\left(\frac{YA}{L}\right)$…….. (ii)
From equations (i) and (ii) we get
$\therefore \text{\hspace{0.17em}}140=\sqrt{\frac{Y×4.9×{10}^{-7}}{0.1×1}}$
$\therefore Y=4×{10}^{9}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$
Therefore, the correct answer is (A).  Register to Get Free Mock Test and Study Material

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