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  • A  0.4 Kg mass is suspended from a wire of negligible mass The length of the wire is  2 m and its cross -sectional area is <img alt="6.2 \times 10^{-8} \mathrm{~m}^2" src="https://entranc

A  0.4 Kg mass is suspended from a wire of negligible mass The length of the wire is  2 m and its cross -sectional area is 6.2 \times 10^{-8} \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 160 rad/s. If the young's modulus of the material of the wire is P \times 10^8 \mathrm{~N} / \mathrm{m}^2 the value of  P

 

Option: 1

3203.23


Option: 2

3303.23


Option: 3

3403.23


Option: 4

3503.23


Answers (1)

best_answer

\frac{F}{A}=\frac{Y \Delta l}{l} \\

F=\frac{A Y}{l} \cdot \Delta l \\

K=\frac{AY}{l}

w=\sqrt{\frac{K}{m}} \\

w=\sqrt{\frac{A Y}{l m}} \\

Y=\frac{\omega^2 m l}{A}

Y=\frac{160 \times 160 \times 0.4 \times 2}{6.2 \times 10^{-8}}=\frac{160 \times 160 \times 8}{62 \times 10^{-8}} \\

Y=\frac{160 \times 160 \times 8 \times 10^8}{62} \\

Y=\frac{204800}{62} \times 10^8 \\

Y=3303.23 \times 10^8 \mathrm{~N} / \mathrm{m}^2 \\

P=3303.23

Posted by

HARSH KANKARIA

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