IIT-JEE 2010 · Paper 1 · Physics Q80

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PhysicsInteger+3Question 80 of 84

A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is $4.9\times10^{-7}\ \mathrm m^2$ . If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s⁻¹. If the Young's modulus of the material of the wire is $n\times10^9$ N m⁻², the value of n is

Your answer

Answer and worked solution

Official answer: 4

Wire as a spring. $k=\dfrac{YA}L$ , so $\omega^2=\dfrac{YA}{mL}$ .

Solve.
$Y=\frac{\omega^2mL}A=\frac{140^2\times0.1\times1}{4.9\times10^{-7}}=\frac{1960}{4.9\times10^{-7}}=4\times10^9\ \mathrm{N\,m^{-2}}.$
So $n=4$ .

Answer: 4

Topics:Hooke's law and elastic moduli Spring-mass systems

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IIT-JEE 2010 · Paper 1, Physics question 80