JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Elasticity & SHM

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 1 0^{- 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 r a d s^{- 1}

. If the Young's modulus of the material of the wire is

n \times 1 0^{9} N m^{- 2}

, the value of n is

Answer: 4

Solution

For a mass suspended from a wire performing SHM, the effective spring constant is:

k = \frac{Y A}{L}

where

Y

is Young's modulus,

A

is cross-sectional area, and

L

is length.

Angular frequency:

ω = \frac{k}{m} = \frac{Y A}{m L}

Given:

ω = 140

rad/s,

m = 0.1

kg,

L = 1

A = 4.9 \times 1 0^{- 7}

m²

140 = \frac{Y \times 4.9 \times 1 0 ^{- 7}}{0.1 \times 1}

14 0^{2} = \frac{Y \times 4.9 \times 1 0 ^{- 7}}{0.1}

19600 = Y \times 4.9 \times 1 0^{- 6}

Y = \frac{19600}{4.9 \times 1 0 ^{- 6}} = \frac{19600}{4.9} \times 1 0^{6} = 4000 \times 1 0^{6} = 4 \times 1 0^{9}

N/m²

Therefore,

n = 4

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer