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 × 10^ -7 ~m^2 . If the mass is pulled a little in the vertically downward direction and

Correct answer: 0 ω= (K)/(m) = (Y A)/(l m) = (n × 10^9)(4.9 × 10^ -7 ) 1 × 0.1 Putting, ω=140 rad ^ -1 in above equation we get, n=4 Answer is 4 .

JEE Advanced 2010Paper 1

JEE Advanced 2010 · Paper 1 · Physics · Mechanical Properties of Solids

A 0.1 ~kg mass is suspended from a wire of negligible mass. The...

0.1 \mathrm{~kg}

mass is suspended from a wire of negligible mass. The length of the wire is

1 \mathrm{~m}

and its cross-sectional area is

4.9 \times 10^{-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 \mathrm{rad} \mathrm{s}^{-1}

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

n \times 10^9 \mathrm{Nm}^{-2}

, the value of

n

\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^9\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}

Putting,

\omega=140 \operatorname{rad} \mathrms^{-1}

in above equation we get, n=4

\therefore

Answer is 4 .