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Physics-

General

Easy

Question

A copper wire of negligible mass, 1 m length and cross-sectional area $10 to the power of negative 6 end exponent$ is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rad $s to the power of negative 1 end exponent. blank$ If the elongation in the wire is $10 to the power of negative 3 end exponent m$ , then the Young’s modulus is

$4 \times 10^{11} N m^{- 2}$
$6 \times 10^{11} N m^{- 2}$
$8 \times 10^{11} N m^{- 2}$
$10 \times 10^{11} N m^{- 2}$

The correct answer is: $4 blank cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent$

$Y equals fraction numerator F l over denominator A increment l end fraction equals blank fraction numerator open parentheses m l omega to the power of 2 end exponent close parentheses l over denominator A blank increment l end fraction o r blank Y equals fraction numerator m l to the power of 2 end exponent omega to the power of 2 end exponent over denominator A increment l end fraction$
Or $Y equals blank fraction numerator 1 cross times 1 cross times 1 cross times 20 cross times 20 over denominator 10 to the power of negative 6 end exponent cross times 10 to the power of negative 3 end exponent end fraction equals 4 blank cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent$

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