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Physics-

General

Easy

Question

A tensile force of $2 cross times 10 to the power of 3 end exponent N$ doubles the length of a rubber band of cross-sectional area $2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent.$ The Young’s modulus of elasticity of the rubber band is

$4 \times 10^{7} N m^{- 2}$
$2 \times 10^{2} N m^{- 2}$
$10^{7} N m^{- 2}$
$0.5 \times 10^{7} N m^{- 2}$

The correct answer is: $10 to the power of 7 end exponent blank N m to the power of negative 2 end exponent$

Strain = $fraction numerator increment l over denominator l end fraction equals fraction numerator l over denominator l end fraction equals 1$
$therefore blank Y equals s t r e s s equals blank fraction numerator 2 blank cross times 10 to the power of 3 end exponent N over denominator 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent end fraction equals 10 to the power of 7 end exponent N m to the power of negative 2 end exponent$

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