A tensile force of $2 \times {10^3}\,{\rm{N}}$ doubles the length of a rubber band of cross-sectional area $2 \times {10^{ - 4}}\,{{\rm{m}}^2}$. The Young’s modulus of elasticity of the rubber band is

${10^7}\,{\rm{N}}{{\rm{m}}^{ - 2}}$

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

$2 \times {10^2}\,{\rm{N}}{{\rm{m}}^{ - 2}}$

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

NEET UG PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Medium

1 of 1

Application +4 | -1

A tensile force of

2 \times {10^3}\,\mathrm{N}

doubles the length of a rubber band of cross-sectional area

2 \times {10^{ - 4}}\,{\mathrm{m}^2}

. The Young’s modulus of elasticity of the rubber band is

View answer and explanation

Correct option: C

{10^7}\,\mathrm{N}{\mathrm{m}^{ - 2}}

Option C is correct. Young's modulus is calculated using the formula Y = (F/A) / (ΔL/L), where the length doubles, meaning ΔL = L.