Q. An area of cross-section of rubber string is $2cm^{2}$. Its length is doubled when stretched with a linear force of $2 \times 10^{5}dynes$. The Young’s modulus of the rubber in $dyne/cm^{2}$ will be
A) $4 \times 10^{5}$
B) $1 \times 10^{5}$
C) $2 \times 10^{5}$
D) ${1 \times 10}^{4}$
Explanation
(b) If length of the wire is doubled then strain = 1
$\therefore Y = Stress = \frac{Force}{Area} = \frac{2 \times 10^{5}}{2} = 10^{5}\frac{dyne}{cm^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.