Q. A metallic ring of radius r and cross-sectional area A is fitted into a wooden circular disc of radius R(R>r). If the Young’s modulus of the material of the ring is Y , the force with which the metal ring expands is
A) $\frac{AYR}{r}$
B) $\frac{AY(R - r)}{r}$
C) $\frac{Y(R - r)}{Ar}$
D) $\frac{YR}{AR}$
Explanation
(b) Young’s modulus Y $= \frac{F}{A} \bullet \frac{L}{l}$
$\therefore$ Force F $= \frac{AYl}{L} = \frac{AY\left\lbrack 2\pi(R - r) \right\rbrack}{2\pi r}$
⟹ $F = \frac{AY(R - r)}{r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.