Q133 / 170 · Manipal (MET) 2021 · Physics

A steel ring of radius $r$ and cross-sectional area $A$ is fitted on to a wooden disc of radius $R(R \gt r)$. If Young's modulus be $Y$, then the force with which the steel ring is expanded, is

  • A
    $A Y \frac{R}{r}$
  • Correct answer
    B
    $A Y\left(\frac{R-r}{r}\right)$
  • C
    $\frac{Y}{A}\left[\frac{R-r}{r}\right]$
  • D
    $\frac{Y r}{A R}$

Solution

Strain $=\frac{\Delta l}{l}=\frac{2 \pi R-2 \pi r}{2 \pi r}=\left(\frac{R-r}{r}\right)$ Stress $=Y \times$ strain $=\frac{F}{A}$ $\therefore \quad F=Y \times A \times$ strain $=A Y \times\left(\frac{R-r}{r}\right)$

Hence, the correct answer is option: 'B'.

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