Question

A steel ring of radius r and cross-section area ‘A’ is fitted on to a wooden disc of radius $R ( R > r )$. If Young's modulus be E, then the force with which the steel ring is expanded is

• A. $A E \frac { R } { r }$
• B. $A E \left( \frac { R - r } { r } \right)$
• C. $\frac { E } { A } \left( \frac { R - r } { A } \right)$
• D. $\frac { E r } { A R }$

Answer 1

Answer: (B)

$A E \left( \frac { R - r } { r } \right)$

Answer 2

Initial length (circumference) of the ring = 2πr

Final length (circumference) of the ring = 2πR

Change in length = 2πR – 2πr.

strain $= \frac { \text { change in length } } { \text { original length } }$ $= \frac { 2 \pi ( R - r ) } { 2 \pi r }$ $= \frac { R - r } { r }$

Now Young's modulus $E = \frac { F / A } { l / L } = \frac { F / A } { ( R - r ) / r }$

∴$F = A E \left( \frac { R - r } { r } \right)$