Question

The length of a steel cylinder is kept constant by applying pressure at its two ends. When the temperature of rod is increased by ${100}^{\circ }$C from its initial temperature, the increase in pressure to be applied at its ends is $(Y_{steel}=2\times {10}^{11}N/m^2,{\alpha }_{steel}=11\times {10}^{-6}{l}^{0}C,1atm={10}^{5}N/m^2.)$

• A. $22\times {10}^{7}atm$
• B. $2.2\times {10}^{3}atm$
• C. zero
• D. $4.3\times {10}^{3}atm$

Answer 1

The correct option is B $2.2\times {10}^{3}atm$

If the rod is allowed to expand, then it will expand by $\Delta l=l\alpha$ $\Delta T$ is due to increase in temperature by $\Delta T$. But as the length of the cylinder is kept constant by applying pressure, a stress is developed in the cylinder. The decrease in length of cylinder due to elasticity is $\Delta l=l\alpha \Delta T$ and a compressive stress will develop in it.

Stress = Excess pressure applied at ends
$=Y\times \frac{\Delta l}{l}Y\alpha \Delta T$
$=2\times {10}^{11}\times 11\times {10}^{-6}\times 100=2.2\times {10}^{3}atm$