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
$(Y_{\text{steel }}=2\times 10^{11}N/m^2$ ${\alpha }_{\text{steel }}=11\times 10^{-6}/{}^{\circ }C,1atm=10^{5}N/m^2.)$

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

Answer 1

Answer: (B)

If the rod is allowed to expand, then it will expand by $\Delta l=l\alpha \Delta T$ 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
$=Y\times \frac{\Delta l}{l}Y\alpha \Delta T$
$=2\times 10^{11}\times 11\times 10^{-6}\times 100$
$=2.2\times 10^{3}atm$