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

• 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
$=Y \times \frac{\Delta l}{l} Y \alpha \Delta T$
$=2 \times 10^{11} \times 11 \times 10^{-6} \times 100$
$=2.2 \times 10^{3} atm$