First slide

Thermal expansion

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^oC from its initial temperature, the increase in pressure to be applied at its ends is $(Y_{steel} = 2 \times 10^{11} N / m^{2}, α_{steel} = 11 \times 10^{- 6} /^{\circ} C, 1 atm = 10^{5} N / m^{2} .)$

Moderate

A

$22 \times 10^{7} atm$
B

$2.2 \times 10^{3} atm$
C

$zero$
D

$4.3 \times 10^{3} atm$

Solution

If the rod is allowed to expand, then it will expand by $Δl = lα$ $ΔT$ due to increase in temperature by $ΔT$ . 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 $Δl = lαΔT$ and a compressive stress will develop in it.
Stress = Excess pressure applied at ends
$\begin{array}{l} ∆ P = Y \times \frac{Δl}{l} = YαΔT \\ = 2 \times 10^{11} \times 11 \times 10^{- 6} \times 100 = 2 .2 \times 10^{3} atm \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App