Question

What force should be applied to the ends of steel rod of a cross sectional area $10c{m}^2$ to prevent it from elongation when heated form $273vK$ to $303k$?
$\alpha$ of steel = ${{10}^{-5}}^{\circ }{C}^{-1},Y=2\times {10}^{11}N{m}^{-2}$

• A. $2\times {10}^{4}N$
• B. $3\times {10}^{4}N$
• C. $6\times {10}^{4}N$
• D. $12\times {10}^{4}N$

Answer 1

Answer: (C)

$6\times {10}^{4}N$

Area A = 10 $c{m}^2=0.01m^2$
Change in temperature = dt = 303 - 273 = 30 [change in temperature in degree kelvin is same as degree centigrade]
Y = stress / strain = (Force / area)/ (increase in length / original length)
= (F/A) / dl /l
Now, $\alpha =dl/\left(ldt\right)$
or, $dl=l\alpha dt$
Putting the value of dl we get,
$Y=F/\left(\alpha dtA\right)$
Or, $F=Y\alpha dtA$
Putting value of $\alpha ={10}^{-5}$ and Y = 2 X ${10}^{11}$
We get force as $6\times {10}^4$ N