Question

What force should be applied to the ends of steel rod of a cross sectional ares $10{cm}^2$ to prevent it from elongation when heated from $273K$ to $303k$?
($\alpha$ to steel ${10}^{-5}{{}^{0}C}^{-1},Y=2\times {10}^{11}{Nm}^{-2}$)

• 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 $A=10c{m}^2=0.01m^2$
Change in temperature $=dt=303-273=30$
[change in temperature in degree kelvin is same as degree centigrade]
$Y=\frac{stress}{strain}=\frac{\frac{force}{area}}{\frac{increaseinlength}{originallength}}=\frac{\frac{F}{A}}{\frac{dl}{l}}$
Now,
$\alpha =\frac{dl}{\left(ldt\right)}$
Or,
$dl=l\alpha dt$
Putting the value of dl we get,
$Y=\frac{F}{\left(\alpha dtA\right)}$
Or,
$F=Y\alpha dtA$
Putting value of $\alpha ={10}^{-5}andY=2\times {10}^{11}$
We get force as $6\times {10}^{4}N$
Hence the option C is the correct option