Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 40 cm respectively. Assuming the load distribution to be uniform, then the compressional strain of each column. $\left(\mathrm{Y}=2\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2\right)$

 

Given,

$\mathrm{mass} =50,000 \mathrm{kg}, \mathrm{Y}=2\times {10}^{11}\mathrm{Pa}$

$\mathrm{Area}={{\mathrm{\pi r}}_1}^2-{{\mathrm{\pi r}}_2}^2=\mathrm{\pi }({(0.4)}^2-{(0.3)}^2) =0.22{\mathrm{m}}^2$

Since there are 4 columns, the stress needed to support the large structure will be divided among them.
​$\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y}\frac{∆l}{l}$

$\frac{\left({\displaystyle \frac{50000}{4}}\right)\mathrm{g}}{0.22}=2\times {10}^{11}\times \frac{∆l}{l}$

$\Rightarrow \frac{∆l}{l}=\frac{500\times 9.8}{4\times 0.22\times 2\times {10}^{11}}=278409\times {10}^{-11}=2.78\times {10}^{-6}$

Hence the correct answer is $2.78\times {10}^{-6}.$