Four identical hollow cylindrical columns of mild steel support a big structure of mass $$50 \times 10^3$$ kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.
[use $$Y = 2.0 \times 10^{11}$$ Pa, $$g = 9.8$$ m s$$^{-2}$$]

The total mass of the structure is given as $$m = 50 \times 10^{3}\ \text{kg}$$. The weight is the gravitational force $$mg$$ acting downward. Because the load is uniformly distributed over four identical columns, each column supports exactly one‐fourth of this weight.

First we calculate the total weight:

$$W_{\text{total}} = mg = \left( 50 \times 10^{3}\ \text{kg} \right)\left( 9.8\ \text{m s}^{-2} \right) = 4.9 \times 10^{5}\ \text{N}.$$

The load per column is therefore

$$F = \frac{W_{\text{total}}}{4} = \frac{4.9 \times 10^{5}\ \text{N}}{4} = 1.225 \times 10^{5}\ \text{N}.$$

Each column is a hollow cylinder with outer radius $$R = 100\ \text{cm} = 1.0\ \text{m}$$ and inner radius $$r = 50\ \text{cm} = 0.50\ \text{m}.$$ The cross-sectional area of a hollow cylinder is

$$A = \pi \left( R^{2} - r^{2} \right).$$

Substituting the radii,

$$A = \pi \left[ (1.0\ \text{m})^{2} - (0.50\ \text{m})^{2} \right] = \pi \left( 1.00 - 0.25 \right)\ \text{m}^{2} = 0.75\pi\ \text{m}^{2}.$$

Numerically,

$$A = 0.75 \times 3.1416\ \text{m}^{2} \approx 2.356\ \text{m}^{2}.$$

Now we compute the compressive stress on each column. By definition,

$$\sigma = \frac{\text{Force}}{\text{Area}} = \frac{F}{A}.$$

Hence,

$$\sigma = \frac{1.225 \times 10^{5}\ \text{N}}{2.356\ \text{m}^{2}} \approx 5.20 \times 10^{4}\ \text{Pa}.$$

The relation between Young’s modulus $$Y$$, stress $$\sigma$$ and longitudinal strain $$\varepsilon$$ is

$$Y = \frac{\sigma}{\varepsilon} \quad\Longrightarrow\quad \varepsilon = \frac{\sigma}{Y}.$$

With $$Y = 2.0 \times 10^{11}\ \text{Pa},$$ we have

$$\varepsilon = \frac{5.20 \times 10^{4}\ \text{Pa}} {2.0 \times 10^{11}\ \text{Pa}} = 2.60 \times 10^{-7}.$$

Thus the compression strain in each steel column is

$$\boxed{2.60 \times 10^{-7}}.$$

Hence, the correct answer is Option B.