The area of cross section of the rope used to lift a load by a crane is $$2.5 \times 10^{-4} \text{ m}^2$$. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be (take $$g = 10 \text{ m s}^{-2}$$)

We need to determine the required area of cross-section of the crane's rope to increase its maximum lifting capacity from 10 metric tons to 25 metric tons.

1. Identify the Working Principle

The maximum lifting capacity of a rope depends on the maximum permissible stress ($$\sigma$$) the material can withstand without breaking. Stress is defined as the restoring force per unit cross-sectional area:

$$\text{Stress } (\sigma) = \frac{\text{Force}}{\text{Area}} = \frac{mg}{A}$$

Since the material of the rope remains the same, the maximum permissible stress stays constant ($$\sigma_1 = \sigma_2$$). Therefore, the lifting load ($$m$$) is directly proportional to the cross-sectional area ($$A$$):

$$\frac{m_1}{A_1} = \frac{m_2}{A_2}$$

2. Substitute the Values and Solve for $$A_2$$

From the problem 

• Initial lifting mass ($$m_1$$) = $$10\text{ metric tons}$$
• Initial area of cross-section ($$A_1$$) = $$2.5 \times 10^{-4}\text{ m}^2$$
• Target lifting mass ($$m_2$$) = $$25\text{ metric tons}$$

Rearranging the proportion formula to isolate the new required area ($$A_2$$):

$$A_2 = A_1 \times \left(\frac{m_2}{m_1}\right)$$

$$A_2 = (2.5 \times 10^{-4}\text{ m}^2) \times \frac{25}{10}$$

$$A_2 = 2.5 \times 10^{-4} \times 2.5 = 6.25 \times 10^{-4}\text{ m}^2$$

Conclusion

The required area of cross-section of the rope should be $$6.25 \times 10^{-4}\text{ m}^2$$, which corresponds to Option A.