The area of cross section of the rope used to lift a load by a crane is $2.5 \times 10^{-4} \mathrm{~m}^{2}$. 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 $g=10 \,m s^{-2}$ )

<p>The relationship between stress (σ), force (F), and area (A) is given by :</p> <p>$\sigma = \frac{F}{A}$</p> <p>In this context, the force is equal to the weight of the load, so we can substitute force with mass (m) times gravity (g) :</p> <p>$F = m \cdot g$</p> <p>From this, we get the formula for the cross-sectional area required to support a given mass :</p> <p>$A = \frac{F}{\sigma} = \frac{m \cdot g}{\sigma}$</p> <p>We can set up a proportionality relationship between the area for 10 metric tons (A₁₀) and the area for 25 metric tons (A₂₅) as follows :</p> <p>$\frac{A_{10}}{A_{25}} = \frac{m_{10}}{m_{25}}$</p> <p>Using the given values :</p> <ul> <li>$A_{10} = 2.5 \times 10^{-4}$ m²,</li> <li>$m_{10} = 10,000$ kg,</li> <li>$m_{25} = 25,000$ kg,</li> </ul> <p>Solving for $A_{25}$ :</p> <p>$$A_{25} = A_{10} \times \left(\frac{m_{25}}{m_{10}}\right) = 2.5 \times 10^{-4} \, \mathrm{m}^{2} \times \left(\frac{25,000 \, \mathrm{kg}}{10,000 \, \mathrm{kg}}\right) = 6.25 \times 10^{-4} \, \mathrm{m}^{2}$$</p> <p>So, Option A $ (6.25 \times 10^{-4} \, \mathrm{m}^{2}) $ is the correct answer.</p>