QJEEM22Ph5bc7: JEE Main 2022 Physics - Class 11 JEEM22Ph5bc7 Solution: The area of cross section of the rope used to lift a load by a…

The relationship between stress (σ), force (F), and area (A) is given by :

$\sigma = \frac{F}{A}$

In this context, the force is equal to the weight of the load, so we can substitute force with mass (m) times gravity (g) :

$F = m \cdot g$

From this, we get the formula for the cross-sectional area required to support a given mass :

$A = \frac{F}{\sigma} = \frac{m \cdot g}{\sigma}$

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 :

$\frac{A_{10}}{A_{25}} = \frac{m_{10}}{m_{25}}$

Using the given values :

• $A_{10} = 2.5 \times 10^{-4}$ m²,
• $m_{10} = 10,000$ kg,
• $m_{25} = 25,000$ kg,

Solving for $A_{25}$ :

$A_{25} = A_{10} \times \left(\frac{m_{25}}{m_{10}}\right) = 2.5 \times 10^{-4} \, \mathrm{m}^{2} \times \left(\frac{25,000 \, \mathrm{kg}}{10,000 \, \mathrm{kg}}\right) = 6.25 \times 10^{-4} \, \mathrm{m}^{2}$

So, Option A $(6.25 \times 10^{-4} \, \mathrm{m}^{2})$ is the correct answer.