Kerala KEAM 5 June 2024 Solved Paper

The weight of a body at a height $h$ above the Earth's surface is given by the formula:$$W = W_0 \left( \frac{R}{R + h} \right)^2$$where:- $W_0$ is the weight at the Earth's surface,- $R$ is the radius of the Earth,- $h$ is the height above the Earth's surface.The weight decreases by 20\% at height $h$, so the new weight $W$ is 80\% of the original weight:$$W = 0.8 W_0$$Substituting into the formula for $W$:$$0.8 W_0 = W_0 \left( \frac{R}{R + h} \right)^2$$Canceling $W_0$ from both sides:$$0.8 = \left( \frac{R}{R + h} \right)^2$$Taking the square root of both sides:$$\sqrt{0.8} = \frac{R}{R + h}$$Simplifying:$$\frac{R}{R + h} = \frac{\sqrt{5}}{2}$$Now, solving for $h$:$$R + h = \frac{2R}{\sqrt{5}}$$$$h = \frac{2R}{\sqrt{5}} - R$$$$h = R \left( \frac{2}{\sqrt{5}} - 1 \right)$$Simplifying further:$$h = R \left( \frac{\sqrt{5}}{2} - 1 \right)$$Thus, the height at which the body loses 20\% of its weight is:$$\boxed{\left( \frac{\sqrt{5}}{2} - 1 \right) R}$$ Quick Tip: The weight of a body decreases with the square of the distance from the center of the Earth. Use the formula $W = W_0 \left( \frac{R}{R + h} \right)^2$ to calculate the change in weight with height.