MHTCET 10 May 2024 Shift 2 Paper

Concept:Total mechanical energy of the body is conserved as it moves under Earth's gravity alone.Explanation:The body starts from rest, so its initial kinetic energy is zero.At distance $R_0$ from Earth's centre, its initial potential energy is $-\frac{GMm}{R_0}$.When it reaches the surface, its distance from the centre is $R$, and its potential energy is $-\frac{GMm}{R}$.Let the speed acquired at the surface be $v$.Using conservation of energy:$$0 + \left(-\frac{GMm}{R_0}\right) = \frac{1}{2}mv^2 + \left(-\frac{GMm}{R}\right)$$Rearranging:$$\frac{1}{2}mv^2 = GMm\left(\frac{1}{R} - \frac{1}{R_0}\right)$$Cancelling $m$ and multiplying by 2:$$v^2 = 2GM\left(\frac{1}{R} - \frac{1}{R_0}\right)$$Taking the square root:$$v = \sqrt{2GM\left(\frac{1}{R} - \frac{1}{R_0}\right)}$$Answer:The correct option is B: $\sqrt{2GM\left(\frac{1}{R} - \frac{1}{R_0}\right)}$.