Conservation of Energy question from MHT CET 2024 (4 May, Shift 2)

Answer: (b)

1. Mass $\propto r^2$, so the roll of radius $R/2$ has mass $\frac{M}{4}$; the unrolled $\frac{3M}{4}$ lies flat on the floor (PE $\approx 0$).
2. Initial PE $= MgR$ (centre at height $R$).
3. Final PE $= \frac{M}{4}g\frac{R}{2} = \frac{MgR}{8}$
4. Change $= MgR - \frac{MgR}{8} = \frac{7}{8}MgR$ (decrease). So the answer is (b).