Q.Two identical water bottles, one empty and the other filled with water, are allowed to roll down an inclined plane. Which one of them reaches the bottom first? Explain your answer.

Step 1. Recall how rolling speed depends on shape, via $K^2/R^2$, for a genuinely rigid rolling body.

For a rigid body of radius of gyration $K$ and radius $R$ rolling without slipping down an incline from height $h$, energy conservation gives its speed at the bottom as

A SMALLER value of $K^2/R^2$ means a LARGER final speed, because less of the total available gravitational PE has to be 'spent' spinning the object up (rotational KE) and more remains available for translational KE. This is exactly why, among genuinely rigid solids released together, a solid sphere ($K^2/R^2=2/5$) beats a disc ($1/2$), which beats a hollow sphere ($2/3$), which beats a ring ($1$) — this is the standard 'racing order' result.

Step 2. Check whether the empty bottle is well modelled as a genuine rigid body.

The EMPTY bottle (essentially a thin hollow shell, close to a hollow cylinder in mass distribution) is a solid object — every part of it is mechanically locked to every other part. When it rolls, its entire mass truly does rotate together with a single well-defined angular velocity $\omega=v/R$, exactly as the rigid-body rolling theory assumes. So its full weight genuinely 'pays into' rotational KE according to its fixed $K^2/R^2$.

Step 3. Check whether the water-filled bottle satisfies the same assumption — and see that it does not.

The water inside the FULL bottle is a liquid, not a rigid solid. As the bottle's shell starts to spin, the water does not instantly (or ever, ideally) get dragged into rotating rigidly along with the shell — internal viscosity is weak and the water's own inertia resists being spun up on the timescale of the roll down a short incline. In the (idealised) limiting case, the water simply translates down the incline along with the bottle's centre of mass but largely does NOT share in the bottle's spin — it behaves almost like a non-rotating mass being carried along for the ride, rather than like a solid shell that must spin at $\omega=v/R$.

Step 4. See the energetic consequence of this difference. …

Rolling-motion energy split, $v=\sqrt{2gh/(1+K^2/R^2)}$, applied to a non- …

• Assuming the heavier (full) bottle must be slower because it 'has more inertia to overcome' — this conflates total mass (which cancels out of the rolling-speed formula entirely) with the SHAPE-dependent ratio $K^2/R^2$, which is what actually matters.
• Treating the water-filled bottle as just a rigid object of some intermediate density and applying the standard rigid-body $K^2/R^2$ table to it directly — the whole point is that the water does NOT rotate rigidly with the shell, so the standard rigid-body formula does not apply to it in the same way. …