A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and a pulley as shown in the figure. Consider PR as a friction-free surface. The velocity of the centre of mass of the wheel when it reaches the bottom Q of the inclined plane PQ will be 1/2√xgh m/s. The value of x is ______. (Round off to the nearest integer.)

The string has a fixed length and runs over the pulley at P, so as the wheel rolls down PQ the block is dragged an equal distance up PR, at the same speed.

Both slopes make the same angle α, so when the wheel has dropped a height h the block has risen the same h.

Energy released: (12-3)gh=9gh — the block takes some of it back.

Energy stored: the wheel both moves and spins, the block only moves.

9gh=1/2(12)v²(1+I/(MR²))+1/2(3)v².

Taking the wheel as a ring, I/(MR²)=1: 9gh=12v²+1.5v²=13.5 v².

v²=(2gh)/3, so v=√(2gh)/3=1/2√(8gh)/3.

Comparing with 1/2√xgh gives x=8/3=2.67, which rounds to 3.

The question never says what shape the wheel is, and it does not have to: a solid disc would give x=(24)/7=3.43, and that rounds to 3 as well.