Given:
Mass of cylinder, M
Radius of cylinder, R
Length of inclined plane, L
Height of inclined plane, h

To find: Speed of centre of mass when cylinder reaches the bottom

Concepts used:
1. Conservation of energy
2. Rolling motion of a solid cylinder

Explanation:

When the cylinder rolls down the inclined plane, there are two types of energies involved, potential energy and kinetic energy. At the top of the inclined plane, the potential energy is maximum and kinetic energy is zero. At the bottom of the inclined plane, the potential energy is zero and kinetic energy is maximum. Assuming no energy is lost due to friction, we can equate the potential energy at the top to the kinetic energy at the bottom.

1. Potential energy at the top:

The potential energy of the cylinder at the top of the inclined plane is given by:

PE = Mgh

where M is the mass of the cylinder, g is the acceleration due to gravity and h is the height of the inclined plane.

2. Kinetic energy at the bottom:

The kinetic energy of the cylinder at the bottom of the inclined plane is given by:

KE = 1/2MVcm^2 + 1/2Iω^2

where Vcm is the velocity of the centre of mass, I is the moment of inertia of the cylinder and ω is the angular velocity of the cylinder.

For a solid cylinder rolling without slipping, the relationship between Vcm and ω is given by:

Vcm = Rω

where R is the radius of the cylinder.

The moment of inertia of a solid cylinder about its centre of mass is given by:

I = 1/2MR^2

Substituting the values of I and Vcm in the expression for KE, we get:

KE = 1/2MVcm^2 + 1/4MR^2ω^2

Substituting ω = Vcm/R, we get:

KE = 3/4MVcm^2

3. Equating potential energy to kinetic energy:

Equating the potential energy at the top to the kinetic energy at the bottom, we get:

Mgh = 3/4MVcm^2

Simplifying, we get:

Vcm = √4/3gh

Therefore, the speed of the centre of mass when the cylinder reaches the bottom is √4/3gh.

Answer: Option (d) √4/3gh.