Consider a solid circular cylinder of radius 2 meters and height 3 meters of uniform density. If the density of the cylinder is ρ kg/meter², then the moment of inertia (in kg meter²) of the cylinder about a diameter of its base is 

Concept:

The moment of inertia (in kg meter2) of the cylinder about a diameter of its base is 

\(I = \frac{1}{4} M r^2 + \frac{1}{3} M h^2\)

Explanation:

Radius of the cylinder, \(r = 2 \, \text{m}\) , Height of the cylinder, \(h = 3 \, \text{m}\), Density of the cylinder, \(\rho \, \text{kg/m}^3\)

\(M = \rho \times \text{Volume of the cylinder}\), where \(V = \pi r^2 h\)

Substituting the given values,

\(V = \pi (2)^2 (3) = 12 \pi \, \text{m}^3\)

Thus,  \(M = \rho \times 12\pi \, \text{kg}\)

\(I = \frac{1}{4} M r^2 + \frac{1}{3} M h^2\)

Now, substitute the values \(r = 2 \, \text{m}\) ,\(h = 3 \, \text{m}\)  and \(M = \rho \times 12\pi \) into the formula

\(I = \frac{1}{4} \times \rho \times 12\pi \times (2)^2 + \frac{1}{3} \times \rho \times 12\pi \times (3)^2\)

⇒\(I = \frac{1}{4} \times 12\pi \times \rho \times 4 + \frac{1}{3} \times 12\pi \times \rho \times 9\)

⇒ \(I = 12\pi \rho + 36\pi \rho\)

⇒ \(I = 48\pi \rho\)

Hence correct option is 1).