Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is \(M,\) radius of its top, \(R\) and height, \(H\), then its moment of inertia about its axis is

Explanation:
Let us derive the moment of inertia of a hollow cone step-by-step using calculus based on the System of Particles and Rotational Motion chapter of the NCERT Class 11 Physics textbook:

Step 1: Understand the Geometry of the Cone
Let the apex (vertex) of the cone be at the origin, pointing downwards. - Slant height of the cone, \(l = \sqrt{R^2 + H^2}\)
- If \(\theta\) is the semi-vertical angle of the cone, then:
\(\tan\theta = \frac{R}{H}\) and \(\cos\theta = \frac{H}{l} = \frac{H}{\sqrt{R^2+H^2}}\)

The total curved surface area of this hollow cone is \(A = \pi R l = \pi R \sqrt{R^2 + H^2}\).
Therefore, the mass per unit area (\(\sigma\)) of the uniform hollow cone is:
\(\sigma = \frac{M}{\pi R \sqrt{R^2 + H^2}}\)

Step 2: Choose a Circular Element
Consider an elemental circular ring of radius \(r\) at a vertical distance \(h\) from the vertex along the axis. The slant width of this element is \(dl\).
From the geometry of the triangle:
\(r = h \tan\theta\)
\(dl = \frac{dh}{\cos\theta}\)

The area of this elemental ring is given by:
\(dA = 2\pi r \cdot dl = 2\pi (h \tan\theta) \frac{dh}{\cos\theta}\)

Step 3: Calculate the Mass of the Element (\(dm\))
Multiply the area of the element by the surface mass density (\(\sigma\)):
\(dm = \sigma \cdot dA = \left(\frac{M}{\pi R \sqrt{R^2 + H^2}}\right) \cdot \left(2\pi h \frac{\tan\theta}{\cos\theta} dh\right)\)
\(dm = \frac{2M \tan\theta}{R \sqrt{R^2 + H^2} \cos\theta} h dh\)

Step 4: Integrate to Find the Total Moment of Inertia (\(I\))
The moment of inertia of a thin ring of radius \(r\) and mass \(dm\) about its central axis is \(dI = (dm)r^2\).
Substituting \(r = h \tan\theta\):
\(dI = (dm)(h \tan\theta)^2 = \left(\frac{2M \tan\theta}{R \sqrt{R^2 + H^2} \cos\theta} h dh\right) \cdot h^2 \tan^2\theta\)
\(dI = \frac{2M \tan^3\theta}{R \sqrt{R^2 + H^2} \cos\theta} h^3 dh\)

Integrating from \(h = 0\) to \(h = H\):
\(I = \int_0^H \frac{2M \tan^3\theta}{R \sqrt{R^2 + H^2} \cos\theta} h^3 dh = \frac{2M \tan^3\theta}{R \sqrt{R^2 + H^2} \cos\theta} \left[ \frac{h^4}{4} \right]_0^H\)
\(I = \frac{2M \tan^3\theta}{R \sqrt{R^2 + H^2} \cos\theta} \cdot \frac{H^4}{4}\)

Step 5: Substitute Trigonometric Values for Simplification
Substitute \(\tan\theta = \frac{R}{H}\) and \(\cos\theta = \frac{H}{\sqrt{R^2+H^2}}\) back into the equation:
\(I = \frac{2M \left(\frac{R^3}{H^3}\right)}{R \sqrt{R^2 + H^2} \left(\frac{H}{\sqrt{R^2 + H^2}}\right)} \cdot \frac{H^4}{4}\)
\(I = \frac{2M R^2}{H^4} \cdot \frac{H^4}{4} = \frac{MR^2}{2}\)

Hence, Option A is the correct choice.