QJEEM19Phfbdd: JEE Main 2019 Physics - Class 11 JEEM19Phfbdd Solution: The moment of inertia of a solid sphere, about an axis parallel …

The correct answer is Option C. Here's why :

The moment of inertia of a solid sphere about an axis passing through its center is given by :

$$I_0=\frac{2}{5}M{R}^2$$

Where :

• $I_0$ is the moment of inertia about the center
• $M$ is the mass of the sphere
• $R$ is the radius of the sphere

Now, using the parallel axis theorem, we can find the moment of inertia about an axis parallel to the diameter and at a distance of 'x' from it :

$$I(x)=I_0+M{x}^2$$

Substituting the value of $I_0$ :

$$I(x)=\frac{2}{5}M{R}^2+M{x}^2$$

This equation represents a parabola, where :

• The coefficient of $x^2$ is positive (M), indicating an upward-opening parabola.
• The y-intercept (the value of $I(x)$ when $x = 0$ ) is $\frac{2}{5}MR^2$ , which is a positive constant.

Therefore, the graph of $I(x)$ vs. $x$ should be an upward-opening parabola, and Option C accurately represents this.