JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 14)

The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly ?
JEE Main 2019 (Online) 12th January Evening Slot Physics - Rotational Motion Question 163 English Option 1
JEE Main 2019 (Online) 12th January Evening Slot Physics - Rotational Motion Question 163 English Option 2
JEE Main 2019 (Online) 12th January Evening Slot Physics - Rotational Motion Question 163 English Option 3
JEE Main 2019 (Online) 12th January Evening Slot Physics - Rotational Motion Question 163 English Option 4

Explanation

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 :

$$I_0 = \frac{2}{5}MR^2$$

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 :

$$I(x) = I_0 + Mx^2$$

Substituting the value of $I_0$ :

$$I(x) = \frac{2}{5}MR^2 + Mx^2$$

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.

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