ICM is the moment of inertia of a circular disc about an axis (CM… — Solution | JEE Physics

NVAEasyJEE 2023Theorems of M.I (Parallel & Perpendicular Axis)

$I_{CM}$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of the disc. $I_{AB}$ is its moment of inertia about an axis $AB$ perpendicular to the plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from the center, where $R$ is the radius of the disc. The ratio of $I_{AB}$ and $I_{CM}$ is $x : 9$ . The value of $x$ is:

A circular disc with central axis CM and a parallel axis AB drawn to the right, separated by a distance 2R/3. Radius R is marked from the center to the left edge of the disc.

Answer

Correct answer:17

Step-by-step solution

Standard Method

Given: $I_{CM}$ is the moment of inertia of a disc about the central axis perpendicular to its plane. The parallel axis $AB$ is at a distance $\frac{2R}{3}$ from the center.

Find: The value of $x$ if $I_{AB} : I_{CM} = x : 9$ .

For a circular disc about its central perpendicular axis,

I_{CM} = \frac{1}{2} mR^2

Using the parallel axis theorem,

I_{AB} = I_{CM} + md^2

where

d = \frac{2R}{3}

So,

I_{AB} = \frac{1}{2} mR^2 + m\left(\frac{2R}{3}\right)^2

I_{AB} = \frac{1}{2} mR^2 + \frac{4}{9} mR^2

I_{AB} = \frac{9}{18} mR^2 + \frac{8}{18} mR^2 = \frac{17}{18} mR^2

Now,

\frac{I_{AB}}{I_{CM}} = \frac{\frac{17}{18} mR^2}{\frac{1}{2} mR^2} = \frac{17}{9}

Thus,

I_{AB} : I_{CM} = 17 : 9

Therefore, the value of $x$ is $17$ .

Ratio Comparison

Given: $I_{CM} = \frac{1}{2}mR^2$ and the new axis is parallel to it at distance $\frac{2R}{3}$ .

Find: The ratio parameter $x$ .

By the parallel axis theorem,

I_{AB} = I_{CM} + m\left(\frac{2R}{3}\right)^2

Substitute $I_{CM} = \frac{1}{2}mR^2$ :

I_{AB} = \frac{1}{2}mR^2 + \frac{4}{9}mR^2

Taking LCM $18$ ,

I_{AB} = \frac{9}{18}mR^2 + \frac{8}{18}mR^2 = \frac{17}{18}mR^2

Now compare with $I_{CM}$ :

I_{CM} = \frac{1}{2}mR^2 = \frac{9}{18}mR^2

Hence,

I_{AB} : I_{CM} = \frac{17}{18}mR^2 : \frac{9}{18}mR^2 = 17 : 9

So the required value is $17$ .

Common mistakes

Using the wrong basic formula for the disc, such as $I = mR^2$ . This is incorrect because for a circular disc about the central axis perpendicular to its plane, the correct expression is $I_{CM} = \frac{1}{2}mR^2$ . Start with the standard disc formula before applying the shift.
Applying the parallel axis theorem with $d = R$ instead of $d = \frac{2R}{3}$ . This is wrong because the axis $AB$ is specifically given at a distance $\frac{2R}{3}$ from the center. Always use the stated separation between the two parallel axes.
Adding the shifted term incorrectly as $m\left(\frac{2R}{3}\right)$ instead of squaring the distance. The theorem requires $md^2$ , not $md$ . First square the offset, then multiply by $m$ .

Practice more Theorems of M.I (Parallel & Perpendicular Axis) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Theorems of M.I (Parallel & Perpendicular Axis)

Answer

Step-by-step solution

Common mistakes

Practice more Theorems of M.I (Parallel & Perpendicular Axis) questions

Related questions