Match List I with List II.
List I
A. Moment of inertia of the rod (length L, mass M, about an axis p

Q: Why is the moment of inertia different when the axis is at the end compared to the center?

Moment of inertia depends on how mass is distributed relative to the axis of rotation. When the axis is at the end, more mass is located further away from the axis compared to when the axis is at the center. Since $I = \int r^2 dm$, the larger distances ($r$) contribute significantly more, resulting in a higher moment of inertia ($ML^2/3$ vs $ML^2/12$).

Q: Can I use the parallel axis theorem to find the moment of inertia about the end?

Absolutely! The parallel axis theorem states $I = I_{CM} + Md^2$. For a rod, $I_{CM} = ML^2/12$. The distance from the center to the end is $d = L/2$. So, $I_{end} = ML^2/12 + M(L/2)^2 = ML^2/12 + ML^2/4 = 4ML^2/12 = ML^2/3$.

Q: In Case C, why does the moment of inertia become $ML^2/3$ even though the axis is at the center?

In Case C, the length of the rod is $2L$. The formula for an axis through the center is $\frac{\text{mass} \times \text{length}^2}{12}$. Substituting the given length, we get $\frac{M(2L)^2}{12} = \frac{M(4L^2)}{12} = \frac{ML^2}{3}$. The increased length drastically increases the moment of inertia.

Q: What happens if both mass and length are doubled, like in Case D?

Both mass and length contribute to the moment of inertia, but length has a squared effect. Doubling the mass doubles the inertia, but doubling the length quadruples it. So, doubling both increases the moment of inertia by a factor of $2 \times 2^2 = 8$.

Q: Is the moment of inertia a vector quantity?

No, moment of inertia is not a vector. It is a tensor quantity, but for fixed-axis rotation problems in introductory physics, it is treated as a scalar. It simply represents the rotational inertia of the body about that specific axis.

Question

Match List I with List II.

List-I:

P. Moment of inertia of the rod (length $L$, mass $M$, about an axis perpendicular to the rod passing through the mid-point)

Q. Moment of inertia of the rod (length $L$, mass $2M$, about an axis perpendicular to the rod passing through one of its end)

R. Moment of inertia of the rod (length $2L$, mass $M$, about an axis perpendicular to the rod passing through its midpoint)

S. Moment of inertia of the rod (length $2L$, mass $2M$, about an axis perpendicular to the rod passing through one of its end)

List-II:

1. $\frac{8ML^2}{3}$

2. $\frac{ML^2}{3}$

3. $\frac{ML^2}{12}$

4. $\frac{2ML^2}{3}$

Sigma Percentile · Accepted Answer

### The Anatomy of Rotational Inertia

Imagine trying to spin a heavy barbell. If you grab it by the center, it's relatively easy to twirl. But if you grab it by one end, it suddenly feels incredibly stubborn and hard to rotate.

This physical intuition is the heart of **Moment of Inertia**. It tells us how difficult it is to change the rotational state of an object. It depends not just on the total mass, but crucially on *how that mass is distributed* relative to the axis of rotation.

In this problem, we are exploring four different variations of a uniform rod. Let's break them down one by one.

### Case A: The Standard Benchmark

We start with the most fundamental case. A rod of mass $M$ and length $L$, rotating about an axis passing through its exact center.

This is our baseline. The mass is distributed symmetrically, and the maximum distance of any mass element from the axis is $L/2$. The standard formula derived via integration is:
$$I_A = \frac{ML^2}{12}$$

This matches perfectly with option **III** in our list.

### Case B: Shifting the Axis

Now, let's look at Case B. The rod now has a mass of $2M$, and the axis has been shifted to one end.

When the axis is at the end, the mass is distributed much further away. The maximum distance is now the full length $L$. The standard formula for an axis at the end is $\frac{	ext{mass} 	imes 	ext{length}^2}{3}$.

Substituting our specific mass $2M$:
$$I_B = \frac{(2M)L^2}{3} = \frac{2ML^2}{3}$$

This result corresponds to option **IV**.

### Case C: Stretching the Rod

In Case C, we return the axis to the center, but we stretch the rod to a length of $2L$. The mass remains $M$.

Because the moment of inertia depends on the *square* of the length, doubling the length has a massive impact. We use our center-axis formula, but we must be careful to substitute $2L$ for the length:
$$I_C = \frac{M(2L)^2}{12}$$

Expanding the square gives us $4L^2$:
$$I_C = \frac{M(4L^2)}{12} = \frac{ML^2}{3}$$

Notice how a rod of length $2L$ rotating about its center has the exact same moment of inertia as a rod of length $L$ (with mass $M$) rotating about its end! This matches option **II**.

### Case D: The Heavyweight Champion

Finally, we have the most extreme case. The mass is doubled to $2M$, the length is doubled to $2L$, and the axis is at the end.

We use the end-axis formula and substitute both new values:
$$I_D = \frac{(2M)(2L)^2}{3}$$

First, square the length: $(2L)^2 = 4L^2$. Then multiply by the mass $2M$:
$$I_D = \frac{(2M)(4L^2)}{3} = \frac{8ML^2}{3}$$

This is the largest moment of inertia of the group, matching option **I**.

### The Grand Conclusion

By carefully applying the foundational formulas and substituting the specific parameters for each case, we have successfully decoded the matrix.

The final matching is $A ightarrow III$, $B ightarrow IV$, $C ightarrow II$, and $D ightarrow I$. Always remember, in rotational mechanics, the position of the axis and the square of the distance are your most critical factors!

Animated Solution for Physics - Rotational Motion: Match List I with List II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

\text{Problem Overview}

\text{Case A: Standard Rod}

\text{Case B: Setup}

\text{Case B: Calculation}

\text{Case C: Setup}

\text{Case C: Calculation}

\text{Case D: Setup}

\text{Case D: Calculation}

\text{Final Matching}

\text{The Way Forward}

The Sigma Insight: Moment of Inertia

The Anatomy of Rotational Inertia

Case A

Case B

Case C

Case D

The Grand Conclusion

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