List-I shows four planar structures made of uniform solid rods each of mass m and length l. In the L

Q: Why is the moment of inertia of a rod at an angle $\theta$ equal to $\frac{ml^2}{3}\sin^2\theta$?

Imagine breaking the rod into tiny point masses. The distance of any point mass at position $x$ from the axis is $r = x\sin\theta$. Integrating $dm \cdot r^2$ from $0$ to $l$ gives $\int_0^l (\frac{m}{l}dx)(x\sin\theta)^2 = \frac{ml^2}{3}\sin^2\theta$.

Q: In Structure Q, why don't we use the $\frac{ml^2}{3}$ formula for the base AB?

The formula $\frac{ml^2}{3}\sin^2\theta$ applies when the axis passes *through* the rod. For the base AB, the axis is parallel to it and does not intersect it. Every point on rod AB is at the exact same perpendicular distance $d$ from the axis, so it acts like a point mass $m$ at distance $d$, giving $I = md^2$.

Q: How did we know the angle is $45^\circ$ for Structure R?

Structure R is a rhombus with $90^\circ$ angles, which makes it a square. The diagonals of a square perfectly bisect the $90^\circ$ corner angles, meaning the angle between any side and the diagonal axis is exactly $45^\circ$.

Q: Does the direction of rotation matter for calculating the moment of inertia?

Not at all! Moment of inertia depends only on the mass distribution relative to the axis of rotation, not on how fast or in what direction the object is spinning.

Q: What if the axis passed through the center of mass of the structures instead?

Then we would use the parallel axis theorem! You would first find the moment of inertia about the center of mass, or use the parallel axis theorem to shift the axis from the ones given in the problem to the center of mass.

Question

List-I shows four planar structures made of uniform solid rods each of mass $m$ and length $l$. In the List-II the possible moment of inertia of these structures about an axis OCO$'$, which lies in the plane of the structures, are given.
Choose the option that describes the correct match between the entries in List-I to those in List-II.

List-I:

P. P

Q. Q

R. R

S. S

List-II:

1. $\frac{5}{4}ml^2$

2. $\frac{1}{6}ml^2$

3. $\frac{1}{12}ml^2$

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

5. $\frac{1}{3}ml^2$

Sigma Percentile · Accepted Answer

### Mastering Moment of Inertia: A Tale of Four Structures

Welcome to a beautiful exploration of rotational mechanics! In this problem, we are tasked with finding the moment of inertia for four distinct planar structures, each composed of uniform solid rods of mass $m$ and length $l$. The beauty of this problem lies in its reliance on a single, powerful fundamental formula.

### The Master Equation

Before we dive into the structures, let's equip ourselves with the master key. The moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through its end and making an angle $	heta$ with the rod is given by:

$$I = \frac{ml^2}{3} \sin^2	heta$$

Additionally, if a rod is perfectly parallel to the axis of rotation at a perpendicular distance $d$, its moment of inertia simplifies to that of a point mass:

$$I = md^2$$

Armed with these two tools, let's conquer each structure one by one.

### Analyzing Structure P

Structure P consists of two rods, $AC$ and $BC$, meeting at a right angle ($90^\circ$) at point $C$. The axis of rotation passes through $C$ and perfectly bisects this angle.

Because the axis bisects the $90^\circ$ angle, both rods make an angle of $	heta = 45^\circ$ with the axis. We simply apply our master formula to both rods and add their contributions:

$$I_P = \frac{ml^2}{3} \sin^2 45^\circ + \frac{ml^2}{3} \sin^2 45^\circ$$

Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$, squaring it gives $\frac{1}{2}$.

$$I_P = 2 	imes \frac{ml^2}{3} \left(\frac{1}{2}ight) = \frac{ml^2}{3}$$

This perfectly matches option (5).

### Analyzing Structure Q

Structure Q is an equilateral triangle $ABC$. The axis passes through the top vertex $C$ and runs parallel to the base $AB$.

For the two slanted rods, $AC$ and $BC$, the geometry of an equilateral triangle dictates that they each make an angle of $	heta = 60^\circ$ with the horizontal axis.

But what about the base $AB$? It doesn't intersect the axis; it runs parallel to it! The perpendicular distance $d$ from the axis to rod $AB$ is simply the height of the equilateral triangle, which is $l \sin 60^\circ = \frac{\sqrt{3}}{2}l$. We use the parallel rod formula $I = md^2$ for this base.

$$I_Q = 2 	imes \left( \frac{ml^2}{3} \sin^2 60^\circ ight) + m \left( \frac{\sqrt{3}}{2}l ight)^2$$

$$I_Q = 2 	imes \frac{ml^2}{3} \left(\frac{3}{4}ight) + m \left(\frac{3}{4}l^2ight)$$

$$I_Q = \frac{ml^2}{2} + \frac{3ml^2}{4} = \frac{5ml^2}{4}$$

This matches option (1).

### Analyzing Structure R

Structure R is a square $ABCD$, and the axis of rotation is its diagonal passing through $A$ and $C$.

One of the elegant properties of a square is that its diagonals perfectly bisect its $90^\circ$ corner angles. This means that all four rods ($AB$, $BC$, $CD$, and $DA$) make exactly a $45^\circ$ angle with the diagonal axis. This symmetry makes our calculation incredibly straightforward!

$$I_R = 4 	imes \left( \frac{ml^2}{3} \sin^2 45^\circ ight)$$

$$I_R = 4 	imes \frac{ml^2}{3} 	imes \frac{1}{2} = \frac{2ml^2}{3}$$

This matches option (4).

### Analyzing Structure S

Finally, Structure S consists of just two rods, $AC$ and $BC$. The axis passes through $C$, and the diagram explicitly shows that each rod makes an angle of $	heta = 30^\circ$ with the axis.

We return to our trusty master formula one last time:

$$I_S = 2 	imes \left( \frac{ml^2}{3} \sin^2 30^\circ ight)$$

Since $\sin 30^\circ = \frac{1}{2}$, squaring it gives $\frac{1}{4}$.

$$I_S = 2 	imes \frac{ml^2}{3} 	imes \frac{1}{4} = \frac{ml^2}{6}$$

This matches option (2).

### Final Conclusion

By systematically applying the fundamental principles of rotational mechanics, we have successfully decoded the moment of inertia for all four structures. The correct matching is **P $ightarrow$ 5, Q $ightarrow$ 1, R $ightarrow$ 4, S $ightarrow$ 2**. Always look for geometric symmetries and parallel axes—they are your best friends in physics!

List-I

List-II

Select Matching Pairs:

Visualized Solution

$I = \frac{m l ^{2}}{3} sin^{2} θ$

$I_{P} = I_{A C} + I_{B C}$

$I_{P} = \frac{m l ^{2}}{3}$

$I_{Q} = I_{A C} + I_{B C} + I_{A B}$

$I_{Q} = \frac{5}{4} m l^{2}$

$I_{R} = 4 \times I_{r o d}$

$I_{R} = \frac{2}{3} m l^{2}$

$I_{S} = I_{A C} + I_{B C}$

$I_{S} = \frac{1}{6} m l^{2}$

\text{Final Match}

The Sigma Insight: Moment of Inertia

Mastering Moment of Inertia

The Master Equation

Analyzing Structure P

Analyzing Structure Q

Analyzing Structure R

Analyzing Structure S

Final Conclusion

Similar Questions

Match List I with List II.

List-I

List-II

Seven identical circular planar discs, each of mass $M$ and radius $R$ are welded symmetrically as shown in the figure. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $P$ is

The linear mass density of a thin rod $A B$ of length $L$ varies from $A$ to $B$ as $λ (x) = λ_{0} (1 + \frac{x}{L})$ , where $x$ is the distance from $A$ . If $M$ is mass of the rod, then its moment of inertia about an axis passing through $A$ and perpendicular to the rod is

Two identical spherical balls of mass $M$ and radius $R$ each are stuck on two ends of a rod of length $2 R$ and mass $M$ (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

Four equal masses, $m$ each are placed at the corners of a square of length ( $l$ ) as shown in the figure. The moment of inertia of the system about an axis passing through $A$ and parallel to $D B$ would be

Four point masses, each of value $m$ , are placed at the corners of a square ABCD of side $l$ . The moment of inertia of this system about an axis passing through A and parallel to BD is

Consider a uniform square plate of side $a$ and mass $m$ . The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

List-I

List-II

Select Matching Pairs:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=3ml2​sin2θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IP​=IAC​+IBC​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IP​=3ml2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IQ​=IAC​+IBC​+IAB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IQ​=45​ml2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IR​=4×Irod​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IR​=32​ml2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IS​=IAC​+IBC​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IS​=61​ml2

\text{Final Match}

The Sigma Insight: Moment of Inertia

Mastering Moment of Inertia

The Master Equation

Analyzing Structure P

Analyzing Structure Q

Analyzing Structure R

Analyzing Structure S

Final Conclusion

Similar Questions

Match List I with List II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

$I = \frac{m l ^{2}}{3} sin^{2} θ$

$I_{P} = I_{A C} + I_{B C}$

$I_{P} = \frac{m l ^{2}}{3}$

$I_{Q} = I_{A C} + I_{B C} + I_{A B}$

$I_{Q} = \frac{5}{4} m l^{2}$

$I_{R} = 4 \times I_{r o d}$

$I_{R} = \frac{2}{3} m l^{2}$

$I_{S} = I_{A C} + I_{B C}$

$I_{S} = \frac{1}{6} m l^{2}$

Consider a uniform square plate of side $a$ and mass $m$ . The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is