Problem 628 - JEE Physics

Question

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

Step-by-Step Solution

Answer

The correct option for this matching exercise is Option D, where the moments of inertia are: X (A ring about its axis) → 4 (MR^2) Y (A uniform circular disc about its axis) → 1 (\(\frac{1}{2} MR^2\)) Z (A solid sphere about any diameter) → 2 (\(\frac{2}{5} MR^2\)) W (A solid sphere about any tangent) → 3 (\(\frac{7}{5} MR^2\))

Step 1: Identifying the moments of inertia for each object in List-I

A Ring about its axis: For a ring with mass (M) and radius (R), the moment of inertia about its axis is given by the formula \(I = MR^2\). In this case, we are looking for the List-II option that matches the formula \(I = MR^2\). A uniform circular disc about its axis: For a uniform circular disc with mass (M) and radius (R), the moment of inertia about its axis is given by the formula \(I = \frac{1}{2} MR^2\). In this case, we are looking for the List-II option that matches the formula \(I = \frac{1}{2} MR^2\). A solid sphere about any diameter: For a solid sphere with mass (M) and radius (R), the moment of inertia about any diameter is given by the formula \(I = \frac{2}{5} MR^2\). In this case, we are looking for the List-II option that matches the formula \(I = \frac{2}{5} MR^2\). A solid sphere about any tangent: For a solid sphere with mass (M) and radius (R), the moment of inertia about any tangent is given by the formula \(I = \frac{7}{5} MR^2\). In this case, we are looking for the List-II option that matches the formula \(I = \frac{7}{5} MR^2\).

Step 2: Matching the formulas for the moments of inertia with the List-II options

A Ring about its axis: We found that the formula is \(I = MR^2\), which matches with option (4). A uniform circular disc about its axis: We found that the formula is \(I = \frac{1}{2} MR^2\), which matches with option (1). A solid sphere about any diameter: We found that the formula is \(I = \frac{2}{5} MR^2\), which matches with option (2). A solid sphere about any tangent: We found that the formula is \(I = \frac{7}{5} MR^2\), which matches with option (3).

Step 3: Identifying the correct option

Using the matched formulas, we can assign the List-II options to the List-I objects: X: A ring about its axis → 4 Y: A uniform circular disc about its axis → 1 Z: A solid sphere about any diameter → 2 W: A solid sphere about any tangent → 3 The correct option is the one that assigns these List-II options to List-I objects: Option B: X: 4 Y: 3 Z: 2 W: 5 Option B does not match our results. Option C: X: 1 Y: 5 Z: 4 W: 3 Option C does not match our results. Option A: X: 2 Y: 1 Z: 3 W: 4 Option A does not match our results. Option D: X: 4 Y: 1 Z: 2 W: 3 Option D matches our results.

Step 4: Conclusion

The correct option for this matching exercise is Option D, where the moments of inertia are: X (A ring about its axis) → 4 (MR^2) Y (A uniform circular disc about its axis) → 1 (\(\frac{1}{2} MR^2\)) Z (A solid sphere about any diameter) → 2 (\(\frac{2}{5} MR^2\)) W (A solid sphere about any tangent) → 3 (\(\frac{7}{5} MR^2\))

Key Concepts

Ring about its axis

The concept of moment of inertia plays a crucial role in understanding rotational motion. When we consider a ring spinning about its axis, the calculation of moment of inertia becomes straightforward. For a ring, which is essentially a thin circular band, all the mass is located at a constant distance from the center, making it uniform. Thus, its moment of inertia is determined by the formula:

• \( I = MR^2 \)

where \( M \) is the mass of the ring and \( R \) is its radius. Utilizing this formula helps in simplifying problems related to rotational dynamics as it provides a clear value for how much resistance the ring offers against changes in its rotational speed. When you encounter problems involving a ring about its axis, remember that the entire mass being at the same distance from the axis makes the calculations intuitive, as you directly apply the basic formula \( MR^2 \). Understanding this helps you connect the dots in physics matching problems by identifying patterns and applying the right formula.

Uniform circular disc

The uniform circular disc is another essential concept in rotational mechanics. Unlike the ring, a disc has its mass distributed throughout its volume, making its moment of inertia calculation slightly different. For a disc rotating around its center, the formula is:

• \( I = \frac{1}{2} MR^2 \)

Here, \( M \) represents the mass of the disc and \( R \) is its radius.
The factor \( \frac{1}{2} \) signifies that only half of the disc's mass effectively contributes to its moment of inertia due to the mass distribution around its center.
This concept is used frequently in physics to describe systems where discs rotate, like wheels or merry-go-rounds.By recognizing the distribution and applying the appropriate formula, you can match the problem requirement in exercises with the correct mathematical expression.

Solid sphere

The concept of a solid sphere is paramount in understanding the dynamics of rotational systems. The moment of inertia for a solid sphere involves more complex calculations due to its three-dimensional mass distribution. For a solid sphere rotating about its diameter, the formula used is:

• \( I = \frac{2}{5} MR^2 \)

In contrast, if the sphere rotates about a tangent, it adopts a different formula:

• \( I = \frac{7}{5} MR^2 \)

In both expressions, \( M \) is the mass and \( R \) is the radius of the sphere. The difference in these values arises from how the mass is positioned relative to the axis of rotation. Each formula accounts for the distribution of mass differently depending on whether the sphere is pivoted about its center or a point on its surface.
Grasping these variations helps in solving matching problems by providing insight into how changes in rotation point affect inertia.

Physics matching problems

Matching problems in physics can be challenging yet insightful for mastering concepts like the moment of inertia. They require identification and correlation of formulas with physical objects and systems.
The key to solving these problems lies in:

• Identifying the object's shape and rotational axis
• Remembering key formulas for various shapes like rings, discs, and spheres
• Applying the correct formula based on given parameters

Such exercises enhance your problem-solving skills by making you apply theoretical knowledge in a practical context. Using these principles effectively not only helps in exams but also in understanding real-world mechanics where rotational dynamics are vital. With practice, solving matching problems can improve your ability to analytically determine moment of inertia for various systems, thus solidifying your physics skills.