The moment of inertia of a uniform rod about a perpendicular axis passing through one of its ends is \(\mathrm{I}_{1}\). The same rod is bent in to a ring and its moment of inertia about a diameter is \(\mathrm{I}_{2}\), Then \(\left[\mathrm{I}_{1} / \mathrm{I}_{2}\right]\) is. \(\\{\mathrm{A}\\}\left(\pi^{2} / 3\right)\) \(\\{B\\}\left(4 \pi^{2} / 3\right)\) \(\\{\mathrm{C}\\}\left(8 \pi^{2} / 3\right)\) \(\\{\mathrm{D}\\}\left(16 \pi^{2} / 3\right)\)

Short Answer

Expert verified
The short answer for the given problem is: \(\frac{I_1}{I_2} = \frac{4\pi^2}{3}\), which corresponds to option B.

Step by step solution

01

Find the Moment of Inertia of the Rod

For a uniform rod of mass \(m\) and length \(L\), the moment of inertia (\(I_1\)) about a perpendicular axis passing through one end is given by: \(I_1 = \frac{1}{3}mL^2\) #Step 2: Moment of Inertia of Ring#
02

Find the Moment of Inertia of the Ring

To find the moment of inertia of the same rod bent into a ring, first, determine the radius (\(R\)) of the resulting ring by dividing the length of the rod by the circumference of the circle: \(R = \frac{L}{2\pi}\) Now, the moment of inertia (\(I_2\)) for a uniform ring of mass \(m\) and radius \(R\) about a diameter is given by: \(I_2 = mR^2\) #Step 3: Substitute the expression for R into \(I_2\)#
03

Replace R with the expression found in Step 2

Use \(R=\frac{L}{2\pi}\) to replace R with an expression involving L in the formula for \(I_2\): \(I_2 = m\left(\frac{L}{2\pi}\right)^2\) Simplify this expression: \(I_2 = \frac{mL^2}{4\pi^2}\) #Step 4: Find the ratio \(\frac{I_1}{I_2}\)#
04

Calculate the ratio of \(I_1\) and \(I_2\)

Divide \(I_1\) by \(I_2\): \(\frac{I_1}{I_2} = \frac{\frac{1}{3}mL^2}{\frac{mL^2}{4\pi^2}}\) Now, cancel \(mL^2\) from the numerator and the denominator: \(\frac{I_1}{I_2} = \frac{1/3}{1/(4\pi^2)} = \frac{1}{3} \cdot (4\pi^2) = \frac{4\pi^2}{3}\) Comparing this result with the given options, we can see that the correct answer is: \\({\mathrm{B}}\\) \(\left(4 \pi^{2} / 3\right)\)

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Most popular questions from this chapter

Statement \(-1\) - Two cylinder one hollow and other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow will reach the bottom of inclined plane first. Statement \(-2-\mathrm{By}\) the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement \(-1\) \\{B \\} Statement \(-1\) is true, statement \(-2\) is true but statement- 2 is not the correct explanation four statement \(-1\). \\{C\\} Statement - 1 is true, statement- 2 is false \\{D \(\\}\) Statement- 2 is false, statement \(-2\) is true

A circular disc \(\mathrm{x}\) of radius \(\mathrm{R}\) is made from an iron plate of thickness \(t\). and another disc \(Y\) of radius \(4 R\) is made from an iron plate of thickness \(t / 4\) then the rotation between the moment of inertia \(\mathrm{I}_{\mathrm{x}}\) and \(\mathrm{I}_{\mathrm{y}}\) is \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{y}}=64 \mathrm{I}_{\mathrm{x}}\) \(\\{B\\} I_{y}=32 I_{x}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{y}}=16 \mathrm{I}_{\mathrm{x}}\) \(\\{\mathrm{D}\\} \mathrm{I}_{\mathrm{y}}=\mathrm{I}_{\mathrm{x}}\)

A cylinder of mass \(5 \mathrm{~kg}\) and radius \(30 \mathrm{~cm}\), and free to rotate about its axis, receives an angular impulse of $3 \mathrm{~kg} \mathrm{M}^{2} \mathrm{~S}^{-1}$ initially followed by a similar impulse after every \(4 \mathrm{sec}\). what is the angular speed of the cylinder 30 sec after initial impulse? The cylinder is at rest initially. \(\\{\mathrm{A}\\} 106.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{B\\} \(206.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{C\\} \(107.6 \mathrm{rad} \mathrm{S}^{-1}\) \\{D \(\\} 207.6 \mathrm{rad} \mathrm{S}^{-1}\)

A particle of mass \(\mathrm{m}\) slides down on inclined plane and reaches the bottom with linear velocity \(\mathrm{V}\). If the same mass is in the form of ring and rolls without slipping down the same inclined plane. Its velocity will be \(\\{\mathrm{A}\\} \mathrm{V}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~V})}\) \(\\{\mathrm{C}\\}(\mathrm{V} / \sqrt{2})\) \(\\{\mathrm{D}\\} 2 \mathrm{~V}\)

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

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