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Chapter 5: Problem 551

Two circular loop \(A \& B\) of radius \(r_{A}\) and \(r_{B}\) respectively are made from a uniform wire. The ratio of their moment of inertia about axes passing through their centres and perpendicular to their planes is \(\left(\mathrm{I}_{\mathrm{B}} / \mathrm{I}_{\mathrm{A}}\right)=8\) then \(\left(\mathrm{r}_{\mathrm{B}} / \mathrm{r}_{\mathrm{A}}\right)\) Ra is equal to... \(\\{\mathrm{A}\\} 2\) \(\\{B\\} 4\) \\{C \(\\}\) \(\\{\mathrm{D}\\} 8\)

Short Answer

Expert verified
The required ratio between the radii \(\left(\frac{r_{B}}{r_{A}}\right)\) is 2, which corresponds to option A.

Step by step solution

01

Write down the moment of inertia for a circular loop.

The moment of inertia (I) of a circular loop can be calculated using the formula: \[I = M r^2\] Here, M is the mass of the loop, and r is the radius of the loop.
02

Set up the equation for the ratio of moments of inertia.

According to the given problem, the ratio of the moments of inertia of loop B to loop A is 8. Therefore, we have: \[\frac{I_B}{I_A} = \frac{M_Br_B^2}{M_Ar_A^2} = 8\]
03

Use the mass of the uniform wire to express M in terms of the radius.

As the wire is uniform, the mass per unit length (m) is the same for both loops. We can express the mass of each loop in terms of its radius and the mass per unit length: \[M_A = m(2 \pi r_A)\] \[M_B = m(2 \pi r_B)\]
04

Substitute the mass expressions into the ratio equation.

Now, replace \(M_A\) and \(M_B\) in the equation for the ratio of moments of inertia: \[\frac{m(2\pi r_B) r_B^2}{m(2\pi r_A) r_A^2} = 8\]
05

Simplify the equation and solve for the ratio of radii.

The constant terms \(m\) and \(2\pi\) cancel out on both sides of the equation, leaving us with: \[\frac{r_B^3}{r_A^3} = 8\] Take the cube root of both sides of the equation to obtain: \[\frac{r_B}{r_A} = \sqrt[3]{8} = 2\] Hence, the required ratio between the radii \(\left(\frac{r_{B}}{r_{A}}\right)\) is 2, which corresponds to option A.

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Most popular questions from this chapter

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

From a circular disc of radius \(\mathrm{R}\) and mass \(9 \mathrm{M}\), a small disc of radius \(\mathrm{R} / 3\) is removed from the disc. The moment of inertia of the remaining portion about an axis perpendicular to the plane of the disc and passing through \(\mathrm{O}\) is.... \(\\{\mathrm{A}\\} 4 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(40 / 9) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(37 / 9) \mathrm{MR}^{2}\)

The M.I. of a body about the given axis is \(1.2 \mathrm{kgm}^{2}\) initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \mathrm{~J}\(. an angular acceleration of \)25 \mathrm{rad} \mathrm{sec}^{2}$ must be applied about that axis for duration of \(\ldots\) \(\\{\mathrm{A}\\} 4 \mathrm{sec}\) \\{B \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)

Particles of \(1 \mathrm{gm}, 1 \mathrm{gm}, 2 \mathrm{gm}, 2 \mathrm{gm}\) are placed at the corners \(A, B, C, D\), respectively of a square of side $6 \mathrm{~cm}$ as shown in figure. Find the distance of centre of mass of the system from geometrical centre of square. (A) \(1 \mathrm{~cm}\) (B) \(2 \mathrm{~cm}\) (C) \(3 \mathrm{~cm}\) (D) \(4 \mathrm{~cm}\)

A circular disc of radius \(R\) is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of \(60 ?\) and released. Its angular velocity when it reaches the equilibrium position will be \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / 3 \mathrm{R})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\) \(\\{\mathrm{C}\\} \sqrt{(2 \mathrm{~g} / \mathrm{R})}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 \mathrm{~g} / \mathrm{R})}\)
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