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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

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 \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)

The ratio of angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of hydrogen atom is ...... \(\\{\mathrm{A}\\} \sqrt{2}\) \(\\{B\\} \sqrt{(1 / 2)}\) \(\\{\mathrm{C}\\}(1 / 2) \quad\\{\mathrm{D}\\} 2\)

Two identical hollow spheres of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is \(\\{\mathrm{A}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(4 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(32 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(34 / 3) \mathrm{MR}^{2}\)

The height of a solid cylinder is four times that of its radius. It is kept vertically at time \(t=0\) on a belt which is moving in the horizontal direction with a velocity \(\mathrm{v}=2.45 \mathrm{t}^{2}\) where \(\mathrm{v}\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) is in second. If the cylinder does not slip, it will topple over a time \(t=\) \(\\{\mathrm{A}\\} 1\) second \(\\{\mathrm{B}\\} 2\) second \\{C \(\\}\) \\} second \(\\{\mathrm{D}\\} 4\) second

Two disc of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio \(1: 3\). The moment of inertia of these disc about the respective axes passing through their centres and perpendicular to their planes will be in the ratio. \(\\{\mathrm{A}\\} 1: 3\) \\{B\\} \(3: 1\) \\{C\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)
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