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Chapter 5: Problem 562

A ring of mass \(\mathrm{M}\) and radius \(\mathrm{r}\) is melted and then molded in to a sphere then the moment of inertia of the sphere will be..... \(\\{\mathrm{A}\\}\) more than that of the ring \\{B \\} Less than that of the ring \(\\{\mathrm{C}\\}\) Equal to that of the ring \\{D\\} None of these

Short Answer

Expert verified
The moment of inertia of the sphere is less than that of the ring. The correct answer is (B) Less than that of the ring.

Step by step solution

01

Calculate the moment of inertia of the ring

The moment of inertia of a ring with mass \(M\) and radius \(r\) is given by the formula: \(I_\text{ring} = Mr^2\) In our case, we have the mass (\(M\)) and radius (\(r\)) of the ring, so we can directly apply this formula: \(I_\text{ring} = M r^2\)
02

Calculate the moment of inertia of the sphere

When the ring is melted and molded into a sphere, its moment of inertia will be given by the formula: \(I_\text{sphere} = \frac{2}{5}M R^2\) Since the mass and radius of the sphere are both equal to the mass and radius of the ring, we can substitute those values: \(I_\text{sphere} = \frac{2}{5}M r^2\)
03

Compare the moments of inertia

Now we have both moments of inertia and can compare them: \(I_\text{ring} = M r^2\) and \(I_\text{sphere} = \frac{2}{5}M r^2\) Start by dividing both sides by \(M r^2\): \(\frac{I_\text{ring}}{M r^2} = 1\) and \(\frac{I_\text{sphere}}{M r^2} = \frac{2}{5}\) Now we compare these two ratios, noticing that \(\frac{2}{5} < 1\) This means that \(I_\text{sphere} = \frac{2}{5}M r^2 < M r^2 = I_\text{ring}\) Hence, the moment of inertia of the sphere is less than that of the ring.
04

Answer

The correct answer is (B) Less than that of the ring.

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Most popular questions from this chapter

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

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\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The frictional force on the disc is \(\\{\mathrm{A}\\}[(\mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{B}\\}[(2 \mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{C}\\} \mathrm{Mg} \sin \theta\) \(\\{\mathrm{D}\\}\) None

Moment of inertia of a sphere of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is \(\mathrm{I}\). keeping mass constant if graph is plotted between \(\mathrm{I}\) and \(\mathrm{R}\) then its form would be.

A thin uniform rod \(A B\) of mass \(M\) and length \(L\) is hinged at one end \(\mathrm{A}\) to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end \(B\) strikes the floor is \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / \mathrm{L})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / \mathrm{L})}\) \(\\{C\\} \sqrt{(3 g / L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(\mathrm{g} / \mathrm{L})}\)
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