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Chapter 5: Problem 561

One solid sphere \(\mathrm{A}\) and another hollow sphere \(\mathrm{B}\) are of the same mass and same outer radii. The moment of inertia about their diameters are respectively \(\mathrm{I}_{\mathrm{A}}\) and \(\mathrm{I}_{\mathrm{B}}\) such that... \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{B}\\} \mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\) $\\{\mathrm{D}\\}\left(\mathrm{I}_{\mathrm{A}} / \mathrm{I}_{\mathrm{B}}\right)=(\mathrm{d} \mathrm{A} / \mathrm{dB})$ (radio of their densities)

Short Answer

Expert verified
The correct answer is Option C: \(\mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\).

Step by step solution

01

Determine the moment of inertia formula for a solid sphere

For a solid sphere (sphere A), the moment of inertia _(I)_ about any diameter can be found using the following formula: \[I_A = \frac{2}{5} m_A R_A^2\] where, - \(I_A\) = moment of inertia of sphere A - \(m_A\) = mass of sphere A - \(R_A\) = radius of sphere A
02

Determine the moment of inertia formula for a hollow sphere

For a hollow sphere (sphere B), the moment of inertia _(I)_ about any diameter can be found using the following formula: \[I_B = \frac{2}{3} m_B R_B^2 \] where, - \(I_B\) = moment of inertia of sphere B - \(m_B\) = mass of sphere B - \(R_B\) = radius of sphere B
03

Use the given information

According to the exercise, both spheres have the same mass and outer radii. So, \(m_A = m_B\) and \(R_A = R_B\). We can replace the mass and radii in the formulas we found in Steps 1 and 2 with the given information.
04

Compare the moment of inertia of both spheres

Now, we need to compare the moment of inertia of both spheres after replacing mass and radii in the formulas from Steps 1 and 2. For sphere A: \[I_A = \frac{2}{5} m_A R_A^2 = \frac{2}{5} m_B R_B^2\] For sphere B: \[I_B =\frac{2}{3} m_B R_B^2\] From the above two equations, it becomes clear that: \[\frac{2}{5} m_B R_B^2 < \frac{2}{3} m_B R_B^2\] Therefore, we can conclude that: \[ \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\] Thus, the correct answer is
05

Option C

\(\mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\).

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Most popular questions from this chapter

Statement \(-1-\) A thin uniform rod \(A B\) of mass \(M\) and length \(\mathrm{L}\) 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 ends \(B\) strikes the floor $\sqrt{(3 g / L)}\( Statement \)-2$ - The angular momentum of the rod about the hinge remains constant throughout its fall to the floor. \(\\{\mathrm{A}\\}\) 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

If distance of the earth becomes three times that of the present distance from the sun then number of days in one year will be .... \(\\{\mathrm{A}\\}[365 \times 3]\) \(\\{\mathrm{B}\\}[365 \times 27]\) \(\\{\mathrm{C}\\}[365 \times(3 \sqrt{3})]\) \(\\{\mathrm{D}\\}[365 /(3 \sqrt{3})]\)

If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(\mathrm{T}\) is..... \(\\{\mathrm{A}\\}\left(\pi \mathrm{MR}^{3} / \mathrm{T}\right)\) \(\\{\mathrm{B}\\}\left(\operatorname{MR}^{2} \pi / \mathrm{T}\right)\) \(\\{C\\}\left(2 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\) \(\\{\mathrm{D}\\}\left(4 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\)

A circular disc of radius \(\mathrm{R}\) is removed from a bigger disc of radius \(2 \mathrm{R}\). such that the circumferences of the disc coincide. The centre of mass of the remaining portion is \(\alpha R\) from the centre of mass of the bigger disc. The value of \(\alpha\) is. \(\\{\mathrm{A}\\} 1 / 2\) \\{B \\} \(1 / 6\) \\{C\\} \(1 / 4\) \(\\{\mathrm{D}\\}[(-1) / 3]\)

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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