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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

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 \(\\{\mathrm{A}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(4 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(32 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(34 / 3) \mathrm{MR}^{2}\)

A constant torque of \(1500 \mathrm{Nm}\) turns a wheel of moment of inertia \(300 \mathrm{~kg} \mathrm{~m}^{2}\) about an axis passing through its centre the angular velocity of the wheel after 3 sec will be.......... $\mathrm{rad} / \mathrm{sec}$ \(\\{\mathrm{A}\\} 5\) \\{B \\} 10 \(\\{C\\} 15\) \(\\{\mathrm{D}\\} 20\)

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 \(\\} 207.6 \mathrm{rad} \mathrm{S}^{-1}\)

Let I be the moment of inertia of a uniform square plate about an axis \(\mathrm{AB}\) that passes through its centre and is parallel to two of its sides \(\mathrm{CD}\) is a line in the plane of the plate that passes through the centre of the plate and makes an angle of \(\theta\) with \(\mathrm{AB}\). The moment of inertia of the plate about the axis \(\mathrm{CD}\) is then equal to.... \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{B\\} I \sin ^{2} \theta\) \(\\{C\\} I \cos ^{2} \theta\) \(\\{\mathrm{D}\\} I \cos ^{2}(\theta / 2)\)

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

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