Answer
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Hint: To solve this question we must know that the moment of inertia of a spherical shell about an axis passing through its center is \[\dfrac{2}{3}M{{R}^{2}}\]. Then we will use the parallel axis theorem to find the moment of inertia about a tangential axis. Parallel axis theorem states that the moment of inertia of a body about an axis parallel to the body passing through its center is equal to the sum of the moment of inertia of the body about the axis passing through the center and product of mass and square of the distance between two points.
Formula used:
\[{{I}_{\text{new}}}={{I}_{c}}+M{{d}^{2}}\]
Complete step by step answer:
Firstly we must draw a diagram for proper understanding of the situation.
Now, we will use parallel axis theorem to determine the moment of inertia about tangential axis by using the expression,
\[{{I}_{\text{new}}}={{I}_{c}}+M{{d}^{2}}\]
Where, \[{{I}_{\text{new}}}\] will be the moment of inertia of the tangential axis.
\[{{I}_{c}}\] is the moment of inertia about the axis through the center which is given by \[\dfrac{2}{3}M{{R}^{2}}\].
\[M\] is the mass of the shell and \[d\] is the distance between two axes which in this case is \[R\].
Now, the moment of inertia about a tangential axis is given as,
\[{{I}_{T}}={{I}_{c}}+M{{d}^{2}}=\dfrac{2}{3}M{{R}^{2}}+M{{R}^{2}}=\dfrac{5}{3}M{{R}^{2}}\]
So, we got the moment of inertia about a tangential axis as \[\dfrac{5}{3}M{{R}^{2}}\].
That means option d is the correct answer.
Note:
We must be aware of taking the moment of inertia of the sphere. The moment of inertia of a solid and hollow sphere are different. A spherical shell is a hollow sphere and the moment of inertia of the hollow sphere about an axis through the center is \[\dfrac{2}{3}M{{R}^{2}}\]. But for a solid sphere, it is \[\dfrac{2}{5}M{{R}^{2}}\]. So we must be very careful while taking moments of inertia for the sphere.
Formula used:
\[{{I}_{\text{new}}}={{I}_{c}}+M{{d}^{2}}\]
Complete step by step answer:
Firstly we must draw a diagram for proper understanding of the situation.
Now, we will use parallel axis theorem to determine the moment of inertia about tangential axis by using the expression,
\[{{I}_{\text{new}}}={{I}_{c}}+M{{d}^{2}}\]
Where, \[{{I}_{\text{new}}}\] will be the moment of inertia of the tangential axis.
\[{{I}_{c}}\] is the moment of inertia about the axis through the center which is given by \[\dfrac{2}{3}M{{R}^{2}}\].
\[M\] is the mass of the shell and \[d\] is the distance between two axes which in this case is \[R\].
Now, the moment of inertia about a tangential axis is given as,
\[{{I}_{T}}={{I}_{c}}+M{{d}^{2}}=\dfrac{2}{3}M{{R}^{2}}+M{{R}^{2}}=\dfrac{5}{3}M{{R}^{2}}\]
So, we got the moment of inertia about a tangential axis as \[\dfrac{5}{3}M{{R}^{2}}\].
That means option d is the correct answer.
Note:
We must be aware of taking the moment of inertia of the sphere. The moment of inertia of a solid and hollow sphere are different. A spherical shell is a hollow sphere and the moment of inertia of the hollow sphere about an axis through the center is \[\dfrac{2}{3}M{{R}^{2}}\]. But for a solid sphere, it is \[\dfrac{2}{5}M{{R}^{2}}\]. So we must be very careful while taking moments of inertia for the sphere.
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