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\[\begin{align}

& A)\dfrac{2}{5}M{{R}^{2}} \\

& B)\dfrac{7}{5}M{{R}^{2}} \\

& C)\dfrac{2}{3}M{{R}^{2}} \\

& D)\dfrac{5}{3}M{{R}^{2}} \\

\end{align}\]

Answer
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\[{{I}_{\text{new}}}={{I}_{c}}+M{{d}^{2}}\]

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 \[\dfrac{5}{3}M{{R}^{2}}\].

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.