Answer
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Hint: As the hollow sphere is rotating with angular frequency ‘$\omega $’, it has rotational kinetic energy. The moment of inertia of the hollow sphere is given in the question. Using this, we can find the rotational kinetic energy and then interpret the relation between it and the heat energy.
Formula used:
${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$
$Q = MS\Delta T$
Complete step by step answer:
The rotational kinetic energy of a given object with moment of inertia, $I$ is given by ${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$.
For the hollow sphere, it is mentioned in the problem that, $I = \dfrac{2}{3}M{R^2}$.
Using the both relations, we have ${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2} \times \dfrac{2}{3}M{R^2} \times {\omega ^2} = \dfrac{1}{3}M{R^2}{\omega ^2}$.
As it is stated that $75\% $ of ${\left( {K.E.} \right)_{rot}}$is changed into heat, $Q$. Hence, $\eqalign{
& Q = 75\% \left( {{{\left( {K.E.} \right)}_{rot}}} \right) \cr
& \Rightarrow Q = \dfrac{3}{4} \times {\left( {K.E.} \right)_{rot}} = \dfrac{3}{4} \times \dfrac{1}{3}M{R^2}{\omega ^2} \cr
& \Rightarrow Q = \dfrac{1}{4}M{R^2}{\omega ^2} \cr} $
But, $Q = MS\Delta T$. Using this formula and obtained relation, we get
$\eqalign{
& Q = MS\Delta T = \dfrac{1}{4}M{R^2}{\omega ^2} \cr
& \Rightarrow \Delta T = \dfrac{{{R^2}{\omega ^2}}}{{4S}} \cr} $
Therefore, the correct option is B.
Note:
Though it is not mentioned in the question about the axis of rotation of the hollow sphere, as the moment of Inertia is given, we calculate the kinetic energy about the given data. But one must always look out for the axis of rotation, as it affects the kinetic energy of the hollow sphere.
Formula used:
${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$
$Q = MS\Delta T$
Complete step by step answer:
The rotational kinetic energy of a given object with moment of inertia, $I$ is given by ${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$.
For the hollow sphere, it is mentioned in the problem that, $I = \dfrac{2}{3}M{R^2}$.
Using the both relations, we have ${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2} \times \dfrac{2}{3}M{R^2} \times {\omega ^2} = \dfrac{1}{3}M{R^2}{\omega ^2}$.
As it is stated that $75\% $ of ${\left( {K.E.} \right)_{rot}}$is changed into heat, $Q$. Hence, $\eqalign{
& Q = 75\% \left( {{{\left( {K.E.} \right)}_{rot}}} \right) \cr
& \Rightarrow Q = \dfrac{3}{4} \times {\left( {K.E.} \right)_{rot}} = \dfrac{3}{4} \times \dfrac{1}{3}M{R^2}{\omega ^2} \cr
& \Rightarrow Q = \dfrac{1}{4}M{R^2}{\omega ^2} \cr} $
But, $Q = MS\Delta T$. Using this formula and obtained relation, we get
$\eqalign{
& Q = MS\Delta T = \dfrac{1}{4}M{R^2}{\omega ^2} \cr
& \Rightarrow \Delta T = \dfrac{{{R^2}{\omega ^2}}}{{4S}} \cr} $
Therefore, the correct option is B.
Note:
Though it is not mentioned in the question about the axis of rotation of the hollow sphere, as the moment of Inertia is given, we calculate the kinetic energy about the given data. But one must always look out for the axis of rotation, as it affects the kinetic energy of the hollow sphere.
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