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# A rigid spherical body is spinning around an axis without any external torque. Due to the rise of temperature its volume increases by 3%. Then the percentage change in its angular speed isA. -2%B. -1%C. -3%D. 1%

Last updated date: 21st Jun 2024
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Answer
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Hint: From the given percentage increase in volume, you could find the ratio of the radii before and after expansion due to the rise in temperature. Absence of external torque directly implies that the angular momentum is conserved. You could substitute the expression for moment of inertia in the conservation relation to get the ratio of angular velocity in terms of the radii. Now substituting the above found relation in here, we could calculate the percentage change in angular velocity.
Formula used:
Moment of inertia of a sphere,
$I=\dfrac{2}{5}M{{R}^{2}}$

Complete answer:
In the question, we are given a rigid spherical body that is spinning around an axis in the absence of any external torque. Its volume increases by 3% as the result of increase in temperature. Using these given information, we have to find the percentage change in angular speed.
Let the radius before expansion be ${{R}_{1}}$ and the radius after the expansion of 3% be ${{R}_{2}}$. We know that the volume of a sphere is given by,
$V=\dfrac{4}{3}\pi {{R}^{3}}$
Then the percentage increase in volume can be given by,
$\dfrac{{{V}_{2}}-{{V}_{1}}}{{{V}_{1}}}\times 100=\dfrac{\dfrac{4}{3}\pi {{R}_{2}}^{3}-\dfrac{4}{3}\pi {{R}_{1}}^{3}}{\dfrac{4}{3}\pi {{R}_{1}}^{3}}\times 100=3$
${{\left( \dfrac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3}}=\left( \dfrac{103}{100} \right)$ ……………………………. (1)
We are also given that external torque is absent in the given case, so, the angular momentum is conserved here, that is,
${{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}$
Also, moment of inertia of a sphere is given by,
$I=\dfrac{2}{5}M{{R}^{2}}$
$\Rightarrow \dfrac{2}{5}M{{R}_{1}}^{2}{{\omega }_{1}}=\dfrac{2}{5}M{{R}_{2}}^{2}{{\omega }_{2}}$
$\Rightarrow \dfrac{{{\omega }_{1}}}{{{\omega }_{2}}}={{\left( \dfrac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}$
But from (1),
$\dfrac{{{R}_{2}}}{{{R}_{1}}}={{\left( \dfrac{103}{100} \right)}^{\dfrac{1}{3}}}$
$\Rightarrow \dfrac{{{\omega }_{1}}}{{{\omega }_{2}}}={{\left( \dfrac{{{R}_{2}}}{{{R}_{1}}} \right)}^{\dfrac{2}{3}}}={{\left( \dfrac{103}{100} \right)}^{\dfrac{2}{3}}}$ ………………………………. (2)
Now the percentage change in angular velocity is given by,
$\dfrac{{{\omega }_{2}}-{{\omega }_{1}}}{{{\omega }_{1}}}\times 100=\left( \dfrac{{{\omega }_{2}}}{{{\omega }_{1}}}-1 \right)\times 100$
Substituting (2) we get,
$\left( {{\left( \dfrac{100}{103} \right)}^{\dfrac{2}{3}}}-1 \right)\times 100=\left( {{\left( 0.97 \right)}^{0.67}}-1 \right)\times 100$
$\therefore \left( 0.98-1 \right)\times 100=-2$
Therefore, we could say that the percentage increase in angular velocity is -2%.

Hence, option A is the right answer.

Note:
We got the percentage change in angular velocity as -2%. The negative sign here indicates that the angular velocity is decreased due to the expansion of the sphere. So, we could conclude that the increase in volume by 3% due to the rise in temperature further results in decrease of angular velocity by 2%.