Answer
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Hint: The radius and thickness of the disc is given. Now find the volume of the disc from there. Now we know that the volume of the disc and sphere is the same as the disc is melted and re-casted into the sphere, then find the moment of inertia for both the bodies.
Complete step-by-step answer:
Radius of the disc =R (given)
Thickness of the disc = R/6(given)
Therefore the volume of the disc will be,
${{V}_{1}}$=$\pi {{R}^{2}}t=\pi {{R}^{2}}\dfrac{R}{6}=\pi \dfrac{{{R}^{3}}}{6}$
Now, let the radius of the sphere be ${R}'$
Therefore the ,volume of the sphere will be
${{V}_{2}}$=$\dfrac{4}{3}\pi {{{R}'}^{3}}$
As the disc is melted and re-casted hence volume will remain the same:
\[\pi \dfrac{{{R}^{3}}}{6}=\dfrac{4}{3}\pi {{{R}'}^{3}}\]
$\begin{align}
& \Rightarrow {{R}^{3}}=8{{({R}')}^{3}} \\
& \Rightarrow R=2{R}' \\
\end{align}$
Now ,let I$_{1}$ be the moment of inertia for disc =$\dfrac{M{{R}^{2}}}{2}$
And ,let I$_{2}$ be moment of inertia for the sphere = $\dfrac{2}{5}M{{({R}')}^{2}}=\dfrac{2}{5}M{{(\dfrac{R}{2})}^{2}}$
${{I}_{2}}={{I}_{1}}(\dfrac{1}{5})$
Therefore option A is the correct answer.
Additional information:
Inertia is a property of matter by virtue of which the body remains at rest or in uniform motion in the same straight line unless acted upon by some external force.
A quantity expressing a body's tendency to resist angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation, is known as moment of inertia.
Note: The metal disc was melted and re-casted hence there will be no change in mass. As the volume of the disc and sphere is the same hence with the volume of the disc, we can figure out the volume of the sphere.
Complete step-by-step answer:
Radius of the disc =R (given)
Thickness of the disc = R/6(given)
Therefore the volume of the disc will be,
${{V}_{1}}$=$\pi {{R}^{2}}t=\pi {{R}^{2}}\dfrac{R}{6}=\pi \dfrac{{{R}^{3}}}{6}$
Now, let the radius of the sphere be ${R}'$
Therefore the ,volume of the sphere will be
${{V}_{2}}$=$\dfrac{4}{3}\pi {{{R}'}^{3}}$
As the disc is melted and re-casted hence volume will remain the same:
\[\pi \dfrac{{{R}^{3}}}{6}=\dfrac{4}{3}\pi {{{R}'}^{3}}\]
$\begin{align}
& \Rightarrow {{R}^{3}}=8{{({R}')}^{3}} \\
& \Rightarrow R=2{R}' \\
\end{align}$
Now ,let I$_{1}$ be the moment of inertia for disc =$\dfrac{M{{R}^{2}}}{2}$
And ,let I$_{2}$ be moment of inertia for the sphere = $\dfrac{2}{5}M{{({R}')}^{2}}=\dfrac{2}{5}M{{(\dfrac{R}{2})}^{2}}$
${{I}_{2}}={{I}_{1}}(\dfrac{1}{5})$
Therefore option A is the correct answer.
Additional information:
Inertia is a property of matter by virtue of which the body remains at rest or in uniform motion in the same straight line unless acted upon by some external force.
A quantity expressing a body's tendency to resist angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation, is known as moment of inertia.
Note: The metal disc was melted and re-casted hence there will be no change in mass. As the volume of the disc and sphere is the same hence with the volume of the disc, we can figure out the volume of the sphere.
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