Find how the volume density of the elastic deformation energy is distributed in a steel rod depending on the distance $r$ from its axis. The length of the rod is equal to $l$, the torsion angle to $\varphi $.
(A) $\dfrac{1}{2}\;\dfrac{{G{\varphi ^2}{r^2}}}{{{l^2}}}$.
(B) $\dfrac{3}{2}\;\dfrac{{G{\varphi ^2}{r^2}}}{{{l^2}}}$
(C) $\dfrac{5}{2}\;\dfrac{{G{\varphi ^2}{r^2}}}{{{l^2}}}$
(D) None of these

Hint:In this question, the concept of the torsion equation is used. The torsion formula is used to calculate the expression for the torsional energy in terms of deformation angle, length, and the radius of the rod.

Complete step by step solution:
We know that elastic energy is the energy stored by a system undergoing deformation. First of all we know that formula for elastic deformation energy in terms of length $l$ and distance $r$of the rod. It is given by
$E = \dfrac{{\pi {r^4}{\varphi ^2}}}{{4l}}......(i)$
Where, $G$is the shear modulus or modulus of rigidity, $r$ is the distance of the steel rod from its axis, $l$ is the length of the rod, and $\varphi $ is the torsion angle.
Then we will differentiate the above equation with respect to $dr$, we get,
$\dfrac{{dE}}{{dr}} = \dfrac{{4G\pi {r^3}{\varphi ^2}}}{{4l}}......(ii)$
Again, As we know that the volume density in terms of distance $r$ and length $l$of the rod is given as,
$V = 2\pi lr......(iii)$
In the next step we will differentiate the above equation with respect to $dr$ and we get, $\dfrac{{dV}}{{dr}} = 2\pi l......(iv)$
Now by dividing equation (ii) by equation (iii) for getting the result in form of $\dfrac{{dE}}{{dV}}$, we get the result as,
$ \Rightarrow \dfrac{{\dfrac{{dE}}{{dr}}}}{{\dfrac{{dV}}{{dr}}}} = \dfrac{{\dfrac{{4G\pi {r^3}{\varphi ^2}}}{{4l}}}}{{2\pi l}}$
After simplification, we get,
$\Rightarrow \dfrac{{dE}}{{dV}} = \dfrac{{G{r^2}{\varphi ^2}}}{{2{l^2}}}......(v)$
So, the answer to the above question is option (A) that is $\dfrac{1}{2}\;\dfrac{{\;\;G{\varphi ^2}{r^2}}}{{{l^2}}}$

Note:
As we know that the basic formula for the torsion energy can be written as,
$E = \dfrac{1}{2}T\phi $
Here, the torque applied is $T$ and the torsion angle is $\phi $.
And we know the torsion equation as,
$\dfrac{T}{J} = \dfrac{{G\phi }}{l}$
Here, $J$ is the polar moment of inertia of the rod and can be written as,
$J = \dfrac{\pi }{2}{r^4}$
Then, torsion equation become,
$\dfrac{T}{{\dfrac{\pi }{2}{r^4}}} = \dfrac{{G\phi }}{l}$
$ \Rightarrow T = \dfrac{{\pi G\phi {r^4}}}{{2l}}$
Now we substitute the value of the torque in the energy equation and get,
$\Rightarrow E = \dfrac{{\pi {r^4}{\varphi ^2}}}{{4l}}$