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**Hint:**The given problem can be solved by using the formula of linear transformation of the temperature with the linear variance of temperature across the length of the rod is assumed to be $\theta $. And also, by using the information which is given in the question, the solution is determined.

**Formula used:**

By using the formula of linear transformation of the temperature in a rod is;

$\dfrac{{{\theta _2} - {\theta _1}}}{L} = \dfrac{{\theta - {\theta _1}}}{l}$

Where, ${\theta _1}$denotes the temperature at the end of point $A$, ${\theta _2}$ denotes the temperature at the end of point $B$, $L$ denotes the length of the rod.

**Complete step by step solution:**

The data given in the problem is;

The temperature at the end $A$ of the rod is, ${\theta _1} = {0^ \circ }\,\,C$.

The temperature at the end $B$ of the rod is, ${\theta _2} = {100^ \circ }\,\,C$.

The length of the rod is given as, $L = 100\,\,cm$.

Distance of the rod from the point $A$ is, $l = 60\,\,cm$

The linear variance of temperature across the length of the rod is assumed. If Heat Transfer is considered as a parallel to the Electric Current model, it will be easier to visualize. The current is equal throughout the rod.

By using the formula of linear transformation of the temperature in a rod is;

$\Rightarrow \dfrac{{{\theta _2} - {\theta _1}}}{L} = \dfrac{{\theta - {\theta _1}}}{l}$

Substitute the values of the temperature of the rod at the point $A$ and $B$, as well as the length of the rod.

$\Rightarrow \dfrac{{{{100}^ \circ } - {0^ \circ }}}{{100}} = \dfrac{{\theta - {0^ \circ }}}{{60}}$

On further simplification in the above equation, then the above equation is written as,

$\Rightarrow \dfrac{{100}}{{100}} = \dfrac{\theta }{{60}}$

By cancelling the same terms in the above equation, then the above equation is written as,

$\Rightarrow 1 = \dfrac{\theta }{{60}}$

By rearranging the terms in the above equation, then the above equation is written as,

$\Rightarrow \theta = {60^ \circ }$

Therefore, The temperature of the rod at a point $60cm$ distant from the end $A$ is $\theta = {60^ \circ }$.

**Hence the option (C), $\theta = {60^ \circ }$ is the correct answer.**

**Note:**When an object is heated or cooled, its length varies by an amount corresponding to the real length and the difference in temperature is known as the linear thermal expansion or it can be called as the linear variance temperature.

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