Question

The V - I graph for a conductor at temperature ${T_1}$ and ${T_2}$ are as shown in the figure. The term $({T_2} - {T_1})$ is proportional to:

A) $\dfrac{{\sin 2\theta }}{{{{\sin }^2}\theta }}$
B) $\dfrac{{\cot 2\theta }}{{{{\sin }^2}\theta }}$
C) $\dfrac{{\cos 2\theta }}{{\sin 2\theta }}$
D) $\dfrac{{\tan 2\theta }}{{{{\sin }^2}\theta }}$

Answer 1

Hint: Recall that the slope of $V - I$ graph gives $\operatorname{R} $ . Revise the trigonometric formulae for the ease of calculations in the question. Also, we must know how the resistance varies with respect to temperature.

Complete step by step solution:
Here we are given a $V - I$ graph.
By Ohm’s law, we know that: $V \propto I$
$ \Rightarrow V = IR$
$ \Rightarrow \dfrac{V}{I} = R$
$ \Rightarrow $ The slope of $V - I$ graph gives $R$
$\therefore $ Consider $\dfrac{{{V_1}}}{{{I_1}}} = {R_1}$ resistance at temperature ${T_1}$ and $\dfrac{{{V_2}}}{{{I_2}}} = {R_2}$ resistance at temperature ${T_2}$ . self-made diagram
But, we also know that
$\tan \theta = \dfrac{{opposite}}{{adjacent}}$
$ \Rightarrow \tan \theta = \dfrac{{{V_1}}}{{{I_1}}}$
$ \Rightarrow {R_1} = \tan \theta $
And ${R_2} = \tan ({90^ \circ } - \theta )$
$ \Rightarrow {R_2} = \cot \theta $

The resistance of a conductor always depends on the temperature. As the temperature increases the resistance of the conductor also increases. For small temperatures, the resistance of the conductor increases linearly with temperature, which is given by the equation:
$R = {R_o}(1 + \alpha T)$
Where $R$ is resistance at temperature $T$ in $Ohms(\Omega )$
${R_o}$ is resistance at absolute temperature in $\Omega $
$T$ is temperature in $Kelvin(K)$
$\alpha $ is temperature coefficient of resistance
$\therefore {R_1} = {R_o}(1 + \alpha {T_1})$ and ${R_2} = {R_o}(1 + \alpha {T_2})$
Now,
${R_2} - {R_1} = {R_o}[1 + \alpha ({T_2} - {T_1})]$
But ${R_2} = \cot \theta $ and ${R_1} = \tan \theta $
Substituting these values in the above equation, we get
$\cot \theta - \tan \theta = {R_o}[1 + \alpha ({T_2} - {T_1})]$
$ \Rightarrow {T_2} - {T_1} \propto \cot \theta - \tan \theta $ $equation(1)$
Now, we need to simply the equation $\cot \theta - \tan \theta $
$\cot \theta - \tan \theta = \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}$
$ \Rightarrow \cot \theta - \tan \theta = \dfrac{{\cos \theta \times \cos \theta - \sin \theta \times \sin \theta }}{{\sin \theta \times \cos \theta }}$
$ \Rightarrow \cot \theta - \tan \theta = \dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{\sin \theta \cos \theta }}$

Now, we can substitute ${\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta $
$ \Rightarrow \cot \theta - \tan \theta = \dfrac{{\cos 2\theta }}{{\sin \theta \cos \theta }}$
Now multiplying the numerator and denominator by $2$ , we get:
$ \Rightarrow \cot \theta - \tan \theta = \dfrac{{2\cos 2\theta }}{{2\sin \theta \cos \theta }}$
We know that $2\sin \theta \cos \theta = \sin 2\theta $
$\therefore \cot \theta - \tan \theta = \dfrac{{2\cos 2\theta }}{{\sin 2\theta }}$
Substituting this value in $equation(1)$
$ \Rightarrow {T_2} - {T_1} \propto \dfrac{{2\cos 2\theta }}{{\sin 2\theta }}$
$ \Rightarrow {T_2} - {T_1} \propto \dfrac{{\cos 2\theta }}{{\sin 2\theta }}$ ($\because 2$ is an integer we can ignore it)

$\therefore $ Option $(C), \dfrac{{\cos 2\theta }}{{\sin 2\theta }}$ is the correct option.

Note: One must know that the resistance of the conductor for small temperatures increases with increase in temperature. One is very likely to forget the trigonometric formulas. Do not confuse or make mistakes in the trigonometric formulas of $\cos 2\theta,\sin 2\theta $.