Current Electricity 1 Question 3

3. In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line.

(2019 Main, 10 April I)

One may conclude that

(a) $R(T)=R _0 e^{-T^{2} / T _0^{2}}$

(b) $R(T)=R _0 e^{T^{2} / T _0^{2}}$

(c) $R(T)=R _0 e^{-T _0^{2} / T^{2}}$

(d) $R(T)=\frac{R _0}{T^{2}}$

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Solution:

  1. From the given graph,

We can say that, $\ln R(T) \propto-\frac{1}{T^{2}}$

Negative sign implies that the slope of the graph is negative.

$$ \begin{array}{ll} \text { or } & \ln R(T)=\text { constant }-\frac{1}{T^{2}} \\ \Rightarrow & \ln R(T)=\frac{\exp (\text { const. })}{\exp \frac{1}{T^{2}}} \\ \Rightarrow & \ln R(T)=R _0 \exp -\frac{T _0^{2}}{T^{2}} \end{array} $$

Alternate Solution

From graph,

$$ \frac{\frac{1}{T^{2}}}{\frac{1}{T _0^{2}}}+\frac{\ln R(T)}{\ln R\left(T _0\right)}=1 $$

$\Rightarrow \quad \ln R(T)=\left[\ln R\left(T _0\right)\right] \cdot 1-\frac{T _0{ }^{2}}{T^{2}}$

or

$$ R(T)=R _0 \exp \frac{-T _0^{2}}{T^{2}} $$



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