Q

# I have a doubt, kindly clarify. 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 staright line.One may conclude that:

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 staright line.

One may conclude that:

• Option 1)

$R(T)=\frac{R_{o}}{T^{2}}$

• Option 2)

$R(T)=R_{o}e^{\frac{-T_{o}^{2}}{T^{2}}}$

• Option 3)

$R(T)=R_{o}e^{\frac{-T^{2}}{T_{o}^{2}}}$

• Option 4)

$R(T)=R_{o}e^{\frac{T^{2}}{T_{o}^{2}}}$

Views

Dependence of Resistance -

i) Length of conductor - $R$ $\alpha$ $l$

ii) Area of cross-section  -  $R \alpha \frac{l}{A}$

iii) Temperature (for conductor)

-

Let   y = mx + c

$\ln R=-m(\frac{1}{T^{2}})+C$

$R=e^{-(\frac{m}{T^{2}})}\cdot e^{C}$

Or

$R=R_oe^{\frac{-m}{T^{2}}}$

where $R_o$ is constant

So, correct option is (2)

which is

$R=R_oe^{\frac{-T_o^{2}}{T^{2}}}$

Option 1)

$R(T)=\frac{R_{o}}{T^{2}}$

Option 2)

$R(T)=R_{o}e^{\frac{-T_{o}^{2}}{T^{2}}}$

Option 3)

$R(T)=R_{o}e^{\frac{-T^{2}}{T_{o}^{2}}}$

Option 4)

$R(T)=R_{o}e^{\frac{T^{2}}{T_{o}^{2}}}$

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