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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}}}

 
Answers (1)
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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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