Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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)=R0/T2 
  • 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)

best_answer

 

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}}}

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main session 2 registration 2025 begins at jeemain.nta.nic.in; documents required, eligibility criteria
anam-khan

JEE Main 2025: BTech mechanical engineering cut-offs for women at NITs remain high despite low enrolment
anam-khan

JEE Main Results 2025: January 28 shift 1 rated ‘toughest’; what is a good score for session 1 exam

Latest Question