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Current Electricity question

2019 · 10 Apr · Shift 1 · Q49
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Current Electricity question

2019 · 10 Apr · Shift 1 · Q49

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
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. One may conclude that : JEE Main 2019 (Online) 10th April Morning Slot Physics - Current Electricity Question 247 English
  1. A
    R(T)=R0e−T2/T02R(T) = {R_0}{e^{ - {T^2}/T_0^2}}R(T)=R0​e−T2/T02​
  2. B
    R(T)=R0T2R(T) = {{{R_0}} \over {{T^2}}}R(T)=T2R0​​
  3. C
    R(T)=R0eT2/T02R(T) = {R_0}{e^{ {T^2}/T_0^2}}R(T)=R0​eT2/T02​
  4. D
    R(T)=R0e−T02/T2R(T) = {R_0}{e^{ - T_0^2/{T^2}}}R(T)=R0​e−T02​/T2
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