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 :

Question

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 : (A) R(T) = (R0/T2)  (B) R(T) = R0 e-T2/T02  (C) R(T) = R0 e-T20/T2  (D) R(T) = R0 eT2/T02

Tardigrade Admin · Accepted Answer

((1/T2)/(1)T20) + (ℓ n(T)/ℓ nR(T0)) = 1 ⇒ ℓ nR(T) =[ℓ nR(T0) ](1- (T20/T2)) ⇒ R(T) =R0e(- (T20/T2))

Q. 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 :

$R (T) = \frac{R _{0}}{T ^{2}}$

$R (T) = R_{0} e^{- T^{2} / T_{0}^{2}}$

$R (T) = R_{0} e^{- T_{0}^{2} / T^{2}}$

$R (T) = R_{0} e^{T^{2} / T_{0}^{2}}$

$\frac{\frac{1}{T ^{2}}}{\frac{1}{T _{0}^{2}}} + \frac{ℓ n ( T )}{ℓ n R ( T _{0} )} = 1$
$\Rightarrow ℓ n R (T) = [ℓ n R (T_{0})] (1 - \frac{T _{0}^{2}}{T ^{2}})$
$\Rightarrow R (T) = R_{0} e^{(- \frac{T _{0}^{2}}{T ^{2}})}$

Q. 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 :

R(T)=T2R0​​

R(T)=R0​e−T2/T02​

R(T)=R0​e−T02​/T2

R(T)=R0​eT2/T02​

T02​1​T21​​+ℓnR(T0​)ℓn(T)​=1 ⇒ℓnR(T)=[ℓnR(T0​)](1−T2T02​​) ⇒R(T)=R0​e(−T2T02​​)

$R (T) = \frac{R _{0}}{T ^{2}}$

$R (T) = R_{0} e^{- T^{2} / T_{0}^{2}}$

$R (T) = R_{0} e^{- T_{0}^{2} / T^{2}}$

$R (T) = R_{0} e^{T^{2} / T_{0}^{2}}$

$\frac{\frac{1}{T ^{2}}}{\frac{1}{T _{0}^{2}}} + \frac{ℓ n ( T )}{ℓ n R ( T _{0} )} = 1$
$\Rightarrow ℓ n R (T) = [ℓ n R (T_{0})] (1 - \frac{T _{0}^{2}}{T ^{2}})$
$\Rightarrow R (T) = R_{0} e^{(- \frac{T _{0}^{2}}{T ^{2}})}$