JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 6)
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 :
_10th_April_Morning_Slot_en_6_1.png)
$$R(T) = {R_0}{e^{ - {T^2}/T_0^2}}$$
$$R(T) = {{{R_0}} \over {{T^2}}}$$
$$R(T) = {R_0}{e^{ {T^2}/T_0^2}}$$
$$R(T) = {R_0}{e^{ - T_0^2/{T^2}}}$$
Explanation
$${{{1 \over {{T^2}}}} \over {{1 \over {T_0^2}}}} + {{\ln \,R\left( T \right)} \over {\ln \,R\left( {{T_o}} \right)}} = 1$$
$$ \Rightarrow $$ ln R(T) = ln R(To) $$\left( {1 - {{T_o^2} \over {{T^2}}}} \right)$$
$$R(T) = {R_o}{e^{ - \left( {{{T_o^2} \over {{T^2}}}} \right)}}$$
$$ \Rightarrow $$ ln R(T) = ln R(To) $$\left( {1 - {{T_o^2} \over {{T^2}}}} \right)$$
$$R(T) = {R_o}{e^{ - \left( {{{T_o^2} \over {{T^2}}}} \right)}}$$
Comments (0)
