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# Suppose that both ends of a rod is kept at a temperature of ${0^ \circ }C$ and the initial temperature distribution is given by $T = {T_0}\sin \left( {\dfrac{{\pi x}}{L}} \right)$ where L is the length of the rod and ‘x’ is measured from the left end of the rod and ${T_0}$ is some constant. Thermal conductivity of rod is k and cross sectional area of rod is A.What is the initial heat current at the centre of the rod?A. 0B. $\dfrac{{kA{T_0}}}{L}$C. $\dfrac{{2kA{T_0}}}{L}$D. $\dfrac{{2kA{T_0}}}{{3L}}$

Last updated date: 22nd Feb 2024
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Hint: Generally when temperature difference is maintained then heat transfers from higher temperature body to the lower temperature body. The amount of heat transferred depends upon the various factors like the temperature difference, thermal resistance of the material.
Formula used:
\eqalign{ & R = \dfrac{L}{{KA}} \cr & H = \dfrac{{\Delta T}}{R} \cr}

Flow of a quantity with time is known as a current. It can be fluid current or heat current or electric current.
In case of fluid current a pressure difference is maintained and that drives the flow of fluid and fluid always flows from high pressure region to the low pressure region and fluid current is governed by fluid resistance too ,whereas in electric current the electric charge flows with time and the voltage difference and electric resistance combined will govern the electric current. Charge flows from higher voltage to lower voltage naturally.
Similarly in thermal current it is governed by temperature difference and thermal resistance.
Thermal resistance of a thermal conductor of length ‘L’ and cross sectional area ‘A’ and thermal conductivity ‘K’ is given by $R = \dfrac{L}{{KA}}$
Heat current is given as $H = \dfrac{{\Delta T}}{R}$ for the length ‘x’ we have
\eqalign{ & H = \dfrac{{\Delta T}}{R} \cr & \Rightarrow H = \dfrac{{kA\Delta T}}{L} \cr & \therefore {H_x} = kA\dfrac{{dT}}{{dx}} \cr}
Here $\dfrac{{dT}}{{dx}}$ is temperature gradient.
\eqalign{ & T = {T_0}\sin \left( {\dfrac{{\pi x}}{L}} \right) \cr & \Rightarrow \dfrac{{dT}}{{dx}} = {T_0}\dfrac{\pi }{L}\cos \left( {\dfrac{{\pi x}}{L}} \right) \cr & at{\text{ }}x = \dfrac{L}{2} \cr & \Rightarrow \dfrac{{dT}}{{dx}} = {T_0}\dfrac{\pi }{L}\cos \left( {\dfrac{{\pi \left( {\dfrac{L}{2}} \right)}}{L}} \right) \cr}
\eqalign{ & \Rightarrow \dfrac{{dT}}{{dx}} = {T_0}\dfrac{\pi }{L}\cos \left( {\dfrac{\pi }{2}} \right) \cr & \therefore \dfrac{{dT}}{{dx}} = 0 \cr}
Since the temperature gradient is zero at the center, heat current also will be zero.

Hence option A is the answer.

Note:
As long as the temperature difference is maintained between the ends of the rods, the heat keeps on flowing. Temperature gradient here between the initial point and the center of the rod is attained as zero. That means the rod initial point and the center of the rod are at the same temperature that is zero degree Celsius.