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# Two diagonally opposite corners of a square loop made of four thin rods of the same material, same dimensions are at temperatures $40^\circ$ and $10^\circ$. If the only heat conduction takes place then the temperature difference between the two other corners will be\begin{align} & \text{A}\text{. 0}{}^\circ \\ & \text{B}\text{. 10}{}^\circ \\ & \text{C}\text{. 25}{}^\circ \\ & \text{D}\text{. 15}{}^\circ \\ \end{align}

Last updated date: 25th Jun 2024
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Hint:Heat transfer occurs due to the temperature difference, and heat flows from higher temperature to lower temperature. If two heat conducting materials, having the same dimensions, are made of the same materials, then temperature also decreases at the same rate. Heat always transfers from high temperature to low temperature. The heat conduction formula can be used to calculate the heat transfer rate across two points, where thermal conductivity constant (k) depends on the material used.

Formula used:
Rate of heat conduction (H) through a slab is given by
$H=\dfrac{kA\Delta T}{\Delta x}$
Where,
A is the cross-sectional area of the slab
$\Delta T$ is the temperature difference between two points
$\Delta x$ is the thickness of the slab
K is the proportionality constant, which is called thermal conductivity of the material.

First, we will draw corresponding diagram:

We are asked to find the temperature difference between points B and D, which is $T_B-T_D$.
From the above formula, we can say that heat conduction rate (H) is directly proportional to cross sectional area (A) and temperature difference $\Delta T$, and H is inversely proportional to slab’s thickness $\Delta x$.
The square loop in the above question is made of the same material, and wire having the same dimensions. So $\Delta x$, A and k will not affect calculation and heat will flow at the same rate through all four wires.
Heat flow through path ADC = Heat flow through path ABC
$\Rightarrow {{H}_{1}}={{H}_{2}}$
But rate of heat conduction (H) through a slab is given by $H=\dfrac{kA\Delta T}{\Delta x}$, so the above equation can be written as,
$\dfrac{kA\left( {{T}_{A}}-{{T}_{D}} \right)}{{{x}_{AD}}}=\dfrac{kA\left( {{T}_{A}}-{{T}_{B}} \right)}{{{x}_{AB}}}$
Now we know,
${{x}_{AD}}={{x}_{AB}}$
So, we get
${{T}_{B}}-{{T}_{D}}=0$
Because heat will flow at the same rate through all four wires, heat will reach point B and point D in the same amount of time. Thus, the temperature difference between B and D will be 0.