Question

The ends of two rods of different materials with their thermal conductivities, radii of cross sections, and lengths all in the ratio 1:2 are maintained at the same temperature difference, if the rate of flow of heat in the larger rod is\[4cal/\sec \], then find the rate of flow of heat in the shorter rod.
A. \[1\,cal/\sec \]
B. \[2\,cal/\sec \]
C. \[8\,cal/\sec \]
D. \[16\,cal/\sec \]

Answer 1

Hint: We must comprehend the rate of heat transfer in order to resolve this problem. The quantity of heat that is moved per unit of time is known as the rate of flow of heat. Here, we will use the heat flow formula to determine the answer.

Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is a cross-sectional area, \[\Delta T\] is the temperature difference between two ends of the metal, L is the length of the metal plate and K is the thermal conductivity.

Complete step by step solution:
The quantity of heat is nothing but how much the heat flow per unit of time in a given material and is given by,
\[Q = KA\dfrac{{\Delta T}}{L}\]
As they have given that, the two rods of different materials have their thermal conductivities, radii of cross sections, and lengths all are in the ratio 1:2 and are maintained at the same temperature difference \[\Delta T\]. That is,
\[\dfrac{{{K_S}}}{{{K_l}}} = \dfrac{{{r_S}}}{{{r_l}}} = \dfrac{{{L_s}}}{{{L_l}}} = \dfrac{1}{2}\]

The rate of flow of heat of the shorter rod and the longer rod is,
\[{\left( {\dfrac{Q}{t}} \right)_S} = \dfrac{{{K_S}\left( {\pi {r_S}^2} \right)\Delta T}}{{{L_S}}}\] and \[{\left( {\dfrac{Q}{t}} \right)_l} = \dfrac{{{K_l}\left( {\pi {r_l}^2} \right)\Delta T}}{{{L_l}}}\]
Then, the ratio of the shorter rod to the longer rod is,
\[\dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_S}}}{{{{\left( {\dfrac{Q}{t}} \right)}_l}}} = \dfrac{{\dfrac{{{K_S}\left( {\pi {r_S}^2} \right)\Delta T}}{{{L_S}}}}}{{\dfrac{{{K_l}\left( {\pi {r_l}^2} \right)\Delta T}}{{{L_l}}}}}\]

Since these two rods are maintained at the same temperature then,
\[\dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_S}}}{{{{\left( {\dfrac{Q}{t}} \right)}_l}}} = \dfrac{{\dfrac{{{K_S}\left( {{r_S}^2} \right)}}{{{L_S}}}}}{{\dfrac{{{K_l}\left( {{r_l}^2} \right)}}{{{L_l}}}}} \\ \]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_S}}}{{{{\left( {\dfrac{Q}{t}} \right)}_l}}} = \dfrac{{{K_S}\left( {{r_S}^2} \right)}}{{{L_S}}} \times \dfrac{{{L_l}}}{{{K_l}\left( {{r_l}^2} \right)}} \\ \]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_S}}}{{{{\left( {\dfrac{Q}{t}} \right)}_l}}} = \dfrac{{{K_S}}}{{{K_l}}} \times \dfrac{{{r_S}^2}}{{{r_l}^2}} \times \dfrac{{{L_l}}}{{{L_S}}}\]

Substitute the value of \[\dfrac{{{K_S}}}{{{K_l}}} = \dfrac{{{r_S}}}{{{r_l}}} = \dfrac{{{L_s}}}{{{L_l}}} = \dfrac{1}{2}\] in above equation, then,
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_S}}}{{{{\left( {\dfrac{Q}{t}} \right)}_l}}} = \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{2}{1} \\ \]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_S}}}{{{{\left( {\dfrac{Q}{t}} \right)}_l}}} = \dfrac{1}{4}\]
Here, they have given the rate of flow of heat in the larger rod is \[4\,cal/\sec \]. Then,
\[\dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_S}}}{4} = \dfrac{1}{4}\]
\[\Rightarrow {\left( {\dfrac{Q}{t}} \right)_S} = \dfrac{4}{4}\]
\[\therefore {\left( {\dfrac{Q}{t}} \right)_S} = 1\]
Therefore, the rate of flow of heat in the shorter rod is \[1cal/\sec \].

Hence, option A is the correct answer.

Note: Here in the given problem do not get confused with the formula for thermal conductivity and the rate of heat transfer. Since thermal conductivity and heat transfer are related to each other. And need to remember what factors the rate of flow of heat depends on.