Question

A cylindrical rod having temperature \[{T_1}\] and ${T_2}$ at its end. The rate of flow of heat ${Q_1}cal{\sec ^{ - 1}}$if all the linear dimensions are double keeping temperature remains constant than rate of flow of heat ${Q_2}$will be:
A. $4{Q_1}$
B. $2{Q_1}$
C. $\dfrac{{{Q_1}}}{4}$
D. $\dfrac{{{Q_1}}}{2}$

Answer 1

Hint: Assuming two cylindrical rods and their temperatures at certain point flow of heat is in calorie per second, now we have to find out the flow of heat in another cylinder keeping the temperature constant, for that we are using Thermal current formula.

Complete step-by-step solution:
Two cylindrical rods having temperatures at its end is ${T_1}$ and${T_2}$,
And their rate of flow of heat is ${Q_1}cal{\sec ^{ - 1}}$
If the linear dimensions are double keeping the temperature remains constant, then the rate of flow of heat is,
For this we are using Thermal current formula, K
That means, $Q = \dfrac{{KA\left( {{T_1} - {T_2}} \right)}}{l}$
Where Q is Thermal current
K is coefficient of thermal current
A is area of cross section
T is the temperature two cylindrical rods is, ${T_1}$and ${T_2}$
L is the length of the cylindrical rod,
For rate of flow of heat ${Q_1}$is
${Q_1} = \dfrac{{KA\left( {{T_1} - {T_2}} \right)}}{l} \to \left( 1 \right)$
Where Area of cross section A is$\pi {r^2}$
By substituting the A value in equation (1) we get,
${Q_1} = \dfrac{{K\pi {r^2}\left( {{T_1} - {T_2}} \right)}}{l}$
The above equation is for flow of heat in ${Q_1}$in linear dimension,
Now we are assuming that linear dimension is doubled then the length and radius r is also doubled then, $l = 2l$and ${r_1} = 2r$ the rate flow of heat for ${Q_2}$ is,
Then we get the thermal current is, ${Q_2}\dfrac{{K\pi {r_1}^2\left( {{T_1} - {T_2}} \right)}}{{2l}}$
The temperature remains constant for rate of flow of heat in ${Q_2}$
$
  {Q_2} = \dfrac{{K\pi {{\left( {2r} \right)}^2}\left( {{T_1} - {T_2}} \right)}}{{2l}} \\
  {Q_2} = 2\left( {\dfrac{{K\pi {r^2}\left( {{T_1} - {T_2}} \right)}}{l}} \right) \\
  {Q_2} = 2{Q_1}cal{\sec ^{ - 1}} \\
 $
In the above equation the rate of flow of heat ${Q_2}$ is double the rate of flow heat to ${Q_1}$
Hence the correct option is B.

Note:From the above solution we have found that the rate of flow of heat is doubled in the linear dimension where the temperature is constant. The rate of flow of heat for ${Q_2}$is doubled the rate of flow of heat to${Q_1}$. Where temperature remains constant the rate of flow heat is $2{Q_1}cal{\sec ^{ - 1}}$.