Question

Rate of heat flow through the cylindrical rod is $Q_1$ . Temperature of the end of rods are $T_1$ and $T_2$ . If all the linear dimensions of the rod become double and temperature remains same, its rate of heat flow is $Q_2$ , then
A. $Q_1=2Q_2$
B. $Q_2=2Q_1$
C. $Q_2=4Q_1$
D. $Q_1=4Q_2$

Answer 1

Hint: Heat conduction, also known as thermal conduction, is the mechanism through which heat is transmitted within a body as a result of particle collisions. Here, the temperature of the end of the rods are given and we have to find the rate of heat flow when its dimensions are changed while keeping the temperature constant. We will do it by using the thermal current formula.

Complete step by step solution:
Let the initial length of the cylinder be $l$ and its radius be $r$ , we know that its rate of heat flow is $Q_1$ and temperature at its ends is $T_1$ and $T_2$ . When $k$ is the thermal conductivity, rate of heat flow through this cylindrical rod will be:
$Q_1={\displaystyle \frac{kA(T_2-T_1)}{l}}$
$\Rightarrow Q_1={\displaystyle \frac{k\pi r^2(T_2-T_1)}{l}}$.........(1)

Now, the linear dimensions are doubled, that is, $R=2r$ and $L=2l$ while the temperature is same, then new rate of heat flow:
$$\begin{gathered} Q_2={\displaystyle \frac{k{A}^{\prime }(T_2-T_1)}{L}} \\ \Rightarrow Q_2={\displaystyle \frac{k\pi R^2(T_2-T_1)}{L}} \\ \Rightarrow Q_2={\displaystyle \frac{k\pi {(2r)}^2(T_2-T_1)}{2l}} \\ \Rightarrow Q_2={\displaystyle \frac{2k\pi r^2(T_2-T_1)}{l}} \end{gathered}$$.........(2)
After dividing the equation (2) by (1), we get;
$Q_2=2Q_1$

Hence, the correct answer is option B.

Note: We note that keeping the temperature constant: when radius is doubled, the rate of heat flow becomes four times when the length is constant. When length is doubled, the rate of heat flow becomes half when radius is constant. When the linear dimensions are doubled, that is, both radius and length is doubled then rate of heat flow becomes two times.