Question

Two identical square rods of metal are welded end to end as shown in figure $\left(a\right)$, $20\text{calories}$ of heat flows through it in $4\text{minutes}$. If the rods are welded as shown in figure $\left(b\right)$, the same amount of heat will flow through the rods in

• A. $1\text{minute}$
• B. $2\text{minutes}$
• C. $4\text{minutes}$
• D. $16\text{minutes}$

Answer 1

The correct option is A $1\text{minute}$

Let $K$ be the thermal conductivity of each rod.
From conduction law,
$\frac{Q}{t}=\frac{K_{eq}A({\theta }_1-{\theta }_2)}{L}$
where,
$K_{eq}=$ equivalent thermal conductivity
$A=$ cross section area of rods
For fig. $\left(a\right)$:
$K_{eq}$ = Harmonic mean of $K_1\&K_2$
$\frac{1}{K_{eq}}=\frac{K_1+K_2}{2K_1{K}_2}=\frac{1}{K}$
$\Rightarrow K_{eq}=K$
for fig. (b):
$K_{eq}=\frac{KA+KA}{A+A}=K$
$\left(\text{AM of}{K}_1\&K_2\right)$
$\because Q_1=Q_2$
$\Rightarrow \frac{K\left(A\right)}{2l}({\theta }_1-{\theta }_2)t_1=\frac{K\left(2A\right)}{l}({\theta }_1-{\theta }_2)t_2$
Given, $t_1=4\text{min}$ & $({\theta }_1-{\theta }_2)={100}^{\circ }\text{C}$
$\Rightarrow t_2=\frac{t_1}{4}=1\text{min}$
Hence, option $\left(a\right)$ is the correct answer.