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Q.43) Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2 K$ while that in the middle has thermal conductivity $K$. The left end of the combination is maintained at temperature $3 T$ and the right end at $T$. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1 / T_2$ is

A) 5/4

B) 3/2

C) 4/3

D) 5/3

 

Answers (1)

best_answer

All rods are identical in size, so heat current $Q \propto K \cdot \Delta T$
- Left rod: $Q \propto 2 K\left(3 T-T_1\right)$
- Middle rod: $Q \propto K\left(T_1-T_2\right)$
- Right rod: $Q \propto 2 K\left(T_2-T\right)$

Since heat flow is same in all rods:

$$
2\left(3 T-T_1\right)=T_1-T_2=2\left(T_2-T\right)
$$


Solve:
- From $2\left(3 T-T_1\right)=T_1-T_2 \rightarrow 6 T-2 T_1=T_1-T_2$

$$
\rightarrow 6 T+T_2=3 T_1
$$

- From $T_1-T_2=2\left(T_2-T\right) \rightarrow T_1=3 T_2-2 T$

Now substitute:

$$
\begin{aligned}
& 6 T+T_2=3\left(3 T_2-2 T\right)=9 T_2-6 T \\
& \rightarrow 12 T=8 T_2 \Rightarrow T_2=\frac{3 T}{2}
\end{aligned}
$$


Now find $T_1=3 T_2-2 T=\frac{9 T}{2}-2 T=\frac{5 T}{2}$
So,

$$
\frac{T_1}{T_2}=\frac{\frac{5 T}{2}}{\frac{3 T}{2}}=\frac{5}{3}
$$

Posted by

Saumya Singh

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