Heat And Thermodynamics Ques 82
- Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod is $4 cm^{2}$. End of copper rod is maintained at $100^{\circ} C$ whereas ends of brass and steel are kept at $0^{\circ} C$. Lengths of the copper, brass and steel rods are 46, 13 and $12 cm$ respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are $0.92,0.26$ and 0.12 in CGS units, respectively. Rate of heat flow through copper rod is
(a) $1.2 cal / s$
(b) $2.4 cal / s$
(c) $4.8 cal / s$
(d) $6.0 cal / s$
(2014 Main)
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Answer:
Correct Answer: 82.(c)
Solution:
Formula:
Series And Parallel Combination Of Rod :
- In thermal conduction, it is found that in steady state the heat current is directly proportional to the area of cross-section $A$ which is proportional to the change in temperature $\left(T _1-T _2\right)$.
Then,
$$ \frac{\Delta Q}{\Delta t}=\frac{K A\left(T _1-T _2\right)}{x} $$
According to thermal conductivity, we get
$$ \begin{aligned} & \text { i.e. } \quad \frac{d Q _1}{d t}=\frac{d Q _2}{d t}+\frac{d Q _3}{d t} \\ & \frac{0.92(100-T)}{46}=\frac{0.26(T-0)}{13}+\frac{0.12(T-0)}{12} \\ & \Rightarrow \quad T=40^{\circ} C \\ & \therefore \quad \frac{d Q _1}{d t}=\frac{0.92 \times 4(100-40)}{40} \end{aligned} $$
$$ =4.8 cal / s $$
BETA
