SATHEE: Heat and Thermodynamics 2 Question 3

Heat and Thermodynamics 2 Question 3

6. The ends $Q$ and $R$ of two thin wires, $P Q$ and $R S$ , are soldered (joined) together. Initially, each of the wire has a length of $1 m 10^{\circ} C$ . Now, the end $P$ is maintained at $10^{\circ} C$ , while the end $S$ is heated and maintained at $400^{\circ} C$ . The system is thermally insulated from its surroundings. If the thermal conductivity of wire $P Q$ is twice that of the wire $R S$ and the coefficient of linear thermal expansion of $P Q$ is $1.2 \times 10^{- 5} K$ $^{- 1}$ , the change in length of the wire $P Q$ is

(2016 Adv.)

(a) $0.78 m m$

(b) $0.90 m m$

(d) $2.34 m m$

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Answer:

Correct Answer: 6. (a)

Solution:

Rate of heat flow from $P$ to $Q$

$\frac{d Q}{d t} = \frac{2 K A (T - 10)}{1}$

Rate of heat flow from $Q$ to $S$

$\frac{d Q}{d t} = \frac{K A (4000 - T)}{1}$

At steady state rate of heat flow is same

$∴ \frac{2 K A (T - 10)}{1} = K A (400 - T)$

or $2 T - 20 = 400 - T$ or $3 T = 420$

$∴ T = 140^{\circ}$

Temperature of junction is $140^{\circ} C$

Temperature at a distance $x$ from end $P$

is $T_{x} = (130 x + 10^{\circ})$

Change in length $d x$ is suppose $d y$

Then, $d y = α d x (T_{x} - 10)$

$\begin{aligned} \int_{0}^{Δ y} d y & = \int_{0}^{1} α d x (130 x + 10 - 10) \\ Δ y & = {[\frac{α x^{2}}{2} \times 130]}_{0}^{1} \\ Δ y & = 1.2 \times 10^{- 5} \times 65 \\ Δ y & = 78.0 \times 10^{- 5} m = 0.78 m m \end{aligned}$

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Heat and Thermodynamics 2 Question 3

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

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महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Heat and Thermodynamics 2 Question 3

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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