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JEE Advance2016
Paper-2
Question
The ends Q and R of two thin wires, PQ and RS , are soldered (Joined) together. Initially each of the wires has a length of 1 m at $10^{\circ}$. Now the end $P$ is maintained at $10^{\circ} \mathrm{C}$, while the end $S$ is heated and maintained at $400^{\circ} \mathrm{C}$. The system is thermally insulted from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of $P Q$ is $1.2 \times 10^{-5} \mathrm{~K}^{-1}$. the change in length of the wire $P Q$ is
Select the correct option:
A
0.78 mm
B
0.90 mm
C
1.56 mm
D
2.34 mm
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Solution
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Question Tags
JEE Advance
Physics
Easy
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