The ends mathrm{Q} and mathrm{R} of two thin | vedclass

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Question

Let the temperature of the junction be $T.$

$	herefore 	ext { rate of heat transfer }=\frac{d Q}{d t}=\frac{2 K A(T-10)}{L}=\frac{K A(400-T)}{L}$

Vedclass · Accepted Answer

$0.78 \mathrm{~mm}$

Vedclass · Answer

Let the temperature of the junction be $T.$

$	herefore 	ext { rate of heat transfer }=\frac{d Q}{d t}=\frac{2 K A(T-10)}{L}=\frac{K A(400-T)}{L}$

$\Rightarrow 2(T-10)=400-T$

$	ext { or, } T =140^{\circ} C$

Now, for the wire $P Q$, let us imagine a small length $\Delta x$ at a distance $x$ from the junction.

$	herefore \frac{\Delta T }{\Delta x }=\frac{140-10}{1}=130$

So, temperature at distance $x :$

$T=10+130 x$

or, $T -10=130 x$

Increase in length of the small element $\Delta x$ is expressed as:

$\frac{d y}{d x}=\alpha \Delta T =\alpha( T -10)$

$	ext { or, } \frac{ dy }{ dx }=\alpha 	imes 130 x$

Integrating both sides, we get:

$\int_0^{\Delta L } dy =13 \circ \alpha \int_0^{ L } xdx$

$\because L =1 m 	ext { (Given) }$

$	ext { or, } \Delta L =\frac{13 \circ ax ^2}{2}=\frac{130 	imes 1.2 	imes 10^{-5} 	imes 1}{0.78} m=0.78 \ mm$

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