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(A) In a steady state of thermal conduction,the temperature gradient along the rod is constant.Let $	heta_A = 100^\circ C$ be the temperature at end

Vedclass · Accepted Answer

$70$

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(A) In a steady state of thermal conduction,the temperature gradient along the rod is constant.Let $	heta_A = 100^\circ C$ be the temperature at end $A$ and $	heta_B = 0^\circ C$ be the temperature at end $B$.The total length of the rod is $L = 20 \, cm$.Let $	heta_x$ be the temperature at a point $x = 6 \, cm$ from end $A$.The temperature gradient is given by $\frac{	heta_A - 	heta_x}{x} = \frac{	heta_A - 	heta_B}{L}$.Substituting the given values:$\frac{100 - 	heta_x}{6} = \frac{100 - 0}{20}$$\frac{100 - 	heta_x}{6} = \frac{100}{20}$$100 - 	heta_x = 5 	imes 6$$100 - 	heta_x = 30$$	heta_x = 100 - 30 = 70^\circ C$.Thus,the temperature at the point is $70^\circ C$.

In a steady state of thermal conduction,the temperatures of the ends $A$ and $B$ of a $20 \, cm$ long rod are $100^\circ C$ and $0^\circ C$ respectively. What will be the temperature of the rod at a point at a distance of $6 \, cm$ from the end $A$ of the rod (in $^\circ C$)?

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