A cylindrical rod having temperature {T_1} an | vedclass

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Question

(b)Rate of heat flow $\left( {\frac{Q}{t}} ight) = \frac{{k\pi {r^2}({	heta _1} - {	heta _2})}}{L} \propto \frac{{{r^2}}}{L}$

$	herefore$ $\fr

Vedclass · Accepted Answer

$2{Q_1}$

Vedclass · Answer

(b)Rate of heat flow $\left( {\frac{Q}{t}} ight) = \frac{{k\pi {r^2}({	heta _1} - {	heta _2})}}{L} \propto \frac{{{r^2}}}{L}$

$	herefore$ $\frac{{{Q_1}}}{{{Q_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} ight)^2}\left( {\frac{{{l_2}}}{{{l_1}}}} ight) = {\left( {\frac{1}{2}} ight)^2} 	imes \left( {\frac{2}{1}} ight) = \frac{1}{2}$

==> ${Q_2} = 2{Q_1}$

A cylindrical rod having temperature ${T_1}$ and ${T_2}$ at its ends. The rate of flow of heat is ${Q_1}$ $cal/sec$. If all the linear dimensions are doubled keeping temperature constant then rate of flow of heat ${Q_2}$ will be

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