A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of…
A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:
Increased 8 times
Doubled
Halved
Unchanged
Solution
Rate of heat i.e., Power developed in the
$
\text { wire }=P=\frac{V^2}{R}
$
Resistance of the wire of length, $L$
$
R_1=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}
$
$\therefore$ Power, $P_1=\frac{V^2}{R_1}$
Resistance of the wire when length is halved i.e., $L / 2$
$
R_2=\frac{\rho \frac{L}{2}}{\pi(2 r)^2}=\frac{\rho L}{\pi 8 r^2}=\frac{R_1}{8}
$
$\therefore$ Power, $P_2=\frac{V}{\frac{R_1}{8}}=\frac{8 V}{R_1}$
or, $\mathrm{P}_2=8 \mathrm{P}_1$ i.e., power increased 8 times of previous or original wire