Question

Masses of three wires of same material are in the ratio $1:3:5$ and their lengths are in the ratio $5:3:1$. If they are connected in series with a battery then the ratio of heats produced in them will be

• A. 5: 9: 5
• B. 25: 9: 1
• C. 125: 15: 1
• D. 25: 15: 9

Answer 1

Answer: (C)

Mass of wire, $m=$ Volume $\times$ Density $=Al\times d$
Resistance of wire, $R=\frac{\rho l}{A}$
or $A=\frac{\rho l}{R}\therefore m=\frac{\rho l}{R}\times ld$
or $R=\frac{\rho l^{2}d}{m}$ i.e., $R\propto \frac{l^2}{m}$
When wires are in series, there will be same current in each wire.
The heat produced $H\propto R$.
$\therefore H_1:H_2:H_3=\frac{l_1^2}{m_1}:\frac{l_2^2}{m_2}:\frac{l_3^2}{m_3}=\frac{5^2}{1}:\frac{3^2}{3}:\frac{1^2}{5}$
$=25:3:\frac{1}{5}=125:15:1$