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 $=A l \times d$
Resistance of wire, $R=\frac{\rho l}{A}$
or $A=\frac{\rho l}{R} \therefore m=\frac{\rho l}{R} \times l d$
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 $H \propto R$.
$\therefore H_{1}: H_{2}: H_{3} =\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}}=\frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5}$
$=25: 3: \frac{1}{5}=125: 15: 1$