Question

The copper wires whose masses are 8 gm and 12 gm have lengths in the ratio 3:4. Find the ratio of their resistances.

• A. 27 : 32
• B. 25 : 32
• C. 32 : 27
• D. 32 : 25

Answer 1

Answer: (A)

Step 1: Given

Masses of the copper wire are $8gand12g$

Ratio of length of wires, $l_1:l_2=3:4$

Step 2 : Formula Used

Resistance $R=\rho \frac{l}{A}$ where,

$\rho =$ resistivity of the material

l = length of the copper wire

A = area of the copper wire

$density\left(d\right)=\frac{mass\left(m\right)}{volume\left(v\right)}$

Step 3: Calculation

Now ($\rho \frac{l}{A}\times \frac{l}{l}=\rho \frac{l\times l}{A\times l})=\rho \frac{l^2}{V}$

where v = volume;

Since $density\left(d\right)=\frac{mass\left(m\right)}{volume\left(v\right)}$

$\Rightarrow$$volume\left(v\right)=\frac{mass\left(m\right)}{density\left(d\right)}$

$\Rightarrow$$R=\rho \frac{l^2}{m}\times d$

For two copper wires, resistivity $\rho$ and the density $d$, remains constant.

So, let resistance of wire 1 be $R_1$, mass 8 gms and length $l_1$

Let resistance of wire 2 be $R_2$, mass 12 gms and length $l_2$

ratio of length of wires:

$l_1:l_2=3:4$

Thus

$\frac{R_1}{R_2}=\frac{\frac{l_1^2}{m_1}}{\frac{l_2^2}{m_1}}$

$\frac{R_1}{R_2}=\frac{l_1^2}{l_2^2}\times \frac{m_2}{m_1}$

$\frac{R_1}{R_2}={\left(\frac{3}{4}\right)}^2\times \frac{12}{8}$

$\frac{R_1}{R_2}=\frac{9}{16}\times \frac{12}{8}$

$\frac{R_1}{R_2}=\frac{27}{32}$

$R_1:R_2=27:32$

Hence, B is the correct option.