Question

Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical resistances are ___ .

• A. 1 : 3 : 5
• B. 5 : 3 : 1
• C. 1 : 15 : 125
• D. 125 : 15 : 1

Answer 1

Answer: (D)

Step 1: Given that:

The ratio of the masses of three wires if copper = $1:3:5$

The ratio of the lengths of the wires = $5:3:1$

Let, $l_1,l_2$ , $l_3$ and $m_1,m_2,m_3$ be the lengths and masses of the three wires resppectively.

Then, $m_1:m_2:m_3=1:3:5$

and, $l_1:l_2:l_3=5:3:1$

Step 2: Formula and concept used:

The volume of wire= Area$\times$ length

If the volume of a wire is V, the area of cross-section is A and the length of the wire is l, then

$V=Al$

And, Mass(m)= Volume(V)$\times$ density(d)

Thus,

$m=Al\times d$

$\Rightarrow Al\times d=m$

$\Rightarrow A=\frac{m}{ld}$

The electric resistance of a wire is given by

$R=\frac{\text{ρ}l}{A}$

Where ρ is the resistivity of the material of the wire.

Putting the value of the area of cross-section , we get

$R=\frac{\text{ρ}l}{\frac{m}{ld}}$

$R=\frac{\text{ρ}{l}^{2}d}{m}$

Step 3: Calculation of the ratio of their resistances:

Let $R_1,R_2$ and $R_3$ be the resistances of the three wires of copper respectively.

As the wires are made of copper the value of resistivity(ρ) and the density of all three wires will be the same.

Thus,

$R_1=\frac{\text{ρ}{\left(l_1\right)}^{2}d}{m_1}$

$R_2=\frac{\text{ρ}{\left(l_2\right)}^{2}d}{m_2}$

And,

$R_3=\frac{\text{ρ}{\left(l_3\right)}^{2}d}{m_3}$

Now,

$\frac{R_1}{R_2}=\frac{\frac{\text{ρ}{\left(l_1\right)}^{2}d}{m_1}}{\frac{\text{ρ}{\left(l_2\right)}^{2}d}{m_2}}$

$\frac{R_1}{R_2}=\frac{{\left(l_1\right)}^2}{{\left(l_1\right)}^2}\times \frac{m_2}{m_1}$

$\frac{R_1}{R_2}={\left(\frac{l_1}{l_2}\right)}^2\times \frac{m_2}{m_1}\frac{R_1}{R_2}=\frac{25}{9}\times \frac{3}{1}$

$\frac{R_1}{R_2}=\frac{25}{3}$

$\frac{R_1}{R_2}=\frac{25\times 5}{3\times 5}$

$\frac{R_1}{R_2}=\frac{125}{15}$.............(1)

And,

$\frac{R_2}{R_3}=\frac{\frac{\text{ρ}{\left(l_2\right)}^{2}d}{m_2}}{\frac{\text{ρ}{\left(l_3\right)}^{2}d}{m_3}}$

$\frac{R_2}{R_3}=\frac{{\left(l_2\right)}^2}{{\left(l_3\right)}^2}\times \frac{m_3}{m_2}$

$\frac{R_2}{R_3}=\frac{9}{1}\times \frac{5}{3}$

$\frac{R_2}{R_3}=\frac{15}{1}$..............(2)

From equation (1) and (2), we get

$R_1:R_2:R_3=125:15:1$

Thus,

The ratio of the resistances of the wires is $125:15:1$ .

Hence option D) 125:15:1 is the correct option.