Question

The ratio of the lengths, densities, masses, and resistivities of two wires A and B are $1:2$, $1:2$, $1:1$ and $1:4$. The ratio of their resistance are:
A) $1:32$
B) $1:16$
C) $8:1$
D) $4:1$

Answer 1

Hint: This is the concept of Ohm’s law. Firstly, we will calculate the area of the wire by using the suitable formula, and then we will calculate the ratio of resistance by putting the value of the area calculated. Here, instead of using $\rho $ for resistivity, we are using $r$ as the resistivity because $\rho $ is used for the density.

Formula used:
The formula for calculating area is given by
$mass = density \times length \times area$
Also, the formula used for resistance is given by
$R = \dfrac{{rl}}{A}$
Where $R$ is the resistance, $r$is the resistivity, $l$ is the length, and $A$ is the area of cross-section.

Complete step by step solution:
Let us first see the terms given in the question.
${l_1}:{l_2} = 1:2$ (Ratio of lengths)
${\rho _1}:{\rho _2} = 1:2$ (Ratio of densities)
${m_1}:{m_2} = 1:1$ (Ratio of masses)
${r_1}:{r_2} = 1:4$ (Ratio of resistivities)
Here, we are not $\rho $ for the resistivity because it is used here for the density.
Now, we know that $mass = density \times length \times area$
Or, $m = \rho \times l \times A$
$\therefore $ $A = \dfrac{{{m_1}}}{{\rho l}}$
Now, we will use the formula of resistance, which is given by
${R_1} = \dfrac{{{r_1}{l_1}}}{{{A_1}}}$
Now, we will put the value of A here,
$\therefore \,\,\,\,\,\,R = \dfrac{{{r_1}{l_1}{\rho _1}{l_1}}}{{{A_1}}}$
$ \Rightarrow \,{R_1} = \dfrac{{{r_1}{\rho _1}{l_1}^2}}{{{m_1}}}$
Similarly, ${R_2} = \dfrac{{{r_2}{\rho _2}{l_2}^2}}{{{m_2}}}$
Therefore, the ratio of ${R_1}$ and ${R_2}$ is given by
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{{{r_1}{\rho _1}{l_1}^2}}{{{m_1}}}}}{{\dfrac{{{r_2}{\rho _2}{l_2}}}{{{m_2}}}}}$
$ \Rightarrow \,\dfrac{{{R_1}}}{{{R_2}}} = (\dfrac{{{r_1}}}{{{r_2}}})(\dfrac{{{\rho _1}}}{{{\rho _2}}})(\dfrac{{{l_1}^2}}{{{l_2}^2}})(\dfrac{{{m_2}}}{{{m_1}}})$
Now, putting the values, we get
$\dfrac{{{R_1}}}{{{R_2}}} = (\dfrac{1}{4})\,(\dfrac{1}{2})\,(\dfrac{1}{4})\,(\dfrac{1}{1})$
$ \Rightarrow \,\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{{32}}$
Therefore, the ratio of the resistance is $1:32$.

Therefore, option (A) is the correct option.

Additional Information:
Now, let us know about the relation of resistance $R$ with length $l$ and area of cross-section $A$.
As we know, the formula of resistance is given by
$R = \dfrac{{\rho l}}{A}$
Now, if we assume $\rho $ as the constant of proportionality, then we will get
$R\,\propto \,l$ And $R\,\propto \,\dfrac{1}{A}$
Therefore, we can say that
The Resistance of a wire increases with an increase in the length of the wire.
The Resistance of a wire decreases with an increase in the area of cross-section of the wire.

Note: As we all know, Ohm’s law is used to find voltage, current, and the resistance of linear electrical circuits. Instead of these uses, Ohm’s law can also be used in daily life. Ohm’s law can be used to control the speed of fans, to operate electrical appliances like iron, kettle, heater, and many more. It is also useful in determining the fuses.