Question

An aluminium rod and a copper rod are taken such that their lengths are same and their resistances are also same. The specific resistance of copper is half that of aluminium, but its density is three times that of aluminium. The ratio of the mass of aluminium rod and that of copper rod will be
$\text{A}\text{. }\dfrac{1}{6}$
$\text{B}\text{. }\dfrac{2}{3}$
$\text{C}\text{. }\dfrac{1}{3}$
D. 6

Answer 1

Hint : Use the formula $R=\rho \dfrac{l}{A}$ and find the ratio of cross sectional areas of both the rods. Then use the formula for density i.e $D=\dfrac{m}{Al}$. Equate the ratio of the densities of both the rods and find the ratio of their masses.
Formula used:
$R=\rho \dfrac{l}{A}$
$D=\dfrac{m}{Al}$

Complete step by step solution :
It is given that there are 2 rods. One rod is made up of aluminium and the other rod is made up of copper. It is said that both the rods are of equal lengths and the resistances of both the rods are also the same.
Let us first understand what is resistance and what is its expression.
Resistance is the ability of a given substance to resist the flow of charges when a potential difference is produced across the given substance. Suppose the substance is a rod of length l and it has a uniform cross sectional area A. Then the resistance of this substance is given as $R=\rho \dfrac{l}{A}$ ….. (i).
Here, $\rho$ is the resistivity of the rod. Resistivity depends on the material of which the rod is made.
When the length and the cross sectional area of the rod are of one unit each, the resistance of the rod is called specific resistance. Then from equation (i), we get that specific resistance is equal to the resistivity of the rod.
The two given rods have the same lengths and resistances. However, they have different cross sectional areas and specific resistances.
Let the cross section areas of aluminium rod and copper rod be ${{A}_{1}}$ and ${{A}_{2}}$ respectively. And let the specific resistances of the rods be ${{\rho }_{1}}$ and ${{\rho }_{2}}$ respectively.
Then we can write the value of R as,
$R={{\rho }_{1}}\dfrac{l}{{{A}_{1}}}$ …. (ii)
And
$R={{\rho }_{2}}\dfrac{l}{{{A}_{2}}}$ …. (iii).
Divide equation (ii) by equation (iii).
$\Rightarrow \dfrac{R}{R}=\dfrac{{{\rho }_{1}}\dfrac{l}{{{A}_{1}}}}{{{\rho }_{2}}\dfrac{l}{{{A}_{2}}}}$
$\Rightarrow 1=\dfrac{{{\rho }_{1}}{{A}_{2}}}{{{\rho }_{2}}{{A}_{1}}}$
$\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}$
It is given that $\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=2$.
Therefore,
$\dfrac{{{A}_{1}}}{{{A}_{2}}}=2$ …. (iv).
Let the densities of the aluminium rod and copper rod be ${{D}_{1}}$ and ${{D}_{2}}$ respectively. And let the masses of the rod be ${{m}_{1}}$ and ${{m}_{2}}$ respectively.
We know that density is equal to mass upon volume.
Therefore, we get
${{D}_{1}}=\dfrac{{{m}_{1}}}{{{A}_{1}}l}$ ….. (v)
And
${{D}_{2}}=\dfrac{{{m}_{2}}}{{{A}_{2}}l}$ …… (vi)
Divide equation (v) by equation (vi).
$\Rightarrow \dfrac{{{D}_{1}}}{{{D}_{2}}}=\dfrac{\dfrac{{{m}_{1}}}{{{A}_{1}}l}}{\dfrac{{{m}_{2}}}{{{A}_{2}}l}}$
$\Rightarrow \dfrac{{{D}_{1}}}{{{D}_{2}}}=\dfrac{{{m}_{1}}{{A}_{2}}}{{{m}_{2}}{{A}_{1}}}$
$\Rightarrow \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{D}_{1}}{{A}_{1}}}{{{D}_{2}}{{A}_{2}}}$
From equation (iv) we know that $\dfrac{{{A}_{1}}}{{{A}_{2}}}=2$ and it is given that $\dfrac{{{D}_{1}}}{{{D}_{2}}}=\dfrac{1}{3}$.
Therefore,
$\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{1}{3}\times 2=\dfrac{2}{3}$
Hence, the correct option is B.

Note : Note that the specific resistance and resistivity are not the same quantities. Only the values of specific resistance and the resistivity are the same for a given substance.
Specific resistance is the resistance of the substance when its length and cross sectional area are of one unit whereas resistivity does not depend on the length and area of the substance. Both the quantities have different dimensional formulas.