Question

Two copper wires whose masses are 8 and 12 gm have lengths in the ratio 3 : 4. Their resistances are in the ratio.
A. 4 : 9
B. 16 : 9
C. 27 : 32
D. 27 : 128

Answer 1

Hint: Use the formula for the resistance of a wire of length l and cross sectional area A. The use the formula for mass density of a body. Find an expression for the ratio of the areas of cross section of the two wires. With this find the ratio of the two resistances.
Formula used:
$R=\rho \dfrac{l}{A}$
$density=\dfrac{mass}{volume}$

Complete answer:
The resistance of a conducting wire is given as $R=\rho \dfrac{l}{A}$,
where $\rho $ is the resistivity of the material of the wire, l is the length of the wire and A is the cross sectional area of the wire.
Here both the wires are made up of copper. Since the material is the same, the resistivity of both the wires will be the same.
Let the length and cross sectional area of first wire with mass ${{m}_{1}}=8gm$ be ${{l}_{1}}$ and ${{A}_{1}}$ respectively.
Let the length and cross sectional area of second wire with mass ${{m}_{2}}=12gm$ be ${{l}_{2}}$ and ${{A}_{2}}$ respectively.
$\Rightarrow {{R}_{1}}=\rho \dfrac{{{l}_{1}}}{{{A}_{1}}}$ …. (i)
And
${{R}_{2}}=\rho \dfrac{{{l}_{2}}}{{{A}_{2}}}$ …. (ii)
Divide (i) by (ii).
$\Rightarrow\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\rho\dfrac{{{l}_{1}}}{{{A}_{1}}}}{\rho\dfrac{{{l}_{2}}}{{{A}_{2}}}}$
$\Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{l}_{1}}{{A}_{2}}}{{{l}_{2}}{{A}_{1}}}$ … (iii).
Since the two wires are made from the same material, the mass density of the two will be the same.
We know that $density=\dfrac{mass}{volume}=\dfrac{m}{V}$.
In his case, ${{V}_{1}}={{A}_{1}}{{l}_{1}}$
And ${{V}_{2}}={{A}_{2}}{{l}_{2}}$.
$\Rightarrow density=\dfrac{{{m}_{1}}}{{{A}_{1}}{{l}_{1}}}=\dfrac{{{m}_{2}}}{{{A}_{2}}{{l}_{2}}}$
$\Rightarrow \dfrac{{{A}_{2}}}{{{A}_{1}}}=\dfrac{{{m}_{2}}{{l}_{1}}}{{{m}_{1}}{{l}_{2}}}$.
Substitute this value in (iii)
$\Rightarrow\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{l}_{1}}}{{{l}_{2}}}\left(\dfrac{{{m}_{2}}{{l}_{1}}}{{{m}_{1}}{{l}_{2}}} \right)=\dfrac{{{m}_{2}}}{{{m}_{1}}}{{\left( \dfrac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}$ …... (iv)
We know that $\dfrac{{{m}_{2}}}{{{m}_{1}}}=\dfrac{12gm}{8gm}=\dfrac{3}{2}$
And $\dfrac{{{l}_{1}}}{{{l}_{2}}}=\dfrac{3}{4}$.
Substitute the values in (iv).
$\Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{3}{2}{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{3}{2}\times \dfrac{9}{16}=\dfrac{27}{32}$.
This means that the ratio of the resistances of the two wires is 27 : 32.

So, the correct answer is “Option C”.

Note:
Note that resistivity is material property. This means that the resistivity of a given conductor only depends on the material of the conductor.
Therefore, conductors of different sizes but the same material have the same resistivity.
Whereas, resistance of a conductor depends on the length and cross sectional area of the conductor along with the material.