Question

We have two wires A and B of same mass and same material. The diameter of the wire A is half of that B. If the resistance of wire A is 24 ohm then the resistance of wire B will be
A. 12 ohms
B. 3.0 ohms
C. 1.5 ohms
D. None of the above

Answer 1

Hint:The resistivity is the property of the material so if different resistors are made of the same material then the resistivity of all the resistors will be equal. The resistance is proportional to the length and inversely proportional to the area of cross-section.

Formula used:
\[R = \dfrac{{\rho l}}{A}\]
where R is the resistance of the wire of length l and cross-sectional area A, \[\rho \] is the resistivity of the material of the wire.

Complete step by step solution:
If the length of the resistor is l and the area of cross-section is A. The wires A and B are made of the same material. So, the resistivity of wire A is equal to the resistivity of the wire B,
\[{\rho _A} = {\rho _B} = \rho \]
The density of wire A is equal to the density of wire B because both are made of the same material. It is given that the mass of wire A is equal to the mass of wire B,
\[{m_A} = {m_B}\]

If the length of the wire A and wire B are \[{l_A}\] and \[{l_B}\] respectively,
\[d\left( {{A_A}{l_A}} \right) = d\left( {{A_B}{l_B}} \right) \\ \]
\[\Rightarrow \dfrac{{{l_A}}}{{{l_B}}} = \dfrac{{{A_B}}}{{{A_A}}} \\ \]
\[\Rightarrow \dfrac{{{l_A}}}{{{l_B}}} = \dfrac{{\pi {{\left( {\dfrac{{{D_B}}}{2}} \right)}^2}}}{{\pi {{\left( {\dfrac{{{D_A}}}{2}} \right)}^2}}} \\ \]
\[\Rightarrow \dfrac{{{l_A}}}{{{l_B}}} = \dfrac{{D_B^2}}{{{{\left( {\dfrac{{{D_B}}}{2}} \right)}^2}}} \\ \]
\[\Rightarrow \dfrac{{{l_A}}}{{{l_B}}} = 4\]

Using the resistance formula, the resistance of the wire A will be,
\[{R_A} = \dfrac{{{\rho _A}{l_A}}}{{{A_A}}}\]
Using the resistance formula, the resistance of the wire B will be,
\[{R_B} = \dfrac{{{\rho _B}{l_B}}}{{{A_B}}}\]
So, the ratio resistances of three wires will be,
\[\dfrac{{{R_A}}}{{{R_B}}} = \dfrac{{\left( {\dfrac{{{\rho _A}{l_A}}}{{{A_A}}}} \right)}}{{\left( {\dfrac{{{\rho _B}{l_B}}}{{{A_B}}}} \right)}}\]
The resistance of wire A is given as 24 ohms, we need to find the resistance of wire B.

On simplifying the ratio, we get
\[\dfrac{{24\Omega }}{{{R_B}}} = \left( {\dfrac{{{l_A}}}{{{l_B}}}} \right) \times \left( {\dfrac{{{A_B}}}{{{A_A}}}} \right) \\ \]
\[\Rightarrow \dfrac{{24\Omega }}{{{R_B}}} = \left( 4 \right) \times \left( 4 \right) \\ \]
\[\therefore {R_B} = \dfrac{{24\Omega }}{{16}} = 1.5\Omega \]
Hence, the required resistance of wire B is 1.5 ohm.

Therefore, the correct option is C.

Note: We should be careful while using the ratio for the resistance. If we have given wires of different metals then the densities and the resistivity of wires would be different.