Question

Masses of 3 wires of the same metal are in the ratio\[1:2:3\] and their lengths are in the ratio \[3:2:1\]. The electrical resistances are in ratio
A. \[1:4:9\]
B. \[9:4:1\]
C. \[1:2:3\]
D. \[27:6:1\]

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 area of cross-section is A then the volume of the resistor is the product of the length and the area of cross-section,
\[V = Al\]
If the mass of the wire is m then,
\[V = \dfrac{m}{d}\]
where d is the density of the material; again a property of the material of the wire.
Using the volume expression,
\[Al = \dfrac{m}{d}\]
\[\Rightarrow A = \dfrac{m}{{ld}}\]
The resistance of the wire will be,
\[R = \dfrac{{\rho l}}{A}\]

Substituting the expression for the area of cross-section,
\[R = \dfrac{{\rho {l^2}d}}{m}\]
So, the ratio resistances of three wires will be,
\[{R_1}:{R_2}:{R_3} = {\left( {\dfrac{{\rho {l^2}d}}{m}} \right)_1}:{\left( {\dfrac{{\rho {l^2}d}}{m}} \right)_2}:{\left( {\dfrac{{\rho {l^2}d}}{m}} \right)_3} \\ \]
As the densities of the wires as well as the resistivity are equal, so the ratio becomes,
\[{R_1}:{R_2}:{R_3} = \dfrac{{l_1^2}}{{{m_1}}}:\dfrac{{l_2^2}}{{{m_2}}}:\dfrac{{l_3^2}}{{{m_3}}} \\ \]
The ratio of the lengths of the wire is given as,
\[{l_1}:{l_2}:{l_3} = 3:2:1\]

So, the ratio of the square of their lengths will be,
\[l_1^2:l_2^2:l_3^2 = {\left( 3 \right)^2}:{\left( 2 \right)^2}:{\left( 1 \right)^2} \\ \]
\[\Rightarrow l_1^2:l_2^2:l_3^2 = 9:4:1\]
The ratio of the masses of the wire is given as,
\[{m_1}:{m_2}:{m_3} = 1:2:3\]
So, the ratio of the inverse of their lengths will be,
\[\dfrac{1}{{{m_1}}}:\dfrac{1}{{{m_2}}}:\dfrac{1}{{{m_3}}} = \dfrac{1}{1}:\dfrac{1}{2}:\dfrac{1}{3}\]

On combining both the ratios, the ratio of the resistances will be,
\[{R_1}:{R_2}:{R_3} = \left( {\dfrac{9}{1}} \right):\left( {\dfrac{4}{2}} \right):\left( {\dfrac{1}{3}} \right) \\ \]
\[\Rightarrow {R_1}:{R_2}:{R_3} = \left( {\dfrac{9}{1} \times 3} \right):\left( {\dfrac{4}{2} \times 3} \right):\left( {\dfrac{1}{3} \times 3} \right) \\ \]
\[\therefore {R_1}:{R_2}:{R_3} = 27:6:1\]
Hence, the required ratio of the resistances is \[27:6:1\].

Therefore, the correct option is D.

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.