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Two different wires, whose specific resistance are in the ratio $2:3$, length $3:4$ and radius of cross section $1:2$. The ratio of their resistances isA. $3:4$B. $16:9$C. $5:6$D. $2:1$

Last updated date: 14th Jul 2024
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Hint: Remember that the ratio of the resistances of two conductors will always be directly proportional to the ration of their lengths and it is inversely proportional to the ratio of their cross sectional areas.

The resistance of a short wire is lesser than the resistance of a long wire because in a long wire the electrons will collide with more ions as they pass through the wire. This makes the relationship between wire length and resistance proportional.

The area of cross section of wire is given by: $A = \pi {r^2}$
The resistance of wire is given by: $R = \dfrac{{\rho l}}{A}$
So, Area of cross section of first wire: ${A_1} = \pi {r_1}^2$
Area of cross section of second wire: ${A_2} = \pi {r_2}^2$
The ratio of areas of two wires will be:
$\dfrac{{{A_1}}}{{{A_2}}} = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2} = \dfrac{1}{4}$
Resistance of first wire: ${R_1} = \dfrac{{{\rho _1}{l_1}}}{{{A_1}}}$
Resistance of second wire: ${R_2} = \dfrac{{{\rho _2}{l_2}}}{{{A_2}}}$
The ratio of resistances of two wires will be:
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{1}{{\dfrac{1}{4}}}$
$\therefore \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{2}{1}$

Therefore, option D is the correct answer.

Note: The resistance of a thin wire will be greater than the resistance of a thick wire because thin wire has lesser electrons to carry current. This makes the relationship between resistance and the area of the cross section of a wire inversely proportional. The resistance that is offered per unit length and unit cross sectional area of that material when a known quantity of voltage is applied at its end is called specific resistance.