Question

Two wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first will be.
A. Twice of the second
B. Half of the second
C. Equal of the second
D. Four times of the second

Answer 1

Hint:For this question we will use the relationship between the resistance, the specific resistance (or the resistivity), the length and the area of cross section of the conductor (here, the wire). We will write the expression of area in terms of the diameter of the wire because we are given the information about diameter of the wire in this question. Then we will substitute these values in the formula for both the cases to get the required answer.

Formula used:
Resistance, $R = \dfrac{{\rho l}}{A}$
where $\rho $ is the specific resistance (or resistivity), $l$ is the length of the conductor (here, wire), and $A$ is the area of the cross section of the wire.
Area of cross section of the wire,
\[A = \pi {r^2}\]
where \[r\] is the radius of the cross section of the wire.
And $r = \dfrac{d}{2}$ where $d$ is the diameter of the cross section of the wire.

Complete step by step solution:
Let the subscripts $1$ and $2$ be used for ${1^{st}}$ and ${2^{nd}}$ wires respectively.
Given: ${l_1} = 2{l_2}$ and ${d_1} = 2{d_2}$ ...(1)
Since, $A \propto {r^2}$ and $r = \dfrac{d}{2} \\ $
This implies that $A \propto {d^2}$ ...(2)
Also, $R = \dfrac{{\rho l}}{A} \\ $ ...(3)
Which means $R \propto l$ and $R \propto \dfrac{1}{{{d^2}}}$ (from equation (2) and (3))
Thus, we get, ${R_1} \propto \dfrac{{{l_1}}}{{{d_1}^2}} \\ $...(4)
And ${R_2} \propto \dfrac{{{l_2}}}{{{d_2}^2}}$ ...(5)

The value of $\rho $ is taken the same because it has the same value for a given material at a given temperature. Substituting equation (1) in equation (4), we get,
${R_1} \propto \dfrac{{2{l_2}}}{{{{\left( {2{d_2}} \right)}^2}}} \\ $
Solving this, we get,
${R_1} \propto \dfrac{{{l_1}}}{{2{d_1}^2}} \\ $ ...(6)
Dividing the equations (5) and (6), we get,
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {\dfrac{{{l_2}}}{{2{d_2}^2}}} \right)}}{{\left( {\dfrac{{{l_2}}}{{{d_2}^2}}} \right)}} \\ $
Thus, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{2} \\ $
This implies that, ${R_1} = \dfrac{1}{2}{R_2}$

Hence, option B is the correct answer.

Note: While solving this question an important point to keep in mind is that the value of resistivity should be taken the same because both the wires are made of the same material. We can also solve this question by equating the value of resistivity obtained by putting the values of the known physical quantities for both the wires in the formula used.