Question

An electric wire of length $l$ and area of cross-section $a$ has a resistance \[R\] ohm. Another wire of the same material having same length and area of cross section \[4a\] has a resistance of
A. \[4R\]
B. $\dfrac{R}{4} \\ $
C. $\dfrac{R}{{16}} \\ $
D. \[16R\]

Answer 1

Hint: In 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 substitute the given 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.

Complete step by step solution:
Let the subscripts $1$ and $2$ be used for ${1^{st}}$ and ${2^{nd}}$ wires respectively.
Given: ${l_1} = {l_2}$ and ${a_2} = 4{a_1}$ where ${l_1} = {l_2} = l$ and ${a_1} = a$ and so ${a_2} = 4a$
Now, we know that,
$R = \dfrac{{\rho l}}{A}$
Also, resistivity (or specific resistance) has the same value for a given material at a given temperature.

So, we can say that resistance is directly proportional to the length of the conductor (here, wire) and inversely proportional to its area of cross section. That is,
$R \propto \dfrac{l}{A} \\ $
This implies that for the first wire,
${R_1} \propto \dfrac{l}{a} \\ $ ...(1)
And since ${R_1} = R$, therefore,
${R_{}} \propto \dfrac{l}{a} \\ $
For the second wire,
${R_2} \propto \dfrac{l}{{4a}}$ ...(2)

Dividing equations (1) and (2), we get,
$\dfrac{{{R_2}}}{R} = \dfrac{{\left( {\dfrac{l}{{4a}}} \right)}}{{\left( {\dfrac{l}{a}} \right)}} \\ $
Solving this, we get,
${R_2} = \dfrac{1}{4}R \\ $
$\therefore {R_2} = \dfrac{R}{4}$

Hence, option B is the correct answer.

Note: A very important point to remember while solving all such questions is that the value of resistivity should be taken the same because both the wires are made of the same material (given in the question). Also we can solve this question by equating the value of resistivity obtained by putting known values in the formula for both the wires.