Question

The specific resistance of a wire is \[\rho \] , its volume is \[3{m^3}\] and its resistance is \[3\Omega \] then its length will be :
A. $\dfrac{3}{{\sqrt \rho }}$
B. $\dfrac{{\sqrt 3 }}{\rho }$
C. $\dfrac{5}{{\sqrt \rho }}$
D. $\dfrac{{\sqrt 5 }}{\rho }$

Answer 1

Hint:In the given question we need to find the length of a conductor in terms of its resistivity. For this we use the formula of volume to find the area of cross section in terms of the length of the conductor. Then we will find the value of length using the relationship between resistance, resistivity, length and area of cross section of the conductor.

Formula Used:
$V = A \times l$ and $R = \dfrac{{\rho l}}{A}$
where $V$ is the volume, $A$ is the area of cross section, $l$ is the length, $R$ is the resistance and $\rho $ is the resistivity of the conductor.

Complete step by step solution:
Given: $V = 3{m^3}$ and $R = 3\Omega $
Now, we know that $V = A \times l$
This implies that, $3 = A \times l$
Therefore, $A = \dfrac{3}{l}$ ...(1)
Now, we know that resistance depends on resistivity, length and area of cross section of the conductor which is represented by the formula,
$R = \dfrac{{\rho l}}{A}$ ...(2)

Substituting equation (1) in equation (2), we get
$R = \dfrac{{\rho l}}{{\left( {\dfrac{3}{l}} \right)}}$
This implies,
$R = \dfrac{{\rho {l^2}}}{3}$
Substitute the value of resistance, we get
$3 = \dfrac{{\rho {l^2}}}{3}$
Solving further we get,
$\rho {l^2} = 9$
This gives,
${l^2} = \dfrac{9}{\rho }$
Thus, we obtain;
$\therefore l = \dfrac{3}{{\sqrt \rho }}$

Hence, option A is the correct answer.

Note: While starting with the solution one must not get confused while using the formula of volume. We just need to use this formula to get the area of the cross section in terms of the length of the conductor so that length is purely in terms of the resistivity of the conductor.