Question

If resistance of the wire formed by 1cc of copper be 2.46ohm. The diameter of the wire is 0.32mm. Then the specific resistance of wire will be:
$A.1.59\times {{10}^{-6}}\Omega cm$
$B.2.32\times {{10}^{-6}}\Omega cm$
$C.3.59\times {{10}^{-6}}\Omega cm$
$D.1.59\times {{10}^{-8}}\Omega cm$

Answer 1

Hint: We have to apply the formula of resistance in terms of specific resistance, length and area to find the correct answer. Specific resistance is defined as the resistance per unit length and unit area of cross-section. We have to use the factors on which the resistance depends to find the correct answer.
Formula used: We will use to following formula to get the correct answer:-
$R=\rho \dfrac{L}{A}$.

Complete step by step solution:
From the given question we have the following parameters:-
Volume of the wire,$V=1cc=1\times {{10}^{-6}}{{m}^{3}}$
Diameter of the wire,$d=0.32\times {{10}^{-3}}m$
Resistance of the wire,$R=2.46\Omega $
We will use the following formula:-
$R=\rho \dfrac{L}{A}$…………………….. $(i)$
Where $\rho $ is specific resistance and $L$ is the length of the conductor. We have to find the specific resistance$(\rho )$. So equation $(i)$ is written as follows:-
$\rho =\dfrac{R\times A}{L}$……………….. $(ii)$
But before putting values we have to find the area of cross section by the use following formula:-
$A=\dfrac{\pi {{d}^{2}}}{4}$
Putting values we get
$A=\dfrac{\pi \times {{\left( 0.32\times {{10}^{-3}} \right)}^{2}}}{4}$
Solving this we get
$A=0.25\times {{10}^{-6}}{{m}^{2}}$.
We have to modify the equation $(ii)$ because length of the wire is not given and volume is given. Therefore multiplying $A$to numerator and denominator both of equation $(ii)$ we get
$\rho =\dfrac{R\times A\times A}{L\times A}$
$\rho =\dfrac{R\times {{A}^{2}}}{V}$
Putting values we get
$\rho =\dfrac{2.46\times {{\left( 0.25\times {{10}^{-6}} \right)}^{2}}}{1\times {{10}^{-6}}}$
Solving further we get
$\rho =1.59\times {{10}^{-8}}\Omega m$
Therefore,$\rho =1.59\times {{10}^{-6}}\Omega cm$.

Hence, option $(A)$ is correct.

Note:
We should not get confused on the fact that both specific resistance and resistivity are the same thing. During our solution we have multiplied the area to the both numerator and denominator to get the equation in desired and easy mode to solve. We can also consider area of cross section as the product of breadth and height and then divide volume by area of cross section to get the length of the conductor and use the formula $\rho =\dfrac{R\times A}{L}$ directly to get the solution.