Question

An electrical cable of copper has just one wire of radius $9\,mm$ . Its resistance is $5\,\Omega $ . This single copper wire of the cable is replaced by $6$ different well insulated copper wires each of radius $3\,mm$ . The total resistance of the cable will now be equal to:
A. $7.5\,\Omega $
B. $45\,\Omega $
C. $90\,\Omega $
D. $270\,\Omega $

Answer 1

Hint:Here we have to apply the concept of electrical resistance and resistivity. An object's electrical resistance is an indicator of its opposition to electric current movement. The characteristic property of the material through which it resists the amount of current through it is resistivity.

Complete step by step answer:
An object's (i.e., a resistor) resistance depends on its form and the material of which it is composed. Resistivity is a material's inherent property which is directly proportional to the overall resistance, an extrinsic quantity that depends on the resistor's length and cross-sectional region.

A given conductor material's electrical resistivity is a measure of how intensely the material resists the passage of electric current through it. This resistivity factor, often referred to as the 'specific electrical resistance', allows the resistance of different types of conductors, regardless of their lengths or cross-sectional areas, to be compared to each other at a given temperature according to their physical properties.

The variables that influence a conductor's resistance in ohms can be described as:
-The resistivity of the material the conductor is made from.
-The complete conductor's duration.
-The conductor's cross-sectional area.
-The conductor's temperature.

Let the resistance of single copper wire be ${R_1}$. If $\rho $ is the resistivity of copper wire, then ${R_1} = \dfrac{{\rho \times 1}}{{{a_1}}}$
${R_1} = \dfrac{{\rho \times 1}}
{{\pi {r_1}^2}}$ ....... (i)
If six wires are substituted for the wire, let the resistance of each wire be ${R_2}$, then:
${R_2} = \dfrac{{\rho \times 1}}{{{A_2}}}$
$\Rightarrow{R_2} = \dfrac{{\rho \times 1}}{{\pi {r_2}^2}}$ ...... (ii)
From equation (i) and (ii), we get:
$
\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{r_2}^2}}{{{r_1}^2}} \\
\Rightarrow\dfrac{5}{{{R_2}}} = {\dfrac{{\left( {3 \times {{10}^3}} \right)}}{{{{\left( {9 \times {{10}^3}} \right)}^2}}}^2} \\
\therefore{R_2} = 45\,\Omega $

Hence, option B is correct.

Note:Here we have to pay attention which is the values for numerator and denominator. Also we have to use standard units and if the units are not in standard form we need to convert them.