Question

In how many parts (equal) a wire of $100\Omega $ be cut so that a resistance of $1\Omega $ is obtained by connecting them in parallel?
A. $10$
B. $5$
C. $100$
D. $50$

Answer 1

Hint:To solve this problem, we need to use two concepts here. First, we will use the relation between resistance and length of the wire as here the wire is to be cut. Second, we will use the formula for equivalent resistance when resistances are connected in parallel connection.

Formulas used:
$R = \rho \dfrac{l}{A}$,
where, $R$ is the resistance, $\rho $ is the resistivity of the material, $l$ is the length of the material and $A$ is the cross-sectional area of the material
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}}$,
where, ${R_{eq}}$ is the equivalent resistance of resistances ${R_1},{R_2},...,{R_n}$ connected parallel.

Complete step by step answer:
We know that $R = \rho \dfrac{l}{A}$. Here, we are cutting one wire, therefore the resistivity and cross-sectional area remain the same.
$
R \propto l \\
\Rightarrow \dfrac{R}{l} = {\text{constant}} \\ $
Let us assume that the wire is cut into n parts. Therefore the length of each part will be $\dfrac{l}{n}$ and let us assume that the resistance of each part is $R'$.
$
\dfrac{R}{l} = \dfrac{{R'}}{{\dfrac{l}{n}}} \\
\Rightarrow R' = \dfrac{R}{n} \\ $
All these parts are connected in parallel connection. Therefore by using the formula
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}}$
We are given the equivalent resistance is $1\Omega $.
$
1 = n\left( {\dfrac{n}{R}} \right) \\
\Rightarrow \dfrac{{{n^2}}}{R} = 1 \\
\Rightarrow {n^2} = R \\ $
We know that resistance of wire is $100\Omega $.
$
{n^2} = 100 \\
\therefore n = 10 $
Thus, a wire of $100\Omega $ should be cut into 10 parts so that a resistance of $1\Omega $ is obtained by connecting them in parallel.

Hence, option A is the right answer.

Note:We have applied the formula $R = \rho \dfrac{l}{A}$. According to this, the resistance is directly proportional to the length of the material which means that the resistance increases as the length increases. Also, the resistance is inversely proportional to the cross=sectional area of the material which means that the resistance decreases as the cross=sectional area increases.