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Question: In how many parts (equal) a wire of \(100\Omega \) be cut so that a resistance of \(1\Omega \) is ob...

In how many parts (equal) a wire of 100Ω100\Omega be cut so that a resistance of 1Ω1\Omega is obtained by connecting them in parallel?
A. 1010
B. 55
C. 100100
D. 5050

Explanation

Solution

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=ρlAR = \rho \dfrac{l}{A},
where, RR is the resistance, ρ\rho is the resistivity of the material, ll is the length of the material and AA is the cross-sectional area of the material
1Req=1R1+1R2+...+1Rn\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}},
where, Req{R_{eq}} is the equivalent resistance of resistances R1,R2,...,Rn{R_1},{R_2},...,{R_n} connected parallel.

Complete step by step answer:
We know that R=ρlAR = \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 ln\dfrac{l}{n} and let us assume that the resistance of each part is RR'.
\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
1Req=1R1+1R2+...+1Rn\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}}
We are given the equivalent resistance is 1Ω1\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 100Ω100\Omega .
n2=100 n=10 {n^2} = 100 \\\ \therefore n = 10
Thus, a wire of 100Ω100\Omega should be cut into 10 parts so that a resistance of 1Ω1\Omega is obtained by connecting them in parallel.

Hence, option A is the right answer.

Note: We have applied the formula R=ρlAR = \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.