Question
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 100Ω be cut so that a resistance of 1Ω is obtained by connecting them in parallel?
A. 10
B. 5
C. 100
D. 50
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=ρAl,
where, R is the resistance, ρ is the resistivity of the material, l is the length of the material and A is the cross-sectional area of the material
Req1=R11+R21+...+Rn1,
where, Req is the equivalent resistance of resistances R1,R2,...,Rn connected parallel.
Complete step by step answer:
We know that R=ρAl. 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 nl 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
Req1=R11+R21+...+Rn1
We are given the equivalent resistance is 1Ω.
1 = n\left( {\dfrac{n}{R}} \right) \\\
\Rightarrow \dfrac{{{n^2}}}{R} = 1 \\\
\Rightarrow {n^2} = R \\\
We know that resistance of wire is 100Ω.
n2=100 ∴n=10
Thus, a wire of 100Ω should be cut into 10 parts so that a resistance of 1Ω is obtained by connecting them in parallel.
Hence, option A is the right answer.
Note: We have applied the formula R=ρAl. 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.