
On folding, a resistance wire of $8\Omega $ doubles up. Find the new resistance of the wire.
A) $0.5\Omega $
B) $1.5\Omega $
C) $2\Omega $
D) $4\Omega $
Answer
580.2k+ views
Hint: On folding, the resistance wire doubled up. This implies that the length of the wire got reduced to its half and the area of the wire doubled.
Formulas used:
The resistance of a wire of length $l$ and area $A$ is given by, $R = \dfrac{{\rho l}}{A}$ where $\rho $ is the resistivity of the material used to make the wire.
Complete step by step answer:
Step 1: List the data given in the question.
The resistance of the wire before folding is $R = 8\Omega $ .
Let $l$ be the length, $A$ be the area of cross-section of the wire and $\rho $ be the resistivity of the material of the wire.
We have the expression for the resistance of the wire, $R = \dfrac{{\rho l}}{A}$ --------- (1).
Step 2: Express the new resistance of the wire on folding.
On folding the new length of the wire becomes half of its original value i.e., $l' = \dfrac{l}{2}$ and the new area of cross-section of the wire will be double of the original cross-sectional area i.e., $A' = 2A$.
Now the new resistance $R'$ can be expressed in terms of its new length and new area as $R' = \dfrac{{\rho l'}}{{A'}}$.
Substituting $l' = \dfrac{l}{2}$ and $A' = 2A$ in the above equation we get, $R' = \dfrac{{\rho \left( {\dfrac{l}{2}} \right)}}{{2A}}$.
Reducing the above equation we get, $R' = \dfrac{{\rho l}}{{4A}}$ -------- (2).
Step 3: Using equations (1) and (2) find the value of the new resistance.
Equation (1) gives $R = \dfrac{{\rho l}}{A}$ and equation (2) gives $R' = \dfrac{{\rho l}}{{4A}}$.
Substituting equation (1) in equation (2) we get, $R' = \dfrac{R}{4}$ .
Given, $R = 8\Omega $ . Then new resistance will be $R' = \dfrac{8}{4} = 2\Omega $ .
Therefore, the new resistance is $2\Omega $.
Note:
On folding the resistance wire, only the dimensions of the wire change, i.e., the length and area of the wire change. The resistivity is a property of the material of the wire. It refers to the ability of the material to conduct electricity. Thus the resistivity remains constant on folding the wire but changes if you change the material of the wire.
Formulas used:
The resistance of a wire of length $l$ and area $A$ is given by, $R = \dfrac{{\rho l}}{A}$ where $\rho $ is the resistivity of the material used to make the wire.
Complete step by step answer:
Step 1: List the data given in the question.
The resistance of the wire before folding is $R = 8\Omega $ .
Let $l$ be the length, $A$ be the area of cross-section of the wire and $\rho $ be the resistivity of the material of the wire.
We have the expression for the resistance of the wire, $R = \dfrac{{\rho l}}{A}$ --------- (1).
Step 2: Express the new resistance of the wire on folding.
On folding the new length of the wire becomes half of its original value i.e., $l' = \dfrac{l}{2}$ and the new area of cross-section of the wire will be double of the original cross-sectional area i.e., $A' = 2A$.
Now the new resistance $R'$ can be expressed in terms of its new length and new area as $R' = \dfrac{{\rho l'}}{{A'}}$.
Substituting $l' = \dfrac{l}{2}$ and $A' = 2A$ in the above equation we get, $R' = \dfrac{{\rho \left( {\dfrac{l}{2}} \right)}}{{2A}}$.
Reducing the above equation we get, $R' = \dfrac{{\rho l}}{{4A}}$ -------- (2).
Step 3: Using equations (1) and (2) find the value of the new resistance.
Equation (1) gives $R = \dfrac{{\rho l}}{A}$ and equation (2) gives $R' = \dfrac{{\rho l}}{{4A}}$.
Substituting equation (1) in equation (2) we get, $R' = \dfrac{R}{4}$ .
Given, $R = 8\Omega $ . Then new resistance will be $R' = \dfrac{8}{4} = 2\Omega $ .
Therefore, the new resistance is $2\Omega $.
Note:
On folding the resistance wire, only the dimensions of the wire change, i.e., the length and area of the wire change. The resistivity is a property of the material of the wire. It refers to the ability of the material to conduct electricity. Thus the resistivity remains constant on folding the wire but changes if you change the material of the wire.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

The computer jargonwwww stands for Aworld wide web class 12 physics CBSE

State the principle of an ac generator and explain class 12 physics CBSE

