
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
510.3k+ 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.
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