
A wire of resistance 1 ohm is stretched to double its length. What is the new resistance?
Answer
568.2k+ views
Hint:
When a wire is stretched to double its length, then the area of the cross section of the wire reduces to half.
The resistance of the wire is directly proportional to its length and inversely proportional to the area of cross section, which is given by the formula \[R = \rho \dfrac{l}{A}\] , where \[\rho \] is the resistivity of the material, \[l\] is the length of the material, and A is the area.
In this question, the resistance of a wire for a length is given,
Complete step by step solution:
Resistance of wire \[R = 1\Omega \]
Let the initial length of the wire be \[l\]
The diameter of the cross section be \[d\]
So the area of the cross section of the wire will be
\[
A = \pi {r^2} \\
= \pi {\left( {\dfrac{d}{2}} \right)^2} \\
\]
The resistance of a wire initially will be
\[
R = \rho \dfrac{l}{A} \\
\rho \dfrac{l}{A} = 1 - - (i) \\
\]
Now it is said that the length of the wire is double, so the new length of will become
\[l' = 2l - - (ii)\]
When the wire is stretched, then its area of cross section reduces by half; hence its new area of cross section becomes
\[
A' = \pi {r^2} \\
= \dfrac{{\pi {{\left( {\dfrac{d}{2}} \right)}^2}}}{2} \\
A' = \dfrac{A}{2} - - (iii) \\
\]
Now let the resistance of the wire when its length is doubled be \[R'\], so this can be written as
\[R' = \rho \dfrac{{l'}}{{A'}} - - (iv)\]
By substituting the new value of length from equation (ii) and equation (iii) in the equation (iv), we get
\[R' = \rho \dfrac{{2l}}{{\dfrac{A}{2}}}\]
This can be written as
\[R' = 4\left( {\rho \dfrac{l}{A}} \right)\]
Since the values of \[\rho \dfrac{l}{A} = 1\], so the resistance of the new will become
\[
R' = 4 \times 1 \\
= 4\Omega \\
\]
Hence the new resistance will be \[ = 4\Omega \]
Note:
Students must note that the resistivity of different materials plays a vital role in selecting the materials to be used for electrical wire also in electronic components such as resistors, integrated circuits, etc.
When a wire is stretched to double its length, then the area of the cross section of the wire reduces to half.
The resistance of the wire is directly proportional to its length and inversely proportional to the area of cross section, which is given by the formula \[R = \rho \dfrac{l}{A}\] , where \[\rho \] is the resistivity of the material, \[l\] is the length of the material, and A is the area.
In this question, the resistance of a wire for a length is given,
Complete step by step solution:
Resistance of wire \[R = 1\Omega \]
Let the initial length of the wire be \[l\]
The diameter of the cross section be \[d\]
So the area of the cross section of the wire will be
\[
A = \pi {r^2} \\
= \pi {\left( {\dfrac{d}{2}} \right)^2} \\
\]
The resistance of a wire initially will be
\[
R = \rho \dfrac{l}{A} \\
\rho \dfrac{l}{A} = 1 - - (i) \\
\]
Now it is said that the length of the wire is double, so the new length of will become
\[l' = 2l - - (ii)\]
When the wire is stretched, then its area of cross section reduces by half; hence its new area of cross section becomes
\[
A' = \pi {r^2} \\
= \dfrac{{\pi {{\left( {\dfrac{d}{2}} \right)}^2}}}{2} \\
A' = \dfrac{A}{2} - - (iii) \\
\]
Now let the resistance of the wire when its length is doubled be \[R'\], so this can be written as
\[R' = \rho \dfrac{{l'}}{{A'}} - - (iv)\]
By substituting the new value of length from equation (ii) and equation (iii) in the equation (iv), we get
\[R' = \rho \dfrac{{2l}}{{\dfrac{A}{2}}}\]
This can be written as
\[R' = 4\left( {\rho \dfrac{l}{A}} \right)\]
Since the values of \[\rho \dfrac{l}{A} = 1\], so the resistance of the new will become
\[
R' = 4 \times 1 \\
= 4\Omega \\
\]
Hence the new resistance will be \[ = 4\Omega \]
Note:
Students must note that the resistivity of different materials plays a vital role in selecting the materials to be used for electrical wire also in electronic components such as resistors, integrated circuits, etc.
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