
The resistance of a wire of uniform diameter $D$ and length $L$ is $R$ . The resistance of another wire of the same material but diameter \[2D\] and length \[4L\;\] will be :
A. $2R$
B. $R$
C. $\dfrac{R}{2}$
D. $\dfrac{R}{4}$
Answer
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Hint:To solve this question we will use the formula that gives the relationship between the resistance, the specific resistance (or the resistivity), the length and the area of cross section of the conductor (here, the wire). We will write the expression of area in terms of the diameter of the wire. Then we will substitute these values in the formula for both the cases to get the required answer.
Formula used:
Resistance,
$R = \dfrac{{\rho l}}{A}$
where $\rho $ is the specific resistance (or resistivity), $l$ is the length of the conductor (here, wire), and $A$ is the area of the cross section of the wire.
Area of cross section of the wire,
\[A = \pi {r^2}\]
where \[r\] is the radius of the cross section of the wire.
And $r = \dfrac{d}{2}$ where $d$ is the diameter of the cross section of the wire.
Complete step by step solution:
We know that, $R = \dfrac{{\rho l}}{A}$
This implies that,
$R = \dfrac{{\rho l}}{{\left( {\dfrac{{\pi {d^2}}}{4}} \right)}} \\ $. . . (1)
Case 1: Diameter of the wire is $D$, Length of the wire is $L$, and the resistance is $R$. Therefore, the resistance of the wire in this case is,
$R = \dfrac{{\rho L}}{A}$
where \[A = \pi {r^2}\]
Since $r = \dfrac{D}{2}$
Therefore,
$R = \dfrac{{\rho L}}{{\left( {\dfrac{{\pi {D^2}}}{4}} \right)}}$ . . . (2)
Case 2: Diameter of the wire is $2D$ , Length of the wire is $4L$ .
Therefore, the resistance of the wire in this case will be,
$R = \dfrac{{\rho \times 4L}}{{\left( {\dfrac{{\pi {{\left( {2D} \right)}^2}}}{4}} \right)}}$ (from equation (1))
Simplifying this we get,
$R = \dfrac{{\rho L}}{{\left( {\dfrac{{\pi {D^2}}}{4}} \right)}}$ which is the same as that in case 1. (From equation (2))
Thus, the resistance of wire when its diameter is $2D$ and its length is $4L$ is $R$ which is equal to the resistance when its diameter is $D$ and its length is $L$ .
Hence, option B is the correct answer.
Note: This question can also be done with a different approach in which we first find the expression of resistivity for both the cases and equate them to get the value of resistance in the second case. This can be done because the value of resistivity is a property of the material and has the same value for a given material at a given temperature.
Formula used:
Resistance,
$R = \dfrac{{\rho l}}{A}$
where $\rho $ is the specific resistance (or resistivity), $l$ is the length of the conductor (here, wire), and $A$ is the area of the cross section of the wire.
Area of cross section of the wire,
\[A = \pi {r^2}\]
where \[r\] is the radius of the cross section of the wire.
And $r = \dfrac{d}{2}$ where $d$ is the diameter of the cross section of the wire.
Complete step by step solution:
We know that, $R = \dfrac{{\rho l}}{A}$
This implies that,
$R = \dfrac{{\rho l}}{{\left( {\dfrac{{\pi {d^2}}}{4}} \right)}} \\ $. . . (1)
Case 1: Diameter of the wire is $D$, Length of the wire is $L$, and the resistance is $R$. Therefore, the resistance of the wire in this case is,
$R = \dfrac{{\rho L}}{A}$
where \[A = \pi {r^2}\]
Since $r = \dfrac{D}{2}$
Therefore,
$R = \dfrac{{\rho L}}{{\left( {\dfrac{{\pi {D^2}}}{4}} \right)}}$ . . . (2)
Case 2: Diameter of the wire is $2D$ , Length of the wire is $4L$ .
Therefore, the resistance of the wire in this case will be,
$R = \dfrac{{\rho \times 4L}}{{\left( {\dfrac{{\pi {{\left( {2D} \right)}^2}}}{4}} \right)}}$ (from equation (1))
Simplifying this we get,
$R = \dfrac{{\rho L}}{{\left( {\dfrac{{\pi {D^2}}}{4}} \right)}}$ which is the same as that in case 1. (From equation (2))
Thus, the resistance of wire when its diameter is $2D$ and its length is $4L$ is $R$ which is equal to the resistance when its diameter is $D$ and its length is $L$ .
Hence, option B is the correct answer.
Note: This question can also be done with a different approach in which we first find the expression of resistivity for both the cases and equate them to get the value of resistance in the second case. This can be done because the value of resistivity is a property of the material and has the same value for a given material at a given temperature.
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