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What is the resistance of a wire if its radius is doubled?
(A) Becomes twice
(B) Becomes one-fourth
(C) Becomes four times
(D) Becomes one-half
Answer
129.9k+ views
Hint: We are asked to find the change in resistance of the wire when one of its parameters is changed. Thus, we will first find the relation of resistance with the parameters of wire.
Formula used
$R = \rho \dfrac{L}{A}$
Where, $R$ is the resistance of the wire, $\rho $ is the resistivity of the material of wire, $L$ is the length of the wire and $A$ is the cross sectional area of the wire.
Step By Step Solution
Firstly,
When the length $L$ of a wire is increased, the atoms of the wire increases and thus the resistance $R$ also increases.
Thus, resistance is directly proportional to the length of the wire.
$R \propto L$
Now,
When the cross sectional area of the wire is increased, the atoms get sparsely separated thus increasing the free space for the charge to flow and in turn decreasing the resistance.
Thus, resistance is inversely proportional to the cross sectional area of the wire.
$R \propto \dfrac{1}{A}$
Then,
Combining both the proportionalities, we get
$R \propto \dfrac{L}{A}$
To remove the proportionality symbol, we introduce a proportionality constant
$R = \rho \dfrac{L}{A}$
Here, $\rho $ is the proportionality constant known as resistivity. Resistivity is the property of a material which refers to the capacity of a material to oppose the flow of a charge.
Now,
According to the question, the radius of the wire is doubled.
Firstly,
The original resistance was,
$R = \rho \dfrac{L}{{\pi {r^2}}} \cdot \cdot \cdot \cdot (1)$
After the increase, the resistance becomes,
\[{R_{New}} = \rho \dfrac{L}{{\pi {{(2r)}^2}}} \Rightarrow {R_{New}} = \dfrac{1}{4}\rho \dfrac{L}{{\pi {r^2}}}\]
Now,
Putting in equation $(1)$ , we get
${R_{New}} = \dfrac{1}{4}R$
Thus, the answer is (B).
Note: We found the answer to be one-fourth as per this particular situation. But if the situation was to double the length, then the new resistance becomes four times of the original resistance.
Formula used
$R = \rho \dfrac{L}{A}$
Where, $R$ is the resistance of the wire, $\rho $ is the resistivity of the material of wire, $L$ is the length of the wire and $A$ is the cross sectional area of the wire.
Step By Step Solution
Firstly,
When the length $L$ of a wire is increased, the atoms of the wire increases and thus the resistance $R$ also increases.
Thus, resistance is directly proportional to the length of the wire.
$R \propto L$
Now,
When the cross sectional area of the wire is increased, the atoms get sparsely separated thus increasing the free space for the charge to flow and in turn decreasing the resistance.
Thus, resistance is inversely proportional to the cross sectional area of the wire.
$R \propto \dfrac{1}{A}$
Then,
Combining both the proportionalities, we get
$R \propto \dfrac{L}{A}$
To remove the proportionality symbol, we introduce a proportionality constant
$R = \rho \dfrac{L}{A}$
Here, $\rho $ is the proportionality constant known as resistivity. Resistivity is the property of a material which refers to the capacity of a material to oppose the flow of a charge.
Now,
According to the question, the radius of the wire is doubled.
Firstly,
The original resistance was,
$R = \rho \dfrac{L}{{\pi {r^2}}} \cdot \cdot \cdot \cdot (1)$
After the increase, the resistance becomes,
\[{R_{New}} = \rho \dfrac{L}{{\pi {{(2r)}^2}}} \Rightarrow {R_{New}} = \dfrac{1}{4}\rho \dfrac{L}{{\pi {r^2}}}\]
Now,
Putting in equation $(1)$ , we get
${R_{New}} = \dfrac{1}{4}R$
Thus, the answer is (B).
Note: We found the answer to be one-fourth as per this particular situation. But if the situation was to double the length, then the new resistance becomes four times of the original resistance.
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