
What length of copper wire of resistivity $ 1.7 \times {10^{ - 8}}\Omega m $ and radius $ 1mm $ is required so that its resistance is $ 1\Omega $ ?
(A) $ 184.7m $
(B) $ 200m $
(C) $ 190.5m $
(D) $ 150m $
Answer
580.2k+ views
Hint
We will use the concept of relation of resistance of a wire to its length and cross sectional area. Then, we will try to equate them.
$ \Rightarrow R = \rho L/A $
Where, $ R $ is the resistance of the wire, $ \rho $ is its resistivity, $ L $ is its length and $ A $ is its cross sectional area.
Complete step by step answer
We know, $ R \propto L $
As if we increase the length of a wire, the resistance increases.
Also, $ R \propto 1/A $
As if we increase the cross sectional area of a wire, the resistance decreases.
Now,Combining both the proportionality and putting in a proportionality constant, we get
$ R = \rho L/A $
Here, the constant $ \rho $ is called the resistivity of the wire.
Now,Given that,
$ \Rightarrow R = 1\Omega $, $ \rho = 1.7 \times {10^{ - 8}}\Omega m $, $ r = 1mm = {10^{ - 3}}m $
Thus, $ A = \pi {r^2} $
Putting in the values and calculating, we get
$ \Rightarrow A = 3.14 \times {10^{ - 6}}{m^2} $
Now, $ R = \rho L/A $
$ \Rightarrow L = RA/\rho $
Thus, Putting in the values,
$ \Rightarrow L = 1\Omega \times 3.14 \times {10^{ - 6}}{m^2}/1.7 \times {10^{ - 8}}m $
After evaluating, we get
$ \Rightarrow L = 1.847 \times {10^2}m $
Thus, the length required is $ 184.7m $ which is option (A).
Additional Information
Resistivity is a property of a material and it differs from material to material. From the name itself our intuition speaks that it is something to do with resistance and it actually is as it the capacity of material to provide opposition or resistance to the charges flowing in the wire or in other words it is referred to how much a wire of certain material can provide resistance to charges or broadly to the current flowing.
Note
If we increase the length of the wire, the stretch of atoms of the material of the wire increases due to which the collisions increases thus the resistance increases. But in the case of cross sectional area, the spacing between the atoms increases reducing the collision possibility and hence reducing the resistance.
We will use the concept of relation of resistance of a wire to its length and cross sectional area. Then, we will try to equate them.
$ \Rightarrow R = \rho L/A $
Where, $ R $ is the resistance of the wire, $ \rho $ is its resistivity, $ L $ is its length and $ A $ is its cross sectional area.
Complete step by step answer
We know, $ R \propto L $
As if we increase the length of a wire, the resistance increases.
Also, $ R \propto 1/A $
As if we increase the cross sectional area of a wire, the resistance decreases.
Now,Combining both the proportionality and putting in a proportionality constant, we get
$ R = \rho L/A $
Here, the constant $ \rho $ is called the resistivity of the wire.
Now,Given that,
$ \Rightarrow R = 1\Omega $, $ \rho = 1.7 \times {10^{ - 8}}\Omega m $, $ r = 1mm = {10^{ - 3}}m $
Thus, $ A = \pi {r^2} $
Putting in the values and calculating, we get
$ \Rightarrow A = 3.14 \times {10^{ - 6}}{m^2} $
Now, $ R = \rho L/A $
$ \Rightarrow L = RA/\rho $
Thus, Putting in the values,
$ \Rightarrow L = 1\Omega \times 3.14 \times {10^{ - 6}}{m^2}/1.7 \times {10^{ - 8}}m $
After evaluating, we get
$ \Rightarrow L = 1.847 \times {10^2}m $
Thus, the length required is $ 184.7m $ which is option (A).
Additional Information
Resistivity is a property of a material and it differs from material to material. From the name itself our intuition speaks that it is something to do with resistance and it actually is as it the capacity of material to provide opposition or resistance to the charges flowing in the wire or in other words it is referred to how much a wire of certain material can provide resistance to charges or broadly to the current flowing.
Note
If we increase the length of the wire, the stretch of atoms of the material of the wire increases due to which the collisions increases thus the resistance increases. But in the case of cross sectional area, the spacing between the atoms increases reducing the collision possibility and hence reducing the resistance.
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