Question

two copper wire A and B of length 30m and 10m have radius 2 cm and 1cm . campare the resistance of the two wire . which will have less resistance.

Answer 1

As both the wires are of same material, therefore their resistivity will be same.
Now, the length of wire A= 30m=L
Radius of wire A=2cm=0.02m
Area of wire=pie*r*r=A
Area of wire A=pie*0.02r*0.02r
= 0.0004r*square pie
=0.0004A
Reaistance of a wire=resistivity*L/A =R
Resistance of wire A= 30L/0.0004A=7500R

Similarly, resistivity of wireB=10L/0.0001A=10000R
So resistance of wire B is more than wire A

Answer 2

Answer:

The resistances of the 2 wires are in the ratio 3:4

Explanation:

Given - The lengths of wires A and B - 30 m and 10 m

             The radii of wires A and B - 2 cm and 1 cm

To find - comparison of resistances between the 2 wires

Formula - $R = \rho \frac{L}{r^2}$

                 where, R = resistance of the wire
                               ρ = resistivity of the wire, a property of its material

                               L = length of the wire
                                r = radius of the wire

Solution -

With the given formula, we calculate the resistances of the 2 wires as follows -

$R_1 = \rho \frac{30}{(4) \times 10^-^4}\\\\R_1 = \rho \frac{30}{0.0004}\\\\R_1 = 75000\; \rho$

Next, we calculate the resistance of the 2nd wire -

$R_2 = \rho \frac{10}{(1) \times 10^-^4}\\\\R_2 = \rho \frac{10}{0.0001}\\\\R_2 = 100000\; \rho$

Now, we compare the 2 resistances as follows. Also, we recollect that because the material of both wires are the same (copper), the resistivities of these wires would be equal.

Thus, $\frac{R_1}{R_2} = \frac{75000 \rho }{100000 \rho} \\\\ \frac{R_1}{R_2} = \frac{3}{4}$

So, we can conclude that the ratios of resistances of these wires are 3:4.

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