A copper wire has a diameter of `0.5 mm` and a resistivity of `1.6 xx 10^-6` ohm cm. How much of this wire would be required to make a 10 ohm coil ? How much does the resistance change if the diameter is doubled ?

Options

A. 122.7 and 1/4
B. 130 and 1/2
C. 120 and 1/5
D. None of these

Answer

A

Explanation

We are given that, diameter of the wire, `D = 0.5 mm = 0.5 xx 10^-3 m`
resistivity of copper, `rho = 1.6 xx 10^-8 ohm m`
required resistance, `R = 10 ohm`
As `R = (rho l)/(A), l = (RA)/(rho) = (R(pi D^2//4))/(rho) = (pi RD^2)/(4 rho) [A = pi r^2 = pi (D//2)^2 = pi D^2//4]`.
or `l = (3.14 xx 10 xx (0.5 xx 10^-3)^2)/(4 xx 1.6 xx 10^-8) m = 122.7 m`
Since, `R = (rho l)/(pi D^2//4) = (4 rho l)/(pi D^2), R prop 1//D^2`. When `D` is doubled, `R` becomes `(1)/(4)` times.