Question

A copper wire has diameter 0.5 mm
and resistivity of 16 x 10-83m.
What will be the length of this
wire to make its resistance
10 P How much does the rls-
instance change if
the diameter
is doubled?​

Answer 1

this is the answer it is an ncert question I hope all will be help full

Answer 2

We know that

Resistance  is given by

$R=\rho \frac{l}{A}$

where

Ris the resistance

l is the length of the wire

A is the area of the wire

and

ρ is the resistivity of the wire

hence

hence substituting the values

$10=16\times 10^{-8}\times \frac{\bold{l}}{\pi (0.5\times 10^{-3})^{2}}\\\frac{10}{16\times 10^{-8}}=\frac{l}{3.14\times 0.25\times 10^{-6}} \\ \bold{l}=\frac{10\times 3.14\times 0.25\times 10^{-6}}{16\times 10^{-8}}\\l=\frac{10\times 3.14 \times 25 \times 10^{(-6-2+8)}}{16} \\l\approx 50 m$

Also we know that

$R∝\frac{1}{A}$

also that A=πr^2

hence

$R∝\frac{1}{\pi r^{2}}\\R∝\frac{1}{r^{2}}$

hence if diameter is doubled i.e, radius is doubled the term in invese becomes 4 times and hence resistance gets reduced by a factor of 4

i.e,

$\frac{R_{1}}{R_{2}}=\frac{4}{1}$ where R1 is the initial value of resistance and R2 is the value after the doubling of diameter