A copper wire has diameter 0.5 mm and resistivity of $1.6 \times {10^{ - 8}}\Omega m$. What will be the length of this wire to make its resistance $10\Omega $ ?

Hint: Assume the length of the wire be l and then use the formula of resistance i.e., $R = \dfrac{{\rho l}}{A}$ and then solve the question.

Step By Step Answer :

Formula used - $R = \dfrac{{\rho l}}{A}$ , $A = \dfrac{{\pi {d^2}}}{4}$

We have given in the question that-

Diameter of the wire is d = 0.5 mm
The resistivity is, $\rho = 1.6 \times {10^{ - 8}}\Omega m$ .
Resistance of the wire $R = 10\Omega $

Now, let us assume the length of the wire needed to make its resistance $10\Omega $ is l.
So, using the formula for resistance i.e., $R = \dfrac{{\rho l}}{A}$

Here, R is the resistance of the wire, $\rho $ is the resistivity, l is the length and A is the area.
Now, since we have the diameter of the wire. So, we can write Area as-

$A = \dfrac{{\pi {d^2}}}{4}$

putting the area value in the above formula, we get-

$R = \dfrac{{\rho l}}{{\dfrac{{\pi {d^2}}}{4}}} = \dfrac{{4\rho l}}{{\pi {d^2}}}$
now further simplifying we get-
$l = \dfrac{{\pi R{d^2}}}{{4\rho }}$
Putting the values of R, d and $\rho $ , we get-
$l = \dfrac{{3.14 \times 10 \times {{(0.5 \times {{10}^{ - 3}})}^2}}}{{4 \times 1.6 \times {{10}^{ - 8}}}}\{ \because 0.5mm = 0.5 \times {10^{ - 3}}m\} $

On solving we get-

$l = 122m$

Therefore, the length of the wire to make the resistance 10 ohm is 122 m.

Note: Whenever such types of questions appear, then always write down the things given in the question. And then as mentioned in the solution, use the formula to find the resistance, and then keeping the area as $A = \dfrac{{\pi {d^2}}}{4}$ (as we have the diameter of the wire) in the formula of resistance, we get the formula for length as $l = \dfrac{{\pi R{d^2}}}{{4\rho }}$ , after keeping the values of the given terms we found the value of length l.