Question

A piece of resistance wire has resistance of $4\Omega$ . Its diameter is doubled. Now its resistance will be :

• A. $8\Omega$
• B. $2\Omega$
• C. $4\Omega$
• D. $1\Omega$

Answer 1

The correct option is D $1\Omega$

Resistance of a piece of wire $=$ 4

$4\Omega =\frac{pL}{A}$

Given diameter of wire is doubled.

Area A $=\frac{\pi d^2}{4}$

Now $A^1=\frac{\pi (2d{)}^2}{4}=4A$

$\therefore$ New resistance $R^1=\frac{pL}{A^1}=\frac{pL}{4A}$

$=\frac{1}{4}\times 4=1\Omega$