Question

A wire of resistance 16 ohm is elongated to double its length find the new resistance?

Answer 1

A wire of resistance , R = 16 ohm

a/c to question, wire is elongated to double of its length. means, volume of wire remains constant.

we know, formula $R=\rho\frac{l}{A}$

or, $R=\rho\frac{l^2}{Al}$

area × length = Al = volume of wire , V

so, $R=\rho\frac{l^2}{V}$

here, $\rho$ is resistivity [ constant for specific wire ] and V is volume , which is also constant here.
so, resistance is directly proportional to square of length.
now, if wire is elongated to double of its length, resistance will be 4 times of its initial value.

so, new resistance of wire = 4 × 16 = 64ohm.

Answer 2

as we know that
RESISTANCE R=
$p \frac{l}{a}$
where p is restivity and l is the length of wire and a is the area
now a*l=l/2* 2a
it means that if length is doubled new area will be reduced to half of it's original value
so new a'=2a while l'=l/2
R'=
$p \frac{l \div 2 }{2a}$
on dividing both
we get r'=16×4=64 ohm