Question

A uniform wire of resistance R is stretched uniformly so that it's length is doubled what is it's new resistance

Answer 1

$\boxed{\bold{\large{GIVEN:}}}$

Before the wire was stretched uniformly so that it's length is doubled.

• A wire of resistance R is given to us.
• The length of the wire is l.
• The resistivity of wire is ρ.
• The cross-sectional area of the wire is A.

After the wire was stretched uniformly so that it's length is doubled.

• Let the new resistance be r.
• The length is doubled. So new length = 2l.
• The resistivity of the wire won't change, so the resistivity remains ρ.
• Since the length is doubled, the cross-sectional area will decrease by 2 times. In other words, the cross-sectional area of the new wire = A/2.

$\boxed{\bold{\large{TO \:\: FIND:}}}$

• New resistance (r).

$\boxed{\bold{\large{SOLUTION:}}}$

We know that the resistance of the wire would be:

$\sf{\red{ R=\rho\dfrac{l}{A} }}$  _______(1)

The length is doubled and the area is decreased by 2 times. By substituting l=2l and A=A/2, the new resistance becomes:

$\sf r=\rho\dfrac{2l}{\dfrac{A}{2} }$

Denominator's denominator goes to the numerator.

$\to \sf r=\rho \dfrac{2 \times 2 \times l}{A}$

$\to \sf r=\rho\dfrac{4l}{A}$

Take out 4 from the equation.

$\to \sf r=4\left(\rho\dfrac{l}{A}\right )$_____(2)

Substitute (1) in (2).

$\boxed{\sf{\blue{r=4 \times R}}}$

NER RESISTANCE IS 4 TIMES MORE THAN THE OLD RESISTANCE.