Question

A PIECE OF WIRE OF RESISTANCE 20 ohm is drawn out so that its length is increased to twice its original length .calcualte the new resistance of the wire.

Answer 1

$\huge \bold{RESISTANCE}$

Resistance is the obstruction offered in the path of the current.

Resistance is directly proportional to length and inversely proportional to area of cross section.

◼So if length is incresses resistance will increase.

◼If area of cross section is increased resistance decreases.

$\underline \bold{ACCORDING \: TO \: QUESTION }$

For a piece of wire,

Resistance R = 20 ohm

Let length be l

Area of cross section be a.

So,

$\boxed{RESISTANCE = \rho \dfrac{Lenght}{Area of cross section }}$

$\boxed{R = \rho \dfrac{l}{a}}$- - - - - (1)

$\underline \bold{GIVEN}$

Length is increased to twice so, new length $l_2 = 2l$

Since ,the lenght is increased by twice area of cross section becomes half

So new area of cross section $a_2 = \dfrac{a}{2}$

So new resistance $R_2$ will be,

$\boxed{RESISTANCE = \rho \frac{Lenght}{Area of cross section }}$

$\boxed{R_2 = \rho \dfrac{2l}{\dfrac{a}{2}}}$

$\boxed{R_2 = \rho \dfrac{4l}{a}}$- - - - - - (2)

On dividing eq(1) by (2)

$\dfrac{R}{R2} = \frac{\rho \dfrac{l}{a}}{\rho \dfrac{4l}{a}}$

$\dfrac{R}{R2} = \dfrac{\cancel{\rho} \dfrac{\cancel{l}}{\cancel{a}}}{\cancel{\rho} \dfrac{4 \cancel{l}}{\cancel{a}}}$

$\dfrac{R}{R2} = \dfrac{1}{4}$

$\huge \boxed {R = \frac{R_2}{4}}$

$\huge \boxed {4R = R_2}$

Hence, the new resistance will be four times the original one.

So, the new resistance will be

▶20 × 4
$\huge \boxed{80 \: ohm}$