Question

3. A piece of wire of resistance 10. ohm is
drawn out so that its length is increased to
three times its original length. Calculate the
new resistance.
Given : R = 1002​

Answer 1

Given: Resistance of wire , `R = 10 Omega`

Length is increased to thrice = 3L

To find : New resistance = ?

Resistance("R") = `("resistivity"(rho) xx "length"(L))/("area"(A))`

`R= (rho L)/(A) implies 10 = (rho L)/(A)`

When length is increased by three times (3L) the area of cross section is reduced by three times `((A)/(3))`.

New length = 3L

New area = `(A)/(3)`

`therefore` resistance R = `(rho. 3L)/((A)/(3)) = 9 ((rhoL)/(A))`

`R = 9.R implies 9 xx10 = 90 Omega`

`R = 90 Omega`

HOPE YOU CAN LEARN FROM THIS

Answer 2

$\huge\tt\red{Q}\tt\pink{U}\tt\blue{E}\tt\green{S}\tt\purple{T}\tt\orange{I}\tt\orange{O}\huge\tt\red{N}$

A piece of wire of resistance 10. ohm is

drawn out so that its length is increased to

three times its original length. Calculate the

new resistance.

Given : R = 1002

$\huge\tt\red{❥}\tt\pink{A}\tt\blue{N}\tt\green{S}\tt\purple{W}\tt\orange{E}\tt\orange{R}\huge\tt\purple{᭄}$

Given:-

Resistance of wire , R=10Ω

Length is increased to thrice = 3L

To find : New resistance = ?

Resistance("R") =

resistivity(ρ)×length(L)

area(A)

Resistance R = 10 Ω

Let l be the length of the wire R ∝ 1

When the length is increased to three times, l’

= 3l

∴ New Resistance R’ ∝ l’ ∝ 3l

∴ RR′=l3l=13RR′=l3l=13

∴ R' = 3R

New resistance = 3 times the original resistance.

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