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
Answers
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
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
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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