A cylindrical wire is stretched to increase its length by 10% . the percentage increase in the resistance of the wire will be. So, new C.S Area of extended wire A2 = Pi (r1)^2 = 0.91 Pi r^2. So, new value of resistance = R1 = Rho.
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Answers
Answer:
Let initial length be ℓ
1
Percentage increment in length =10%
So, Final length, ℓ
2
=ℓ
1
+
100
10
×ℓ
1
=
10
11 ℓ
1
Step 2: Finding Initial and Final Resistances
When the wire is stretched, its volume V remains the same.
We know that, Resistance R=
A
ρ ℓ
=
A ℓ
ρ ℓ
2
=
V
ρ ℓ
2
So, R
1
=
V
ρ ℓ
1
2
....(1)
and R
2
=
V
ρ ℓ
2
2
....(2)
Step 3: Divide equation (1) by equation (2)
R
2
R
1
=(
ℓ
2
ℓ
1
)
2
⇒R
2
=(
ℓ
1
ℓ
2
)
2
R
1
=(
10
11
)
2
R
1
Step 4: Percentage change in resistance
% Change =
R
1
R
2
−R
1
×100
=
1
(
10
11
)
2
−1
×100
=21%
Answer:
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