Cylindrical copper rod has a resistance of 10 ohm .The rod is reformed to twice it's original length without any change in volume.Find the resistance of the newly formed cylinder.
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Answered by
2
resistance =40 ohm
Explanation:
as we increase length twice by keeping volume constant
i.e)suppose new radius is R so
pie×r×r×h=pie×R×R×2h
r^2=2R^2
so R=r/root(2)
so previous cross section area = 2× new cross section area
now we know that R(resistance)=rho×length/cross section area
now put values we get
new resistance =4×previous resistance
=4×10=40 ohm
Answered by
0
The resistance of rod before reformation
R
1
=R=
πr
1
2
ρl
1
[∵R=
A
ρl
=
πr
2
ρl
]
Now the rod is reformed such that
l
2
=
2
l
1
∴πr
1
2
l
1
=πr
2
2
l
2
(∵ Volume remains constant)
or
r
2
2
r
1
2
=
l
1
l
2
..(i)
Now the resistance of the rod after reformation
R
2
=
πr
2
2
ρl
2
∴
R
2
R
1
=
πr
1
2
ρl
1
/
πr
2
2
ρl
2
=
l
2
l
1
×
r
1
2
r
2
2
or
R
2
R
1
=
l
2
l
1
×
l
2
l
1
=(
l
2
l
1
)
2
=(2)
2
(using (i))
∴R
2
=
4
R
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