If 18 meters of high-resistance wire with a diameter of 1.2 mm has a resistance of 10 ohms, what is the resistance of 27 m of the same type of wire with a diameter of 1.5 mm?

R
2
=
9.61

O
h
m
s
Explanation:
l
1
=
18
m

r
1
=
1.2
2
=
0.6
m
m
=
6
⋅
10
−
4
m

R
1
=
10

O
h
m
s
R
1
=
τ
⋅
l
π
⋅
r
2
1
R
1
=
τ
⋅
18
π
⋅
36
⋅
10
−
8
10
=
τ
⋅
18
36
⋅
π
⋅
10
−
8
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
l
2
=
27
m

r
2
=
1
⋅
5
2
=
0.75
⋅
10
−
3
=
7.5
⋅
10
−
4
m
R
2
=
τ
⋅
l
2
π
⋅
r
2
2
R
2
=
τ
⋅
27
π
⋅
56.2
⋅
10
−
8
R
2
=
τ
⋅
27
56.2
π
⋅
10
−
8
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
.
.
10
R
2
=
τ
⋅
18
36
⋅
π
⋅
10
−
8
⋅
56.2
⋅
π
⋅
10
−
8
τ
⋅
27
10
R
2
=
18
⋅
56.2
36
⋅
27
10
R
2
=
56.2
54
R
2
=
540
56.2
R
2
=
9.61

O
h
m
s