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Problem
Problem
Find the percentage error in specific resistance given by
ρ
=
π
r
2
R
l
\rho = \frac{\pi r^2 R}{l}
ρ
=
l
π
r
2
R
where
r
r
r
is the radius having value
(
0.2
±
0.02
)
cm
(0.2 \pm 0.02) \text{ cm}
(
0.2
±
0.02
)
cm
,
R
R
R
is the resistance of
(
60
±
2
)
ohm
(60 \pm 2) \text{ ohm}
(
60
±
2
)
ohm
and
l
l
l
is the length of
(
150
±
0.1
)
cm
(150 \pm 0.1) \text{ cm}
(
150
±
0.1
)
cm
.
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