ii) The period of oscillation of a simple pendulum is given by the relation T = 20 2010 The measured values of length L and the period of oscillations for 50 oscillations are (20.0 + 0.1) cm and (45 + 1)s respective- ly. Find the absolute error in determination of g.
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Correct option is
D
5%
Here, T=2π
g
L
Squaring both sides, we get, T
2
=
g
4π
2
L
org=
T
2
4π
2
L
The relative error in g is,
g
Δg
=
L
ΔL
+2
T
ΔT
Here, T=
n
t
and ΔT=
n
Δt
∴
T
ΔT
=
t
Δt
The errors in both L and t are the least count errors.
∴
g
Δg
=
10
0.1
+2(
50
1
)=0.01+0.04=0.05
The percentage error in g is
g
Δg
×100=
L
ΔL
×100+2(
T
ΔT
)×100=[
L
ΔL
+2(
T
ΔT
)]×100=0.05×100=5%
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