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Standard XII
Physics
Multiplication & Division of Errors
In an experim...
Question
In an experiment, the refractive index of glass was observed to be
1.45
,
1.56
,
1.54
,
1.44
,
1.54
,
1.44
,
1.54
and
1.53
Calculate percentage error.
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Solution
Refractive index
=
μ
μ
1
=
1.45
μ
5
=
1.54
μ
2
=
1.46
μ
6
=
1.44
μ
3
=
1.54
μ
7
=
1.54
μ
4
=
1.44
μ
8
=
1.53
u
average
=
1.45
+
1.46
+
1.54
+
1.44
+
1.54
+
1.44
+
1.54
+
1.53
8
¯
u
=
11.94
8
=
1.4925
=
1.50
Δ
μ
=
¯
μ
−
μ
Δ
μ
1
=
1.50
−
1.45
=
0.05
Δ
μ
2
=
1.50
−
1.46
=
0.04
Δ
μ
3
=
1.50
−
1.54
=
−
0.04
Δ
μ
4
=
1.50
−
1.44
=
0.06
Δ
μ
5
=
1.50
−
1.5
4
=
−
0.04
Δ
μ
6
=
1.50
−
1.44
=
0.06
Δ
μ
7
=
1.50
−
1.54
=
−
0.04
Δ
μ
8
=
1.50
−
1.53
=
−
0.03
Mean absolute error
=
Δ
¯
μ
=
0.05
+
0.04
+
0.04
+
0.06
+
0.04
+
0.03
+
0.06
+
0.04
8
=
0.36
8
=
0.045
=
0.04
.
Fractional Error
=
Δ
¯
μ
¯
μ
=
0.04
1.50
=
0.03
∵
Error
=
Δ
¯
μ
¯
μ
=
0.03
×
100
%
=
3
%
⟶
Ans
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Similar questions
Q.
In an experiment, the refractive index of glass was observed to be
1.45
,
1.56
,
1.54
,
1.44
,
1.54
,
1.44
,
1.54
and
1.53
Calculate fractional error.
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In an experiment, the refractive index of glass was observed to be
1.45
,
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,
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1.44
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and
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In an experiment, the refractive index of glass was observed to be
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,
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,
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. Calculate relative error.
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In an experiment refractive index of glass was observed to be
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