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IIT JEE Advanced 2008 Paper 1
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Section:
Physics
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© examsnet.com
Question : 5 of 69
Marks:
+1
,
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Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.
Least count for length = 0.1 cm.
Least count for time = 0.1 s.
Student
Length of pendulum (cm)
Number of oscillations (n)
Total time for (n) oscillations (s)
Time period (s)
I
64.0
8
128.0
16.0
II
64.0
4
64.0
16.0
III
20.0
4
36.0
9.0
If
E
I
E_I
E
I
,
E
I
I
E_{II}
E
II
and
E
I
I
I
E_{III}
E
III
are the percentage errors in g, that is,
(
Δ
g
g
×
100
)
\left(\frac{\Delta g}{g} \times 100\right)
(
g
Δ
g
×
100
)
for students I, II, and III, respectively, then
[JEE Adv 2008 P1]
E
1
E_1
E
1
= 0
E
1
E_1
E
1
is minimum
E
1
E_1
E
1
=
E
n
E_n
E
n
E
n
E_n
E
n
is maximum
Validate
Solution:
Student-I:
E
I
E_I
E
I
=
Δ
g
g
\frac{\Delta g}{g}
g
Δ
g
× 100 =
(
0.1
64.0
+
2
×
0.1
128
)
\left(\frac{0.1}{64.0} + \frac{2 \times 0.1}{128}\right)
(
64.0
0.1
+
128
2
×
0.1
)
× 100 =
5
16
\frac{5}{16}
16
5
%
Student-II:
E
I
I
E_{II}
E
II
=
Δ
g
g
\frac{\Delta g}{g}
g
Δ
g
× 100 =
(
0.1
64.0
+
2
×
0.21
64
)
\left(\frac{0.1}{64.0} + 2 \times \frac{0.21}{64}\right)
(
64.0
0.1
+
2
×
64
0.21
)
× 100 =
15
32
\frac{15}{32}
32
15
%
Student-III :
E
I
I
I
E_{III}
E
III
:
Δ
g
g
\frac{\Delta g}{g}
g
Δ
g
× 100 =
(
0.1
20.0
+
2
×
0.1
36
)
\left(\frac{0.1}{20.0} + \frac{2 \times 0.1}{36}\right)
(
20.0
0.1
+
36
2
×
0.1
)
× 100 =
19
18
\frac{19}{18}
18
19
Therefore, the percentage error in g for
E
1
E_1
E
1
is minimum
© examsnet.com
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