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JEE Main 24 Feb 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 6
Total: 90
The period of oscillation of a simple pendulum is
T
=
2
π
√
L
g
.
Measured value of
L
is
1.0
m
from metre scale having a minimum division of
1
m
m
and time of one complete oscillation is
1.95
s
measured from stopwatch of
0.01
s
resolution. The percentage error in the determination of
g
will be
[24 Feb 2021 Shift 2]
1.13
%
1.03
%
1.33
%
1.30
%
Validate
Solution:
Given,
T
=
2
π
√
L
g
.
.
.
(i)
where, time period,
T
=
1.95
s
Length of string,
I
=
1
m
Acceleration due to gravity
=
g
Error in time period,
∆
T
=
0.01
s
=
10
−
2
s
Error in length,
∆
L
=
1
m
m
=
1
×
10
−
3
m
Squaring Eq. (i) on both sides, we get
T
2
=
4
π
2
L
g
g
=
4
π
2
L
T
2
⇒
∆
g
g
=
∆
L
L
+
2
∆
T
T
=
10
−
3
1
+
2
×
10
−
2
1.95
=
10
−
3
+
1.025
×
10
−
2
=
10
−
3
+
10.25
×
10
−
3
=
11.25
×
10
−
3
∵
∆
g
∕
g
×
100
=
11.25
×
10
−
3
×
10
2
=
1.125
%
=
1.13
%
© examsnet.com
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