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JEE Advanced Model Paper 2 with solutions for online practice
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© examsnet.com
Question : 3
Total: 54
Consider a simple pendulum of a very large length L. For small amplitudes, its time period is T. Given, radius of earth is
R
e
= 6400 km, then choose the correct option(s).
T = 2π
√
1
g
(
1
L
+
1
R
e
)
When L =
R
e
, T = 59.8min
When L ➔ ∞ , T = 84.6min
When L =
R
e
, T = 69.8min
Validate
Solution:
ln the figure, F = -mg
s
i
n
(
θ
1
+
θ
2
)
for small amplitude sin
(
θ
1
+
θ
2
)
=
θ
1
+
θ
2
∴ F = -mg
(
θ
1
+
θ
2
)
F = - mg (sin
θ
1
+ sin
θ
2
) = -mg
(
x
L
+
x
R
e
)
= -mg
(
1
L
+
1
R
e
)
x
∴ a = - g
(
1
L
+
1
R
e
)
x
Comparing with a = -
ω
2
x ,
ω =
√
g
(
1
L
+
1
R
e
)
∴ T = 2π
√
1
g
(
1
L
+
1
R
e
)
When L =
R
e
i.e = L = 6400 ×
10
3
m
We get T = 59.8 min
When L → ∞ ,
R
e
L
= 0
Therefore , T = 2π
√
R
e
g
= 84.6 min
© examsnet.com
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