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MHT CET 2018 Physics Solved Paper
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© examsnet.com
Question : 1
Total: 50
The path length of oscillation of simple pendulum of length 1 meter is 16 cm. Its maximum velocity is (g =
π
2
m
∕
s
2
)
2π cm/s
4π cm/s
8π cm/s
16π cm/s
Validate
Solution:
(c) Given: Path length =16 cm
∴ Amplitude a =
16
2
= 8cm
Time period T = 2π
√
l
g
= 2π
√
1
π
2
= 2π ×
1
π
= 2s
Maximum velocity
V
max
= aω
= a x
2
π
T
= 8 x
2
π
2
= 8π cm/s
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