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JEE Main Physics Class 11 Oscillations Part 2 Questions
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© examsnet.com
Question : 8
Total: 76
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle
(
θ
)
of thread deflection in the extreme position will be :
[27-Jan-2024 Shift 2]
tan
−
1
(
√
2
)
2
tan
−
1
(
1
2
)
tan
−
1
(
1
2
)
2
tan
−
1
(
1
√
5
)
Validate
Solution:
👈: Video Solution
Loss in kinetic energy
=
Gain in potential energy
⇒
1
2
m
v
2
=
mg
ℓ
(
1
−
cos
θ
)
⇒
v
2
ℓ
=
2
g
(
1
−
cos
θ
)
Acceleration at lowest point
=
−
v
2
ℓ
Acceleration at extreme point
=
g
s
i
n
θ
Hence,
v
2
ℓ
=
g
s
i
n
θ
∴
s
i
n
θ
=
2
(
1
−
cos
θ
)
⇒
tan
θ
2
=
1
2
⇒
θ
=
2
tan
−
1
(
1
2
)
© examsnet.com
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