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JEE Mains 25-Jan-2023 Shift 1 Solved Paper
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© examsnet.com
Question : 5
Total: 90
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is
100
g
. The time period of the motion of the particle will be (approximately)
(take
g
=
10
ms
−
2
,radius of earth
=
6400
‌
km
)
[25-Jan-2023 Shift 1]
24 hours
1 hour 24 minutes
1 hour 40 minutes
12 hours
Validate
Solution:
Let at some time particle is at a distance
x
from centre of Earth, then at that position field
E
=
‌
GM
R
3
x
∴
Acceleration of particle
→
a
=
−
‌
GM
R
3
→
x
‌
⇒
ω
=
√
‌
GM
R
3
=
√
‌
g
R
‌
‌
Now
‌
T
=
‌
2
Ï€
ω
=
2
Ï€
√
‌
R
g
‌
⇒
T
=
2
×
3.14
×
√
‌
6400
×
10
3
10
‌
=
2
×
3.14
×
800
s
e
c
≈
1
‌
hour
‌
24
‌
minutes
‌
© examsnet.com
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