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BITSAT 2023 Shift 2 Solved Paper
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© examsnet.com
Question : 11
Total: 130
Two identical particles each of mass '
m
' go round a circle of radius
a
under the action of their mutual gravitational attraction. The angular speed of each particle will be:
√
G
m
2
a
3
√
G
m
8
a
3
√
G
m
4
a
3
√
Gm
a
3
Validate
Solution:
👈: Video Solution
Radius of circle,
r
=
a
Centrifugal force,
F
=
m
ω
2
r
⇒
Gmm
(
2
a
)
2
=
m
ω
2
a
(
∵
F
=
GM
M
1
M
2
d
)
Here
M
1
=
m
;
M
2
=
m
and
d
=
2
a
⇒
angular speed,
ω
=
√
G
m
4
a
3
© examsnet.com
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