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JEE Main Physics Class 11 Gravitation Part 1 Questions
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© examsnet.com
Question : 49
Total: 100
Four identical particles of mass M are located at the corners of a square of side ‘a’. What should be their speed if each of them revolves under the influence of others’ gravitational field in a circular orbit circumscribing the square ?
[8 April 2019 I]
1.35
√
G
M
a
1.16
√
G
M
a
1.21
√
G
M
a
1.41
√
G
M
a
Validate
Solution:
A
C
=
a
√
2
∵
r
=
A
C
2
=
a
√
2
2
=
a
√
2
Resultant force on the body
B
=
G
M
2
a
2
^
i
+
G
M
2
a
2
^
j
+
G
M
2
(
a
√
2
)
2
(
cos
45
°
^
i
+
sin
45
°
^
j
)
⇒
|
F
|
=
G
M
2
a
2
(
√
2
)
+
G
M
2
2
a
2
M
v
2
r
=
Resultant force towards centre
∴
M
v
2
(
a
√
2
)
=
G
M
2
a
2
(
√
2
+
1
2
)
⇒
v
2
=
G
M
a
(
1
+
1
2
√
2
)
⇒
v
=
√
G
M
a
(
1
+
1
2
√
2
)
=
1.16
√
G
M
a
© examsnet.com
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