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JEE Main 26 Feb 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 1
Total: 90
Find the gravitational force of attraction between the ring and sphere as shown in the figure, where the plane of the ring is perpendicular to the line joining the centres. If
√
8
R
is the distance between the centres of a ring (of mass
m
) and a sphere (of mass
M
), where both have equal radius
R
.
[26 Feb 2021 Shift 1]
√
8
9
⋅
G
m
M
R
2
√
2
3
⋅
G
M
m
R
2
1
3
√
8
⋅
G
M
m
R
2
√
8
27
⋅
G
m
M
R
2
Validate
Solution:
Given, distance between centre of ring and sphere,
d
=
√
8
R
Since, gravitational field at the axis of ring,
E
=
G
m
d
(
d
2
+
R
2
)
3
∕
2
Here,
G
is the gravitational constant.
⇒
E
=
G
m
R
√
8
(
8
R
2
+
R
2
)
3
∕
2
=
G
m
R
√
8
(
3
R
)
3
⇒
E
=
G
m
R
√
8
27
R
3
=
G
m
√
8
27
R
2
Force between ring and sphere,
F
=
M
E
. . . (i)
Substituting the value of
E
in Eq. (i), we get
F
=
√
8
27
G
m
M
R
2
© examsnet.com
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