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JEE Mains 06-Sep-2020 Shift 2 Solved Paper
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© examsnet.com
Question : 1
Total: 75
PHYSICS
Two planets have masses
M
and
16
M
and their radii are a and 2a, respectively. The separation between the centers of the planets is 10a. A body of mass
m
is fired from the surface of the larger planet towards the smaller planet along the line joining their centers. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :
√
GM
2
ma
3
2
√
GM
a
4
√
GM
a
2
√
GM
a
Validate
Solution:
G
M
x
2
=
G
(
16
M
)
(
10
a
−
x
)
2
1
x
=
4
(
10
a
−
x
)
⇒
4
x
=
10
a
−
x
⇒
x
=
2
a
....(i)
COME
−
G
M
m
8
a
−
G
(
16
M
)
m
2
a
+
K
E
=
−
G
M
m
2
a
−
G
(
16
M
)
m
8
a
K
E
=
G
M
m
[
1
8
a
+
16
2
a
−
1
2
a
−
16
8
a
]
K
E
=
G
M
m
[
1
+
64
−
4
−
16
8
a
]
1
2
m
v
2
=
G
M
m
[
45
8
a
]
V
=
√
90
G
M
8
a
V
=
3
2
√
G
M
a
© examsnet.com
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