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JEE Main 7 Jan 2020 Shift 1 Solved Paper
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© examsnet.com
Question : 16
Total: 75
A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass
m
10
so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth):
[7-Jan-2020 Shift 1]
m
20
(
u
−
√
2
G
M
3
R
)
2
5
m
(
u
2
−
119
200
G
M
R
)
3
m
8
(
u
+
√
5
G
M
6
R
)
2
m
20
(
u
2
+
113
200
G
M
R
)
Validate
Solution:
Applying energy conservation
K
i
+
U
i
=
K
f
+
U
f
1
2
m
u
2
+
(
−
G
M
m
R
)
=
1
2
m
v
2
−
G
M
m
2
R
v
=
√
u
2
−
G
M
R
...(i)
By momentum conservation, we have
m
10
v
T
=
9
m
10
√
G
M
2
R
....(ii)
&
m
10
v
r
=
m
v
⇒
m
10
v
r
=
m
√
u
2
−
G
M
R
....(iii)
Kinetic energy of rocket
=
1
2
m
(
v
T
2
+
v
r
2
)
=
m
20
(
81
G
M
2
R
+
100
u
2
−
100
G
M
R
)
=
m
20
(
100
u
2
−
119
G
M
2
R
)
=
5
m
(
u
2
−
119
G
M
200
R
)
© examsnet.com
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