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JEE Main 25 Feb 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 3
Total: 90
Two satellites
A
and
B
of masses
200
k
g
and
400
k
g
are revolving around the Earth at height of
600
k
m
and
1600
k
m
, respectively. If
T
A
and
T
B
are the time periods of
A
and
B
respectively, then the value of
T
B
−
T
A
is
(Given, radius of Earth
=
6400
k
m
, mass of Earth
=
6
×
10
24
k
g
)
[25 Feb 2021 Shift 1]
1.33
×
10
3
s
3.33
×
10
2
s
4.24
×
10
3
s
4.24
×
10
2
s
Validate
Solution:
Given,
M
A
=
200
k
g
,
M
B
=
400
k
g
,
H
A
=
600
k
m
,
H
B
=
1600
k
m
and
R
A
=
R
E
+
H
A
=
6400
+
600
=
7000
k
m
R
B
=
R
E
+
H
B
=
6400
+
1600
=
8000
k
m
Let
T
A
,
T
B
,
ω
A
,
ω
B
,
R
A
and
R
B
be the time period, angular frequencies and radii of satellites
A
and
B
, respectively. Force on satellite
A
,
F
A
=
m
A
ω
A
2
R
A
=
G
M
m
A
R
A
2
⇒
ω
A
2
=
G
M
R
A
3
but,
ω
A
=
2
π
T
A
∴
(
2
π
T
A
)
2
=
G
M
R
A
3
⇒
T
A
=
√
4
π
2
R
A
3
G
M
Similarly,
T
B
=
√
4
π
2
R
B
3
G
M
∴
T
B
−
T
A
=
√
4
π
2
G
M
(
√
R
B
3
−
√
R
A
3
)
=
2
π
√
G
M
[
√
(
8
×
10
3
)
3
−
√
(
7
×
10
3
)
3
]
=
2
π
×
10
9
√
G
M
(
8
√
8
−
7
√
7
)
=
2
π
×
10
9
√
6.67
×
10
−
11
×
6
×
10
24
(
4.107
)
=
2
π
√
4
×
10
14
×
10
9
(
4.107
)
=
π
×
10
2
×
4.107
=
12.9
×
10
2
s
=
1.33
×
10
3
s
© examsnet.com
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