Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
CBSE
CBSE Question Papers
NCERT Books
NCERT Exemplar Books
NCERT Study Notes
CBSE Study Concepts
CBSE Class 10 Solutions
CBSE Class 12 Solutions
NCERT Text Book Class 11 Solutions
NCERT Text Book Class 12 Solutions
ICSE Class 10 Papers
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
Gravitation
Show Para
Hide Para
Section:
Physics
Share question:
© examsnet.com
Question : 5
Total: 21
Two spherical stars
A
and
B
have densities
ρ
A
and
ρ
B
, respectively.
A
and
B
have the same radius, and their masses
M
A
and
M
B
are related by
M
B
=
2
M
A
. Due to an interaction process, star
A
loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains
ρ
A
. The entire mass lost by
A
is deposited as a thick spherical shell on
B
with the density of the shell being
ρ
A
. If
v
A
and
v
B
are the escape velocities from
A
and
B
after the interaction process, the ratio
v
B
v
A
=
√
10
n
15
1
∕
3
. The value of
n
is
[JEE Adv 2022 P1]
Your Answer:
Validate
Solution:
Given
R
A
=
R
B
=
R
M
B
=
2
M
A
Calculation of escape velocity for A:
Radius of remaining star
=
R
A
2
.
Mass of remaining star
=
ρ
A
4
3
π
R
A
3
8
=
M
A
8
−
GM
A
∕
B
R
A
∕
2
+
1
2
m
v
A
2
=
0
⇒
v
A
=
√
2
GM
A
∕
B
R
A
∕
2
=
√
GM
A
2
R
Calculation of escape velocity for B
Mass collectedover
B
=
7
8
M
A
Let the radius of
B
becomes
r
.
∴
4
3
π
(
r
3
−
R
B
3
)
ρ
A
=
7
8
ρ
A
4
3
π
R
A
3
⇒
π
3
=
7
8
R
A
3
+
R
B
3
=
(
15
)
1
∕
3
R
2
∴
V
B
2
2
=
23
GM
A
8
×
15
1
∕
3
R
2
=
23
GM
A
4
×
15
1
∕
3
R
∴
V
B
=
√
23
GM
A
2
×
15
1
∕
3
R
∴
V
B
V
A
=
√
23
15
1
∕
3
=
√
10
×
2.30
15
1
∕
3
n
=
2.30
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Prev Question
Next Question
Similar Free Exams:
AIEEE Previous Papers
BITSAT Exam Previous Papers
JEE Advanced Model Papers
JEE Advanced Previous Papers
JEE Mains Chapters Wise Previous Papers
JEE Mains Model Papers
JEE Mains Previous Papers
VITEEE Previous Papers