Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A uniform ring of mass  and radius <img alt="\mathrm{r}" src="https://entrancecorner.oncodecogs.c

A uniform ring of mass \mathrm{m} and radius \mathrm{r} is placed directly above a uniform sphere of mass m and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance \mathrm{r\sqrt3} as shown in the figure. The gravitational force exerted by the sphere on the ring will be

Option: 1

\frac{\mathrm{G} \mathrm{Mm}}{8 \mathrm{r}^2}

 


Option: 2

\frac{\mathrm{G} \mathrm{Mm}}{4 \mathrm{r}^2}
 


Option: 3

\sqrt{3} \frac{\mathrm{G} \mathrm{Mm}}{8 \mathrm{r}^2}
 


Option: 4

\mathrm{\frac{\mathrm{G} \mathrm{Mm}}{8 \mathrm{r}^2 \sqrt{3}}}


Answers (1)

best_answer

Let the field of the ring at the centre of the sphere be

\mathrm{E}=\frac{\mathrm{GM}}{(2 \mathrm{r})^2} \cos \alpha=\frac{\mathrm{GM}}{4 \mathrm{r}^2} \times \frac{\sqrt{3 \mathrm{r}}}{2 \mathrm{r}}

Force on the sphere of mass \mathrm{M}

\mathrm{ \mathrm{F}=\mathrm{ME}=\frac{\mathrm{GMm} \sqrt{3}}{8 \mathrm{r}^2} }

Hence option 3 is correct.
 

Posted by

Irshad Anwar

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

'PM Modi personally supervising the situation': Centre tells SC in NEET case
anam-khan

NEET Paper Leak Row: NTA informs SC of security overhaul, institutional reforms
anam-khan

Re-NEET 2026 City Intimation Slip LIVE: NTA MBBS admit card soon; direct link, login details, exam date

Latest Question