“God used
beautiful mathematics in creating the world”.
– P. A. M. Dirac
The following question involving gravitation and circular motion appeared in JEE (Main) 2014 question paper:
Four particles, each of mass M
and equidistant from each other, move along a circle of radius R under the action of their mutual
gravitational attraction. The speed of each particle is
(1) √(GM/R)
(2) √(2√2 GM/R)
(3) √[(GM/R)(1+2√2)]
(4) (½)√[(GM/R)(1+2√2)]
The answer
is (½)√[(GM/R)(1+2√2)] given in option (4).
This
question was worked out in the post dated 6th January 2013 on this
site. The statement of the question is slightly different; but the essential
points are the same. Click here to see the solution (Question No. 2 in the
post)
You can access all questions on gravitation on
this site by clicking on the label ‘gravitation’ below this post or by trying a
search for ‘gravitation’ using the search box provided on this page.
Use the ‘older posts’ tab to access all the posts in this section.
Use the ‘older posts’ tab to access all the posts in this section.
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