A uniform ring of mass m and radius 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 r3 as shown in the figure. The gravitational force exerted by the sphere on the ring will be
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
GMm8r2
b
GMm4r2
c
3GMm8r2
d
GMm8r33
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
dF=GMdm4r2F=ΣdFcosθ=ΣGMdm4r2cosθ=GM4r2×3r2rΣdm=3GMm8r2Alternative solution:The gravitational field due to the ring at a distance 3r is given byE=Gm(3r)r2+(3r)23/2 or E=3Gm8r2The required force is EM, i.e., (3Gm)M/8r2