0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# A uniform sphere of mass M and radius R exerts a force F on a small mass m situated at a distance of 2R from the centre O of the sphere. A spherical portion of diameter R is cut from the sphere as shown in the figure. The force of attraction between the remaining part of the sphere and the mass m will be:

A

7F9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

2F3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

4F9

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

F3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 7F9 The force of attraction between the complete sphere and mass m is: F=GmM(2R)2=GmM4R2 Mass of complete sphere is: M=4π3R3×ρ where ρ is the density of the sphere. [Density =MassVolume] Mass of the cut-out portion is m∘=4π3(R2)3×ρ=M8 Now, the distance between the centre of the cut out portion and mass m: 2R−R2=3R2 Hence, the force of attraction between the cut out portion and mass m is: f=Gm∘m(3R2)2=G(M8)m9R24 =GmM4R2×29=2F9 Therefore, the force of attraction between the remaining part of the sphere and mass m: F−f=F−2F9=7F9

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Field Due to Continuous Bodies - 1
PHYSICS
Watch in App
Join BYJU'S Learning Program