TS EAMCET 19-July-2022 Shift 1 Solved Paper

The force of attraction between the complete sphere of mass M and particle of mass m is
F=‌

GmM

4R2

....(i)
Mass of the complete sphere, M=‌

4

3

πR3ρ
where, ρ is density of sphere.
Mass of the cut out portion, m′=‌

4

3

π(‌

R

2

)3ρ=‌

M

8

Now, the distance between the centre of the cut out portion and mass m,
=2R−‌

R

2

=‌

3R

2

Hence, the force of attraction between the cut out portion and mass ˙m is
F′‌=‌

Gm′M

(‌ 3R 2 )2

=‌

G( M 8 )m

9R2 4

‌=‌

GmM

4R2

×‌

2

9

=‌

2

9

F  [from Eq (i)]
Thus, the force of attraction between the remaining part of the sphere and mass
m=F−F′=F−‌

2

9

F=‌

7

9

F