KCET 2021 Physics Solved Paper

According to the question, the arrangement of the masses is as shown below, Gravitational force between two masses is given as
F=

Gm1m2

r2

where, G= gravitational constant and r= distance between them.
From the given values, we can say that force between masses at A and G.
= Force between masses at B and G
⇒ FA=

GmAmG

AG2

and FB=

GmBmG

BG2

or FA=FB=

G×8×2

12

=16G ...(i)
[∵ Given, m1=mA=mB=8‌kg,m2=mG=2‌kg,,AG=BG=r=1m]
From the figure, resultant of FA and FB is given as
FAB=√FA2+FB2+2FAFB‌cos‌θ
=√2FA2+2FA2‌cos‌120°
=√FA2=√(26G)2=16G [from Eq. (i)]
⇒FAB=FA=FB=16G
For resultant force on 2‌kg body to be zero, 4‌kg body should be placed at a certain distance from G such that
FCG=−FAE
⇒ |FCG|=|FAB|
or

G×mCmG

x2

=G(16)
where, x= distance between 4‌kg and 2‌kg body.
Here, mC=4‌kg
⇒

G×4×2

x2

=G(16)
⇒ x2=

1

2

⇒ x=

1

√2

m
This means, 4‌kg body should be placed at point P on line CG such that PG=

1

√2

m.