NDA General Science Physics Questions Part 3

When the substances are mixed in equal volumes, then
4=

Vρ1+Vρ2

2V

⇒ρ1+ρ2=2×4...(i)
When the two substances are mixed in equal masses,then

m

ρ1

+

m

ρ2

=

2m

3

∴

1

ρ1

+

1

ρ2

=

2

3

....(2)
or

ρ1+ρ2

ρ1ρ2

=

2

3

or

8

ρ1ρ2

=

2

3

or ρ1ρ2=12...(3)
Now ρ1−ρ2=[(ρ1+ρ2)2−4ρ1ρ2]

1

2

=[64−48]

1

2

=4.....(4)
Solving equation (3) and (4), we get
ρ1=6
and ρ2=2