Question

A body of density 0.8 g cm floating on a liquid has 80% of its volume submerged in a liquid. Find
the density of the liquid.
​

Answer 1

Answer:

Let the volume of the block of wood be

V

c

m

3

and its density be

d

w

g

c

m

−

3

So the weight of the block

=

V

d

w

g

dyne, where g is the acceleration due to gravity

=

980

c

m

s

−

2

The block floats in liquid of density

0.8

g

c

m

−

3

with

1

4

t

h

of its volume submerged.So the upward buoyant force acting on the block is the weight of displaced liquid

=

1

4

V

×

0.8

×

g

dyne.

Hence by cindition of floatation

V

×

d

w

×

g

=

1

4

×

V

×

0.8

×

g

⇒

d

w

=

0.2

g

c

m

-3

,

Now let the density of oil be

d

o

g

c

m

-3

The block floats in oil with 60% of its volume submerged.So the buoyant force balancing the weight of the block is the weight of displaced oil =

60

%

×

V

×

d

o

×

g

dyne

Now applying the condition of floatation we get

60

%

×

V

×

d

o

×

g

=

V

×

d

w

×

g

⇒

60

100

×

V

×

d

o

×

g

=

V

×

0.2

×

g

⇒

d

o

=

0.2

×

10

6

=

1

3

=

0.33

g

c

m

−

3