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Sagot :
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
Explanation:
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