In this illustration, we'll analyze air cavities
in a gold bar. we are given that a gold bar
of mass 108 gram weighs 1 0 2 gram in water.
and if in the gold bar some air cavities are
left at the time of manufacturing we are required
to find the volume of these air cavities.
given that the specific gravity of gold to
be 19 point 3. so in solution here, if we
consider. that, volume of, air cavities. be
v c. then, we can write total volume. of,
gold bar. is, total volume of bar we can write
as in air it is weighing 1 0 2 gram. in water
and in air it weighs 1 0 8 grams so this is
1 0 8, divided by 19 point 3 times ro of water
as specific gravity of gold to be 19 point
3, the, density of gold can be taken as 19
point 3 ro w. plus, volume of cavities. and,
we can write, in water. buoyant force. on
gold bar. is, you can see in air it weighs
1 0 8 gram in water it weighs 1 0 2 gram,
so it weighs. 1 0 2 gram in water so buoyant
force on it will be equal and to 6 gram of
water. so here we can write buoyant force
on gold bar. is, v, total volume of the gold
bar multiplied by ro of water, multiplied
by g. and this should be equal to. the difference
of. the 2 weights, which we are measuring,
that can be written as 6 multiplied by g,
where 6 is the, difference in the weight of,
the gold bar in water as well as in air. so
in this situation here we can take in c g
s unit ro w is, equal to 1. taking ro w is
equal to 1 gram per centimeter cube. we use.
in this situation this is 1 0 8, divided by
19 point 3. plus, v c. and here g gets canceled
out, should be equal to 6. so here simplifying
we are getting the value of v c is equal to
zero point 4 0 4 centimeter cube, that is
a result of this problem.
