In this illustration we'll discuss about copper
impurity in gold piece. we are given that
a piece of gold weighing 36 grams in air weighs
34 gram in water, if their is some copper
impurity in gold we are required to find the
amount of copper in it. given that the specific
gravity of gold is 19 point 3 and that of
copper is 8 point 9. so, if we consider that,
the amount. of copper. in gold. to be. x gram.
this implies, mass of, gold will be, this
can be given as 36 minus, x gram because,
in air it is weighing 36 gram in total, which
includes some copper impurity. so if x is
the mass of copper, then the mass of gold
will be actually 36 minus x. so here we can
write the, volume of, whole piece. is, this
volume we can write, x gram is the mass of
copper so its volume will be x divided by.
8 point 9 times ro of water which is the density
of, copper, plus this is 36 minus x divided
by. 19 point 3 times the density of water.
as these are the specific gravity and ro w
is the, water density. and if we solve this
problem in c g s units we can take ro w to
be 1. so here we can write, in water. force
of, buoyancy. on, piece. is, this will be
v, ro water g and that should be equal to
2 gee. because the difference of the 2 weights
in air and water is 2 gram only. here g gets
cancelled out and we can substitute the value
of this volume over here. so this gives x
by 8 point 9. plus 36 minus x divided by 19
point 3. is equal to 2. and if we cross multiply
and simplify. this relation gives us 19 point
3 x plus, 8 point. 9, multiplied by 36. minus
8 point 9 x. is equal to 2 into, 8 point 9
into, 19 point 3. so here this is gives us
10 point 4. x. is equal to 23 point 1 4 which
gives us the value of x is 23 point 1 4. divided
by 10 point 4 that is 2 point 2 2 5 gram.
that is the result of this problem.
