in this example we are given that a cube of
wood supporting 200 gram mass. just floats
in water in fully submerged state. and it
is also given that when the mass is removed
cube rises by 2 centimeter above water level.
and we are required to find the size of cube
and we are also given with density of water
to be thousand kg per meter cube. now in this
solution if we draw the initial situation
we can state . this is a cube. which is of
size l by l, and , it is supporting . a block
of mass m , where we are given that m is zero
point two kilograms. and right now it is in
fully submerged state so if it is water, then
the cube is fully is submerged state. so you
can state the total . weight of, cube plus
, this mass. in this situation, this balanced
by , the total buoyant force by the water.
so in this situation we can simply write for.
iquilibrium. of the system. we use, here we
can easily use. the weight of this wooden
block can be written as , l cube is its volume
multiplied by density of wood. multiplied
by g this is the total weight of this, cube,
plus the weight of this mass can be written
as zero point two g. and the total weight
of this system is balanced by the buoyant
force due to the weight of water displace
by the cube. so it can be written as, l cube
multiplied by density of water i am taking
it as ro multiplied by g and, after this mass
is removed you can simply state, that cube
will become lighter and it’ll be raised
up . so we can say after the mass is, removed.
the cube will rise , above a water level say
by some distance h. so in this situation we
can say if this distance is h , still the
submerged part of cube, below the water level
can be written as, l minus h. so buoyancy
is only due to the submerged part of, the
object because this is the, only volume which
is displacing the liquid . so in this situation
we can simply state after. removal of. block.
here we use. in this situation, again we can
state, the total weight of, this block is
balance by the buoyant force. so it can be
written as l cube density of wood into g,
must be balance by the buoyant force which
is of weight of liquid displace so here the
volume of liquid displaced is, l minus h multiplied
by l square into density of water into g,
now . in this expression g we can cancel here
also we can cancel g . now, if we substitute
the values after opening it, we can simply
see what we are getting here l cube density
of wood can be written as. l cube density
of water . minus zero point two is equal to
this l cube density of water . minus, h , l
square, ro . so in this situation this term
also gets cancelled out. this negative sign
also can be cancelled, so in this situation
, we can see l square is equal to zero point
two, divided by h ro. as it is given that
when the block is removed. the level of cube
will rise by two centimeter so here this h
can be written as zero point zero two. so
when we substitute the values this point two
divided by zero point zero two multiplied
by, thousand. so on solving this, we’ll
get the result to be zero point zero one.
so if l square is this l can be written as
zero point one meter . which is equal to ten
centimeter that is the answer to this problem
as, we are just require to find the dimensions
of the cube we are required to find the size
of cube so it is a cube of size ten by ten
by ten centimeter.
