let us discuss one of the most important application
on archimedes principle that is apparent weight
of a body submerged in a liquid. to understand
it lets first consider a spring balance . through
which on the hook. we, hang a body of mass
m, say this is spring balance s-p. and through
its hook, a body , weight of mass m is suspended.
in this situation we know that , the spring
balance will have a tendency to read , the
tension in the connecting string in kilogram
force. so in this situation, the block will
be experiencing the weight mg. and the tension
in this situation for the iquilibrium we can
write t is equal to mg. so we can simply state
if, mass m. is hanging. through a spring balance
s-p. we can simply write, its reading will
be. reading is always given as tension divided
by g in kilogram force . and here tension
is mg so this will be m k g of that is the
tension , in the spring which is given as.
as its reading. now see what will happen if
the mass is submerged in a liquid. if i re-draw
the situation in such a way , that again the
spring balance is their. and through its hook
. the same mass is suspended. but, the difference
is . now in this situation the mass is submerged
in a container . which is filled with a liquid
of density ro l. in this situation i take
the density of liquid to be ro l. and let
us consider density of this mass should be
ro as this is the density of solid. now in
this situation, if we wish to find out. the
volume of, weight. which is suspended . this
volume can be written as mass by density of
solid . now in this situation, obviously when
it is submerged, a tension will be acting,
in the connecting string of spring balance.
as well as it’ll experience a buoyant force
in upward direction and its weight mg in downward
direction. so if we draw the free body diagram
of this mass, which is submerged in liquid
as well as hanging through the spring balance
s-p. so this can be written mg. in downward
direction, and in upward direction the forces
can be written as tension plus the buoyant
force. and buoyant force we know . buoyant
force . on block . can always be written as
this, v ro of liquid into g. and the volume
we can directly calculate like this. and for
the iquilibrium of object, we can also write
in this situation. for iquilibrium. of weight.
which is suspended. we can write in this situation
t + f b is equal to mg. or in this situation
tension will be mg minus f-b we can directly
state as buoyancy force is acting in upward
direction . tension will be less. and if tension
is less certainly the reading of weighing
balance will also be less. that’s why we
say , when an object is submerged in a fluid.
its apparent weight decreases. so in this
situation we can write it mg minus f-b f-b
can be written as v ro of liquid g and v can
be given as m by ro s. so it is mg multiplied
by ro l upon ro s this is the expression we
are getting . so here the value of tension
will be mg . one minus, ro l upon ro s. one
point you need to keep this in your mind as
this tension is written as. apparent weight
of object. and if we wish to find out the
reading. we know well that reading is just
t by g. as we have calculated the initial
reading to be m kilogram force, here t by
g when we calculate the reading will be m
multiplied by one minus ro l by ro s. in kilogram
force that is the reading of weighing balance.
so always remember whenever a body of mass
is submerged in a liquid. then within its
submerged state its apparent weight is written
as mg one minus ro l by ro s this is also
very useful relation will use in different
applications. numerical and different kind
of illustrations also.
