here in this illustration, we’ll discuss
about the resistance of a conductor with varying cross section. here you can see the figure
shows a conductor of length l. and having a cross section which is varying. and, the
radius of cross section varies linearly from, radius ay to b. and the resistivity of material
given as ro, and assuming b minus ay that means, their is not much variation. in the
radii compare to this length. we are required to find the resistance of this conductor.
now in the solution here we can write. for this linear variation if we consider at a
distance x, an elemental disc. which is of width d x. and for this elemental disc if
the radius is taken as r we can write. the radius of. elemental disc. shown is. this
r we can calculate by, similarity of triangles like this distance is x and the whole distance
is l. so here i can write r minus ay by x is equals to b minus ay by l. using which
we can write the value of r is equal to, ay plus, b minus ay by l. multiplied by x. and
this relation can also be directly written, by linear relationship from ay to b, with
distance x. so if this is the radius here we can calculate the resistance, of. elemental
disc is. this can be written as d r this is a resistance of this disc. of which the area
can be taken as pie r square and width is d x so this is ro, d x by area is, pie r square
so this will be, ro d x by. pie, ay plus b minus, ay by l multiplied by x, whole square.
and from end ay to b all such elemental disc can be considered to be connected is series,
so. we can write here the total resistance. of. conductor. here it can be given as r.
which is integration of d r and that can be written as, integration of ro d x by. pie.
ay plus b minus, ay by l, multiplied by x whole square. and this can be integrated from
zero to l. so if we integrate it you can see the result we are getting is. here this ro
by pie can be taken as a constant. and 1 by ay plus b minus ay by l, x whole square when
it is integrated it will be minus. this is 1 by, ay plus b minus ay by, l. x. multiplied
by this, l by b minus, ay. and we apply the limits from zero to l. so we substitute the
limits you can see what we are getting this is ro l upon pie. b minus ay this can be.
taken out and inside when we substitute the value l. so this l can gets cancelled out
it is ay plus b minus ay is minus 1 by b and when we substitute zero. this will be, minus
1 by ay so negative will become plus so this will be 1 by ay minus, 1 by b. so, if we further
simplify it you can see the result we are getting is ro l upon, here b minus ay also
gets cancelled out this is pie, ay b. so that is the resistance of this conductor and the
final result of this problem.