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.

Sir thank you so much for helping students like me you are a true hero

respected sir

how can we come to know that we have to make a relation between r and the parameters a and b

Thank you sir,it really helped

sir in this case when a=0 we will get resistance as infinite.. does that mean if we connect it in a circuit current doesnt flow… and how resistance = infinite is possible… i am getting the same infinite answer incase of sphere also.. and zero resistance for cylinder… what is the meaning of this…( all have radius R and height h and for sphere h=2R ( this is understood))

sir a cone of base radius R and height h and cylinder of base radius R and height h and sphere of radius R and ( height h = 2R this is understood ) which of these have more electrical resistance towards current…

Sir, What would have been the case if (b-a) is not <<l ?

sir if we want to find resistance between slant heights of frustum then , while writing area for element can we assume it as a cylinder .. if no then what Will be the approach …… also sir pls try to upload videos on experimental section asap. .. thanks for your great effort

sir which formula did u use for integration sir please reply sir.

sir at 2:50 if b–a<<l then before integration if we state that (b-a/l)

x is also<<1 then using binomial approximation if we bring 2 (the power) inside the brackets and we get the denominator as pie(a+2x(b-a)/l) ..it gives a complete different answer with log termSir how we decide they are in series and parallel sir example is recent jee question in which aluminium and iron are taken and the equivalent resistance was asked then how we judge it is in series or parallel?

sir how is dr of disc is pl/A pls prove it

Sir god bless you! I have been struggling with exactly this problem!

Sir can u tell which module we should watch first in thermal effect of current or advance illustration…

If we do it in terms of area we get

P*L÷(pi(a^2 – b^2)* ln(a^2÷b^2)

Is this answer wrong

Sir i m dropper and my focus is jee mains January attempt..n advance..in which course of u i should focus…jr. medium or advanced

Sir disc has three surface two circle, and one cylindrical since it length is dx by neglecting cylindrical surface the area will be 2pir^2 , please say the reason why you have considered pir^2 as area

HELLO SIR..BIG FAN…

DONT TAKE IT OTHERWISE..BUT I HAVE A SHORT METHOD TO SOLVE THIS PROBLEM….IT IS VERY HASTLLE METHOD..