very complicated ansln((a+b)/a)----------- (whole divided by)2pi[L^2+(b-a)^2][] stands for sqr root
na.
pendi bhaiya....i am not able to solve this ques...plz help me....
ln(3a/a+b)----------- (whole divided by)2pi[L^2+(b-a)^2]if this is wrong then plz telll me is the curved surface area of frustrum l(r+R) ; where l is slant ht???
[] stands for sqr root
ankush...u can urself derive that...i just did....it is the same....only that u have to multiply that by pi..my answer...(2log2)/pi[h^2 + (b-a)^2]
the answer of the previous hollow frustum problem is L/6pi ,and i am 100 percent confident....... ANUJ SIR PLEASE TELL THE CORRECT ANSWER .........
let me give you a few hints:1)assume the current flow to be perpendicular to the axis of symmetry of the hollow frustum..2)try to visualize it as a combination of cylinders.
This comment has been removed by the author.
to the previous problem, your ans. was right.this hint was for this question.
IS THE ANSWER IS ln2/2piL
did no one else get this answer???cmon people..its a parallel combination..
please post the answer if my answer is wrong
looking at ur hints.......do u mean we take a cylinder whose centre is at a distance x from smaller basewhose radii area+(b-a)x/landa+b/2+(b-a)x/l and integrate??but it wud involve integration of dx/(ln x)
amritpal's last ans. was correct.to integrate, you need the resistance of a hollow cylinder b/w inner and outer radii..ankush, inner radius is a/2+(b/2-a/2)x/Louter radius is twice this..maybe i commented on the inner radius of the left end somewhere...
yahoo my answer is correct, really tough problem....
tough???? yes it was tough.
very complicated ans
ReplyDeleteln((a+b)/a)
----------- (whole divided by)
2pi[L^2+(b-a)^2]
[] stands for sqr root
na.
ReplyDeletependi bhaiya....
ReplyDeletei am not able to solve this ques...
plz help me....
ln(3a/a+b)
ReplyDelete----------- (whole divided by)
2pi[L^2+(b-a)^2]
if this is wrong then plz telll me
is the curved surface area of frustrum l(r+R) ; where l is slant ht???
[] stands for sqr root
ReplyDeleteankush...u can urself derive that...i just did....it is the same....only that u have to multiply that by pi..
ReplyDeletemy answer...
(2log2)/pi[h^2 + (b-a)^2]
the answer of the previous hollow frustum problem is L/6pi ,and i am 100 percent confident....... ANUJ SIR PLEASE TELL THE CORRECT ANSWER .........
ReplyDeletelet me give you a few hints:
ReplyDelete1)assume the current flow to be perpendicular to the axis of symmetry of the hollow frustum..
2)try to visualize it as a combination of cylinders.
This comment has been removed by the author.
ReplyDeleteto the previous problem, your ans. was right.
ReplyDeletethis hint was for this question.
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteIS THE ANSWER IS ln2/2piL
ReplyDeletedid no one else get this answer???
ReplyDeletecmon people..its a parallel combination..
please post the answer if my answer is wrong
ReplyDeleteThis comment has been removed by the author.
ReplyDeletelooking at ur hints.......
ReplyDeletedo u mean we take a cylinder whose centre is at a distance x from smaller base
whose radii are
a+(b-a)x/l
and
a+b/2+(b-a)x/l
and integrate??
but it wud involve integration of dx/(ln x)
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteamritpal's last ans. was correct.
ReplyDeleteto integrate, you need the resistance of a hollow cylinder b/w inner and outer radii..
ankush, inner radius is a/2+(b/2-a/2)x/L
outer radius is twice this..
maybe i commented on the inner radius of the left end somewhere...
yahoo my answer is correct, really tough problem....
ReplyDeletetough???? yes it was tough.
ReplyDelete