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nopes.......ans is 1/3
n plz i want a hint or a soln
need a quick reply
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hi here is my ques.......... if the points P1, P2, P3..... are the middle pts. of line segments AB, P1B, P2B........respectively nd particles of masses m, m/2, m/2^2........are placed respectively on these pts. if G is the mass-center of so placed infinite particles nd vector BG = P vector BA then find P.
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2nd one...............
l x y z l
l x^2 y^2 z^2 l
ly+z z+x x+y l
r1>r1+r3 nd take (x+y+z) common
(x+y+z) l 1 1 1 l
l x^2 y^2 z^2 l
ly+z z+x x+y l
c1>c1-c2 nd c2>c2-c3
u ll finally get............
(x+y+z)(x-y)(y-z)(z-x)
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1st one.................
determinant of l a^2 2ab b^2l
l b^2 a^2 2abl
l 2ab b^2 a^2l
c1>c1+c2+c3 nd then take (a+b)^2 common frm c1
(a+b)^2 l 1 2ab b^2 l
l 1 a^2 2ab l
l 1 b^2 a^2 l
r1> r1 - r2
r2> r2 - r3
(a+b)^2 l 0 2ab-a^2 b^2-2ab l
l 0 a^2-b^2 2ab-a^2 l
l 1 b^2 a^2 l
nw expand the determinant along c1
u ll get...................
(a+b)^2 ( a^4 + b^4 + 3 a^2b^2 - 2a^3 b - 2ab^3)
= (a^3 + b^3)^2
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hi i tried it like diis......a sin^-1 x - b cos^-1 x = c.....(1)
we knw dat sin^-1 x + cos^-1 x = pi/2.......(2)
using (2) n solving (1).........
u ll get the value of sin^-1 x nd cos^-1 x
n finally the desired ans
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hi lolita........wot i think is......since dimensions r the same soo R is proportional to rho. wires r connected in parallel.....n if equivalent resisitivity is denoted by rho then...........1/rho = 1/rho1 + 1/rho2
sooo....v ll get rho = rho1.rho2/(rho1+rho2)
sorry if i m wrong......but i think dis is the answer :)
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tan(x+2x)=(tanx + tan2x)/(1-tanx tan2x)
solving it..we get....................
tanx.tan2x.tan3x=tan3x - tanx - tan2x
soo nw....
differentiation of (tan3x - tanx - tan2x)
= 3sec^2 3x - sec^2 x - 2sec^2 2x
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hi try it like dis................
integral of root{ (cosx-cos^3 x)/(1-cos^3 x) }
" { (cosx sin^2 x)/(1-cos^3 x) }
integral of { root(cosx/1-cos^3x) * sinx dx}
put cosx=t >>>>> sinx dx = dt
integral of { root(t/1-t^3) dt}
integral of { t^1/2/ (1- (t^3/2)^2)^1/2 dt
nw put t^3/2=z
3/2 t^1/2 dt = dz >>>>>>>>>>>> t^1/2 dt = 2/3 dz
sooo....finally we get........ integral of{ 2/3 dz/1-z^2 }
nw u ll get the ans :)
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thanx a lottt 4 the info. krish :)
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hi.....here is the soln.
sin10 sin50 sin70 = sin10 sin(90-40) sin(90-20)
= sin10 cos40 cos20
mul 2cos10 in num. nd deno.....ull get.......
= sin20 cos20 cos40/2cos10
= sin40 cos40/4cos10
=sin80/8cos10
=sin(90-10)/8cos10
=1/8
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hi..........lim x>pi (1-sinx/2)/cosx/2.(cosx/4 - sinx/4)
mul. num n deno. by (1+ sinx/2)(cosx/4+sinx/4).....nd on solving ull get....
lim x> pi 1/(1+sinx/2) = 1/2
hope u got it :)
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hi wot i think is.......level of mppet is nt much.....nd it is easy 2 score a gud %age here....just mugg up all the formulaes nd start doing previous years papers.............nd dnt b tensed.......one thing......if u get 60%or 70% in mppet u really wont get a decent college...u ve 2 fight 4 10%marks.........if u get 80% marks then surely u ll get gsit or mits.........n its nt at all tuff.........soo just concentrate on ur studies.....make ur concepts clear........n u ll easily get thru it.
:)
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hi gayathri i think if u use "by parts" in dis question then it will b tedious......aditya's soln is perfect.........always remember ilate cant b applied newhere.......by parts is aaplied only where the question can b reduced in a form where one part is easy to integrate nd the other is easy to differentiate
nw in dis case if u apply ilate prefernce can b given to neone....but u ve to reduce sin^3 x or cos^5 x in a form dat can b integrated.....nd it will bcum unnecessary lengthy..........sooo....where 2 use wot depends on the question.....i hope u r getting wot i wanna say
thanx
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u can go 4 VMC's or brilliant's correspondence
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hi nunoxic....i m explaining u via examples.....hope it helps
1st example : suppose a block is kept on a fixed inclined plane whose inclination is "Q"
sooo....in dis case the forces dat will act are.....normal force wich is perpendicular to the inclined plane...........nd weight of the block. so in dis case weight will b resolved. so wcosQ will act downwards nd perpendicular to plane nd wsinQ parallel to plane
( draw the fig nd it will b clear). suppose here ne force F is acting on the bloch in upward direction of the incline, then we ll get the eqn as.........
N = wcosQ
F=wsinQ
now the 2nd example: suppose a block of mass 'm' is kept on an inclined plane of mass 'M' nd a force F is acting on clined plane wich is parallel to the ground { here also draw the fig. n u ll b clear}
soo...nw in dis case we ve to resolve the normal force as the normal force is acting perpendicular to the incline but it is making some angle with the force F
soooo..........ovr all it depends upon the case wot 2 resolve n wot nt
hope u got it
n if dere's ne more problem u can ask me :)
thanx
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hi abhinav........sinQ + cosQ can also b written as root(( sinQ + cosQ)^2)= root(1- sin2Q) or as elessar solved
soo...plz do make it clear wot do u exactly wanna ask :)
thanq
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i too think theta is neither a scalar quantity nor a vector quantity.
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Vriti Edustore
