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i got a seat in eee thr...but i returned it..primarily becoz of the environment over thr is everythng but studying.....alrite if placement n infra. is gud.....but guys the fees they ask is not at all reasonable!!!!...if u top in ne coll(provided its A grade) u'll b picked up by the same companies!!.....nw i'm trying mumbai thro CAP rounds........my verdict go if its a dire scenerio.....
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x^x=t
xlogx=logt
diff. both sides wrt to x
and using product rule in lhs
x.1/x +logx= 1/t dt/dx
subst. t
we have
d x^x/dt= x^x(1+logx)
rhs is the ans....
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hi there
this is really easy
well we have binomial theorem and in a kind of expansion of
(y-a)^n +(y+a)^n
when we expand each term using binomial then the even terms get cut whereas the odd terms get doubled.
(y-a)^n -(y+a)^n
in this case it is opposite i.e. the odd get cut and even double
hence the shortcut
cheersss
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here is another method,
the expression can be written as
z^3-z'=0,........................1
we know that zz'=|z|^2
putting in one,
we have ,
z^4-|z|^2=0
or
z^2= |z|
this condition is only true
when z is unimodular i.e.
0
1
-1
i
-i
|i|=1
therefore 5 sol.
do rate ma efforts
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now see, there are 8 squares horizontally and 8 vertically there fore 8^2= 64 now, assuming a small square of unit length we can group two squares (vert. and hor.) hence giving 7 new squares vertically and hor. total = 7*7=49 now if we gp. 3 we get 6*6=36 THINK VIRTUALLY.... AND SO ON TILL WE GET 1 WHOLE SQUARE HENCE TOTAL IS 64 +49 +36...............1=204 square of no.s i dont have time, hence could not explain(sorry people) do rate ma efforts.........
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i plan to renew this topic,
so contribute PEOPLE!!!
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did neone check this out????????
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see we can easily find the acc. of the bodies by using the kine. eq
2g-T=2a
and
T-g=a
solving them we get
a=g/3
as initial velocity is 0,
vel(v) after 1 sec of motion is,
v=at=g/3 m/sec
now for the string to be tight again the distances moved by the two masses should be equal........RIGHT
for 1 kg APPLYING SIGN CONVENTIONS
x=vt-1/2 g t^2.....1
for 2 kg velocity is 0 as it is stopped
x=1/2gt^2.............2
equating the two,
vt= gt^2
v/g=t...................v=g/3 proved
hence t=1/3.sec.
plss rate ma efforts
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hi tanvi,
some days ago i posed quite a similar Q out there..
i have fr long maintained a top 5,6 rank in my coaching (could have been better had i worked on chemistry)
last time i thought i had done well but it went down to 20+,
what went wrong???
u see,the Q in the paper are somewhat discussed in the class,some even are same!!!, while some are directly picked from
a book, slightly tisting
u see at this point the studen t "might hve(by luck of course)"
referred that book or remember that ans....hence without thinking marking the ans..
cheating of course is another pain ,though THEY MAY EMERGE WINNERS AT THE END OF day,
BUT u know and i know that we worked fair and on the day we write our iit this is what is gonna be imp.
results of coaching are no doubt imp. but there is more to success than topping them
they are seldom true,and never ever BUILD CONFIDENCE THROUGH THEM..
just keep working hard and care damn of what others willl think
cheers
NICK
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hi there,
see let the first term of one A.P. be a
and others be
a
now given
2a +(n-1)d/2a+(n-1)d= 5n+4/9n+6.........right???/...................1
now we have to take th ration of their m term
which is,
a +(m-1)d/a'+(m-1)d
now if multiply this by 2, we have
2a +(2m-2)d/2a'+(2m-2)d
NOW U SEE WE CAN SUBSTITUTE
n=2m-1 THIS IS BECAUSE THEN WE CAN MAKE USE OF EQ. 1....
here m=18
therefore,n=35
putting this in the RHS OF 1
we have
the ratio as, 179/321
which i suppose should be the ans....
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u see,as they are the roots,
theta=@
cosec@+cot@=-b/a.................1
cosec@*cot@=c/a..............2
putting 1 in 2, we get
and simplifying we get,
acosec@^2 +c +bcosec=0
therefore,
cosec @= -b+- (b^2 -4ac)^1/2 /2ac
also ,similarly we get
cot@=-b+- (b^2 -4ac)^1/2 /2ac
as cot@ cannot be equal to cosec@,
let cosec@ be
-b+ (b^2 -4ac)^1/2 /2ac................4
and
cot@ be
-b- (b^2 -4ac)^1/2 /2ac...............3
we also have,
cosec@^2=cot@^2+1
putting 3 and 4 above and simplifying (very easily we get,
a^4-b^4 + 4b^2ac=0 which is (d)
cosec@ and cot@ cannot be equal as
cosec@^2=1+cot@^2
if they are equal then
1=0, which is wrong
cheers and do rate....
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u see,
as time taken by car is less than using
distance/velocity= time
500/p +10= 500/q......1
500q +pq=500p
substituting ,p(or q)
from
p=q+25.......2
we get a quadratic solving which we get,
p=50
and
q=25
hence
p=2q i.e. (b)
do rate....
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that has to be CONSTANT VELOCITY NOT CONSTANT ACC.
......
CHECK IT OUT
NINADDROXX
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well see,
time taken by the truck to cover this distance is
58.8/14.7=4 sec.
now for an observer on ground it happens to be a simple projectile motion.......IMAGINE...
ucos@t=58.8.......1
usin@t- gt^2/2=0
as displacement is 0(he catches the ball)
therefore
usin@=gt/2........2
dividing 1 by 2 we get
tan@=4/3(after subst. the values...of course)
hence @=53 degrees
cos@=3/5
therefore from 1
u=14.7*5/3= 24.5m/SEC
cheers and DO RATE.......
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fine work here is another
let w1 be the angular velocity of small sphere ,
w2 be the angular velocity of big sphere.
now using the centre of mass eq.
w1R=w2(5R-R) (think over this people....).......................1
now writing the energy eq.
1/2 Iw1^2 +1/2mv1^2 +mg(5R-R)(1-cos@)= k
here w1 is the angular velocity of small body
v1 is the linear velocity
@ is theta
k is a constant as energy cannot be destroyed
now substituting values
I=2/5mR^2
w1=w2(5R-R)/R^2....................from 1
v1=w2(5R-R) ( as it is W.R.T. the big sphere)
putting the values in the above expression and simplifying it,
we have
1/5 (w2)^2(4R)^2 + 1/2(w2)^2(4R)^2 +g(4R)-g4Rcos@=k
now as this is conservative force ,d/dt of the expression will be 0........
hence differentiating(very easily)
(g)(4R) will become 0
w2^2 will be 2w2 a' where a' is angular acc.
sin@==@(very small)
.....
the final eq. is
28/5a'=g@
a'=5g/28R @
there fore
T=2pi root 28R/5g
which is the ans... try using the diiff. method whenever possible....
rate ma effortsssssssssssssss
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let
S =1+1/2-1/4+1/8...............(a)
2S=2+1-1/2+1/4-1/8...............(b)
(a) +(b)
3S=2+1+1 rest of the terms will get cut,as it is till INFINITY
NO TERMS SHALL REMAIN UNCUT as infinity is
undefined
therefore
S=4/3
which i think should be the ans.......
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the stone is thrown vertically upwards
this mean it WILL COME COME DOWN WHEN IT REACHES ITS MAX. HEIGHT (from the top of tower)
here final velocity will also be 0
let the distance travelled be "l"
-2gl=-u^2
therefore 2gl=u^2
on its ways down,from l to the top of tower
its initial velcoity is 0 and final can be found v out by,
2gl=v^2
therefore v=u,
this means the velocity of going up from the tower is same as it is to come down,
hence THE DISTANCE TRAVELLED FROM HERE IS SAME AS THAT OF THE HEIGHT OF THE TOWER.....
for the ans.
let the height be s
2gs=9u^2-u^2
s=4u^2/g.....
plss rate meeeeeeeeeee
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ok this sounds easy but i am messing up the signs.. there is plank of mass m1 which lies on a frictionless surface,a solid sphere of mass m2 lies on it and rolls without slipping. a constant force F is applied to the plank. find the individual acc. of the bodies....
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can anybody out there verify me......???
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