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ans 5..
3rd option is right...
select 5 games whose results r wrongly predicted 10c5
then 4 wrong prediction u hv 2 options(as one is correct result out of three)
so total wayz 10c5. 2^5
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ans. 4 atmost 1 box is empty means either no enpty or 1 empty
no empty box no of wayz.. 3!=6
1 box empty... 3C1.2c1 3c2 1c1=18...
3c1 to select empty box... 2c1 to select box wid 2 balls... 3c2 to select the 2 balln 1c1 to put remainin ball in remaini box
total 24
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ans 3 ..
5 animals can fit into 4 cages... so no of ways is 5c4 . 4! .7!=604800
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ans 2. 2 circles can intersect at atmost 2 points .... max no of pts can be obtained if no 3 circles intersect at one point...
no. of all possible pairs of circles 8C2
max
no. of points of intersection 2. 8c2=56
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ans1. the ans is
nC2 =66
n(n-1)=132
or n =12
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say there is a positive charge particle q in a external electric field E
a force f (< qE) is applied on the charge opposite to the field...
the net force on the charge is qE -f along E....
here work done by f is -ve
since displacement of charge is opposite to f
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nishant is right....
auxillary circle of hyperbola is x^2+y^2=a^2 ie circle hvin transverse axis as diameter
normal chord means a chord tat is normal to the curve....
there r infinite chords which can be drawn from a pt
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ans is a
1 atm = 101325 Pa
1 bar = 100000 Pa
exact
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hey... the 2 nd question should be how much charge is flown n not current k....
initially
Ceq=c/2
V1=V2=V/2
charge on each cap
q=cv/2
finally..
Ceq' = kc.c/kc+c=kc/k+1
charge on each cap is
q'=kcv/k+1
v1=q'/kc=v/k+1
so ratio V1 initial: V1 final=v/2/v/k+1=k+1/2....
total charge flown= qfinal- qini=
kcv/k+1 -cv/2
=(k-1)cv/2(k+1)
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hey dude pole n polar is a hyge theory to type...u can read it from SL loney....
however equation of polar wrt a curve is equation of chord of contact wrt pt h,k ....where h,k is called pole of the polar...
{h,k here may be inside or outside the circle}
so now...
let pole be h,k
eq of polar becomes hx+ky-3/4(h+x)+5/2(y+k)-7=0.....(1)
1) is same as given line 9x+y=28
so comparin the 2 eqs u can get the ans...
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HEY...
imagine 3D X,Y,Z ...8 octet r formed ...right...so let the corner wid charge be origin...other seven cubes r imaginary....
see all 8 cubes as a whole....it is a cube wid side double each single cube...
this is a cube wid charge in the centre....so flux from each face of larger cube is q/6
now each face of original cube is 1/4 of the larger one...n symmetrical to...
so from each face of cube flux is 1/4 into q/6
which gives the ans as q /.24
(rem the flux is q/24 on the 3 face not in contact wid the charge not the other 3)
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thanx!!
its nice to be appreciated!!!
hehehe
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hey karthik u r wrong... in such prob we never take g to be different!
the height difference is negligible as compared to radius of earth!
so this a prob of circular motion ...
rem whenever a car travels at a curved path...it must hv centripetal acc....
it may or may not hv tangential acc...
so direction of total acc. on a curved path can be either
1...towards the centre ie only centripetal( wen tangential acc is zero...as in this ques n uniform circular motion)
2...vector sum of tangential n centipetal acc... wen tangential acc is present...
thus car travellin on a curved path can never hv total acc in tangential direction...
(on a st. line car has tangential acc only)
at point one...say normal reaction is N1 and car is travellin on a part wid radius r
so mg-N1= centripetal force at one =mv^2/r......(!)
similarly at pt 2 radius be r"
mg -N2=mv^2/r"....(2)
from the fig....
curvature of one > curvature of 2
implies r<r"
or mv^2/r > mv^2/r"
from (1) n (2)..
mg-n1 > mg -n2
or -n1> -n2
n2>n1
so ans is n2>n1
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hey..wat is the 1st question? then..do we hv to fine no. of wayz of doin it?????
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see which ever graph lies above the other along y axis is max in tat region..
- <x<0 y=x^2 is above y=x
i.e value of x^2 >x in this region
similarly in 0<x<1
y=x is above y=x^2
so x>x^2 in tjis region so max{x,x^2} in region 0<x<1 is x
in the last case 1<x<
max{x,x^2} is x^2 as its graph is above graph of x so the ans is the graph wid red dots
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hey koel u right on your part...but....
it is true a>b does not neccessarily imply a^2>b^2
but if both a>0 n b>0
we can square the inequation widout changin the sign of inequality ....
generally in such questions of inequalities...
we assume x>0 y>0 z>0...don be confused...
this information will be provided in the question....
so x+y+z>0 n same wid x^2+y^2=z^2>0
so we can square the inequality...
so alwayz remember if u r squarin the inequality tat both sides r >0
also if both sides r less than 0,..... u can square n change the inequality sign....
remember squarin means expandin the domain so we have to take care.....
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soo...nobody wid hard luck!!!!!! gr8....
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ans is infinity....
the form is infinity/infinity..
so we can apply l hospital...
 [x ] [ infinity] 2^x/x^2
usin l hospital
=
[x ][infinity ] 2^x ln2/2x
again l hospital
[ ][ ] 2^x ln2 .ln2/2
now puttin x tendin to infinity
we get infinity....
also rem alwayz
[x ][infinity ] 2^x/x^n where n is any integer ans is alwayz infinity...because on usin l hospital n times u will get
[x ][infinity ] 2^x.(ln2)^n/n(n-1)....1
whish is infnity...
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this is a question of multinomial theorum...
'A' can occur zero to m times
ie x^0+x^1+x^2......x^m
similarly 4 rest letters...
so total no. of combinations of a, n,s,h,u tat can occur in m bags is...
coefficient of x^m in (x^0+x^2...+x^m)^5
calcucate it,,,,say it is p
now to form anshu... all letters should be atleast once
so we omit x^0
so no. of such cases
is coefficient of x^m in
( x^1+x^2+x^3...+x^m)^5
say this comes out to be q
so prob is1-q/p
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y isn't anybody helpin me??????????????????????????/
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Vriti Edustore
