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1) since +ve plate of 1st capacitor is connected to - ve plate of second capacitor
they are in series
eq capacitanceC'=C1C2/C1+C2

=5C/3
now since they are in series
eq pot diff V'=V1+V2
V'=2V+V=3V

initial energy=( C1V1^2)/2 + (C2V2^2)/2
(8+1)/2 CV^2=4.5CV^2

final energy= (C'V'^2)/2
=3CV^2

so loss in energy=(4.5-3)CV^2
=1.5CV^2

0???????

thanks ........was looking for something like this!!!!!!!

congrats buddy!!!!!!!!!!!!!!!!!!

is it 4

since magnetic field far away in both the cases wud be 0

1)V^{_cm=_{m1v1+m2v2/(m1+m2)}}

_{^{vel of heavy block=14m/s vel of lighter block=0}}

_{^{so Vcm=14*5/(5+2)}}

_^=10m/s

_^(b)

_{^{2) vel of block 1 wrt com=14-10=4m/s}}

_{^{vel of block 2 wrt com=10-0=10m/s}}

_{^{diff in mag=10-4=6m/s}}

_^(a)

_{^{3) total kinetic energy=1/2 m1v1^2 + 1/2 m2v2^2}}

_{^{here v1 and v2 are vel wrt to com}}

_{^{so KE =100+40=140J}}

_^(c)

thanks guys my doubt is cleared now!!!!!!

the ans shud be 1/4

the last 2 digits shud be divisible by 4
no of possible ways are 25 (24 + 1(for 00)
in a normal no there last 2 digits can occupied by any of 10 digits(inc 0)

so prob=25/(10*10)
=1/4

shud be 10!!
no of ways = 5!/(3! * 2!)=10

my ans is (A)

here we have to consider both forces acting on the same side of the block

net force in horizontal direc=P+Qsin@

for vertical direc
mg+Qcos@=N
for equilibrium

force in hor direc=u N

P+Qsin@=u (mg+Qcos@)
u= P+Qsin@/(mg +Qcos@)......................u is coeff of friction

my ans is (c) 16 J

x=t^3/3
v=dx/dt=t^2 dx=(t^2)dt

a=dv/dt=2t
F=2 x a =4t

W=integral(Fdx)
=integral(4t x t^2 x dt) within the limits 0to 2

=t^4

=2^4

=16J

@raulrag009

my pleasure

of course.....u just need to identify ur weaknesses
bmat tests are logical..........so most probably u know most of the topics but are not able to do the ques correctly

u must be able to do them at home
so there r only 2 possibilities
1) due to lack of time u are not able to comlplete the questions and as a result end up missing the sitters
2)u are not able to solve ques during that time
as u r not used to sitting for 3hrs at a strech .......especially during the second paper

i ll suggest u look up the previous bmat papers and try to solve the ques ....ull know the weaknesses

u must have practice of sitting for 3hrs at a strech

hope ull do better in the test on 17th!!!

1 b
2 a
3 b
5
a-q
b-p
c-pr
d-rs

6 abc

sorry man not very satisfactory

still thanks for trying!!!!!

OKAY SO HERE'S THE SOL
I CANT MAKE THE DIAG(THEY HAVE CONSIRED SHM AS REFLEC ON CIRCLE)
SO ILL JUST TYPE THE SOL

the phase diff b/w the particles
phi=pi/6 +pi/4 =5pi/12

th will collide when the perpendicular s drawn on the the diameter coincide.this will happen when any of the particle rotates an angle

theta =phi +phi' (on the figure they ,ve shown phi''=5pi/24 (dont know how)

=5pi/12 + 5pi/24=5pi/8
or wt =5pi/8
t= 5pi/8 x T/2pi

=5T/16'