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Ishan here I am giving solution of first two parts. For other two parts even I am not very clear so i don't want to say anything right now.

Let us call the particle which is projected as A and the other particle as B.
Initially uxA (initial velocity of A along x axis)=ucos(60)=10sqrt(3)*cos(60)=5sqrt(3)
Initially uyA (initial velocity of A along y axis)=usin(60)=10sqrt(3)*sin(60)=15 m/s.

Displacement os A in 2 sec
xA=uxA*t=10*sqrt(3)
yA=uyA*t-(1/2)gt^2=10 m
At this stage string will becomes taut. So length of string =sqrt( 10^2 +(10*sqrt(3)^2)
=20m

At this point the string makes an angle of taninverse(10/(10*sqrt(3))=+30 degree with horizontal

Velocity of A just before this is
vxA(2) = uxA = 5sqrt(3) m/s
vyA(2) = uyA-gt = 15-10*2=-5m/s
So velocity of A is 10 m/s making an angle of -30degree with horizontal,

So velocity makes an angle of 60 degree with direction of string. It will have two componets. 5 m/s along string and 5sqrt(3) m/s perpendicular to string. A will continue to move with 5sqrt(3) m/s perpendicular to string as string cannot exert force perpendicular to it. But velocity along string will decrease with coresponding increase in velocity of B.
Let after string becomes taut velocity of A and B along string is v.
Momentum along string just before t=2sec = 4*5=20kgm/s
Momentum along string justsftere t=2sec =(4+ 4)*v=20kgm/s
v = 2.5 m/s

Good answer arnydude12

The angular momenum of first sphere is along y-axis and of the second is along z-axis. There is no particle in the system which has an angular momentum about x-axis, so net momentum about x-axis is zero.

The level of water will fall and the best explanation comes from kghedriu. Enjoy

Ashish I am just answering your first question. For others please repost them as fresh question. You can't express length of a bent wire and sphere as one vector but a plane area can be expressed as a vector

Q2) Depends velocity in which direction you take as positive

If you take velocity of north as positive

Vm/g = Vm/t+Vt/g = (-18) + (54) = 36 km/hr

If you take velocity of south as positive

Vm/g = Vm/t+Vt/g = (18) + (-54) = -36 km/hr

In both the cases velocity of monkey wrt ground is 36 km/hr towards north

PS: How do you get this 18 Km/hr and what is the use of train B

HI Trisha I answered it in some other post today only. Its 4*omega. Please check for the same

Umang consider the small elemet where you have taken the area between two circles of radius r and r+dr. And then you think that the area of this small element is 2*pi*r*dr. There are two mistakes. First this figure will not be a cylinder (Remember that the radius at top and bottom of cylinder is equal). Second the height of this cylinder is not dr (As taken by you)
Because of these mistakes you are not getting right answers. By the way Subs answer is correct.

Nick I have already answered this question. Please check problem number 1844. Please copy following link and paste it in your browser's address
http://www.goiit.com/posts/list/1844.htm

And answer for first question is Ph-CH-(CH3)2

Hi Aman how do you attach a mass of M/2 to both the parts. It should also be a function of soda in the can.

Ayush here comes my answer. I am answering just part (d) and all the answers will come from there. Let at any time the cylinder contains soda upto a height x.

Mass of can=M

center of mass of can = H/2

Mass of soda upto level x = mx/H

center of mass os soda = x/2

Center of mass of whole system (h)=(MH/2+mx^2/(2H))/(M+mx/H)

for part a) and b) put x=H and x=0 and you get h+H/2

for c) minimize x by diffrentiating h

It will execute SHM but Aman has done it is not clear to me. Anyhow here is my solution. Let the distance of midpoint (C) of given cord from earth's center (O) is a. Let at any time mass is at a point P which is at a distance x from C.

OP=sqrt(a^2+x^2)

g'=(g/R)*OP

The component of g' along chord = g'*x/sqrt(a^2+x^2)

=gx/R

acceleration is

x and is along mean position so motion is SHM

Let A and B be points on intermediate cylinder which touch inner and outer cylinder at P and Q respectively. Let V be velocity of center of intermediate cylinder and

_i be its
Velocity of A=velocity of P

V-R

_i=

R

Velocity of B=velocity of Q

V+R

_i=9

R

Solving V=5

R

_i = 4

Tension is an internal force between ring and block. If you consider ring block system the external forces are force of gravity on block and normal reaction on ring which are both vertical.

HI Ishan, question is not clear. If the mass is held at P and P is fixed then why will it move