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m(dv/dt) = Fext + urel (dm/dt)
This is Merchersky's equation for a variable mass system
urel = Relative velocity of the mass being gained/ejected w.r.t the system
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Don't apply the approximation at the start. Get the complete ratio, then apply h<<R.
The required ratio
g0(1-h/R) / g0/(1+h/R)2
= (1-h/R) (1+h/R)2
= (1-(h/R)2 ) (1+h/R)
As h<<R, (h/R)2<<1
So the required ratio is  (1+h/R)
Hence linear variation
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If the particle is above earth,
g = g0 (1-2h/R)
mg = mg0 (1-2h/R)
W = W0 (1-2h/R)
Where W represents weight.
dW/dh = -2W0/R
which is independent of h.
If the particle is inside the earth's surface
g = g0 (1-h/R)
W = W0 (1-h/R)
dW/dh = -W0h/R
which is independent of h.
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The exact expression for the areal velocity is L/2
where  is the reduced mass of the system.
Considering the mass of satellite megligible w.r.t that of earth = M, mass of earth
Hence, areal velocity is L/2M
Note that the expression isindependent of m. So areal velocity varies as m0
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The self potential energy of a system is defined as
U =(1/2)  V dm
where V is the potential energy of the rest of the system at the location of dm.
Let dm be an elementary mass on the surface of the sphere. Then
V = -GM/R
U =(1/2)  V dm = =(1/2) (-GM/R) dm = -GM/2Rdm = -GM2/2R
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Check the third step. There should be a 2 there
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Plz make sure u didn't forget to include the limits of integration in Q2
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Well I went way wrong with that. Sorry dude ... the occasional slip !
The number of ways in which each element in the domain can be linked to either of the two elements in the co-domain = 2n
Now, there are two possibilities that we must overrule to make it onto, that no element in the co-domain is devoid of a pre-image. The above linkings will have 2 such cases, when either one of the elements in the codomain is unlinked.
So total number of onto functions = 2n -2
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Excellent solution Vinu
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For the ball thrown up
-h = ut1 - (1/2)gt12
h = -ut1 + (1/2)gt12 --------(1)
For the ball thrown down
-h =- ut2 - (1/2)gt22
h = ut2 + (1/2)gt22 --------------(2)
For the freely falling ball
-h = - (1/2)gt2 -----------(3)
Multiply (1) by t2 & (2) by t1 and add
h(t1 + t2) = (1/2)gt1t2 ((t1 + t2)
Cancelling (t1 + t2) and substituting the value of h from (3)
(1/)gt2 = (1/2)gt1t2
t =  t1t2
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Start from rest.
Keep applying the force and move in a closed path.
If the force is conservative, it won't do any work & by work energy theorem, there will not be any change in KE. So when you reach the starting point your particle will again be at rest.
If this doesn't happen, the force in non conservative.
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Sneha has decided that she won't rate me!!!
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The answer given in the book seems to be incorrect. My approach is a bit different from urs.
The left portion of the string is a variable mass system in which mass is entering.
By Merchersky's equation,
m(dv/dt) = Fext + urel (dm/dt) = Fext + urel (dm/dx)(dx/dt)
As the left portion of the chain is in equilibrium,
0 = {T-(m/L)[(L/2)+(x/2)]}j +(- 2gxj)(m/L)( 2gx)
T = (mg/L)[(L/2) + (5x/2)]
= (W/2)[1+(5x/L)]
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