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adiabatic expansion .

apply p^( gamma - 1 ) / T ^ ( gamma ) = const

See that it is an isobaric expansion .

so , dQ = nCp dT

again dQ/dt = KA ( Ts - T ) /X ;

combining these two eqns , we get

dT/ ( Ts - T ) ={ KA/( n Cp X ) } dt

integrating ( taking final temp as Tf )

Ts - Tf / ( Ts - T ) = exp ( KA t / X n Cp )

from this reln we find Tf

Then from PVf = n RTf

we find Vf

again from PV i = n RT ( as given )

we find Vi

Now we find increase in volume ( Vf - Vi )

distance moved = ( Vf - Vi ) / A

assume root of the form p + sqrt ( q )

put it for x

Now equate the rational & irrational part .

Or ,

Show the discriminant to be a perfect square ( a, b , c are rational )

Oh !!!!!!!!!!!!!!!!!!!! Extremely sorry !!!!!!!!!!!!!!!!!

What u have to do is to introduce corriolis force . ( = 2mv )

Then u will get the ans .

Centrifugal force will not do .

Got it ?

because of conservation of charge .

centrifugal force = m ^2 R ( R = radius at equator )

= m v

= W v /g ( since W = mg )

Now the centrifugal force may add to the wt of the body or may decrease it depending on the direction of velocity of the body .

So the apparent wt is = W ( 1 + or - v/g ) ( proved )

apply pv power gamma = const

so , p1 : p2 = 3^ ( gamma ) =6.26

gamma = 1.67 for monoatomic gas

See ,the air in the pipe is relatively confined but the air in the atmosphere is open . This feature distinguishes thetwo media . And whenever there is a distinction betwn media there is a reflection.

The reflection occurs at the every jn ( where there is a discontinuity in the medum) . Here u get the jn open air (atmos ) & closed air ( inside the pipe ) .

These two are completely different medium . Hence we get a reflected wave .

Oh !!!! everything is fine .

Torque applied wrt the bottom most pt =

T = ( R + r ) 2t = dL / dt

so L = ( R + r ) t^2

See ,the distance betwn the bottom most point & the point of application of force is ( R + r ) . Hence the factor ( R + r ) . I think u have now got it .

Here L denote the angular momentum sought

Its polar eqn is r = exp ( k

)

or log r = k

T = ( R + r ) 2t

= dL/ dt

So L = ( R+ r ) t^2

Suggesting u two methods

1 >

find dx/ dt , dy/ dt first

& then find dy/dx from dy/dt / dx/ dt

2>

see we get y^2/3 + x^2/3 = 1/ (cos 2t ) ^1/3 .............(1 )

and y^2/3 - x^2/3 = ( cos 2t ) ^2/3 ...............( 2 )

eliminate cos 2t from ( 1 ) & ( 2 ) to get a reln between y & x

Now differentiate it wrt x to get dy/dx

do u know what

s in the expn mean exactly ?

They are std deviations with probability fn mod

^2

Now the RHS of the h -principle depends on how this deviation or uncertainity is defined . Diff authors have defined it differently so there is a ' uncertainity ' in the uncertainity principle .

the last term isa 2 arctan (x )

write log sin x to the base cos x as ln sin x/ ln cos x etc

Now differentiate as per standard rules.