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abvegto

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Everything posted by abvegto

  1. yes we have met before although i dunno where... thats where iam having troubles...i cant figure out where this inernal energy has gone?....iam aware that this process is diff from adiabatic expansion wch is an isentropic....where entropy increses due to increase in volume then simultaneously entropy decreases due to decrease in temp..therefore keeping the entropy const... in adiabatic process s(v ,T) = 0 = Ds = ds/dv T Dv + ds/dT v DT such that ds/dv = -ds / dT in the throtlling process...if hydrogen is taken as the gas...after the expansion its temp. increases and its volume also increases thus s(v , T) is positive i.e there is an increase in entropy => that the ability of hydrogen to do work has decreased => that internal energy of the hydrogen has also descreased ..(?)..but this contradicts to the fact that the temp of the gas has increased inspite of the decrease in its internal energy?.... may be iam not clear....but i cannot put things together here.....
  2. @abhay....u are basically violating the law of conservation of energy....moreover...from the laws of thermodynamics ..it is impossible to get 100% efficiency even if u neglect friction, let alone more than 100%....
  3. in the throtling process or porous plug experiment, the gas either cools (most) or is heated (hydrogen helium) when they pass from a region of high pressure to a region of low pressure. i do not understand how a gas can be heated via this expansion?...becoz we know that expansion => increase in volume => decrease in internal energy wch should => fall in temp always...but for hydrogen the temp. increses even after the decrease in its internal energy...how is this possible?
  4. the question itself does not contain sufficient info....like how does the densities of different gases vary with altitude...for example oxygen is heavier thus applies more pressure...as we go up the atmoshere...the % of 02 decreases and it is replaced possibly by nitrogen which apllies less pressure...therefore the pressure at 4km depends on the composition of gases which is surely different from that at 2km.....so, i guess u r wrong...not sure though...
  5. i guess the answer is 1....cos(-t) = cos t and sin(-t) = - sin t
  6. here is a representation of the carnot cycle in the pv diagram(y=p x=v)... we kno that entropy is defined as the measure of randomness of a system..... from the veiw point of kinetic theory the entropy change during an adiabatic process is zero since q = 0( bc or da)for an ideal gas......but during the process there is an increase in the randomness of the system i.e there is an increase in the volume of the system.. does this increase in volume account for any change in entropy for a real gas? if yes then how to measure it?... D U = (dU/dT v) dT + (dU/dv T) dV.....in case of a real gas can we write the change in entropy as ds = dU/dT?.... plz help...
  7. yes....it is in accord with law of conservation of energy.......when light enters any medium its electric field interacts with those of the electrons of the medium and hence the electric field is disturbed .....and hence the speed "appears" to have been altered w.r.t our frame... but in that frame of reference( i.e w.r.t the medium itself )... light still travels at c...
  8. becoz ds vector = r dy (like s = r x) so i have the freedom of using either 'rx' (r dy) or 's'(ds) since they represent one and the same thing...
  9. well....the attraction can be zero at infinite distances.. f = gm1m2/ r now put r = infinity and we get f =Gm1m2 1/ infinity = 0 but this is just mathematically....u may well ask the value of infinity(there should be)...and that value will be the limit for this formula and probably for the range of gravitation itself..
  10. Lepton ...Consider a planet which moves from A to B in a small time dT...under polar co- ordinate system let the radius vector be r vector from the origin o. that is OA = Ob = rvector angle AOX = y angle BOX = y+dy therefore change in angle is dy now let the planet move from A to B in small time dT....let it cover a distance dSvector ...here dA ( the area covered by the planet in time dT) = 1/2 rvector x dSvector dividing by dT .... dA/dT = 1/2 rvector x dS/dT ..... dA/dT =1/2 rvector x vvector multiplying and dividing by 'm'.. the mass of the planet .... dA/dT = 1/2m (r x mv) .... dA/dT = 1/2m (r x p)............. 1 differentiating 1 again with T we get ..... d2A/dT2 = 1/2m [ (dr/dt x p) x (r x dp/dt)] ......d2A/dT2 = 1/2 m {[ v x (mv)] x [(r x f)]} ....................= 1/2 m (r x f).........2 but since the gravitational force is an example for a central force therefore force acts along the direction of r...that is sin theta = 0 therefore r x f = 0 therefore 2 becomes ....d2A/dT2 = 0 => dA /dT = CONSTANT hence the second law is proved MY QUESTION IS....IN MY EXAM I WROTE THIS PROOF AND I GOT 0 MARKS!..I CANT UNDERSTAND WHERE DID I GO WRONG... OF COURSE THE BOOKISH PROOF IS DEFFERENT....CAN U GUYS PLEASE PIN POINT MY MISTAKE?... and sorry for my english...
  11. take tanx = sinx / cosx ...then simplify then rationalize
  12. ...Consider a planet which moves from A to B in a small time dT...under polar co- ordinate system let the radius vector be r vector from the origin o. that is OA = Ob = rvector angle AOX = y angle BOX = y+dy therefore change in angle is dy now let the planet move from A to B in small time dT....let it cover a distance dSvector ...here dA ( the area covered by the planet in time dT) = 1/2 rvector x dSvector dividing by dT .... dA/dT = 1/2 rvector x dS/dT ..... dA/dT =1/2 rvector x vvector multiplying and dividing by 'm'.. the mass of the planet .... dA/dT = 1/2m (r x mv) .... dA/dT = 1/2m (r x p)............. 1 differentiating 1 again with T we get ..... d2A/dT2 = 1/2m [ (dr/dt x p) x (r x dp/dt)] ......d2A/dT2 = 1/2 m {[ v x (mv)] x [(r x f)]} ....................= 1/2 m (r x f).........2 but since the gravitational force is an example for a central force therefore force acts along the direction of r...that is sin theta = 0 therefore r x f = 0 therefore 2 becomes ....d2A/dT2 = 0 => dA /dT = CONSTANT hence the second law is proved MY QUESTION IS....IN MY EXAM I WROTE THIS PROOF AND I GOT 0 MARKS!..I CANT UNDERSTAND WHERE DID I GO WRONG... OF COURSE THE BOOKISH PROOF IS DEFFERENT....CAN U GUYS PLEASE PIN POINT MY MISTAKE?... and sorry for my english...
  13. In an experiment to study photocurrent( cathode ray tube?..)..connecred to a battery ..let us consider 2 cases CASE 1 ----a particular matallic emmitter is taken( work function = 4 ev)..a suitable frequency of light is made to fall(energy = 6 ev say..)..intensity being constant(say 3 electrons r emmitted).. and no potential is applied... the energy of the electrons will be 2 ev ,1.9 ev, 2.1ev say..( theoritically all electrons hav energy = 2ev).. CASE 2 ----now keeping all as constants except potential wch is made negative( say 1 v >> stopping potential)..then energies of the electrons changes to--- ...1 ev, 0.9 ev and 1.1 ev.. now my question is...whether tje current in the circuit changes?..if so how?..is it too neglogible?... if the current changes then if the negative potential of 1 v is also kept constan and suppose the frequency is increased ...to say 12 ev..the enrgies of the electrons increases beyond 2 ev...now does the current increase?.... i hav referred many books wch say that negative potential effects the current but frequency doesnt,. once it is above the threshold level....pls clarify ..thanx...
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