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One mole of an ideal gas is contained in a perfectly insulating cylinder. Initially adiabatic piston of unit are and unit mass is in equilibrium. Now a block of same mass is kept gently on piston as shown in figure. (Take lambda=1.5,g=10m//s^(2) and neglect friction and assume change in temperature in whole gas is simultaneous) Depth upto which (piston+block) will move before coming to rest again is : |
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Answer» `(sqrt(5)-1)/(2)m` `2mgh=(4)/(2)RDeltaT` `DeltaT=(10h)/(R )` …..(i) Initially, `(mg)/(A)(A.l)=RT_(1)` …(ii) and finally, `P_("max")A(l-h)=RT_(2)` …..(III) (iii)-(ii) `P_(m)(1-h)-10=10h` …..(iv) By equation of adiabatic PROCESS `PV=` constant `(mg)/(A)(Al)^(gamma)=P_("max"){A(l-h)}^(gamma)` `10=P_(m)(1-h)^(gamma)` Put this `P_(m)` in equation (iv) `10(1-h)^(1-gamma)-10=10h` `(1-h)^(1-gamma)-1=h` `(1-h)^(1-gamma)=(h+1)` `(1)/(sqrt(1-h))=(1+h)rArr(1)/((1-h))=(1+h)^(2)` `h(h^(2)+h-1)=0` `h=(sqrt(5)-1)/(2)m` and `P_(n)=(10)/((1-(sqrt(5)-1)/(2))^(3//2))=(20sqrt(2))/((3-sqrt(5))^(3//2))N//m^(2)` 
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