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)

Maximum pressure of gas during its motion:

Options

A. `(20sqrt(2))/((5-sqrt(3))^(3//2))N//m^(2)`
B. `(20sqrt(2))/((3+sqrt(5))^(1//2))N//m^(2)`
C. `(20sqrt(2))/((3+sqrt(5))^(1//2))N//m^(2)`
D. `(20sqrt(2))/((3-sqrt(5))^(3//2))N//m^(2)`

Explanation

By energy conservation.
`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)`