1.

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V_(0) and its pressure is P_(0). The piston is slightly displaced from the quilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency :

Answer»

<P>`(1)/(2pi)(V_(0)MP_(0))/(A^(2)GAMMA)`
`(1)/(2pi)sqrt((A^(2)gammaP_(0))/(MV_(0)))`
`(1)/(2pi)sqrt((MV_(0))/(A gammaP_(0)))`
`(1)/(2pi)(A gammaP_(0))/(V_(0)M)`

Solution :
FBD of piston at equilibrium
`impliesP_("atm")A+mg=P_(0)A""…(1)`
FBD of piston when piston is pushed down a distance x.
`P_("atm")+mg-(P_(0)+dP)A=m""(d^(2)x)/(dt^(2))""...(2)`
Process is adiabatic `impliesPV^(gamma)=Cimplies-dP=(gammaPdV)/(V)`
Using 1, 2, we get `f=(1)/(2pi)sqrt((A^(2)gammaP_(0))/(MV_(0)))`
So, correct choice is (b).


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