A fixed thermally conducting cylinder has a radius `R` and height `L_(0)`. The cylinder is open at its bottom and has a smaller hole at its top. A piston of mass `M` is held at a distance `L` from the top surface, as shown in the figure. The atmospheric pressure is `P_(0)`.

The piston is now pulled out slowly and held at a distance `2L` from the top. The pressure in the cylinder between its top and the piston will then be

Options

A. `rho g (L_0 -R)^2 + P_0 (L_0 -H) + L_0 P_0 = 0`
B. `rho g (L_0 -R)^2 - P_0 (L_0 -H) - L_0 P_0 = 0`
C. `rho g(L_0 - H)^2 + P_0 (L_0 - H) - L_0 P_0 = 0`
D. `rho g (L_0 - H)^2 - P_0 (L_0 - H) + L_0 P_0 = 0`

Answer

C

Explanation

`P_1 = P_2`
`P_0 + rhog(L_0 - H )=P" ...(i)"`
Now, applying `P_1 V_1 = P_2 V_2` for the air inside the cylinder, we have
`P_0 (L_0) = p(L_0 - H) " "p=(p_0 L_0)/(L_0 - H)`
Substituting in Eq. (i), we have `P_0 + rhog (L_0 -H) =(P_0 L_0)/(L_0 -H)`
`rArr " "rhog(L_0 - H)^2 + p_0 (L_0 -H) - L_0 P_0 =0`
Therefore, option (C) is correct.