Two vertical cylinders are connected by a small tube at the bottom. It contains a gas at constant temperature . Initially the pistons are located at the same height. The diameters of the two cylinders are different . Outside the cylinder the space is vaccum . Gravitational acceleration is `h`,`h_(0) = 20cm` , `m_(1) = 2kg "and" m_(2) =1kg`. The pistons are initially in equilibrium. If the masses of the piston are interchanged find the separation between the two pistons when they are again in equilibrium. Assume constant temperature .

Answer

`30`cm

Explanation

For equilibrium `(m_(1)g)/(A_(1)) + rhogh_(0) = (m_(2)g)/(A_(2)) + rhogh_(0)`
`m_(1) = 2kg,m_(2)= 1kg implies (2)/(A_(1)) = (1)/(A_(2))`
For final equilibrium `(m_(1)g)/(A_(2)) gt (m_(2)g)/(A_(1))`
`therefore m_(1)` block will fall down
For constant temperature forces and pressure being constant `V_("initial") = V_("final")`
`h_(0)A_(1) + h_(0)A_(2) = h_(1)A_(1) + h_(2)A_(2) implies h_(1) = 30` cm & `h_(2) = 0`