For the system shown in the figure, the cylinder on left has a mass (M) of 25 kg, cross-sectional area 20 cm2 and thickness 5 cm. It is connected to a spring of spring constant 1880 N/m. The piston on the right has mass m(= 5 kg) and cross-sectional area 4 cm2. Initially heights of water column in both cylinders are equal and pistons are in equilibrium. The minimum mass m (in kg) to be kept on m so that water just spills out from the left is (g = 10 m/s2)
Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.
Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA
Kx + P0 A' + Mg = P1 A'
...(i)
P0 A + mg = P1 A
...(ii)
(i) – (ii)
Solving, x = 0
So, spring is in natural length initially.
Let mass put on 5 kg be m
For water to just spill out, y = 5cm
as volume is conserved, y × A' = x × A
y × 20 = x × 4
x = 5y = 25 cm
h = x + y = 30 cm = level difference between two masses
For equilibirum
Solving, m = 2