As shown in the figure two cylindrical vessels A and B are interconnected . Vessel A contains water up to 2m height and vessel B contains kerosene . Liquids are separated by movable, airtight disc C . If height of kerosene is to be maintained at 2m , calculate the mass to be placed on the piston kept in vessel B. Also calulate the force acting on disc C due to this mass. Area of piston `=100cm^(2)`, area of disc
`C=10cm^(2)` , Density of water `=10^(3)kgm^(-3)`, specific density of kerosene `=0.8`.

Explanation

Area of piston `A_(1)=100cm^(2)=10^(-2)m^(2)`
Area of disc `A_(2)=10cm^(2)=10^(3)m^(2)`
Density of water `rho_(w)=10^(3)kgm^(-3)`
`therefore` Specific density of kerosene
`=("Density of kerosene")/("Density of water")`
`=0.8`
`therefore` Density of kerosene `rho_(k)`
`=0.8xx` density of water
`=0.8xx10^(3)`
`=800kgm^(-3)`
If height of kerosene is maintained at 2m ,
Pressure of water column `=(mg)/(A_(1))+` Pressure of kerosene column.
`thereforeh rho_(w)g=hrho_(k)g+(mg)/(A_(1))`
`therefore2xx10^(3)=2xx800+(m)/(10^(-2))`
`therefore2000-1600=(m)/(10^(-2))`
`thereforem=400xx10^(-2)`
`thereforem=4kg`
Now pressure due to mass m is transmitted undiminished to disc C . So, pressure due to 4kg mass
`=("Force on disc C")/("Area of disc C")`
`therefore(mg)/(A_(1))=(F_(C))/(A_(2))`
`thereforeF_(C)=mg((A_(2))/(A_(1)))=4xx9.8xx((10^(-3))/(10^(-2)))`
`F_(C)=3.92N`