A uniform solid cylinder of density `0.8g//cm^(3)` floats in equilibrium in a combination of two non mixin liquids `A` and `B` with its axis vertical. The densities of liquids `A` and `B` are `0.7g//cm^/(3)` and `1.2g//cm^(3)`, respectively. The height of liquid `A` is `h_(A)=1.2cm.` The length of the part of the cylinder with liquid `B` is `h_(B)=0.8 cm.`

a. Find the total force exerted by liquid `A` on the cylinder.
Find `h` the length of part of the cylinder in air.
c.The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid `A` and is then released. Find the aceleration of the cylinder immediately just after it is released.

Answer

(i) 0 (ii) 0.25 (iii) ` (10)/(6) `

Explanation

(i) 0 (ii) 0.25 (iii) ` (10)/(6) `
(i) Liquid A is applying the hydrostatic force on cylinder from all the sides. So, net force is zero.
(ii) In equilibrium weight of cylinder = Net upthrust on the cylinder let s be the area of cross-section of the cylinder, then weight ` = (s) (h + h_A+ h _B) rho _ ("cylinder") g ` and upthrust on the cylinder= upthrust due to liquid A + upthrust due to liquid B = ` sh _A rho _ A g + sh _B rho _B g `
Equation these two, ` s(h+ h_A + h _B) rho _("cylinder") g = sg (h_A rho A + h_B rho _B ) `
or ` (h+ h _A + h_B ) rho _("cylinder" ) = h_A rho_A + h_B rho _B `
Substituting,
` h_A= 1.2 cm, h_B = 0.8 cm and rho _A = 0.7 g //cm ^3 , rho _B = 1.2 g// cm ^3 and rho _("cylinder") = 0.8 g//cm ^3`
In the above equation , we get h = 0.25 cm
(iii) Net upward force = extra upthrust ` = shrho _B g `
` therefore ` Net acceleration ` a = ("force")/("mass of cylinder")`
or ` a = (shrho _B g)/( s (h + h_A + h _B) rho _("cylinder") ) or a = (sh rho _B g )/( (h + h _A + h _B) rho_("cylinder")) `
Substituting the values of ` h, h_Ah_B, rho _B and rho _("cylinder")`, We get ` a = (g)/(6) ` (upwards )