Properties of Matter 2 Question 19

19. A uniform solid cylinder of density 0.8g/cm3 floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7g/cm3 and 1.2g/cm3, respectively. The height of liquid A is hA=1.2cm. The length of the part of the cylinder immersed in liquid B is hB=0.8cm.

(2002, 5M)

(a) Find the total force exerted by liquid A on the cylinder.

(b) Find h, the length of the 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 acceleration of the cylinder immediately after it is released.

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Answer:

Correct Answer: 19. (a) Zero (b) 0.25 cm (c) g / 6

Solution:

  1. (a) Liquid A is applying the hydrostatic force on cylinder from all the sides. So, net force is zero.

(b) 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+hA+hB)ρcylinder g

and upthrust on the cylinder

= upthrust due to liquid A+ upthrust due to liquid B

=shAρAg+shBρBg

Equating these two,

s(h+hA+hB)ρcylinder g=sg(hAρA+hBρB)

or (h+hA+hB)ρcylinder =hAρA+hBρB

Substituting,

hA=1.2cm,hB=0.8cm and ρA=0.7g/cm3

ρB=1.2g/cm3 and ρcylinder =0.8g/cm3

In the above equation, we get h=0.25cm

(c) Net upward force = extra upthrust =shρBg

Net acceleration a= Force  Mass of cylinder 

 or a=shρBgs(h+hA+hB)ρcylinder  or a=hρBg(h+hA+hB)ρcylinder 

Substituting the values of h,hAhB,ρB and ρcylinder , we get, a=g6(upwards)



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