Question

A cube made of material having a density of 0.9 x 100 kg/m floats between water and a liquid of
density 0.7 x 10 kg/m", which is immiscible with water. What part of the cube is immersed in
water?

Answer 1

Explanation:

In the end, the cube floats only. So according to the law of floatation, its weight = total upthrust

Total upthrust = upthrust provided by liquid + upthrust provided by water

= weight of liquid displaced + weight of water displaced

=> Vdg = Vd(liquid)g + volume of cube submerged x d(water)g ( V is the total volume of solid)

ATQ;

900Vg = 70Vg + 1000*volume of cube submerged*g

=> 0.83Vg = volume of solid submerged*g

=> volume of solid submerged / V = 0.83 or 83/100

=> Volume of solid submerged = 0.83 of total V

Hence, 0.83 of the total volume is immersed.

If we were to calculate how much % of volume is submerged,

= volume of solid submerged / volume of solid x 100

= 0.83 x 100

= 83%

Hence