A cylindrical ice block is floating in water. 10% of its total volume is outside water. Kerosene oil (relative density = 0.8) is poured slowly on top of water in the container. Assume that the oil does not mix with water. Height of the ice cylinder is H.

(a) What is the thickness of kerosene layer above the water when 20% of the volume of the ice block is above the water surface?
(b) Find the ratio of volume of ice block in kerosene to its volume in water after the kerosene layer rises above the top surface of ice and the block gets completely submerged. Neglect any melting of ice

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$\frac{H}{4}, 2 : 3$

$\frac{H}{6}, 1 : 2$

$\frac{H}{8}, 1 : 1$

$\frac{H}{5}, 3 : 4$

answer is A.

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Detailed Solution

(a) Let A = area of cross section of the ice cylinder
H = height of the cylinder
F₁ = force on top surface due to atmospheric pressure (P_o)
F₂ = force on the bottom surface due to water pressure
W = weight of the cylinder
For equilibrium -

$\begin{array}{l} F_{2} - F_{1} = W \\ \Rightarrow [P_{o} + ρ_{k} gx + ρ_{w} g (0 .8 H)] A - P_{o} A = {AHρ}_{icee} g \\ \frac{ρ_{k}}{ρ_{w}} x + 0 .8 H = \frac{ρ_{ike}}{ρ_{w}} H \\ 0 .8 x + 0 .8 H = 0 .9 H \\ x = \frac{H}{8} \end{array}$

$(b)$ Buoyancy = W
Wt. of water displaced + wt. of kerosene displaced = W
$0 .8 {HAρ}_{w} g + {xAρ}_{k} g = {HAρ}_{ice} g$

Using Archimedes principle
${Ah}_{1} ρ_{k} g + {Ah}_{2} ρ_{w} g = A (h_{1} + h_{2}) ρ_{ice} g Solving, h_{1} = h_{2}$