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A cubic block of wood of side 10cm floats at interface between an oil and water with its lower surface 2cm below the interface.the height of the oil and water columns are 10cm each. Density of oil is o.8g/cc.the mass of the block is

A cubic block of wood of side 10cm floats at interface between an oil and water with its lower surface 2cm below the interface.the height of the oil and water columns are 10cm each. Density of oil is o.8g/cc.the mass of the block is

Grade:11

2 Answers

Arun
25750 Points
6 years ago

Let us assume the pressure at the bottom of the block to be P1 and the top to be P2. The density of oil be d1 and density of water  be d2. So we know that

P1 = P2 + (d1)(g)(10-h) + (d2)(g)(h) where h stands for the height of block inside water.

P1 - P2 = 0.8(g)(10-2) + 1(g)(2) = 8.4(g)

We know that P = F/A and hence F = PA = 8.4(g) x area.

Also we know that weight of block is mg = (density)(volume)(g) since { Density = Mass/Volume} and also volume = area x height.

8.4(g) x area = (density)(area)(height)(g)

8.4 = density x height { height = 10 cm }

density = 0.84 g/cc

mass = volume x density = 0.84 x 1000 = 840 g

Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
Dear Student

Let assume the pressure at the bottom of the block to be P1 and the top to be P2. The density of oil be d1 and density of water be d2. So we know that

P1 = P2 + (d1)(g)(10-h) + (d2)(g)(h) where h stands for the height of block inside water.

P1 - P2 = 0.8(g)(10-2) + 1(g)(2) = 8.4(g)

We know that P = F/A and hence F = PA = 8.4(g) x area.

Also we know that weight of block is mg = (density)(volume)(g) since { Density = Mass/Volume} and also volume = area x height.

8.4(g) x area = (density)(area)(height)(g)

8.4 = density x height { height = 10 cm }

density = 0.84 g/cc

mass = volume x density = 0.84 x 1000 = 840 g

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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