Grade 11Mechanics

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

Varshith

8 Years agoGrade 11

2 Answers

Arun

8 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

6 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