Two liquids of densities ρ ₁ an ρ ₂ (ρ ₂ = 2ρ ₁) are filled up behind a square wall of side 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing)

Solution

Physics sir · 4 steps

Find the pressure at M, N and O first; each part's force is its average pressure times area.

1. Force on a strip. Pressure rises linearly with depth, so the force on a vertical rectangle is the average of the pressures at its top and bottom edges times its area. Each part of the wall is 10 m × 5 m = 50 m². Atmospheric pressure acts on both sides of the wall, so we leave it out.

   Point	Gauge pressure	Why
   M (top)	0	free surface
   N (interface)	5ρ₁ g	5 m of liquid 1 above
   O (bottom)	5ρ₁ g + 5ρ₂ g = 15ρ₁ g	both layers press down, with ρ₂ = 2ρ₁

2. Upper part MN. Average pressure = (0 + 5ρ₁ g)/2 = 2.5ρ₁ g, so F_(MN) = 2.5ρ₁ g × 50 = 125ρ₁ g.
3. Lower part NO. Average pressure = (5ρ₁ g + 15ρ₁ g)/2 = 10ρ₁ g, so F_(NO) = 10ρ₁ g × 50 = 500ρ₁ g.
4. So. F_(MN)/F_(NO) = 125/500 = 1/4, which is option (c).

The idea to keep. For layered liquids on a wall, find the pressure at every boundary first. A lower layer carries the full weight of every layer above it. Then force = average pressure times area.

Answer from the NTA final answer key. Solution by PrepWiser.