Physics 2005 JAMB Past Questions - Question 40: Two liquids L1 and L2 are contained in a U-tube. The height and the density of L1 are 8 cm and 10-³ kgm-³ respectively. If the density  is 800 kgm-³, its height measured from the same level is

Correct Answer

D

Explanation

To solve this problem, we can use the principles of hydrostatics and the concept of pressure in a fluid column. The pressure at any point in a fluid at rest is given by the hydrostatic pressure formula:

P = ρ * g * h

Where:
P = Pressure
ρ = Density of the fluid
g = Acceleration due to gravity
h = Height of the fluid column

Now, in the U-tube, the pressure on both sides must be equal since they are at the same level (same height). Therefore, we can set up the equation:

For liquid L1:
P1 = ρ1 * g * h1

For liquid L2:
P2 = ρ2 * g * h2

Since P1 = P2, we can equate the two equations:

ρ1 * g * h1 = ρ2 * g * h2

Now, let's solve for h2, the height of liquid L2:

h2 = (ρ1 * h1) / ρ2

Given:
Density of L1 (ρ1) = 10^(-3) kg/m³
Height of L1 (h1) = 8 cm = 0.08 m
Density of L2 (ρ2) = 800 kg/m³

Now, plug in these values to find h2:

h2 = (10^(-3) kg/m³ * 0.08 m) / (800 kg/m³)

h2 = (8 * 10^(-5) m³) / (800 kg/m³)

h2 = (8 * 10^(-5) m³) / (8 * 10^2 kg/m³)

h2 = 10^(-7) m / (10^2) m

h2 = 10^(-9) m

So, the height of liquid L2 measured from the same level is 10^(-9) meters or 1 nanometer.