Physics 2015 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 10³kgm-³ respectively. If the density of L2 is 800 kgm-³, its height measured from the same level is

Correct Answer

D

Explanation

To determine the height of liquid L2 in the U-tube, we can use the principle of hydrostatic pressure. The pressure at any point in a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.

Since the two liquids are in equilibrium, the pressure at the same level in the U-tube must be the same. Therefore, we can equate the pressures of L1 and L2 at the same level:

P(L1) = P(L2)

ρ(L1) * g * h(L1) = ρ(L2) * g * h(L2)

Given that the height of L1 is 8 cm and its density is 10³ kg/m³, and the density of L2 is 800 kg/m³, we can substitute these values into the equation:

10³ * 9.8 * 0.08 = 800 * 9.8 * h(L2)

Simplifying the equation:

784 = 7840 * h(L2)

Dividing both sides by 7840:

h(L2) = 784 / 7840

h(L2) = 0.1 m

Therefore, the height of liquid L2 measured from the same level is 0.1 meters or 10 cm.