Manometer with two liquids

GCE O-level 5076 Science (Physics) 2015 November Paper 2 Question 4(c)

Question:

How would you find the density of liquid B? I believe most of you would find the pressure at point X and then equate it with the pressure at point Y. Have you ever wonder why this method works? At least I do.

Let us investigate by considering the pressure at points P and Q. Point P is at the junction between air and liquid A. Point Q is at the same horizontal level in liquid B.

Pressure at P = Atmospheric pressure

Pressure at Q = Atmospheric pressure + Pressure due to the column of liquid above Q

In other words, the pressure at A and at B are different. Hence, we cannot solve the question by equating the pressure at P and pressure at Q. This begs the following question: What's so special about points X and Y?

Explanation:

As the liquids are at rest (i.e. in equilibrium), the pressure at the bottom of the manometer due to the left column must be equal to the pressure due to the right column.

Pressure due to left liquid column = Pressure due to right liquid column.

For points X and Y,

Pressure at X + Pressure due to water column below X = Pressure at Y + Pressure due to water column below Y.

Since Pressure due to water column below X = Pressure due to water column below Y, Pressure at X = Pressure at Y.

For point A and B,
Pressure at P + Pressure due liquid column between P and X + Pressure due to water column below X = Pressure at Q + Pressure due liquid column between Q and Y + Pressure due to water column below Y.

Since Pressure due to water column below X = Pressure due to water column below Y,
Pressure at P + Pressure due liquid column between P and X = Pressure at Q + Pressure due liquid column between Q and Y.

Since liquid A and liquid B have different densities, pressure due liquid column between P and X is NOT equal to pressure due liquid column between Q and Y even though both liquid columns have the same vertical height (i.e. PX = QY). Hence, pressure at P is NOT equal to pressure at Q.

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O-level Science (Physics) Suggested Answers for Electromagnetic Spectrum

N05/2/3
(a) ultra-violet radiation [1], radio waves [1]
(b) number of complete waves produced per second [1]
(c) speed of wave = 3.0 × 10^8 m/s [1]
(d) transverse wave OR can travel through a vacuum [1]

N06/2/5a

different frequencies or wavelengths [1]

N08/2/2

(a) infra-red radiation [1]
(b) microwaves [1]
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O-level Science (Physics) N04/2/4bii

An athlete runs a 100 m race in 12.5 s. Explain why, for part of the race, the athlete must have been running faster than the speed calculated in (a).

GCE O-level 5052 Science (Physics) 2004 November Paper 2 Question 1(b)

Answer: The runner accelerates from rest.

Proof: Consider the area under an arbitrary speed-time graph that starts from rest and the area under another speed-time graph that is travelling at a constant speed, which is equal to the maximum speed of the first graph. Both graphs stop at time t. Thus, the area under the first speed-time graph  = average speed x t while the area under the second speed-time graph = maximum speed x t. Since first area < second area, hence average speed < maximum speed.