Which formula could be correct for the speed v of ocean waves in terms of the density ρ of seawater, the acceleration of free fall g, the depth h of the ocean and the wavelength λ?

Question 14

Which formula could be correct for the speed v of ocean waves in terms of the density ρ of seawater, the acceleration of free fall g, the depth h of the ocean and the wavelength λ?

A v = √gλ                     B v = √(g/h)                 C v = √(ρgh)                D v = √(g/ρ)

Reference: Past Exam Paper – June 2007 Paper 1 Q3

Solution:

Answer: A.

Unit of velocity: m s^-1

Consider the overall unit of the options provided.

The unit of the correct option should have the units of velocity.

Note that square root means ‘a power/index of ½ or 0.5’.

Unit of √(gλ): [ms^-2 m]^0.5 = [m²s^-2]^0.5 = m s^-1 

Unit of √(g/h): [ms^-2 m^-1]^0.5 = [s^-2]^0.5 = s^-1

Unit of √(ρgh): [kgm^-3 ms^-2 m]^0.5 = [kg m^-1 s^-2]^0.5 = kg^0.5 m^-0.5 s^-1

Unit of √(g/ρ): [ms^-2 kg^-1m³]^0.5 = [m⁴ s^-2 kg^-1]^0.5 = m² s^-1 kg^-0.5

Noise reduction headphones actively produce their own sound waves in order to cancel out external sound waves.

Question 33

Noise reduction headphones actively produce their own sound waves in order to cancel out external sound waves.

A microphone in the headphones receives waves of one frequency. A loudspeaker in the headphones then produces a wave of that frequency but of a different phase.

What is the phase difference between the external sound wave and the wave produced by the loudspeaker in the headphones?

A 90°                          B 180°                         C 270°                         D 360°

Reference: Past Exam Paper – June 2013 Paper 11 Q29

Solution:

Answer: B.

The noise can be cancel out by destructive interference (using the principle of superposition).

By producing a wave of the same frequency and amplitude but with a phase difference of 180° (this is the condition for destructive interference), the loudspeaker can cancel the noise.

A phase difference of 360° means that the waves are completely in phase (there is no phase difference between them.). Constructive interference occurs causing the amplitude of the noise to become greater.

Air is trapped inside a glass bulb which is immersed in water and attached to a U-tube containing mercury.

Question 40

Air is trapped inside a glass bulb which is immersed in water and attached to a U-tube containing mercury. The densities of water and mercury are ρw and ρm respectively. The surface of the water is open to the atmosphere where atmospheric pressure is P.

The acceleration of free fall is g.

What is the pressure of the air in the glass bulb?

A P + gρwhw + gρmhm

B P + gρwhw – gρmhm

C gρwhw + gρmhm

D gρwhw – gρmhm

Reference: Past Exam Paper – November 2017 Paper 12 Q15

Solution:

Answer: B.

The pressure is the same at the same level in a liquid.

Pressure is always due to what is above the level being considered.

Consider the level of mercury on the left-hand side of the U-tube. It is at a depth of h_w below the surface of water. Let this level be O.

At this level,

Pressure on left-had side = Pressure on right-hand side

Again, remember that pressure is always due to what is ABOVE that level.

On the left-hand side, we have water of depth h_w above level O. And above the water surface, we have the atmospheric pressure P acting on the level O.

Pressure on left-hand side = due to water of depth h_w + due to atmospheric pressure

Pressure on left-hand side = h_w ρ_w g + P

On the right-hand side, we have a height hm of mercury of level O. In addition to the mercury, the air inside the glass bulb also exerts a pressure (P_air).

Pressure on right-hand side = due to mercury of height h_m + due to air

Pressure on right-hand side = h_m ρ_m g + P_air

At level O,

Pressure on left-hand side = Pressure on right-hand side

h_w ρ_w g + P = h_m ρ_m g + P_air

P_air = h_w ρ_w g + P – h_m ρ_m g