The warning signal on an ambulance has a frequency of 600 Hz. The speed of sound is 330 m s-1. The ambulance is travelling with a constant velocity of 25 m s-1 towards an observer.

Question 4

The warning signal on an ambulance has a frequency of 600 Hz. The speed of sound is 330 m s^-1.

The ambulance is travelling with a constant velocity of 25 m s^-1 towards an observer.

Which overall change in observed frequency takes place between the times at which the

ambulance is a long way behind the observer and when it is a long way in front of the observer?

A 49 Hz                       B 84 Hz                       C 91 Hz                       D 98 Hz

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q26

Solution:

Answer: C.

The equation for the Doppler effect is as follows (given in the list of formula):

f_o = f_s v / (v ± v_s)

where  fo is the observed frequency

            fs is the frequency of the source

            v is the speed of sound

            v_s is the speed of motion of the source (ambulance)

When the ambulance moves towards the observer, the observed wavelength decreases and so, the observed frequency increases. The negative sign is used.

When approaching the observer (initial position of ambulance),

Observed frequency f_o1 = f_s v / (v – v_s) = 600 × 330 / (330 – 25) = 649 Hz

When the ambulance moves away from the observer, the observed wavelength increases and so, the observed frequency decreases. The positive sign is used.

When receding the observer (final position of ambulance),

Observed frequency f_o2 = f_s v / (v + v_s) = 600 × 330 / (330 + 25) = 558 Hz

Overall change in observed frequency = 649 – 558 = 91 Hz

The diagram shows a pump called a hydraulic ram. In one such pump the long approach pipe holds 500 kg of water.

Question 10

The diagram shows a pump called a hydraulic ram.

In one such pump the long approach pipe holds 500 kg of water. A valve shuts when the speed of this water reaches 2.0 m s^-1 and the kinetic energy of this water is used to lift a small quantity of water by a height of 15 m.

The efficiency of the pump is 10%.

Which mass of water could be lifted 15 m?

A 0.15 kg                     B 0.68 kg                     C 1.5 kg                      D 6.8 kg

Reference: Past Exam Paper – June 2015 Paper 11 Q18

Solution:

Answer: B.

This question is on the conservation of energy. Energy is being transformed from one form to another.

The kinetic energy of the water is being converted into gravitational potential energy of the water.

However, since the efficiency of the pump is only 10%, only 10% of the kinetic energy is converted into GPE.

10 % of Kinetic energy of water = GPE of water lifted

0.10 × ½ × m × v² = m_l g h     where m_l is the mass of water lifted

0.10 × ½ × 500 × 2² = m_l × 9.81 × 15 

m_l = (0.10 × ½ × 500 × 2²) / (9.81 × 15) = 0.68 kg

The diagram shows a d.c. circuit. What is the resistance between the points P and Q due to the resistance network?

Question 10

The diagram shows a d.c. circuit.

What is the resistance between the points P and Q due to the resistance network?

A 0.47 Ω          B 2.1 Ω           C 3.0 Ω           D 21 Ω

Reference: Past Exam Paper – June 2011 Paper 12 Q38

Solution:

Answer: C.

We need to simplify the resistances.

Consider the two 6 Ω resistors in parallel. Let their combined resistance be R₁.

1 / R₁ = 1/6 + 1/6

R₁ = [1/6 + 1/6]^-1 = 3 Ω

This 3 Ω combined resistance is in series with the 3 Ω resistor in the lower branch.

Combined resistance in the lower branch = 3 + 3 = 6 Ω

Equivalent resistance between the points P and Q,

R_T = [1/6 + 1/(3+3)]^-1 = 3 Ω