Which statement about electromagnetic radiation is correct?

 Question 26

Which statement about electromagnetic radiation is correct?

A Waves of wavelength 5 × 10^-9 m are high-energy gamma rays.

B Waves of wavelength 3 × 10^-8 m are ultra-violet waves.

C Waves of wavelength 5 × 10^-7 m are infra-red waves.

D Waves of wavelength 9 × 10^-7 m are light waves.

 

 

 

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q22

 

 

 

Solution:

Answer: B.

Estimates for the different components of the electromagnetic spectrum should be known at A-Level.

 

All EM waves travel at the same speed of 3.0×10⁸ m s^-1.

Gamma rays have the highest frequencies, followed by X-rays and UV.

 

v = fλ

 

For the given wavelength of UV waves,

f = v / λ = 3.0×10⁸ / 3.0×10^-8 = 1×10¹⁶ Hz

 

This is a correct estimate for the frequency of UV waves.

A man standing next to a stationary train hears sound of frequency 400 Hz emitted from the train’s horn.

Question 25

A man standing next to a stationary train hears sound of frequency 400 Hz emitted from the train’s horn. The train then moves directly away from the man and sounds its horn when it has a speed of 50 m s^-1. The speed of sound is 340 m s^-1.

What is the difference in frequency of the sound heard by the man on the two occasions?

A 51 Hz                       B 69 Hz                       C 349 Hz                     D 469 Hz

Reference: Past Exam Paper – November 2016 Paper 12 Q25

Solution:

Answer: A.

The formula 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 (train)

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

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

When stationary, the sound from the train has a frequency of 400 Hz. This is the frequency of the source, f_s (= 400 Hz).

Speed of motion of the train, v_s = 50 m s^-1

Speed of sound, v = 340 m s^-1

For a receding source (going away),

f_o = f_s v / (v + v_s)

f_o = 400 × 340 / (340 + 50)

f_o = 348.7 = 349 Hz

Difference in frequency = 400 – 349 = 51 Hz

When sound travels through air, the air particles vibrate. A graph of displacement against time for a single air particle is shown.

Question 24

When sound travels through air, the air particles vibrate. A graph of displacement against time for a single air particle is shown.

Which graph best shows how the kinetic energy of the air particle varies with time?

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q23

Solution:

Answer: D.

Kinetic energy = ½ mv²

A graph of displacement against time for a single air particle is shown. 

The gradient of the displacement-time graph gives the velocity of the air particle at that point in time. This is done by calculating the gradient of the tangent at that point.

The gradient (and hence, velocity) is found to be zero at the maximum displacement (the tangent is horizontal) and maximum when the displacement is zero (the tangent is steepest).

Note that the velocity is NOT zero when the displacement is zero. At the points where displacement = 0, we need to draw a tangent to obtain the velocity. It can clearly be observed then that the line is steepest at these points – that is, the velocity is greatest when displacement = 0.

Thus, at time = 0, T and 2T the velocity is zero and hence kinetic energy is zero. [A and C incorrect]

But, between time = 0 and T or between time = T and 2T the displacement is zero (in case) twice. So, the velocity (and kinetic energy) reaches its maximum value 2 times in each of the 2 intervals. [B is incorrect]