A cathode-ray oscilloscope (c.r.o.) is used to determine the frequency of a sound wave. The diagram shows the waveform on the screen.

Question 6

A cathode-ray oscilloscope (c.r.o.) is used to determine the frequency of a sound wave.

The diagram shows the waveform on the screen.

The time-base setting is 5.0 ms / div.

What is the frequency of the sound wave?

A 57 Hz                       B 71 Hz                       C 114 Hz                     D 143 Hz

Reference: Past Exam Paper – June 2018 Paper 13 Q22

Solution:

Answer: B.

The time-base setting is 5.0 ms / div.

That is, 1 div represents 5.0 mV.

To have a correct measurement, we need to find positions where the waves occupy a ‘whole’ number of divisions.

2.5 waves occupy 7 divisions.

2.5 T - - > 7 divisions

2.5 T = 7 × 5.0

T = 7 × 5.0 / 2.5

Period T = 14 ms = 14×10^-3 s

Frequency = 1 / T = 1 / (14×10^-3)

 
Frequency = 71 Hz

An astronaut on the Moon, where there is no air resistance, throws a ball. The ball’s initial velocity has a vertical component of 8.00 m s-1 and a horizontal component of 4.00 m s-1, as shown.

Question 19

An astronaut on the Moon, where there is no air resistance, throws a ball. The ball’s initial velocity has a vertical component of 8.00 m s^-1 and a horizontal component of 4.00 m s^-1, as shown.

The acceleration of free fall on the Moon is 1.62 m s^-2.

What will be the speed of the ball 9.00 s after being thrown?

A 6.6 m s^-1                        B 7.7 m s^-1                        C 10.6 m s^-1                     D 14.6 m s^-1

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q7

Solution:

Answer: B.

The weight of the ball would affect the vertical component of the velocity as it acts downwards, towards the surface of the Moon.

The absence of air resistance indicates that the horizontal component of velocity is NOT affected during the motion. This component remains constant even after 9 s.

Consider the vertical component.

Take the upward direction to be positive.

Initial velocity, u = + 8.00 m s^-1

Acceleration, a = – 1.62 m s^-2 (as it acts downwards)

Time t = 9.00 s

Final velocity = v

v = u + at = 8 + (–1.62 × 9)

v = - 6.58 m s^-1 (it is downwards)

So, after 9.00 s,

Horizontal component = 4.00 m s^-1

Vertical component = (-) 6.58 m s^-1

The speed v can be found by applying Pythagoras’ theorem.

v² = 4² + 6.58²

v = 7.7 m s^-1

A charged oil drop of mass m, with n excess electrons, is held stationary in the uniform electric field between two horizontal plates separated by a distance d.

Question 8

A charged oil drop of mass m, with n excess electrons, is held stationary in the uniform electric field between two horizontal plates separated by a distance d.

The voltage between the plates is V, the elementary charge is e and the acceleration of free fall is g.

What is the value of n ?

A eV / mgd                  B mgd / eV                  C meV / gd                  D gd / meV

Reference: Past Exam Paper – June 2015 Paper 12 Q30

Solution:

Answer: B.

The oil drop has an excess of electrons, so it is negatively charged.

Charge of electron = n × e     

where e is the charge of an electron and n is the excess electrons

The oil drop would be attracted to the positive plate, which is the upper one. So, the electric force on the electron is upwards.

The weight of the oil drop acts downwards.

Since the oil drop is stationary, the resultant force on it is zero.

(Upward) Electric force = (Downward) Weight

Electric field strength (E) × charge (n×e) = mg

E × n×e = mg                         

[but E = V / d]

(V/d) × n×e = mg

n = mgd / eV