• Physics 9702 Doubts | Help Page 221

Question 1052: [Electric field]

There is a potential difference between a pair of parallel plates.

Which values of potential difference and separation of the plates will produce an electric field strength of the greatest value?

potential difference                 separation

A                     2V                               2d

B                     2V                               d / 2

C                     V / 2                            2d

D                     V / 2                            d / 2

Reference: Past Exam Paper – June 2009 Paper 1 Q28

Solution 1052:

Answer: B.

Electric field strength, E = potential difference / separation

Choice A: E = 2V / 2d = V/d

Choice B: E = 2V / (d/2) = 4 (V/d)                 [greatest value]

Choice C: E = (V/2) / (2d) = ¼ (V/d)

Choice D: E = (V/2) / (d/2) = V/d

Question 1053: [Waves > Interference]

Two light sources produce visible interference fringes only in certain circumstances.

Which condition enables visible interference fringes to be formed?

A using a white light source

B using incoherent sources

C using one light source which is polarised at right angles to light from the other source

D using sources from which the light does not overlap

Reference: Past Exam Paper – November 2011 Paper 12 Q28

Solution 1053:

Answer: A.

White light will produce fringes since it consists of different colors at different frequencies.

Sources should be coherent (have constant phase difference and same frequency) for interference. In the same way, the lights should overlap, causing constructive and destructive interference. It is in this way that fringes are formed.

Using one light source which is polarised at right angles to light from the other source will not cause interference. They should be in the same plane.

Question 1054: [Turning effects of forces]

A uniform metre rule is pivoted at the 34.0 cm mark, as shown.

The rule balances when a 64 g mass is hung from the 4.0 cm mark.

What is the mass of the metre rule?

A 38 g                                     B 44 g                         C 120 g                                   D 136 g

Reference: Past Exam Paper – June 2015 Paper 11 Q15

Solution 1054:

Answer: C.

A uniform metre rule would have mass (M) at the centre, that is, the 50.0cm mark.

Moment = Force × perpendicular distance from line of action to pivot

In this case, the weights involved would cause the moments. The distances should be taken form the pivot, which is at the 34.0cm mark.

For equilibrium, clockwise moment = anticlockwise moment

(M×g) × (50.0 – 34.0) = (64×g) × (34.0 – 4.0)

16M = 64 × 30

Mass M = 120g

Question 1055: [Waves > Intensity]

The intensity of a progressive wave is proportional to the square of the amplitude of the wave. It is also proportional to the square of the frequency.

The variation with time t of displacement x of particles in a medium, when two progressive waves P and Q pass separately through the medium, are shown on the graphs.

The intensity of wave P is I₀.

What is the intensity of wave Q?

A ½ I₀                         B I₀                              C 8 I₀                           D 16 I₀

Reference: Past Exam Paper – November 2005 Paper 1 Q24

Solution 1055:

Answer: B.

Intensity I ∝ (Amplitude A)²

Intensity I ∝ (Frequency f)²

Intensity I = kA²f²      where k is a constant

Intensity of wave P = I₀

Amplitude A_P of wave P = x₀

Frequency f_P of wave P = 1 / T = 1 / t₀

Intensity of wave P = I₀ = kA_P²f_P²

Let the intensity of wave Q = I_Q

Amplitude of wave Q = 2x₀ = 2A_P

Frequency of wave Q = 1 / T = 1 / 2t₀ = 0.5f_P

Intensity I_Q = k (2A_P)² (0.5f_P)² = k(4A_P²) (0.25f_P²) = kA_P²f_P² = I₀