• Physics 9702 Doubts | Help Page 132

Question 660: [Waves]

(a)

(i) Define, for a wave,

1. wavelength λ

2. frequency, f

(ii) Use definitions to deduce relationship between λ, f and the speed v of the wave.

(b) Plane waves on the surface of water are represented by Fig.1 at one particular instant of time.

Waves have frequency 2.5 Hz.

Determine, for waves,

(i) amplitude

(ii) speed

(iii) phase difference between points A and B

(c) Wave in (b) was produced in a ripple tank. Describe briefly, with aid of a sketch diagram, how the wave may be observed.

Reference: Past Exam Paper – November 2013 Paper 23 Q5

Solution 660:

(a)

(i)

1. The wavelength of a wave is the minimum distance between 2 points moving in phase

OR distance between neighboring or consecutive peaks or troughs

OR wavelength is the distance moved by a wavefront in time T or one oscillation/cycle or period (of source)

2. The frequency of a wave is the number of wavefronts per unit time OR number of oscillations per unit time or oscillations/time

(ii)

Speed = distance / time = wavelength / time period   [M1]

            = λ / T = λf      [A0]

(b)

(i) Amplitude = 8.0 /2 = 4.0mm

(ii)

Frequency, f = 2.5 Hz

Speed = λ / T = λf

(From diagram given,) 3.75 λ = 18cm

Wavelength λ = 18 / 3.75 = 4.8cm

Speed = λf = 4.8 x 10^-2 x 2.5 = 12 x 10^-2 ms^-1

(iii) 180⁰ or π rad

(c)

Light and screen and correct positions above and below ripple tank

Strobe or video camera {placed in a position such that an image of the waves in the tank and the wave patterns on the viewing screen can be taken. So, the camera may be placed to the side of the tank, some distance away.}

Question 661: [Current of Electricity > Resistance]

(a) Two resistors, each of resistance R, are connected first in series and then in parallel.

Show that ratio of

combined resistance of resistors connected in series to combined resistance of resistors connected in parallel

is equal to 4.

(b) Variation with potential difference V of current I in a lamp shown in Fig.1. 

Calculate resistance of the lamp for a potential difference across the lamp of 1.5V

(c) Two lamps, each having the I-V characteristic shown in Fig.1, are connected first in series and then in parallel with battery of e.m.f. 3.0V and negligible internal resistance.

Complete table of Fig.2 for the lamps connected to battery

            pd across each lamp / V          resistance of each lamp / Ω     combined resistance

series              

parallel            

(d)

(i) Use data from completed Fig.2 to calculate ratio of

combined resistance of lamps connected in series to combined resistance of lamps connected in parallel

(ii) Ratios in (a) and (d)(i) are not equal.

By reference to Fig.1, state and explain qualitatively the change in resistance of a lamp as the potential difference is changed

Reference: Past Exam Paper – November 2009 Paper 22 Q6

Solution 661:

(a)

In series, total resistance = 2R

In parallel, total resistance = ([1/R + 1/R]^-1) = ½ R

Ratio = 2R / (½R) = 4

(b)

(from graph,) At p.d. = 1.5V, current = 0.10A

Resistance = V/I = 1.5 / 0.1 = 15Ω

(c)

            pd across each lamp / V          resistance of each lamp / Ω     combined resistance

series               1.5                                           15                                            30

parallel             3.0                                           20                                            10

(d)

(i) Ratio = (30/10) = 3

(ii)

The resistance increases as the potential difference increases.

{The gradient of the I-V graph is seen to decrease as the value of V increases. That is, as V increases, the current I increases but by a smaller amount than it would have increased at lower values of V. This means that the resistance (which restricts current) is increasing as V increases.}

The increasing potential difference increases the current.

{Even if the amount by which the current increases becomes smaller at higher values of V, the current is still increasing as V increases. This is shown by the gradient of the graph being positive.}

The current increases non-linearly, so the resistance increases.

Question 662: [Waves > Superposition]

(a) State principle of superposition.

(b) Coherent light of wavelength 590 nm is incident normally on a double slit, as shown in Fig.1.

Separation of the slits A and B is 1.4 mm.

Interference fringes are observed on screen placed parallel to the plane of the double slit.

Distance between the screen and the double slit is 2.6 m.

At point P on the screen, path difference is zero for light arriving at P from the slits A and B.

(i) Determine separation of bright fringes on the screen near to point P.

(ii) Variation with time of the displacement x of the light wave arriving at point P on the screen from slit A and from slit B is shown in Fig. 6.2a and Fig. 6.2b respectively.

1. State phase difference between waves forming the dark fringe on the screen that is next to point P.

2. Determine ratio

intensity of light at a bright fringe / intensity of light at a dark fringe

Reference: Past Exam Paper – November 2010 Paper 23 Q6

Solution 662:

(a) The principle of superposition states that when two (or more) waves meet (at a point), the (resultant) displacement is the (vector) sum of the individual displacements of the waves.

(b)

(i)

For double slits: λ = ax / D

590 × 10^–9 = (1.4 × 10^–3 × x) / 2.6

Separation of fringes, x = 1.1 mm

(ii).

1. Phase difference = 180° (allow π if rad stated)

{For a dark fringe, destructive interference must occur. (The first) Destructive interference occurs when the phase difference is 180°.}

2.

{For a bright fringe, we add the amplitudes (maximum displacement) of the wave from slit A and B. For a dark fringe, we subtract their amplitudes.}

At the maximum (bright fringe), the amplitude is (2 + 1.4 =) 3.4 units and at the minimum, the amplitude is (2.0 – 1.4 =) 0.6 units

{Intensity is proportional to the square of the amplitude.}

Intensity ~ amplitude² allow I ~ a²

Ratio = 3.4² / 0.6² = 32

Question 663: [Current of Electricity]

Circuit used to measure the power transfer from a battery is shown in Fig.1. Power is transferred to a variable resistor of resistance R.

Battery has an electromotive force (e.m.f.) E and an internal resistance r. There is a potential difference (p.d.) V across R. Current in the circuit is І.

(a) By reference to the circuit shown in Fig.1, distinguish between the definitions of e.m.f. and p.d.

(b) Using Kirchhoff’s second law, determine an expression for current І in the circuit.

(c) Variation with current І of the p.d. V across R is shown in Fig.2.

Use Fig.2 to determine

(i) e.m.f. E,

(ii) internal resistance r.

(d)

(i) Using data from Fig.2, calculate power transferred to R for a current of 1.6 A.

(ii) Use answers from (c)(i) and (d)(i) to calculate the efficiency of the battery for a current of 1.6 A.

Reference: Past Exam Paper – November 2012 Paper 23 Q4

Solution 663:

(a)

e.m.f. is the amount of chemical energy converted to electrical energy per unit charge (by the battery).

p.d. is the amount of electrical energy converted to thermal energy per unit charge (by the resistor).

(b)

{Total resistance in circuit = R + r}

E = I (R +r) or I = E / (R +r) (any subject)                         

(c)

(i) e.m.f. E = 5.8 V

(ii)

{The resistance R can be obtained by R = V / I by choosing a point on the graph. Consider point (1.0, 4.0).

Resistance R = V / I = 4.0 / 1.0 = 4Ω}

{E = I (R + r) = IR + Ir

When I = 1.0A, V = IR = 4.0V.

5.8 = 4 + (1.0×r)}

e.g. 5.8 = 4 + (1.0 × r)

Internal resistance r = 1.8 Ω

(d)

(i) Power P = VI = 2.9 × 1.6 = 4.6 (4.64) W

(ii)

Power from battery (= IE) = 1.6 × 5.8 = 9.28 W

{P = VI = 4.64W as calculated in part (i)}

Efficiency = (4.64 / 9.28) × 100 = 50 %

OR

Efficiency = VI / EI (= V / E)

{From graph, when I = 1.6A, V = 2.9V}

Efficiency = (2.9 / 5.8) × 100 = 50%