# 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)
180

^{0}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 ~ amplitude

^{2}allow I ~ a^{2}
Ratio = 3.4

^{2}/ 0.6^{2}= 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%

Here are the remaining questions for now:

ReplyDelete23/O/N/11 Q.6(c)

23/M/J/12 Q.2(c)

21/O/N/12 Q.6(c)

22/O/N/12 Q.3(b)

21/M/J/13 Q.6(c)(d)

22/O/N/13 Q.3(c)(ii)

For 23/O/N/11 Q.6(c), check solution 665 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-133.html

I still don't get why the resistance of each lamp connected in parallel is 20 ohm.

ReplyDeleteshouldn't it be 15 ohm?

The resistance of the lamp is not constant, but depends on the p.d. across it (as given from the graph). So, we should identify from the graph the corresponding current flowing when the p.d. is 1.5 and/or 3.0. The resistance can then be calculated.

DeleteIn solution 663 c(i), how do we obtain the value of E

ReplyDeleteWhen no current flows in the circuit, the voltmeter can be considered to be connected in parallel to the battery. So, the voltmeter would give the e.m.f. of the battery.

Delete