Monday, April 13, 2015

Physics 9702 Doubts | Help Page 111

  • Physics 9702 Doubts | Help Page 111

Question 566: [Waves]
(a) Source of sound has frequency f. Sound of wavelength λ is produced by the source.
(i) State
1. what is meant by frequency of the source,
2. distance moved, in terms of λ, by a wavefront during n oscillations of the source.
(ii) Use answers in (i) to deduce an expression for speed v of the wave in terms of f and λ.

(b) Waveform of a sound wave produced on screen of a cathode-ray oscilloscope (c.r.o.) is shown in Fig.

Time-base setting of the c.r.o. is 2.0 ms cm–1.
(i) Determine frequency of the sound wave.
(ii) A second sound wave has same frequency as that calculated in (i). Amplitude of the two waves is the same but phase difference between them is 90°.

Reference: Past Exam Paper – June 2010 Paper 23 Q5

Solution 566:
1. The frequency of the source is the number of oscillations produced by the source per unit time

{For 1 oscillation, the distance moved by a wavefront is λ}
Distance moved during n oscillations = nλ

Speed v = distance / time = (nλ) / t
But n / t = frequency f            hence Speed v= fλ

If there are f oscillations per unit time, so (fλ) is the distance per unit time.
The distance per unit time is the speed v, so v = fλ

{1 period corresponds to 3 squares (3cm) on the screen. The time-base setting is such that 1cm represents 2ms.}
1.0 period is 3 × 2 = 6.0 ms
Frequency (= 1/ T) = 1 / (6 × 10–3) = 170 Hz

{1 complete wave (1 wavelength) corresponds to 360°. So, 90° would correspond to a quarter of a wavelength. So, the new wave should be displaced from the one shown by a distance equal to λ/4. Or, in terms of time, this is displaced by a time of T/4}

The wave should be drawn (with approx. same amplitude and) with correct phase difference

Question 567: [Matter > Density]
Diagram shows the atoms of a substance with atoms at the corners of a cube. Average separation of the atoms at a particular temperature is 15 nm.

When the temperature changes so that average separation becomes 17 nm, by which factor will the density of the substance change?
A 0.61                                     B 0.69                         C 0.78                         D 0.88

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q19

Solution 567:
Answer: B.
Density = mass / volume
The mass M is independent of changes in temperature,

Initial volume = 15 x 15 x 15 = 153 nm3
Initial density ρi = M / (153)

Final volume = 17 x 17 x 17 = 173 nm3
Final density ρf = M / (173)

Let the factor by which the density of the substance changes = y
ρf = y (ρi)
Factor y = ρf / ρi = [M / (173)] / [M / (153)] = 153 / 173 = 0.69

Question 568: [Nuclear Physics]
U++ is doubly-ionised uranium atom. Uranium atom has a nucleon number of 235 and a proton number of 92.
In a simple model of the atom, how many particles are in this ionised atom?
A 235                          B 325                          C 327                          D 329

Reference: Past Exam Paper – June 2014 Paper 11 Q39

Solution 568:
Answer: B.
The nucleus of the uranium atom contains 235 nucleons (this includes the number of neutrons and protons).

A neutral atom would have 92 electrons (equal to the number of proton so that the atom is neutral), but this atom is doubly-ionised (2 electrons have been removed) and positive, so it must have 90 electrons.

There are (235 nucleons + 90 electrons =) 325 particles in total.

Question 569: [Kinematics]
Small steel ball falls freely under gravity after being released from rest.
Which graph best represents variation of the height h of the ball with time t?

Reference: Past Exam Paper – June 2010 Paper 12 Q8

Solution 569:
Answer: B.
Since the ball is released from rest, its height initially has greatest value. [D is incorrect]

The gradient of the graph represents the velocity of the ball.
At rest, the velocity is zero, so the gradient should be zero at t = 0. [C is incorrect]

Due to gravity, the ball accelerates as it falls. So, its speed increases with time. This is represented by a change in gradient in the graph. [A is incorrect]


  1. How is the gradient zero in C?

    1. at t=0, gradient is NOT zero. that's why C is incorrect.

  2. I am actually so confused between C and B and i dont get it :(

    1. I already made a detailed explanation on gradient before.

      Try to combine the explanation here along with the notes on gradient at

      If you still do not understand, let me know. Also, tell me what actually you are confused about, then. But I believe you should understand after reading the notes on gradient.

    2. so in B, at time t=0, the gradient is infinite where it has to be Zero?

    3. So in C, at time t=0 the gradient is infinite, when it has to be zero as it is released from rest.

    4. no, in B, a tangent drawn at t=0 will be a horizontal line. a horizontal line has a gradient = 0, not infinite

  3. I cant get your explanation of option A in solition 569...
    If speed is increasing with time then why B is answer its gradient also decreasing

    1. B is not decreasing. It is negative but increasing.

      For more details about gradient, I recommend


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