# 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:__**(a)**

(i)

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

2.

{For 1 oscillation, the
distance moved by a wavefront is λ}

Distance moved during n oscillations
= nλ

(ii)

EITHER

Speed v = distance / time = (nλ) / t

But n / t = frequency f hence Speed v= fλ

OR

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

**(b)**

(i)

{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
(ii)

{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 = 15

^{3}nm

^{3}

Initial density ρ

_{i}= M / (15

^{3})

Final volume = 17 x 17 x 17 = 17

^{3}nm

^{3}

Final density ρ

_{f}= M / (17

^{3})

Let the factor by which the density of the substance changes = y

ρ

_{f}= y (ρ

_{i})

Factor y = ρ

_{f}/ ρ

_{i}= [M / (17

^{3})] / [M / (15

^{3})] = 15

^{3}/ 17

^{3}= 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]

How is the gradient zero in C?

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

DeleteI am actually so confused between C and B and i dont get it :(

ReplyDeletewhich question?

Delete569 please

DeleteI already made a detailed explanation on gradient before.

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

http://physics-ref.blogspot.com/2015/03/physics-graphs-basics-2-nature-of.html

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.

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

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

Deleteno, in B, a tangent drawn at t=0 will be a horizontal line. a horizontal line has a gradient = 0, not infinite

DeleteI cant get your explanation of option A in solition 569...

ReplyDeleteIf speed is increasing with time then why B is answer its gradient also decreasing

B is not decreasing. It is negative but increasing.

DeleteFor more details about gradient, I recommend

http://physics-ref.blogspot.com/2015/03/physics-graphs-basics-2-nature-of.html