# Physics 9702 Doubts | Help Page 171

__Question 855: [Waves > Stationary waves]__
A stationary wave is set up a
stretched string, as shown.

Which statement about points on the
string is correct?

A Point Q vibrates with the largest
amplitude.

B Points P and R vibrate in phase.

C Point S is an antinode.

D The horizontal distance between R
and S is half the wavelength.

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q29*

__Solution 855:__**Answer: B.**

Points P and R (they both move up) vibrate
in phase while (P and Q) or (Q and R) are out of phase.

Point R vibrates with the largest
amplitude (R is an antinode).

Point S is a node (zero amplitude).

The horizontal distance between R
and S is a quarter of the wavelength.

__Question 856: [Current of Electricity]__
Diagram shows a fixed resistor and a
light-dependent resistor (LDR) in series with a constant low-voltage supply.

When LDR is in the dark, the fixed
resistor and the LDR have the same value of resistance.

Light is shone on the LDR.

What happens to the potential
differences across the two components?

p.d.
across resistor p.d. across LDR

A decreased
increased

B increased
decreased

C no
change increased

D no
change decreased

**Reference:**

*Past Exam Paper – June 2011 Paper 12 Q37*

__Solution 856:__**Answer: B.**

The resistance of the LDR decreases
as the light intensity on it increases. Thus, in the dark, the resistance of
the LDR was greater.

Ohm’s law: V = IR

When light is shone on it, the resistance
of the LDR decreases. So, the p.d. across the LDR decreases.

From Kirchhoff’s law, the sum of
p.d.’s in a circuit is equal to the e.m.f. in the circuit. Thus, the p.d. across
the fixed resistor increases such that the sum of p.d.’s is still equal to the e.m.f.

__Question 857: [Radioactivity]__
The isotope

^{222}_{86}Rn decays in a sequence of emissions to form isotope^{206}_{82}Pb. At each stage of decay sequence, it emits either an Î±-particle or a Î²-particle.
What is the number of stages in the
decay sequence?

A 4 B
8 C 16 D 20

**Reference:**

*Past Exam Paper – November 2014 Paper 11 & 12 Q39*

__Solution 857:__**Answer: B.**

An Î±-particle is

^{4}_{2}Î± (mass number = 4 and atomic number = 2) and a Î²-particle is^{0}_{-1}Î² (mass number = 0 and atomic number = -1).
Difference in mass number A between

^{222}_{86}Rn and^{206}_{82}Pb = 222 – 206 = 16
Since a Î²-particle does not
contribute anything to the mass number, this difference in mass number is
solely due to the Î±-particles.

Number of Î±-particles = 16 / 4 = 4

Number of stages liberating an
Î±-particle = 4

Increase in atomic number due to Î±-particles
= 4 × 2 = 8

Let the number of Î²-particles be x.

Consider the initial and final atomic
numbers, which should be equal.

86 = 82 +8 + x(-1)

x = -4 / -1 = 4

Thus, 4 Î²-particles are liberated,
each one in a stage. There are 4 stages here too.

Total number of stages = 4 + 4 = 8

__Question 858: [Waves > Diffraction]__
Diagram 1 shows ripple tank
experiment in which plane waves are diffracted through a narrow slit in a metal
sheet.

Diagram 2 shows same tank with a
slit of greater width.

In each case, the pattern of the
waves incident on slit and the emergent pattern are shown.

Which action would cause the waves
in diagram 1 to be diffracted less and so produce an emergent pattern closer to
that shown in diagram 2?

A increasing the frequency of
vibration of the bar

B increasing the speed of the waves
by making the water in the tank deeper

C reducing the amplitude of
vibration of the bar

D reducing the length of the vibrating
bar

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q24 & Paper 12 Q23 & Paper 13 Q25*

__Solution 858:__**Answer: A.**

The amount
of diffraction is less as the curvature of the diffracted wavefront decreases
(diffraction is more in diagram 1).

During the
diffraction of plane wavefronts through an opening, neither the frequency nor
wavelength (given by the distance between the wavefronts) changes. The
frequency only depends on the source of the vibrating bar (e.g. a motor).

The amount
of diffraction (the sharpness of the bending) depends on the wavelength. Diffraction
increases with increasing wavelength and vice versa.

The closer
the obstacle/opening/slit is to the wave's wavelength, the greater the amount
of diffraction. When the slit is equal to the wavelength, maximum diffraction
is seen and the waves spread out greatly, with the wavefronts being almost
semicircular.

If the
slit is larger than the wavelength, diffraction is less. The water wave passes
through the gap and does not spread out much on the other side. See diagram 2.
Actually, when the wavelength is smaller than the slit, diffraction may not be noticeable.

We want
the waves in diagram 1 (with a smaller slit) to be diffracted less. So, the
difference between the slit and the wavelength should be large, resulting in
the diffraction to be less. Since the slit is itself small, we should be
producing waves of even smaller wavelength.

Speed v =
fÎ»

By
increasing the frequency f, the wavelength is decreased (while the speed is
constant), causing the relative size of the opening/slit to be larger which
results in less diffraction. [A is correct]

As the
depth of water in the tank is increased, the wavelength and thus the speed
increases. A wavelength smaller than the slit causes less diffraction while a
wavelength larger than the slit would result in relatively more diffraction. [B is incorrect]

The amplitude is the maximum
displacement of the water molecule from its equilibrium position. The amplitude
gives us an idea of the energy of the wave, but does not affect the wavelength
and diffraction. [C is incorrect]

Reducing the length of the vibrating
bar does not affect the experiment since it is the wave (in the water) that
moves. [D is incorrect]

Hi

ReplyDeleteFor question 858

Why B is incorrect? The speed of the water wave when the depth of the tank increases. So therefore wavelength increases as frequency constant. This means more diffraction. So therefore B is not correct. Is my understanding correct?

I'm confused now. In my O Levels, we learn that the speed of wave increases when the depth of the water tank increases.

DeleteSorry what I mean is the speed of wave increases as the depth of the water in the water tank increases.

DeleteSorry, you are right. There may have been some mistakes in the explanation. It is now been corrected. Read through it again, it should be clear now.

DeleteI understand the explanation now. Thanks admin :)

Delete