# Physics 9702 Doubts | Help Page 114

__Question 578: [Wave-Particle Duality]__
Electrons,
travelling at speed v in vacuum, are incident on a very thin carbon film, as illustrated
in Fig.1.

Emergent
electrons are incident on a fluorescent screen.

A
series of concentric rings is observed on screen.

**(a)**Suggest why the observed rings provide evidence for wave nature of particles.

**(b)**Initial speed of the electrons is increased. State and explain effect, if any, on radii of the rings observed on the screen.

**(c)**A proton and an electron are each accelerated from rest through same potential difference.

Determine
the ratio

**of**de Broglie wavelength of the proton**to**de Broglie wavelength of the electron**Reference:**

*Past Exam Paper – November 2013 Paper 41 & 42 Q7*

__Solution 578:__**(a)**

*Either*

If
light passes through suitable film/cork dust, etc, diffraction
occurs and a similar pattern is observed.

*Or*

Concentric
circles are evidence of diffraction, which is a wave property.

**(b)**

An
increase in speed also causes the momentum to increase. Since λ = h/p, λ decreases.
Hence the radii decrease.

*Or*

As
speed increases, energy also increases. λ = h / (2Em)

^{½}. Hence, the radii of the rings decrease.**(c)**

Since
the potential difference is the same, the electron and the proton will have the
same (kinetic) energy.

{Kinetic energy E = ½ mv

^{2}. Momentum p = mv. Speed v = p/m. Replacing v = p/m in the equation for kinetic energy,
E = ½ m (p/m)

^{2}= p^{2}/ 2m [this equation can be used directly without derivation]
From the above equation, p = (2Em)

^{½}}
Either
E = p

^{2}/2m or p = (2Em)^{½}
Since
λ is inversely proportional to momentum, p

Ratio
= λ

_{p}/ λ_{e}= p_{e}/p_{p}= (m_{e}/m_{p})^{½}= {(9.1x10^{-31})/(1.67x10^{-27})}^{½}= 2.3x10^{-2}

__Question 579: [Electric field + Vectors]__
An
electron enters the space between two parallel charged plates with initial
velocity u.

While
in the electric field, its direction changes by θ and it emerges with velocity
v.

What
is the relation between v and u?

A
v = u / cos θ B v = u cos θ C v = u / sin θ D v = u sin θ

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q30*

__Solution 579:__**Answer: A.**

The
initial velocity u is not affected by the electric field since it is
horizontal. Therefore, upon emerging from the parallel plates, the horizontal
component of the speed of the electron is still u.

Thus,
we have a right-angled triangle with the hypotenuse v and the adjacent side to
the angle θ being u.

cos
θ = u / v giving v = u / cos θ

__Question 580: [Ideal Gas]__**(a)**Volume of an ideal gas in a cylinder is 1.80 × 10

^{–3}m

^{3}at a pressure of 2.60 × 10

^{5}Pa and temperature of 297 K, as illustrated in Fig.1.

Thermal energy required to raise
temperature by 1.00 K of 1.00 mol of the gas at constant volume is 12.5 J.

Gas is heated at constant volume
such that internal energy of the gas increases by 95.0 J.

(i) Calculate

1. amount of gas, in mol, in
cylinder

2. rise in temperature of the gas:

(ii) Use answer in (i) part 2 to
show that final pressure of the gas in the cylinder is 2.95 × 10

^{5}Pa.**(b)**Gas is now allowed to expand. No thermal energy enters or leaves the gas. Gas does 120J of work when expanding against the external pressure.

State and explain whether final
temperature of the gas is above or below 297K.

**Reference:**

*Past Exam Paper – June 2013 Paper 42 Q2*

__Solution 580:__
(i)

1.

pV = nRT

2.60x10

^{5}x1.80x10^{-3}= n x 8.31 x 297
Number of moles, n = 0.19mol

2.

Δq = ncΔT

95.0 = 0.190 x 12.5 x ΔT

ΔT = 40K

(ii)

p/T = constant

(2.6x10

^{5}) / 297 = p / (297+40)
Pressure p = 2.95x10

^{5}Pa**(b)**

{No thermal energy enters
or leaves the gas (ΔQ = 0). Gas does 120J of work when expanding against the
external pressure (ΔW = +120J).

Principle of conservation
of energy:

ΔQ = ΔU + ΔW

ΔQ is the amount of
heat/energy (+ve when heat is given TO the system and –ve when heat is taken
FROM system)

ΔU is the change in
internal energy (+ve when internal energy of system increases and –ve when
internal energy of system decreases. A change in internal energy is shown by a
change in temperature) – this is the energy of the gas itself

ΔW is the work done (+ve
when external work is done BY system and –ve when external work is done ON
system. If the system expands, it does external work against external forces)}

Replacing the energy values,

ΔU = – ΔW = –120J.

As ΔU is negative, the internal
energy decreases. But internal energy is the sum of KE and PE of the gas
molecules, the so, KE of the molecules decreases. Thus, the temperature lowers.

{Or if we consider that
the initially, the gas was heated at constant volume such that the internal
energy of the gas increases by 95.0 J. Now, it is given that the gas does 120J
of work when expanding against the external pressure. Some may take to energy
of the gas that allows it only to expand to be 120 – 95 = 25J. But, the above method
is better sine we are dealing with different part of the question.}

__Question 581: [Work, Energy and Power]__

Trains supply coal to a power
station. Table below gives quantities describing the operation of the power
station.

symbol unit

power station output P
W

number of trains per day N

mass of coal on a train M
kg

energy from 1 kg of coal J
J

number of seconds in one day S

Which expression gives efficiency of
the power station?

A PS / NMJ B PSN / MJ C
NMJ / PS D NM / PSJ

**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q22*

__Solution 581:__**Answer: A.**

Firstly, efficiency is a ratio of 2
similar appropriate quantities. So, the numerator and the denominator must have
same units.

Here, the efficiency can be given in
terms of the energy.

Efficiency = Output energy per day /
Total energy available per day [Only A is correct]

Power = Energy / Time.

Energy = Power × Time

Output energy = Power station output
×
Number of seconds in one day

Output energy = PS

Total energy available per day =
Number of trains per day × Mass of coal on a train × Energy from 1kg of coal

Total energy available per day = NMJ

Efficiency = PS / NMJ

__Question 582: [Simple harmonic motion]__
Needle of a sewing machine is made
to oscillate vertically through a total distance of 22mm, as shown in Fig.1.

Oscillations are simple harmonic
with a frequency of 4.5Hz.

The cloth that is being sewn is positioned
8.0mm below point of the needle when the needle is at its maximum height.

**(a)**State what is meant by simple harmonic motion

**(b)**Displacement y of the point of the needle represented by the equation

y = a cos(ωt)

(i) Suggest position of the point of
the needle at time t = 0

(ii) Determine the values of

1. a

2. ω

**(c)**Calculate, for the point of the needle,

(i) maximum speed

(ii) speed as it moves downwards
through the cloth

**Reference:**

*Past Exam Paper – November 2008 Paper 4 Q3*

__Solution 582:__**(a)**For simple harmonic motion, the acceleration / force is (directly) proportional to the displacement and EITHER is directed towards the fixed point OR the acceleration and displacement are in opposite directions.

**(b)**

(i) Maximum / minimum height / 8mm
above the cloth / 14mm below cloth

(ii)

1. {a =
amplitude = total distance between maximum and minimum heights / 2}

a = 11mm

2. ω
= 2πf = 2π(4.5) = 28.3rads

^{-1}**(c)**

(i) Maximum speed, v = ωa = 28.3 x 11.x10

^{-3}= 0.31ms^{-1}
(ii)

{For simple harmonic
motion, the speed at a displacement y from the equilibrium position is given by
Speed, v = ω √(a

^{2}– y^{2})}
Speed, v = ω √(a

^{2}– y^{2})
{The figure shows the
point of the needle when it is at its maximum height – it is 8mm above the
cloth. The amplitude of the oscillation is 11mm. We need the displacement of
the needle when the points of the needle is at the same level as the cloth.

For the needle to reach
this position, it needs to travel 8mm. But remember, the needle was previously
at its maximum height – that is, at a displacement of 11mm from its equilibrium
position. It has now moved down by 8mm. So, its displacement from the
equilibrium position is now 11 – 8 = 3mm}

Displacement, y (= 11 – 8) = 3mm

Speed v = 28.3 x 10

^{-3}√(11^{2}– 3^{2}) = 0.30ms^{-1}

__Question 583: [Pressure]__
An object, immersed in liquid in a
tank, experiences an upthrust.

What is physical reason for this
upthrust?

A The density of the body differs
from that of the liquid.

B The density of the liquid
increases with depth.

C The pressure in the liquid
increases with depth.

D The value of g in the liquid
increases with depth.

**Reference:**

*Past Exam Paper – June 2010 Paper 12 Q13*

__Solution 583:__**Answer: C.**

An object, immersed in a liquid in a
tank, experiences an upthrust.

Choices B and D are obviously
incorrect. The density ρ of the liquid is constant. Similarly, the value of g is a constant
at a specific location.

The physical reason for this
upthrust is actually due to the difference in pressures at the top and at the
bottom of the object.

Pressure P = hρg

As stated above, ρ and g are constant in this situation.

As the depth h increases, the
pressure increases. So, the pressure at the bottom of the object (which is
deeper) is greater than the pressure at the top of the object (the depth is
less). This results in a resultant pressure upwards. This pressure acts on the
surface area of the bottom of the object. This is the upthrust.

Upthrust is a force. Pressure =
Force / Area

Can you please explain the solution for 582 b) ii)

ReplyDeleteWhy is the amplitude in this case equal to {a = amplitude = total distance between maximum and minimum heights / 2}?

Shouldn't it be the maximum amplitude?

We don’t know the height of the equilibrium position, but we know that at the maximum height, the point of the needle is at a distance equal to the amplitude from the equilibrium position. Similarly, at the minimum height, the point of the needle is at a distance equal to the amplitude from the equilibrium position. Thus, the distance between the maximum and minimum heights is twice the amplitude.

Deletecan you please explain with a diagram

DeleteIt would take too much time to draw a diagram in softcopy.

DeleteJust imagine a wave. The crest is the maximum displacement and the trough is the minimum displacement.

But the distance between the crest and the trough would be equal to twice the amplitude.

oh I got it now!

DeleteThank you so much!!

thank you!

Delete