• 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 mv². Momentum p = mv. Speed v = p/m. Replacing v = p/m in the equation for kinetic energy,

E = ½ m (p/m)² = p² / 2m        [this equation can be used directly without derivation]

From the above equation, p = (2Em)^½}

Either E = p²/2m         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 m³ at a pressure of 2.60 × 10⁵ 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 × 10⁵ 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:

Go to
The volume of an ideal gas in a cylinder is 1.80 × 10-3 m3 at a pressure of 2.60 × 105 Pa and a temperature of 297 K, as illustrated in Fig. 2.1.

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 √(a² – y²)}

Speed, v = ω √(a² – y²)

{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 √(11² – 3²) = 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