Sunday, May 17, 2015

Physics 9702 Doubts | Help Page 144

  • Physics 9702 Doubts | Help Page 144

Question 710: [Nuclear Physics]
The controlled reaction between deuterium (21H) and tritium (31H) has involved ongoing research for many years. Reaction may be summarised as
21H       +          31H      - - - >   42He    +          10n       +          Q
where Q = 17.7 MeV.
Binding energies per nucleon are shown in Fig.1.
binding energy per nucleon / MeV
21H                               1.12
10n                                -
42He                             7.07

(a) Suggest why binding energy per nucleon for neutron is not quoted.

(b) Calculate mass defect, in kg, of a helium 42He nucleus.

(i) State name of the type of reaction illustrated by this nuclear equation.
(ii) Determine binding energy per nucleon, in MeV, of tritium (31H).

Reference: Past Exam Paper – November 2009 Paper 41 Q8

Solution 710:
(a) The neutron is a single nucleon / particle

{A 42He nucleus contains 4 nucleons. So, we need 4 times the binding energy per nucleon. To convert the energy from MeV to J, we multiply by 1.6 × 10-13.}
Binding energy = 4 × 7.07 × 1.6 × 10-13 = 4.52 × 10-12 J
Binding energy = c2 Δm
4.52 × 10-12 = (3.0 × 108)2 × Δm
Mass defect, Δm = 5.03 × 10-29 kg

(i) Fusion

{We need the binding energy per nucleon. Tritium contains 3 nucleons and deuterium contains 2 nucleons. Values given in table are for BE per nucleon, thus we need to multiply by the corresponding number of nucleons present in the element
Left hand side of equation, total binding energy = 2(1.12) + 3x
where x is binding energy per nucleon for tritium.

Helium contains 4 nucleons. Q is the energy released in the reaction, so it should be subtracted. Binding energy is the energy binding (keeping together) the nucleons (protons and neutrons) in the nucleus. So, binding energies should be added, while the energy released Q (which is not the energy keeping the nucleons together in the nucleus) should be reduced from the total binding energy.
From the law of conservation of energy, the overall energy should be the same on both sides of the equation. But here, we are dealing with binding energies.
Right hand side of equation, total binding energy = 4(7.07) – 17.7}

(2 × 1.12) + 3x = 28.28 – 17.7
Solving gives
x = 2.78MeV per nucleon

Question 711: [Gravitation]
(a) State Newton’s law of gravitation.

(b) Earth and the Moon may be considered to be isolated in space with their masses concentrated at their centres.
Orbit of the Moon around the Earth is circular with a radius of 3.84 × 105 km. Period of the orbit is 27.3 days.
Show that
(i) angular speed of the Moon in its orbit around the Earth is 2.66 × 10–6 rad s–1,
(ii) mass of the Earth is 6.0 × 1024 kg.

(c) Mass of the Moon is 7.4 × 1022 kg.
(i) Using data from (b), determine gravitational force between the Earth and the Moon.
(ii) Tidal action on Earth’s surface causes the radius of the orbit of the Moon to increase by 4.0 cm each year.
Use answer in (i) to determine the change, in one year, of the gravitational potential energy of the Moon. Explain your working.

Reference: Past Exam Paper – June 2012 Paper 42 Q1

Solution 711:
(a) Newton’s law of gravitation states that the force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation, provided that the separation is much larger than the size of the masses.

Angular speed, ω [= 2π / T] = 2π / (27.3×24×3600) [= 2π / (2.36×106)]
Angular speed, ω = 2.66×10-6rads-1

(Gravitational force provides centripetal force)
GMm/r2 = mω2r           or GMm/r2 = mv2/r
GM = ω2r3       or GM = v2r    
M = (3.84×105 ×103)3 × (2.66×10-6)2 / (6.67×10-11)   
Mass of Earth, M = 6.0×1024kg

Gravitational force, F = GMm / r2 = (6.67×10-11) (6.0×1024) (7.4×1022) / (3.84×108)2
Gravitational force, F = 2.0×1020N

ΔEp = F Δx where F can be considered as being constant since the increase x (= 4.0cm) is very much smaller than distance r (radius of the orbit)                                       
{For large changes in the distance, the value of the gravitational field strength would change (the same case as distances far from the Earth). Thus F would change. But if x is very small compared to r, then the force F may be considered to be constant.}
ΔEp = 2.0×1020 × 4.0×10-2 = 8.0×1018J

{The following calculation require a higher degrees of accuracy, or an answer equal to zero would be obtained. It is advised to use the method above.}
ΔEp = GMm/r1 – GMm/r2      
r1 = 3.84×108               r2 = (3.84×108) + 4.0×10-2  
Correct substitution   
1/r1 – 1/r2 = 2.7127×10-19
ΔEp = GMm (1/r1 – 1/r2) = (6.67×10-11) (6.0×1024) (7.4×1022) × 2.7127×10-19
ΔEp = 8.0×1018J

Question 712: [Current of Electricity]
Battery of e.m.f. 12 V and internal resistance 2.0 Ω is connected in series with an ammeter of negligible resistance and an external resistor. External resistors of various different values are used.

Which combination of current and resistor value is not correct?
current / A       external resistor value / Ω
A         1.0                               10
B         1.2                               8
C         1.5                               6
D         1.8                               4

Reference: Past Exam Paper – November 2010 Paper 12 Q33

Solution 712:
Answer: D.
We need to identify which combination is NOT correct.
Total resistance in circuit = r + R = 2 + R

From Ohm’s law: V = I (r+R)
12 = I (2+R)
Current I = 12 / (2+R)

Choice A: when R = 10, I = 1A.
Choice B: when R = 8, I = 1.2A.
Choice C: when R = 6, I = 1.5A.
Choice D: when R = 4, I = 2A.

Question 713: [Electromagnetism]
Charged particle of mass m and charge –q is travelling through a vacuum at constant speed v.
It enters a uniform magnetic field of flux density B. Initial angle between the direction of motion of the particle and the direction of the magnetic field is 90°.
(a) Explain why path of the particle in the magnetic field is the arc of a circle.

(b) Radius of the arc in (a) is r.
Show that ratio q/m for the particle is given by the expression
q / m = v / Br.

(c) Initial speed v of the particle is 2.0 × 107 m s–1. Magnetic flux density B is 2.5 × 10–3 T.
Radius r of the arc in the magnetic field is 4.5 cm.
(i) Use these data to calculate ration q/m.
(ii) Path of the negatively-charged particle before it enters the magnetic field is shown in Fig.1.
Direction of the magnetic field is into the plane of the paper.
On Fig.1, sketch path of the particle in the magnetic field and as it emerges from the field.

Reference: Past Exam Paper – November 2013 Paper 41 & 42 Q6

Solution 713:
(a) The force due to the magnetic field on the particle is constant and is always normal to the direction of motion of the particle. This force provides the centripetal force, causing the path to be the arc of a circle.

mv2/r = Bqv
hence, q/m = v / Br

v = 2.0x107ms-1
B = 2.5x10-3T
R = 4.5x10-2m
q/m = v /Br = (2.0x107) / (2.5x10-3 x 4.5x10-2) = 1.8x1011Ckg-1

In the field, the sketch should be a curved path, of constant radius, in direction towards the bottom of the plane.
As it emerges, the path is tangent to the curved path (on entering and on leaving the field).

Question 714: [Electric field]
Small charged metal sphere is situated in an earthed metal box. Fig.1 illustrates the electric field between the sphere and the metal box.

(a) By reference to Fig.1, state and explain
(i) whether sphere is positively or negatively charged
(ii) why it appears as if charge on sphere is concentrated at centre of sphere

(b) On Fig.1, draw arrow to show direction of force on a stationary electron situated at point A

(c) Radius r of the sphere is 2.4 cm. Magnitude of the charge q on the sphere is 0.76 nC.
(i) Use expression
V = Q / 4πϵor
to calculate a value for magnitude of potential V at surface of the sphere

(ii) State sign of the charge induced on the inside of the metal box. Hence explain whether actual magnitude of potential will be greater or smaller than the value calculated in (i).

(d) A lead sphere is placed in a lead box in free space, in a similar arrangement to that shown in Fig.1. Explain why it is not possible for the gravitational field to have a similar shape to that of electric field.

Reference: Past Exam Paper – November 2007 Paper 4 Q4

Solution 714:
(i) EITHER The lines are directed away from the sphere OR Lines go from positive to negative OR The line shows the direction of the force on a positive charge.
So, the sphere is positively charged.  

EITHER All the lines (appear to) radiate from the centre
OR All the lines are normal to the surface of the sphere.

(b) The arrow is tangent to the curve (at A) in the correct position and direction (towards sphere)

(i) V = (0.76x10-9) / [4π (8.85x10-12) (0.024)] = 285V

(ii) {Detailed explanations for this part of the question is available as Solution 270 at Physics 9702 Doubts | Help Page 44 -}

A negative charge is induced on (the inside of) the box. The formula applies to isolated (point) charge OR Less work is done in moving the test charge from infinity. So the potential is lower.

(d) EITHER Gravitational field is always attractive OR The field lines must be directed towards both the box and the sphere.           
{Unlike electric force (like charges attract and unlike charges repel), gravitational force is always attractive, that is, it attracts a mass towards the object. A gravitational force never repels a mass.
So, the lead sphere would attract the lead box towards it (the sphere) and the lead box would attract the lead sphere towards it (the lead box). So, the direction of the field lines should be drawn both towards the box and the sphere.}


  1. For 11/M/J/11 Q.9, check solution 715 at


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