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.
(c)
(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
(b)
{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
× Î”m
Mass defect, Δm = 5.03 × 10-29
kg
(c)
(i) Fusion
(ii)
{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.
(b)
(i)
Angular speed, ω [= 2Ï€ / T] = 2Ï€ / (27.3×24×3600) [= 2Ï€ / (2.36×106)]
Angular
speed, ω = 2.66×10-6rads-1
(ii)
(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
(c)
(i)
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
(ii)
EITHER
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
OR
{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.
(b)
mv2/r
= Bqv
hence, q/m = v
/ Br
(c)
(i)
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
(ii)
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:
(a)
(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.
(ii)
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)
(c)
(i) V
= (0.76x10-9) / [4Ï€ (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 - http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-44.html}
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.}
11/M/J/11 Q.9,10,23,24
ReplyDeleteFor 11/M/J/11 Q.9, check solution 715 at
ReplyDeletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-145.html
Q710)c)ii) i dont understand why we subtract Q from the the total binding energy on the product side?
ReplyDeletethe energy Q is used for other purposes (such as to heat water to produce electricity).
Deleteit is NOT used to keep the nucleus together. so, it should be excluded from our calculations