# Physics 9702 Doubts | Help Page 144

__Question 710: [Nuclear Physics]__
The controlled reaction between
deuterium (

^{2}_{1}H) and tritium (^{3}_{1}H) has involved ongoing research for many years. Reaction may be summarised as^{2}

_{1}H +

^{3}

_{1}H - - - >

^{4}

_{2}He +

^{1}

_{0}n + Q

where Q = 17.7 MeV.

Binding energies per nucleon are
shown in Fig.1.

binding energy per nucleon / MeV

^{2}

_{1}H 1.12

^{1}

_{0}n -

^{4}

_{2}He 7.07

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

**(b)**Calculate mass defect, in kg, of a helium

^{4}

_{2}He 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 (

^{3}_{1}H).**Reference:**

*Past Exam Paper – November 2009 Paper 41 Q8*

__Solution 710:__**(a)**The neutron is a single nucleon / particle

**(b)**

{A

^{4}_{2}He 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 = c

^{2}Î”m
4.52 × 10

^{-12}= (3.0 × 10^{8})^{2}× Î”m
Mass defect, Î”m = 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 × 10

^{5}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 × 10

^{24}kg.**(c)**Mass of the Moon is 7.4 × 10

^{22}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, Ï‰ [= 2Ï€ / T] = 2Ï€ / (27.3×24×3600) [= 2Ï€ / (2.36×10

^{6})]
Angular
speed, Ï‰ = 2.66×10

^{-6}rads^{-1}
(ii)

(Gravitational force
provides centripetal force)

GMm/r

^{2}= mÏ‰^{2}r or GMm/r^{2}= mv^{2}/r
GM = Ï‰

^{2}r^{3}or GM = v^{2}r
M = (3.84×10

^{5}×10^{3})^{3}× (2.66×10^{-6})^{2}/ (6.67×10^{-11})
Mass of Earth, M = 6.0×10

^{24}kg**(c)**

(i)

Gravitational force, F = GMm / r

^{2}= (6.67×10^{-11}) (6.0×10^{24}) (7.4×10^{22}) / (3.84×10^{8})^{2}
Gravitational force, F = 2.0×10

^{20}N
(ii)

EITHER

Î”Ep = F Î”x 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.}

Î”Ep = 2.0×10

^{20}× 4.0×10^{-2}= 8.0×10^{18}J
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.}

Î”Ep = GMm/r

_{1}– GMm/r_{2}
r

_{1}= 3.84×10^{8}r_{2}= (3.84×10^{8}) + 4.0×10^{-2}
Correct substitution

1/r

_{1}– 1/r_{2}= 2.7127×10^{-19}
Î”Ep = GMm (1/r

_{1}– 1/r_{2}) = (6.67×10^{-11}) (6.0×10^{24}) (7.4×10^{22}) × 2.7127×10^{-19}
Î”Ep = 8.0×10

^{18}J

__Question 712: [Current of Electricity]__
Battery of e.m.f. 12 V and internal
resistance 2.0 Î© 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 / Î©

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 × 10

^{7}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)**

mv

^{2}/r = Bqv
hence, q/m = v
/ Br

**(c)**

(i)

v = 2.0x10

^{7}ms^{-1}
B = 2.5x10

^{-3}T
R = 4.5x10

^{-2}m
q/m = v /Br =
(2.0x10

^{7}) / (2.5x10^{-3}x 4.5x10^{-2}) = 1.8x10^{11}Ckg^{-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__(on entering and on leaving the field).**curved path**

__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 / 4Ï€Ïµ

_{o}r
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

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