Physics 9702 Doubts | Help Page 143
Question 705: [Quantum
Physics]
(a) State what is meant by de Broglie wavelength.
(b) An electron is accelerated from rest in a vacuum through potential
difference of 4.7 kV.
(i) Calculate de Broglie wavelength
of the accelerated electron.
(ii) By reference to answer in (i),
suggest why such electrons may assist with an understanding of crystal
structure.
Reference: Past Exam Paper – November 2012 Paper 43 Q7
Solution 705:
(a) The de Broglie wavelength is the wavelength associated with a
particle that is moving
(b)
(i)
Kinetic
energy {= VQ} = 1.6 × 10–19 × 4700 =
7.52 × 10–16 J
EITHER
energy = p2/2m OR EK
= ½ mv2 and p = mv
Momentum
p = √(7.52×10–16 × 2 × 9.1×10–31) = 3.7 × 10–23
N s
Wavelength
λ = h / p = (6.63 × 10–34) / (3.7 × 10–23) = 1.8 × 10–11
m
(ii) The wavelength is about the
separation of atoms, so they can be used in (electron) diffraction.
{The separation of atoms
can be identified by diffraction of the electrons.}
Question 706: [Current
of Electricity > Capacitance]
(a)
(i) Define capacitance.
(ii) A capacitor is made of two
metal plates, insulated from one another, as shown in Fig.1.
Explain why capacitor is said to
store energy but not charge.
(b) Three uncharged capacitors X, Y and Z, each of capacitance 12 μF,
are connected as shown in Fig.2.
Potential difference of 9.0 V is
applied between points A and B.
(i) Calculate combined capacitance
of the capacitors X, Y and Z.
(ii) Explain why, when potential
difference of 9.0 V is applied, the charge on one plate of capacitor X is 72
μC.
(iii) Determine
1. potential difference across
capacitor X
2. charge on one plate of capacitor
Y
Reference: Past Exam Paper – November 2012 Paper 41 & 42 Q5
Solution 706:
(a)
(i) Capacitance is the ratio of
charge and potential (difference) / voltage.
(ii) On each of the two plates, the capacitor has
equal magnitudes of (+)ve and (–)ve charge. The total charge on the capacitor
is therefore zero, and so, the capacitor does not store charge. Work needs to
be done to separate the (+)ve and (-)ve charges on the plates. So, the
capacitor stores energy to achieve this.
(b)
(i)
Capacitance of Y and Z together is
24μF.
1/C = (1/24) + (1/12)
C = 8.0μF
(ii) (Some
discussion as to why all charge of one sign on one plate of X.)
As the p.d. is applied, the positive
charge would flow from positive terminal to each one plate of the capacitor,
which would then become positively charged.
Q = CV = 8.0x10-6
x 9.0 = 72μC
(iii) Determine
1. V = Q / C = (8.0x10-6)
/ (12x10-6) = 6.0V
2.
Either Q = CV = 12x10-6 x
3.0 or charge is shared between Y and Z (9V – 6V)
Charge = 36μC.
Question 707: [Waves]
Microwave transmitter emits waves
towards a metal plate. Waves strike the plate and are reflected back along
their original path.
A microwave detector is moved along
line PT.
Points P, Q, R, S and T are the
positions where minima of intensity are observed. These points are found to be
15 mm apart.
What is the frequency of the
microwaves?
A 5.0 GHz B 6.7 GHz C
10 GHz D 20 GHz
Reference: Past Exam Paper – November 2010 Paper 12 Q26
Solution 707:
Answer:
C. A stationary wave is formed. The distance between 2 consecutive minima gives half the wavelength.
Wavelength λ = 2(15) = 30mm = 0.03m
Speed v = f λ
Frequency, f = v / λ = (3×108) / 0.03 = 1×1010Hz = 10GHz
Question 708: [Electromagnetism]
(a) State what is meant by a magnetic field.
(b) Charged particle of mass m and charge +q is travelling with
velocity v in a vacuum. It enters a region of uniform magnetic field of flux
density B, as shown in Fig.1.
Magnetic field is normal to the
direction of motion of the particle. Path of the particle in the field is the
arc of a circle of radius r.
(i) Explain why path of the particle
in the field is the arc of a circle.
(ii) Show that radius r is given by
the expression
r = mv / Bq
(c) A thin metal foil is placed in magnetic field in (b).
A second charged particle enters
region of the magnetic field. It loses kinetic energy as it passes through the
foil. The particle follows the path shown in Fig.2.
(i) On Fig.2, mark with an arrow the
direction of travel of the particle.
(ii) Path of the particle has
different radii on each side of the foil.
The radii are 7.4 cm and 5.7 cm.
Determine the ratio
final momentum of particle / initial momentum of particle
for the particle as it passes
through the foil.
Reference: Past Exam Paper – June 2011 Paper 41 Q5
Solution 708:
Go toA charged particle of mass m and charge +q is travelling with velocity v in a vacuum. It enters a region of uniform magnetic field of flux density B, as shown in Fig. 5.1.
Question 709: [Gravitation]
Spherical planet has mass M and
radius R.
The planet may be assumed to be
isolated in space and to have its mass concentrated at centre.
Planet spins on its axis with
angular speed ω, as illustrated in Fig.1.
Small object of mass m rests on
equator of the planet. Surface of planet exerts a normal reaction force on the mass.
(a) State formulae,
in terms of M, m, R and ω for
(i) gravitational force between planet
and object
(ii) centripetal force required for
circular motion of the small mass
(iii) normal reaction exerted by
planet on the mass
(b)
(i) Explain why normal reaction on the
mass will have different values at equator and at the poles
(ii) Radius of planet is 6.4×105m.
It completes 1 revolution in 8.6×104s. Calculate magnitude of centripetal acceleration at
1. equator
2. one of the poles
(c) Suggest 2 factors that could, in case of a real planet, cause
variations in the acceleration of free fall at its surface
Reference: Past Exam Paper – November 2008 Paper 4 Q1
Solution 709:
(a)
(i) F = GMm / R2
(ii) F = mRω2
(iii) Reaction force = (GMm / R2)
– mRω2
{The force of gravity is
attractive, so it attracts the mass towards the planet. The normal reaction
opposes the gravitational force and thus acts outwards, away from the planet.
Now, the resultant of these 2 forces provides the centripetal force.
Gravitational force –
Normal reaction = Centripetal force
Normal reaction =
Gravitational force – Centripetal force}
(b)
(i)
EITHER The value of R in the
expression Rω2 varies
OR mRω2 is no longer
parallel to GMm /R2 / normal to surface
and becomes smaller as the object
approaches a pole / is zero at pole
{The R in the expression GMm/R2
and the R in the expression mRω2 are different. In the first one, R
is the radius of the planet (considered to be spherical). In the second equation,
R is the perpendicular distance of the mass from the axis of spin of the
planet. This axis passes through the centre of the sphere. At the equator, R is
the same in both cases. However, at the poles, the ‘perpendicular distance of
the mass from the axis of spin of the planet’ is actually zero. At the top of
the sphere (pole), the distance of the surface from the axis is zero.}
(ii)
1.
{Acceleration = Rω2 and ω = 2πf}
Acceleration = (6.4x105)
x (2Ï€ /
{8.6x104})2 = 0.034ms-2
2. Acceleration = 0
(c) Choose any 2:
The ‘radius’ of the planet varies
The density of the planet is not
constant
The spinning of the planet
Nearby planets / stars
12/O/N/10 Q.33
ReplyDelete11/M/J/11 Q.9,10
For 12/O/N/10 Q.33, go to
Deletehttp://physics-ref.blogspot.com/2014/11/9702-november-2010-paper-12-worked.html
can you please explain oct/nov 2012, variant -11, (p-1) , Q.21? :)
ReplyDeleteCheck solution 803 at
Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-161.html
41/M/J/11 Q.5(c)(i) Can you please show the path of the particle?
ReplyDeleteFor Q.708 Please show the path of the particle in part (c)(i)
ReplyDeleteThe path is already shown in the diagram - it's the curved line. You only need to insert the head of an arrow to indicate its direction. This is from the bottom to the top along the path.
DeleteThere could only be 2 directions - either the one stated above, or its reverse - both are along the path shown - the curved line.
why is the direction from top to bottom, and not its reverse?
DeleteFrom Fig 5.2, we can observe that the path is NOT the arc of a circle as in Fig 5.1.
DeleteWe are 2 options for the direction of the path: either from right to left or from bottom to top of the diagram.
As the charged particle passes through the coil, it loses some kinetic energy. Its speed v decreases and so, the radius of the path also decreases (since r = mv / Bq). In the diagram, this reduction can be observed as we pass through the foil from the bottom to top and NOT from right to left along the path.
Solution 705 (b)(ii)
ReplyDeleteCould you please explain the answer because I'm not sure what the question or answer is trying to specify...
By accelerating the electrons through different p.d.`s, electrons with different de Broglie wavelength are produced. Now, if one of these electrons is diffracted by a crystal, it means that the value of its de Broglie wavelength (this can be calculated as above) is similar to the separation of the atoms (this is inferred due to the diffraction taking place). Thus, diffraction allows us to predict the separation of the atoms in the crystal.
DeleteWhy when mrw^2 =0 then normal reaction =0? Isn't the normal reaction at the pole= GMm/r^2 - mRw^2? So the normal reaction will be equal to the gravitational force isn't it
ReplyDeletecould you read again. It seems that the explanation is already available above - (b)(i) in red.
Deletewhy is the acceleration 0 at the axis??
ReplyDeleteAcceleration = Rω^2 where R is the perpendicular distance from the axis. At the poles, the value of R is zero. So, the acceleration = 0
DeletePls try to reply asap....
ReplyDeleteForce of gravitation is towards the centre and F(centripetal) is also centre directed so how ????? F(normal) becomes F(gravity)-F(centripetal).
Normal reaction is balancing both the forces by acting in opposite direction....
or Fn should have been Fn= Fg + Fc
Consider 2 forces: 10N and 7N both opposing each other.
DeleteTHe resultant would be 3N in the same direction as the 10N force.
SImilarly, the centripetal is the resultant of the 2 other forces and there is no problem for it to be in the same direction as one of them. Here the centripetal force is in the same direction as the force of gravity.
2015 may/june 9702/42 ...question number 5?
ReplyDeletewhere can i find the answer to may-june, 2015, paper 41 question 6 b (ii) ?
ReplyDeletego to
Deletehttp://physics-ref.blogspot.com/2017/12/9702-june-2015-paper-41-43-worked.html
can you please help me understand why for question 708 solution (c) (i) above why the answer is 'from bottom to top'? [june 2011 paper 41 question 5 c(i)]..
ReplyDeletethank you
the explanation has been updated
Delete