9702 November 2007 Paper 4 Worked Solutions | A-Level Physics
Paper 4
SECTION A
Question 1
(a)
(i)
A radian is the angle subtended at
the centre of a circle such that the arc is equal in length to the radius.
(ii)
Why 1 complete revolution is
equivalent to angular displacement of 2π
rad:
Length of arc = rθ and for one
revolution, the length of arc formed (= circumference) = 2πr. So, θ = 2πr / r =
2π
(b)
Elastic cord has unextended length
of 13.0cm. One end of cord attached to fixed point C. Small mass of weight 5.0N
is hung from free end of cord. Cord extends to length of 14.8cm, as shown.
Cord and mass are now made to rotate
at constant angular speed ω in vertical
plane about point C. When cord is vertical and above C, its length is the
unextended length of 13.0cm, as shown.
(i)
Show that angular speed ω of cord
and mass = 8.7rads-1:
EITHER The weight provides / equals
the centripetal force OR The acceleration of free fall is the centripetal
acceleration.
(a = rω2)
9.8 = 0.13ω2
ω = 8.7 rads-1
(ii)
Cord and mass rotate so that cord is
vertically below C, as shown. Length L of cord, assuming it obeys Hooke’s law:
{When the cord and mass is vertically below C, the centripetal force {FC} is provided by the resultant force from the force in the cord [tension in the cord {T} – which is upwards] and the weight [W - which is downwards]. (T should be greater than W so that the mass remains attached to that cord) So, FC = T – W giving T = W + FC}
{When the cord and mass is vertically below C, the centripetal force {FC} is provided by the resultant force from the force in the cord [tension in the cord {T} – which is upwards] and the weight [W - which is downwards]. (T should be greater than W so that the mass remains attached to that cord) So, FC = T – W giving T = W + FC}
Force in the cord = Weight +
Centripetal force
{To know the force in the cord, the spring constant, which is always constant should be found first. Hooke’s law: Force, F = kx. Extension, x = L – 13 and spring constant, k = F /x. When F = 5.0N, x = 14.8 – 13 = 1.8cm, k = spring constant [this always constant] = 5.0 / 1.8}
{To know the force in the cord, the spring constant, which is always constant should be found first. Hooke’s law: Force, F = kx. Extension, x = L – 13 and spring constant, k = F /x. When F = 5.0N, x = 14.8 – 13 = 1.8cm, k = spring constant [this always constant] = 5.0 / 1.8}
Force in the cord = (L – 13) x 5/1.8 or force constant = 5.0 / 1.8
{Force constant mentioned above =
spring constant}
{Weight = 5.0N, Centripetal force =
m rω2. Since only the weight
is given, we need to find the mass to calculate the centripetal force. Mass, m = weight / g = 5.0 / 9.8}
{Force in the cord = Weight +
Centripetal force}
(L – 13) x 5/1.8 = 5.0 + (5/9.8) (L
x 10-2) (8.72)
Length of cord, L = 17.2cm
Question 2
(a)
Amount of 1.00mol of Helium-4 gas
contained in cylinder at pressure of 1.02x105Pa and temperature of
27oC.
(i)
Volume of gas in cylinder:
pV = nRT
(T = 273 + 27 = 300)
V = (nRT/p =) (8.31 x 300) /
(1.02x105) = 0.0244m3
(ii)
Show that average separation of gas
atoms in cylinder is approximately 3.4x10-9m:
Volume occupied by 1 atom = 0.0244 /
(6.02x1023) = 4.06x10-26m3
Separation ≈ (4.06x10-26)1/3
= 3.44x10-9m
(b)
(i)
Gravitational force between 2
Helium-4 atoms that are separated by distance of 3.4x10-9m:
Force = GMm / r2 =
[(6.67x10-11) (4{1.66x10-27})2] / (3.44x10-9)2
= 2.49x10-46N
(ii)
Ratio of weight of a Helium-4 atom to
gravitational force between 2 Helium-4 atoms that with separation 3.4x10-9m:
Ratio = [4(1.66x10-27) x
9.8] / (2.49x10-46) = 2.6x1020
(c)
Comment on answer to (b)(ii) with
reference to 1of the assumptions of kinetic theory of gases:
One assumption of the kinetic theory
of gases is that the forces between the atoms are negligible.
Comment: e.g
The ratio shows the gravitational
force to be very small
The force is very much less than the
weight
If there are forces, they are not
gravitational forces.
Question 3
{Detailed explanations for this question is available as Solution 1125 at Physics 9702 Doubts | Help Page 243 - http://physics-ref.blogspot.com/2016/10/physics-9702-doubts-help-page-243.html}
Question 4
{Detailed explanations for this question is available as Solution 714 at Physics 9702 Doubts | Help Page 144 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-144.html}
Question 5
A capacitor is charged to a potential difference of 15 V and then connected in series with a switch, a resistor of resistance 12 kΩ and a sensitive ammeter, as shown in Fig. 5.1.
Question 6
A straight conductor carrying a current I is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig.6.1. ...
Question 7
{Detailed explanations for this question is available as Solution 1029 at Physics 9702 Doubts | Help Page 215 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-215.html}
SECTION B
Question 8
Fig. 8.1 shows a circuit incorporating an ideal operational amplifier (op-amp).
Question 9
(a)
Acoustic impedance is the product of
density (of the medium) and the speed of sound (in the medium).
(b)
Why acoustic impedance is important
when considering reflection of ultrasound at boundary between 2 media:
The difference in acoustic impedance
determines the fraction of incident intensity that is reflected / amount of reflection.
(c)
Principles behind use of ultrasound
to obtain diagnostic information about structures within body:
A pulse of ultrasound (which
is directed into the body) is reflected at the boundary (between the tissues).
(The reflected pulse is) detected and processed. The time for return of the
echo gives (information on the) depth and the amount of reflection gives
information on the tissue structures.
Question 10
Fig. 10.1 shows the variation with frequency f of the power P of a radio signal.
Question 11
In cellular phone network, country
is divided into a number of cells, each with its own base station. Fig shows a
number of these base stations and their connection to cellular exchange.
(a)
Explain why country is divided into
a number of cells:
The carrier frequencies can be
re-used (simultaneously without interference) so that the number of handsets
possible is increased.
OR
Any sensible e.g. UHF used, so ‘line
of sight’
(b)
What happens at base station and
cellular exchange when mobile phone handset is switched on, before call is made?:
The handset sends out an (identifying)
signal which is communicated by the base stations to (the computer at) the
exchange. The computer selects a base station with the strongest signal
and allocates a (carrier) frequency.
asalamualaikum.
ReplyDeleteCan you please explain Q3 part b(ii)? How do you find the amplitude using the graph and/or otherwise?
Wa`alaikum-us-salaam.
DeleteI have added some details. See it you understand now
04/m/j/07 q4c(ii) pls
ReplyDeletedetails have been added for that part
Deletes06 qp 4, question 8 b ii) v(velocity)= electric potential/ magnetic force, as the velocity increases shouldn't the magnetic field decrease ;since, they are inversely proportional? meaning the E should increase as the velocity increases but why is that not the case?
ReplyDeleteIt's explained as solution 654 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-130.html
M/J 07 Q3(b) and 4(b) and (c) please. Thank you :)
ReplyDeleteFor question 3, see solution 26 at
Deletehttp://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-4.html
For question 4, see solution 699 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-141.html
Can you explain why is the answer for question 4(d.) ? MJ 2007
ReplyDeleteSee question 699 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-141.html
Can you explain ON2007 q4(d)? Thank you.**
ReplyDeleteThe solution has been updated
DeleteMJ 2007 7(b) ,can you explain what's the function of resistor and how to draw the graph? Thank you :) thank you
ReplyDeleteCheck question 106 at
Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-20.html
May Allah bless you for your kindness!!
ReplyDeleteCan u explain question 6bii)
ReplyDeleteExplanation has been updated
DeleteCan you explain Q5 b2) because im getting a different answer and i dont understand how to calculate the area like explained in marking scheme. I solved it using trapezium rule, and by calculating the area of trapeziums normally.
ReplyDeleteExplanation updated.
DeleteIt can work with trapezia. Make sure you follow what each cm represents as explained.
Errors may occur due to overestimation or underestimation
PLZ EXPLAIN THE REASONS FOR THE TICKS AND CROSSES IN 0N/07 Q.8(part B)
ReplyDeletePLEASE MAKE A COMPLETE ILLUSTRATED DIAGRAM FOR ON/07 Q.8 P(part C).. THANKSSS
ReplyDeleteexplanation updated
DeleteCan you explain Q 10 part c by giving a graphical diagram
ReplyDeletecan you give a graphical diagram of solution of Question 10 part c
ReplyDeletediagram has been added
DeleteMay Allah (SWT) give u success in this life and the Hereafter...Ameen!
ReplyDelete