Tuesday, November 25, 2014

Physics 9702 Doubts | Help Page 24

  • Physics 9702 Doubts | Help Page 24

Question 129: [Waves > Stationary waves]
Basic principle of note production in a horn is to set up stationary wave in an air column.

For any note produced by horn, a node is formed at the mouthpiece and an antinode is formed at bell. Frequency of lowest note is 75 Hz.
What are the frequencies of next two higher notes for this air column?

Reference: Past Exam Paper – June 2011 Paper 11 Q25 & June 2014 Paper 11 Q27

Solution 129:
Answer: D.
As stated from the question, for any note produced by the horn, a node is formed at the mouthpiece and an antinode is formed at the bell.

The lowest note consists of only a node at the mouthpiece and an anti-node at the bell, forming ¼ of a wavelength. The frequency of this lowest note is 75Hz.

The first higher note consists of a node at the mouthpiece, followed by an anti-node, then a node and finally an anti-node at the ball, forming ¾ of a wavelength.
So, its frequency is 3 times that of the lowest note (that is, for the same distance {the length of the horn}, the wave formed is 3 times greater {3/4 is 3(1/4) and ¼ of the wavelength is formed in the lowest note}). Frequency = 3(75) = 225Hz.

Similarly, the second higher note would formed 5/4 of a wavelength, with a frequency = 5(75) = 375Hz

Question 130: [Power > Hydroelectric power]
Turbine at hydroelectric power station is situated 30 m below the level of surface of a large lake. Water passes through turbine at a rate of 340 m3 per minute.
Overall efficiency of turbine and generator system is 90%.
What is the output power of power station? (Density of water is 1000 kg m–3.)
A 0.15 MW                 B 1.5 MW                   C 1.7 MW                   D 90 MW

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q19

Solution 130:
Answer: B.
Work done by fluid = PΔV where P is the pressure (constant) and ΔV is the change in volume.
If ΔV is positive, W.D is +ve, so the fluid does work. If ΔV is negative, W.D is also –ve, meaning work is done on the fluid.

Power is the rate of doing work.
Power = Work done / time = (PΔV)/t = P (ΔV/t)
(ΔV/t) is the flow rate (= 340m3 / min)

In 1min (60sec), a volume of 340m3 passes through the turbine
In 1sec, a volume of (340 / 60 =) [17/3] m3 passes through the turbine

Flow rate, ΔV/t = 340m3 / min = (17/3) m3s-1
Pressure, P = hρg = 30 (1000) (9.81)

Output power = Efficiency x Power = 0.9 x P x (ΔV/t)
Output power = 0.9 x hρg x (ΔV/t)
Output power = 0.9 x [30 (1000) (9.81)] x [17/3] = 1.5 x 106W = 1.5 MW

Flow rate = (Av) – the units are consistent = 340m3/min = (17/3) m3/sec
Output Power = 0.9 x WD/time
[Power = Work Done / time = Force x velocity]

Output Power = 0.9 x Fv
[Force = Pressure x Area] Output Power = 0.9 x PAv
[Pressure = hρg] Output Power = 0.9 x hρg(Av) = 1500930 = 1.5 x 106W = 1.5 MW

Potential energy = mgh
Mass = Density x Volume

Input Power = Energy / time = mgh / t = (m/t) gh

Mass per unit time = Density x (Volume per unit time)
Mass per unit time = 1000 x (340)
{Convert 340m3 / min to (17/3) m3/sec as shown before}

Output Power = Efficiency x Input Power
Output Power = 0.9 x [1000 x 17/3] x 9.81 x 30 = 1.5MW

Question 131: [Pressure > Work done > Gas]
Gas is enclosed inside cylinder which is fitted with frictionless piston.

Initially, gas has volume V1 and is in equilibrium with an external pressure p. Gas is then heated slowly so that it expands, pushing piston back until the volume of the gas has increased to V2.
How much work is done by gas during this expansion?
A p(V2 – V1)               B ½ p(V2 – V1)           C p(V2 + V1)               D ½ p(V2 + V1)

Reference: Past Exam Paper – June 2013 Paper 12 Q18

Solution 131:
Answer: A.
Increase in volume, ΔV = (V2 – V1)
Work done by gas during expansion = pΔV = p(V2 – V1)
{The above formula for work done on/by gas is given in the formula list}

Question 132: [Waves > Stationary Waves]
Diagram shows two tubes.

Tubes are identical except tube X is closed at its lower end while tube Y is open at its lower end. Both tubes have open upper ends.
Tuning fork placed above tube X causes resonance of air at frequency f. No resonance is found at any lower frequency than f with tube X.
Which tuning fork will produce resonance when placed just above tube Y?
A a fork of frequency f / 2
B a fork of frequency 2f / 3
C a fork of frequency 3f / 2
D a fork of frequency 2f

Reference: Past Exam Paper – November 2010 Paper 12 Q25

Solution 132:
Answer: D.
For tube X, a node is present at lower (closed) end while for tube Y, an antinode is present at the lower (open) end. For both tubes, antinodes are present at the upper open end.

For closed tube X, resonant frequency are at fc = nv / 4L
where v: speed of wave, L: length of tube and n = 1, 3, 5, …

For open tube Y, resonant frequency are at fo = nv / 2L
where n = 1, 2, 3, …

{Note that the above formulae may be derived by studying the wave formed at each frequency. As explained in Question 129 at Physics 9702 Doubts | Help Page 24 (above), at the lowest frequency, a quarter of a wavelength will be formed in the length L of tube X. So, L = λ / 4 giving λ = 4L. Speed v = fλ . Frequency f = v / λ = v / 4L.  A similar reasoning may be used to find the frequency for an open tube at the lowest frequency. The same is done at higher frequencies. The general formulae are given above.}

So, the resonant frequency for the open tube Y, fo = 2fc for n =1.

Question 133: [Matter > Elastic Deformation]
Spring is stretched over range within which elastic deformation occurs. Its spring constant is 3.0 N cm–1.
Which row, for stated applied force, gives the correct extension and strain energy?

Reference: Past Exam Paper – June 2013 Paper 13 Q 21

Solution 133:
Answer: D.
Hooke’s law: Force F = ke      where k is the spring constant and e is the extension.
Spring constant, k = 3.0Ncm-1 = 300Nm-1  

Extension = F / k
Strain energy = ½ Fe

Consider A:
Force = 3.0N and extension = 3 / 300 = 0.01m
Strain energy = 0.5 (3.0) (0.01) = 0.015J = 15mJ                   [A is incorrect]

Consider B:
Force = 6.0N and extension = 6 / 300 = 0.02m
Strain energy = 0.5 (6.0) (0.02) = 0.06J = 60mJ                     [B is incorrect]

Consider C:
Force = 12.0N and extension = 12 / 300 = 0.04m                  [C is incorrect]

Consider D:
Force = 24.0N and extension = 24 / 300 = 0.08m
Strain energy = 0.5 (24.0) (0.08) = 0.96J = 960mJ

Question 134: [Forces > Moment]
Diving board of length 5.0 m is hinged at one end and supported 2.0 m from this end by spring of spring constant 10 kN m–1. Child of mass 40 kg stands at far end of the board.

What is the extra compression of the spring caused by child standing on the end of board?
A 1.0 cm                     B 1.6 cm                      C 9.8 cm                      D 16 cm

Reference: Past Exam Paper – November 2013 Paper 13 Q15

Solution 134:
Answer: C.
Both the child and the spring cause a moment, at a distance from the hinge, on the board.
For equilibrium of the board (for the board to stay horizontal), the torque of the child (anticlockwise moment produced by child) should be equal to the torque of the spring (clockwise moment produced by spring).

Let the force by the spring = F
(40 x 9.81) x 5.0 = F x 2.0
Spring force, F = 981N

Hooke’s law: F = ke
Spring constant, k = 10kNm-1 = 10000Nm-1
Extension = F / k = 981 / 10000 = 0.0981m = 9.8cm

Question 135: [Electric Field > Non-uniform electric field]
Diagram shows non-uniform electric field near a positively charged and a negatively charged sphere.
Four electrons, A, B, C and D, are shown at different positions in field.
On which electron is direction of the force on the electron shown correctly?

Reference: Past Exam Paper – November 2011 Paper 12 Q31

Solution 135:
Answer: A.
The direction of electric field lines is the direction of the force on a positive charge. That is, the direction is from (away) a positive charge towards a negative charge.
An electron, which has a negative charge, will be attracted towards the positive sphere [B incorrect], NOT towards the negative charge [D incorrect] since like charges repel.

In a non-uniform electric field, the direction of the electric field on a charge at a specific point is given by tangent to the electric field line at that point in the electric field. [C incorrect since it is not a tangent]  
If the electric field was uniform, then C would have been correct.

Question 136: [Electric Field > Electric Potential energy]
Positive charge experiences force F when placed at point X in a uniform electric field.
Charge is then moved from point X to point Y.
Distances r and s are shown on diagram.

What is the change in potential energy of charge?
A decreases by Fs
B increases by Fs
C decreases by Fr
D increases by Fr

Reference: Past Exam Paper – June 2008 Paper 1 Q16 & November 2013 Paper 13 Q18

Solution 136:
Answer: A.
Electric field direction is (away) from positive to negative. The positive charge loses potential energy when moving in the direction of the electric field.
(This is similar to the gravitational potential energy. The gravitational field is usually downward, but a mass would have a higher potential energy when its height is greater – height is in opposite direction to the direction of the gravitational field)

Work done = Force x distance moved in direction of force
Since the distance moved in the direction of the electric field is s, the potential energy decreases by Fs.

Question 137: [Kinematics > Projectile motion]
Bullet is fired horizontally with speed v from rifle. For short time t after leaving rifle, the only force affecting its motion is gravity. Acceleration of free fall is g.
Which expression gives the value of the ratio of the horizontal distance travelled in time t to the vertical distance travelled in time t?
A vt / g                        B v / gt                                    C 2vt / g                      D 2v / gt

Reference: Past Exam Paper – June 2011 Paper 12 Q 6

Solution 137:
Answer: D.
The acceleration of free fall acts (vertically) downwards. It does not affect the horizontal motion.
Horizontal distance travelled in time t, s = vt

Since the bullet is fired horizontally, it’s vertical component of velocity is initially zero.  
Vertical distance travelled in time t, s = ut + ½ at2 = 0 + ½ gt2 = ½ gt2

Ratio = vt / (½gt2) = 2v / gt


  1. can explain june 2014 paper 12 question 26 please? thank you

    1. Details for the solution have been added at the respective worked solutions page. Check it there

  2. can you explain june2010 paper11 question 34?

    1. See solution 600 at

    2. Is there a way I can just look at questions about a certain topic like [Electric Field > Electric Potential energy] for example?

  3. Is there a way to just see questions about a certain topic like [Electric Field > Electric Potential energy] for example?

    1. Not yet, but I've been think about this too.
      But if you know from which year a question is, you may search for it at

  4. Hi admin, can u do worked solutions for 2015 m/j paper 11, 12 ,and 13 pls????? I realised the papers get harder and harder each year, 2015 was so tough. Ur worked solutions will help me a lot!!! PLSS Thank you!

    1. You need to specify the questions, one by one. try them first, and let me know why questions are you are difficulties

  5. Hi,
    can u help me with these please.I found them from a topical past exam question book.Since i dont know what the reference means i will note it just as it is in the book
    J93/III/3 (aii only)
    N94/III/3 (bii only)
    Thanks in advance :)

    1. see the questions and let me know on which topic they are.
      + try to type a few lines of the questions or completely.

    2. Ive finally got to know what they mean and i have got some more question .some questions i dont understand part of them only so i am noting that between brackets.
      June 1985 paper 1 question 17
      June 1993 paper3 question 3 (part aii only)
      nov 1994 paper3 question 3 (part biii only)
      june 2007 paper 1 question 33
      These are all questions on current n electricity .Hope u can help me with these .Thanks for ur awesome worked solutions.Your work is trully praiseworthy.

    3. For June 2007 P1, go to

      For June 1985 paper 1 question 17, see solution 1073 at

  6. Can you do for nov 2007 paper1 question 35 please?Thanks

    1. See solution 770 at

  7. Can you publish worked solution of Summer 12, qp11 question 12 about the man rising up. i dont get the same answer as marscheme. thank you, your page is a saviour

    1. See solution 178 at

  8. is tyhere such sites for A level mathematics!! I just loved this one!

    1. Don't know. Maybe I'll think about adding maths later.

  9. An Olympic athelete of mass 80kg completes in a 100m race.
    What is the best estimate of his mean kinetic energy during the race? (october November 2011 paper 11)

    Thanks alot for looking into this problem, Please solve it as soon as possible.

    1. Consider solution 752 at

  10. Q134, the diving board should have mass but why weight is not considered?

    1. the weight has been calculated. (40 x 9.81)

    2. isn't that the weight of child?

    3. it may be neglected as no information is given in the quesiton

  11. Thank you very much, whoever made this page. I've got my physics paper 1 three days from now and I've got no words to tell how much these explanations to MCQs helped me. Thanks a million for making things easier for us and keep up the good work! :)

  12. In solution 135, why would C be correct if it was a uniform field instead of non-uniform one?

    1. the electron would be directly attracted to the positive charge


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