Wednesday, February 11, 2015

Physics 9702 Doubts | Help Page 65

  • Physics 9702 Doubts | Help Page 65

Question 362: [Nuclear Physics > Radioactive decay]
In 2002, two-proton radioactive decay of isotope of iron, 4526 Fe, was observed.
What could be resulting product?
A 4326Fe                       B 4324Cr                       C 4524Cr                       D 4728Ni

Reference: Past Exam Paper – June 2014 Paper 11 Q38

Solution 362:
Answer: B.
A proton radioactive decay of an isotope is a decay emitting a proton.

The isotope of Fe is 4526Fe.
A proton is 11p.

Since it is a two-proton radioactive decay, 2 protons are emitted.

4526Fe               - - - >               X         +          2 11p
The proton number and mass number should be equal on both sides of the equation. So, the resulting product should have
proton number: 26 – 2(1) = 24
Mass number is 45 - 2(1) = 43 since mass number of each proton is one.

Question 363: [Matter > Hooke’s law]
Wire stretches 8 mm under a load of 60 N.
A second wire of same material, with half the diameter and a quarter of the original length of the first wire, is stretched by same load.
Assuming that Hooke’s law is obeyed, what is extension of this wire?
A 1 mm                       B 4 mm                       C 8 mm                       D 16 mm

Reference: Past Exam Paper – June 2012 Paper 12 Q25

Solution 363:
Answer: C.
Hooke’s law: F = kx

The extension x does not depend on the diameter and area of the wire.
As the wire is of the same material, the spring constant, k is also the same.

So, the wire will have the same extension since the same load is used.

Question 364: [Current of Electricity > Kirchhoff’s laws]
Diagram shows circuit with four voltmeter readings V, V1, V2 and V3.

Which equation relating voltmeter readings must be true?
A V = V1 + V2 + V3
B V + V1 = V2 + V3
C V3 = 2(V2)
D V – V1 = V3

Reference: Past Exam Paper – June 2005 Paper 1 Q36 & June 2012 Paper 12 Q35

Solution 364:
Answer: D.
The voltmeters are used to give the potential difference across the resistors. The voltmeter reading V gives the e.m.f of the battery.

The resistors connection in the circuit shown forms 2 different loops (the voltmeters are not considered).
Loop 1: battery – resistor R (in parallel) – resistor R
Loop 2: battery – resistor 2R (in parallel) – resistor R  

From Kirchhoff’s laws, the sum of p.d. in any loop should be equal to the e.m.f in the circuit.

V = V2 + V1
V = V3 + V1                (This can be written as V – V1 = V3)

Question 365: [Simple Harmonic Motion]
A mass at the end of a helical spring is given a vertical displacement of 3.0cm from its rest position and then released. If the subsequent motion is simple harmonic with a period of 2.0s, find the distance covered by the mass in (i) the first 1.0s (ii) the first 0.75s.

Solution 365:
The equation describing the displacement x (from rest) of a particle undergoing simple harmonic motion with respect to time t is given by
x = x0 sin (ωt)
where x0 is the amplitude (maximum displacement) (= 3.0cm) and ω is the angular frequency of the simple harmonic motion.

But initially, the mass is at a displacement of 3.0cm. The equation for simple harmonic motion that should be use here is thus
x = x0 cos (ωt)

Since the mass is displaced from rest by 3.0cm, the amplitude (maximum displacement) x0 is 3.0cm. The mass cannot go beyond a displacement of 3.0cm (from the law of conservation of energy).

The period is given to be 2.0s.
Angular frequency ω = 2π /T = 2π / 2 = π

So, the equation can be simplified as
x = 3 cos (πt)

Since the motion is simple harmonic, the displacement varies from – 3.0cm to + 3.0cm. However, the distance travelled for a specific time (which is a scalar) may be greater.

The question says that the period is 2.0s. For a mass starting at a maximum displacement in one direction, the period is the time taken for the mass to move to the rest position, then the maximum displacement in the other direction, then back to the rest position and finally back to the maximum displacement it initially was. The displacement is zero but the distance travelled is 3.0 + 3.0 + 3.0 + 3.0 = 12.0cm.

Since we need to find the distance travelled for times of 1.0s and 0.75s, it would be less than 12.0cm.

When time t = 1.0s,
Displacement x = 3.0 cos (π(1.0)) = – 3.0cm

This displacement is to the opposite direction (than the mass was initially) from rest position.  So, the mass needs to go to the rest position first. This requires it to travel a distance of 3.0cm.
Total distance travelled for a time of 1.0s = 3.0 + 3.0 = 6.0cm

When time t = 0.75s,
Displacement x = 3.0 cos (π(0.75)) = – 2.12cm
Total distance travelled for a time of 0.75s = 3.0 + 2.12 = 5.12cm


  1. Replies
    1. Q5 is explained at

    2. For Q6, check q 391 at

  2. Replies
    1. Check at

  3. for solution 362 here, you said that mass number is unchanged since mass number of each proton is zero and you said the answer is B but in answer B the mass number is changed to 43. I dont understand

    1. Sorry, this was a mistake. a proton is 1 1p. I corrected it above.

  4. in solution 364, how do the 2 equations give V – V1 = V3?

    1. Consider only the 2nd equation and bring V1 on the left hand side of the equation.

    2. so we consider the 2nd equation since the first equation isn't available in the options?

  5. Replies
    1. For Q9, see solution 947 at

  6. Replies
    1. For Q33, see solution 935 at

      For Q35, see solution 770 at

  7. The solution and answer to Q.363 is wrong. The actual answer is B. It is to be calculated using the Young Modulus equation.
    For the 1st wire:
    Young Modulus = 60 X L/8 X A = 7.5 L/A
    The second wire is made of the same material so we can equate this Young Modulus equation there. It also has 4 times more area, is twice the original length and is stretched by the same load, thus for the second wire:
    Young Modulus = 60 X 2L/4A X e
    7.5 L/A = 60 X 2L/4A X e
    e = 60 X 2/7.5 X 4 = 4mm
    You guys are doing a great job and this site proves to be really helpful in this painful phase of CIEs. Well, we are humans, bound to mistakes, and 1 wrong solution in a thousand right ones doesn't make that much of a difference. Hope you update the solution above. Thanks.

    1. THanks, but actually you made a few mistakes. See details for the correct work below.

      Young modulus = stress / strain = (F/A) / (e/l) = Fl / Ae
      Since the wires are of the same material, the young modulus of the 2 wires are the same.
      For 1st wire: Young modulus = 60l / 8A
      Area depends on (diameter d)^2. When the diameter is halved, the area becomes A/4. The original length of the first wire is l/4.
      For 2nd wire: Young modulus = 60(l/4) / (A/4)e = 60l / Ae

      Equating, 60l / Ae = 60l / 8A giving e = 8mm

    2. The official CIE answer is 4mm. Please do correct it.

    3. I verified again and it's C


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