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 (Ï€t)
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 (Ï€(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 (Ï€(0.75)) = – 2.12cm
Total distance travelled for a time of 0.75s = 3.0 + 2.12 = 5.12cm
21/O/N/10 Q.5(b),Q.6(a)(ii)
ReplyDeleteQ5 is explained at
Deletehttp://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-66.html
For Q6, check q 391 at
Deletehttp://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html
22/O/N/10 Q.2(b)
ReplyDeleteCheck at
Deletehttp://physics-ref.blogspot.com/2014/09/9702-november-2010-paper-22-worked.html
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
ReplyDeleteSorry, this was a mistake. a proton is 1 1p. I corrected it above.
Deletein solution 364, how do the 2 equations give V – V1 = V3?
ReplyDeleteConsider only the 2nd equation and bring V1 on the left hand side of the equation.
Deleteso we consider the 2nd equation since the first equation isn't available in the options?
DeleteYes
Deleteq9 and q20 june 2005
ReplyDeleteFor Q9, see solution 947 at
Deletehttp://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-195.html
q35, 33 and 34 nov 2007
ReplyDeleteFor Q33, see solution 935 at
Deletehttp://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-192.html
For Q35, see solution 770 at
http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-155.html
The solution and answer to Q.363 is wrong. The actual answer is B. It is to be calculated using the Young Modulus equation.
ReplyDeleteFor 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.
THanks, but actually you made a few mistakes. See details for the correct work below.
DeleteYoung 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
The official CIE answer is 4mm. Please do correct it.
DeleteI verified again and it's C
DeleteAssalamualaekum everyone, there are two questions in past papers one with answer=8mm and other has the answer=4mm. But the question#363 has the correct answer=8mm.
DeleteAOA;you are doing a fantastic job.Need some help in Q5 M-j/2012/Paper 12.
ReplyDeleteWslm. GO to
Deletehttps://physics-ref.blogspot.com/2014/06/9702-june-2012-paper-12-worked.html