Thursday, February 5, 2015

Physics 9702 Doubts | Help Page 61

  • Physics 9702 Doubts | Help Page 61

Question 345: [Radioactive Decay]
The nuclide 22688Ra occurs in appreciable quantities in a mineral of great geological age despite the fact that its half-life is only a very small fraction of the age of the mineral. This is because
A the radioactive decay of 22688Ra is exponential and the activity can never become zero.
B 22688Ra is constantly being formed by the decay of a longer-lived isotope.
C the mineral is constantly being subjected to neutron bombardment from the centre of the Earth, forming more 22688Ra.
D the concentration of 22688Ra is constantly being regenerated by radioactive fall-out from the atmosphere.
E the rock surrounding the 22688Ra slows down the escape of the alpha-particles and gamma-rays.

Reference: Past Exam Paper – N79 / II / 36 & J82 / II / 38 & J87 / I / 27

Solution 345:
Answer: B.
The correct answer should be reasonable and relevant, based on the simple theories learned in the course.

Even if choice A may be true, the 22688Ra would be present in only a scarce amount.

The only correct and reasonable choice is that the 22688Ra is constantly being formed by the decay of a longer-lived isotope.

Question 346: [Nuclear Physics]
The isotope 22688Ra decays into 22286Rn with the emission of an α-particle and a γ-ray photon of frequency υ.
The principle of conservation of energy is expressed in the decay equation
A 88 = 86 + 2
B 226 = 222 + 4
C [mRa – mRn] c2 = hυ
D ½ mRn uRn2 = hυ + ½ mα uα2
E [mRa – (mRn + mα] c2 = hυ + ½ mα uα2 + ½ mRn uRn2

Reference: Past Exam Paper – N80 / II / 39

Solution 346:
Answer: E.
The equation for the decay is
22688Ra             - - - >   22286Rn             +          42α        +          γ

Choice A is an equation for the proton number while choice B equates the mass number without considering the γ-ray. [A and B are incorrect]

The 22688Ra, 22286Rn and 42α all have some masses while the γ-ray, which is an EM wave, has no mass.

After the decay, the 22286Rn and 42α emitted would move with some energy – kinetic energy. Their kinetic energy would be provided by the mass defect between the parent nucleus (22688Ra) and the daughter nucleus (22286Rn) along with the alpha particle emitted.

The energy due to this mass defect is [mRa – (mRn + mα] c2

This energy is converted to the kinetic energy of the 22286Rn and 42α. Part of that energy is also used to liberate the γ-ray photon.

Energy of γ-ray photon = hυ
Kinetic energy of 22286Rn = ½ mRn uRn2
Kinetic energy of 42α = ½ mα uα2

Only choice E accounts for all these energies.

Question 347: [Current of Electricity > Kirchhoff’s law]
Diagram shows a circuit in which the battery has negligible internal resistance.

What is the value of the current I?
A 1.0 A                       B 1.6 A                       C 2.0 A                       D 3.0 A

Reference: Past Exam Paper – November 2003 Paper 1 Q31

Solution 347:
Answer: A.
Total resistance in circuit = [1/6 + 1/3]-1 + 2 = 2 + 2 = 4Ω

Total current supplied by battery = V / R = 12 / 4 = 3.0A

2 loops can be seen in the circuit.
Loop 1: battery – 6.0 Ω resistor – 2.0 Ω resistor
Loop 2: battery – 3.0 Ω resistor – 2.0 Ω resistor

From Kirchhoff’s law, the sum of potential difference across the components in each loop is the same (and equal to 12V here).

Consider loop 1:
The total current of 3.0A flows through the 2.0 Ω resistor.
p.d. across 2.0 Ω resistor = 3.0 (2.0) = 6.0V

Therefore, the p.d. across the 6.0 Ω resistor = 12 – 6 = 6.0V
Current I through the 6.0 Ω resistor = V / R = 6.0 / 6.0 = 1.0 A

The total current of 3.0 A would split at the junction of the 2 resistors in parallel. From Ohm’s law (V = IR), the current is inversely proportional to the resistance. That is, when the resistance is big, the current is small.
Since the resistance 6.0 Ω resistor is twice that of the 3.0 Ω resistor, the current I through the 6.0 Ω resistor would be half that through the 3.0 Ω resistor.

So, current in 6.0 Ω resistor = 1.0A and current in 3.0 Ω resistor = 2.0A.
Total current = 1.0 + 2.0 = 3.0A

Question 348: [Electromagnetism > Oscillations]
Aluminium sheet is suspended from an oscillator by means of a spring, as illustrated.

Electromagnet is placed a short distance from centre of the aluminium sheet.
Electromagnet is switched off and frequency f of oscillation of the oscillator is gradually increased from a low value. Variation with frequency f of amplitude a of vibration of the sheet is shown.

A peak on graph appears at frequency f0.
(a) Explain why there is a peak at frequency f0.

(b) Electromagnet is now switched on and frequency of the oscillator is again gradually increased from a low value. On Fig, draw a line to show variation with frequency f of the amplitude a of vibration of the sheet.

(c) Frequency of oscillator is now maintained at a constant value. Amplitude of vibration is found to decrease when current in the electromagnet is switched on.
Use laws of electromagnetic induction to explain this observation.

Reference: Past Exam Paper – June 2003 Paper 4 Q3

Solution 348:
(a) The frequency f0 is the natural frequency of the spring (system). This frequency is at the driver frequency.
{So, resonance occurs.}

(b) The line drawn should have the amplitude less (than the graph shown) at all frequencies. The peak should be flatter with the peak being at f0 or slightly below f0.

(c) The (aluminium) sheet cuts the magnetic flux/field, so currents / e.m.f. are induced in the (metal) sheet. These current dissipate energy. So, less energy is available for the oscillations. So, the amplitude is smaller.

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