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Question 1044: [Radioactivity]

(a) An isotope of an element is radioactive. Explain what is meant by radioactive decay. 

(b) At time t, a sample of a radioactive isotope contains N nuclei. In a short time Δt, the number of nuclei that decay is ΔN.

State expressions, in terms of the symbols t, Δt, N and ΔN for

(i) the number of undecayed nuclei at time (t + Δt),

(ii) the mean activity of the sample during the time interval Δt,

(iii) the probability of decay of a nucleus during the time interval Δt,

(iv) the decay constant.

(c) The variation with time t of the activity A of a sample of a radioactive isotope is shown in Fig. 9.1.

The radioactive isotope decays to form a stable isotope S. At time t = 0, there are no nuclei of S in the sample.

On the axes of Fig. 9.2, sketch a graph to show the variation with time t of the number n of nuclei of S in the sample.

Reference: Past Exam Paper – June 2015 Paper 41 & 43 Q9

Solution 1044:

(a) During radioactive decay, a nucleus/nuclei emits spontaneously/randomly α-particles, β-particles, or γ-ray photons

(b)

(i) N – ΔN

{Number of undecayed nuclei = Total number of nuclei (N) – number of nuclei that decay (ΔN)}

(ii) ΔN / Δt

{Activity is the number of nuclei that decay per unit time.}

(iii) ΔN / N

{Probability of decay of a nucleus = number of nuclei that decay (ΔN) / total number of nuclei (N)}

(iv) ΔN / NΔt

{ΔN / Δt = (–) λN       where λ is the decay constant

λ = ΔN / NΔt}

(c)

graph: a smooth curve in correct direction starting at (0,0)

n at 2t_½ is 1.5 times that at t_½ (± 2 mm)

{At time t = 0, there are no nuclei of S in the sample. (0 , 0)

At time = t_½, half of the radioactive sample (=N/2) has decayed. Let’s say this produces an amount n = 2 of isotope S. This is point (t_½, 2).

Now, only half of the radioactive sample remains. At time = 2t_½, half of the remaining radioactive sample decays (=N/4). Compared to the previous case, only half the amount has decayed, so this produces half the amount of isotope S (= 2/2 = 1). Total amount of isotope S present at time 2t_½ is n = 2 + 1 = 3. This is point (2t_½, 3).

Similarly, at time = 3t_½, the amount of isotope S produced is half of the previous amount (= ½ = 0.5). Total amount of isotope S present at time 3t_½ is n = 2 + 1 + 0.5 = 3.5. This is point (3t_½, 3.5)}

Question 1045: [Newtonian Mechanics > Forces > Centre of gravity]

What is the centre of gravity of an object?

A the geometrical centre of the object

B the point about which the total torque is zero

C the point at which the weight of the object may be considered to act

D the point through which gravity acts

Reference: Past Exam Paper – June 2005 Paper 1 Q12 & November 2009 Paper 11 Q11 & Paper 12 Q10

Solution 1045:

Answer: C.

The centre of gravity of an object is the point at which the weight of the object may be considered to act.

The geometrical centre of the object may not always be the centre of gravity (e.g. if the object is not uniform).

Question 1046: [Electromagnetism]

Two long, straight, current-carrying conductors, PQ and XY, are held a constant distance apart, as shown in Fig. 6.1.

The conductors each carry the same magnitude current in the same direction.

A plan view from above the conductors is shown in Fig. 6.2.

(a) On Fig. 6.2 draw arrows, one in each case, to show the direction of

(i) the magnetic field at Q due to the current in wire XY (label this arrow B),

(ii) the force at Q as a result of the magnetic field due to the current in wire XY (label this arrow F).

(b) (i) State Newton’s third law of motion.

(ii) Use this law and your answer in (a)(ii) to state the direction of the force on wire XY.

(c) The magnetic flux density B at a distance d from a long straight wire carrying a current I is given by

B = 2.0 × 10^–7 × I / d

Use this expression to explain why, under normal circumstances, wires carrying alternating current are not seen to vibrate. Make reasonable estimates of the magnitudes of the quantities involved. 

Reference: Past Exam Paper – June 2006 Paper 4 Q6

Solution 1046:

(a)

(i) arrow B in correct direction (down the page)

{Direction of magnetic field is given by the right-hand grip rule. Current is out of paper, so the magnetic field is anticlockwise. At Q, this is down the page.}

(ii) arrow F in correct direction (towards Y) B1

{The direction of the force can be obtained by Fleming’s left-hand rule. Thumb: force, forefinger: field and middle finger: current}

(b)

(i) When two bodies interact, the force on one body is equal but opposite in direction to the force on the other body.

(ii) The direction is opposite to that in (a)(ii) {towards Q}

(c)

Consider a current of I = 0.5A. The magnetic flux density at a distance d = 1.0 cm (= 0.01m) is

B = 2.0 × 10^–7 × 0.5 / 0.01 = 0.0001 T

The force is given by F = BIL

The force between the wires is small compared to the weight of wire

Question 1047: [Moments > Equilibrium]

A rigid uniform beam is pivoted horizontally at its midpoint.

Different vertical forces are applied to different positions on the beam.

In which diagram is the beam in equilibrium?

 

Reference: Past Exam Paper – November 2012 Paper 12 Q17

Solution 1047:

Answer: C.

For equilibrium, clockwise moment = anti-clockwise moment

Choice A: both forces causes an anti-clockwise moment. This is not equilibrium, instead the beam would rotate.

Choice B:

Clockwise moment = 5 × 10 = 50 Ncm

Anti-clockwise moment = 4 × 15 = 60 Ncm  (they are not equal)

Choice C: [correct]

Clockwise moment = 5 × 14 = 70 Ncm

Anti-clockwise moment = 2(5) + 4(15) =70 Ncm