# Physics 9702 Doubts | Help Page 4

__Question 22: [Radioactivity]__**(a)**Sample of radioactive isotope contains N nuclei at time t. At time (t + Δt), it contains (N – ΔN) nuclei of isotope. For period Δt, state, in terms of N, ΔN and Δt,

(i) Mean activity of sample

(ii) Probability of decay of nucleus

**(b)**Cobalt-60 source having half-life of 5.27 years is calibrated and found to have activity of 3.50 × 10

^{5}Bq. Uncertainty in calibration is ±2%. Calculate length of time, in days, after calibration has been made, for stated activity of 3.50 × 10

^{5}Bq to have maximum possible error of 10%.

**Reference:**

*Past Exam Paper – June 2009 Paper 4 Q9*

__Solution 22:__**(a)**

(i)

{Activity is the rate of
disintegration OR the number of nuclei decaying per unit time}

Mean activity of sample = ΔN / Δt

(ii)

Probability of decay of nucleus = ΔN
/ N

**(b)**

{The calibration of the
initial activity A

_{o}contains an uncertainty of 2%.
[Activity is the rate of
disintegration OR the number of nuclei decaying per unit time] Activity
decreases with time as the source decays. So,

__the change in activity with time tells us about the amount of nuclei decayed during that time__.
The calibration for the activity
A, at time t, contains an uncertainty of 10%. So, the change in activity ΔA (= A

_{o}– A) corresponds to the increase in error of 10 – 2 = 8%. As stated before, this percentage also tells us about the amount of nuclei that have decayed. So, the source must have decayed by 8%}The source must decay by 8%

{Activity, A at time t = A

_{o}exp(-λt) and the decay constant, λ = ln2 / T_{½}where T_{½}is the half-life.}
Activity, A = A

_{o}exp(- (ln2) t / T_{½}) or A / A_{o}= 1 / (2^{t/T})
{For the OR case above, -
(ln2) t / T

_{½}= ln 2^{-t/T½}. So, exp(ln 2^{-t/T½}) = 2^{-t/T½}= 1 / (2^{t/T½}).The source has decayed by 8%. The initial activity (which represents the initial amount of nuclei in the source) corresponds to 100%. Therefore, the activity A (which represents the amount of nuclei at time t), at time t, corresponds to 92%. A / A

_{o}= 92 / 100 = 0.92}

0.92 = exp(- (ln2) t / 5.27) or 0.92 = 1 / (2

^{t/5.27})

Time, t = 0.634 years = 230 days

__Question 23: [Electromagnetism > Lenz’s law]__
You are provided with coil of wire,
bar magnet and sensitive ammeter. Outline experiment to verify Lenz’s law.

**Reference:**

*Past Exam Paper – June 2009 Paper 4 Q7*

__Solution 23:__
A coil is connected in series with a
meter [sensitive ammeter - galvanometer] (do not allow inclusion of a cell). A

__known__pole (of a magnet) is pushed into the coil. The current__direction__(not reading) is observed. The (induced) field / field from the coil repels the magnet.
EITHER State the rule to determine
the direction of the magnetic coil {Right hand grip
rule – curled fingers indicate the direction of the current and the thumb,
which is perpendicular to the curled finger shows the direction of the field}
OR Reversing the magnet direction gives an opposite deflection on the meter.

(Lenz’s law:)The direction of the
induced current is (gives rise to effects) such as to oppose the change
producing it.

(The diagram was taken from the
internet. Credit goes to its original owner.)

__Question 24: [Radioactive decay > Momentum]__
Francium-208 is radioactive and
emits α-particles with kinetic energy of 1.07 × 10

^{–12}J to form nuclei of astatine, as illustrated.**(a)**Nature of α-particle:

**(b)**Show that initial speed of α-particle after decay of francium nucleus is approximately 1.8 × 107ms

^{–1}:

**(c)**

(i) State principle of conservation
of linear momentum:

(ii) Francium-208 nucleus is stationary
before decay. Estimate speed of astatine nucleus immediately after decay:

**(d)**Close examination of decay of francium nucleus indicates that astatine nucleus and α-particle are not ejected exactly in opposite directions. Suggest explanation for this observation:

**Reference:**

*Past Exam Paper – November 2006 Paper 2 Q3*

__Solution 24:__**(a)**

An α-particle is a helium nucleus OR
contains 2 protons and 2 neutrons

**(b)**

{Mass of α-particle = 4u. 1u = 1.66x10

^{-27}kg}
Kinetic energy of α-particle = ½mv

^{2}
½ [4 (1.66x10

^{-27})] v^{2}= 1.07x10^{-12}
Initial speed of α-particle, v = 1.8x10

^{7}ms^{-1}**(c)**

(i)

The principle of conservation of
linear momentum states that the sum of momenta (in any direction) is constant /
total momentum is constant in a closed system / no external force

(ii)

{Before decay,
francium-209 nucleus is stationary – speed = 0 and hence, momentum (=mv) = 0}

Momentum of francium (= 0) =
momentum of α-particle + momentum of astatine

{momentum of astatine =
(-) momentum of α-particle}

204(v) = 4 (1.8x10

^{7})
Speed of astatine nucleus, v =
3.5x10

^{5}ms^{-1}**(d)**

{They are not ejected
exactly in opposite directions – this means that the sum of momenta is not
constant. So, there should be another particle / photon that would account for
the missing momentum}

Another particle / photon is emitted
{simultaneously} at an angle to the direction of
the α-particle

__Question 25: [Waves > Interference > Double Slits]__**(a)**State three conditions required for maxima to be formed in interference pattern produced by two sources of microwaves.

**(b)**Microwave source M emits microwaves of frequency 12 GHz. Show that wavelength of microwaves is 0.025 m.

**(c)**Two slits S

_{1}and S

_{2}are placed in front of microwave source M described in (b), as shown. Distances S

_{1}O and S

_{2}O are equal. Microwave detector is moved from O to P.

Distance S

_{1}P is 0.75 m and distance S_{2}P is 0.90 m. Microwave detector gives maximum reading at O. State variation in readings on microwave detector as it is moved slowly along line from O to P.**(d)**Microwave source M is replaced by source of coherent light. State two changes that must be made to slits in Fig. 5.1 in order to observe interference pattern.

**Reference:**

*Past Exam Paper – June 2013 Paper 21 Q5*

__Solution 25:__**(a)**

Choose any 3:

The waves should overlap / meet /
superpose

The waves must be coherent /
constant phase difference (not constant λ or frequency)

The path difference = 0, λ, 2 λ OR phase difference = 0, 2π, 4π

The y must have the same direction
of polarization / unpolarised

**(b)**

Wavelength, λ = v / f

Frequency, f = 12x10

^{9}Hz
Wavelength, λ = (3x10

^{8}) / (12x10^{9}) = 0.025m**(c)**

{Maxima are formed when
path difference = n λ where n = 0, 1, 2, 3, … At O, the maxima corresponds to
n = 0. The path difference between S

_{1}P and S_{2}P = 0.90 – 0.75 = 0.15m. This corresponds to n = 0.15 / 0.025 = 6. Therefore, including the maxima at O, there is a total of 7 maxima from O to P [excluding the maxima at O and P, there is 5 maxima between O and P].
Minima are found when path
difference = 0.5λ,
1.5λ, 2.5λ, … that is, midway between 2
maxima. So between 2 consecutive maxima, there is a minima. Since there is a
total of 7 maxima, 6 minima are found are present between O and P.}

There is a maximum at P and

__several__minima or maxima between O and P. There are 5 maxima / 6 minima between O and P OR 7 maxima / 6 minima including O and P.**(d)**

Microwave source M is replaced by
source of coherent light. State two changes that must be made to slits in Fig.
5.1 in order to observe interference pattern:

The slits must be made narrower

The slits must be put closer
together

__Question 26: [Electric field > Electric field strength]__
Two charged points A and B are
separated by distance of 6.0 cm, as shown.

Variation with distance d from A of
electric field strength E along line AB is shown.

Electron is emitted with negligible
speed from A and travels along AB.

**(a)**Relation between electric field strength E and potential V.

**(b)**Area below line of the graph of Fig represents potential difference between A and B. Use Fig to determine potential difference between A and B.

**(c)**Use answer to (b) to calculate speed of electron as it reaches point B.

**(d)**

(i) Use Fig to determine value of d
at which electron has maximum acceleration.

(ii) Without any further
calculation, describe variation with distance d of acceleration of electron.

**Reference:**

*Past Exam Paper – June 2007 Paper 4 Q3*

__Solution 26:__**(a)**

The electric field strength is equal
to the negative potential gradient

{E = - V / d}

**(b)**

Area under the line of graph = 21.2
cm

^{2}± 0.4 cm^{2}
{On the x-axis (the
position axis), 10 squares represent 2cm of the quantity d. But if we measure
with a ruler, 1cm on the graph paper corresponds to 5 squares. So, 1cm (the
length measured with ruler) on the x-axis represents [(2/10) x 5 =] 1.0cm of
the quantity d. In metre this is 1.0 × 10

^{–2}m of the quantity d.
Similarly, on the y-axis,
10 squares represent 5kVm

^{-1}of the quantity E. Again, if we measure with a ruler, 1cm on the graph paper corresponds to 5 squares. So, 1cm (the length measured with ruler) on the y-axis represents [(5/10) x 5 =] 2.5 kVm^{-1}of the quantity E.}From the graph, 1.0 cm

^{2}represents ([1.0 × 10

^{–2}] × [2.5 × 10

^{3}] =) 25 V

Potential difference between A and B
(= 21.2 x 25) = 530 V

**(c)**

Kinetic energy of electron, ½ mv

^{2}= qV
½ (9.1x10

^{-31}) v^{2}= (1.6x10^{-19}) (530)
Speed of electron, v = 1.37x10

^{7}ms^{-1}**(d)**

(i)

{Electric force, F = (Electric
field strength, E) x (charge , q). F = ma. So, force and hence, acceleration
are maximum when E is maximum}

Electron has maximum acceleration at
d = 0

(ii)

The acceleration of the electron
decreases then it increases.

Some quantitative analysis (e.g. it
is minimum at 4.0cm)

__Question 27: [Electric field > Electric potential]__**(a)**Define electric potential at a point.

**(b)**Two isolated point charges A and B are separated by distance of 30.0 cm, as shown.

Charge at A is + 3.6 × 10

^{–9}C. Variation with distance x from A along AB of potential V is shown.
(i) State value of x at which
potential is zero.

(ii) Use answer in (i) to determine
charge at B.

**(c)**Small test charge is now moved along line AB in (b) from x = 5.0 cm to x = 27 cm. Explain value of x at which force on test charge will be maximum.

**Reference:**

*Past Exam Paper – June 2008 Paper 4 Q4*

__Solution 27:__**(a)**

Electric potential at a point is
defined as the work done moving unit positive charge from infinity to the point

**(b)**

(i)

Potential is zero at x = 18cm

(ii)

{At x = 18cm (this
position is defined by the question), the potential is zero. So, at position x
= 18cm, the sum of potential due to A and B = 0.

Note that the position is
taken to be 18cm relative to A. The same position is at a distance = 30 – 18 =
12cm from B.

When calculating the
potential, the distance from the point charge being considered is taken, not
the values describing the positions which are only defined for convenience in
the question.}

V

_{A}+ V_{B}= 0 {Sum of potential at x = 18cm is zero}
{Potential, V due to point
where charge is q, at distance x from that point = q / 4πϵ

_{o}x. From B, the distance is 30 – 18 = 12cm}
[(3.6x10

^{-9}) / 4πϵ_{o}(18x10^{-2})] + [q / 4πϵ_{o}(12x10^{-2})] = 0
Charge, q at B = - 2.4x10

^{-9}C**(c)**

Electric field strength = (-)
gradient of graph {potential gradient}

{E = - V / d and electric
force, F = Eq}

Electric force = (charge, q) x
gradient {potential gradient}

OR Electric force is proportional to
gradient

So, electric force is largest at x =
27cm {gradient is maximum there}

__Question 28: [Electric field]__**(a)**Two parallel metal plates P and Q are situated 8.0 cm apart in air, as shown. Plate Q is earthed and plate P is maintained at potential of +160 V.

(i) On Fig, draw lines to represent
electric field in region between plates

(ii) Show that magnitude of electric
field between plates is 2.0 × 10

^{3}Vm^{–1}.**(b)**Dust particle suspended in air between plates. Particle has charges of +1.2 × 10

^{–15}C and –1.2 × 10

^{–15}C near its ends. Charges may be considered to be point charges separated by distance of 2.5 mm, as shown. Particle makes angle of 35° with direction of electric field.

(i) On Fig, draw arrows to show
direction of force on each charge due to electric field.

(ii) Calculate magnitude of force on
each charge due to electric field.

(iii) Calculate magnitude of couple
acting on particle.

(iv) Describe the subsequent motion
of particle in electric field.

**Reference:**

*Past Exam Paper – June 2005 Paper 2 Q6*

__Solution 28:__

**(a)**

(i)

The lines are normal to the plates
and have equal spacing (at least 4 lines). The direction is from (+) plate to
earthed plate

(ii)

Electric field strength, E (= V / d)
= 160 / 0.08 = 2.0x10

^{3}Vm^{-1}**(b)**

(i)

{Electric field is from
positive to negative. So, the field line shows the direction towards a negative
charge. Positive charges are attracted to negatives charges and vice versa. So,
the electric field line shows the direction of the force on a positive charge}

Arrows should have correct
directions {force on the +ve (upper) charge is to the
right [in same direction as field line] and the force on the –ve (lower) charge
is to the left (since –ve charges are attracted to +ve). Note that both
directions are horizontal} with the line of action of the arrows passing
through the charges {each of the arrows drawn should
begin at the respective charges}

(ii)

{Electric field strength,
E is shown (in previous part of the question) to be 2.0x10

^{3}Vm^{-1}}
{Both charges have the
same magnitude, so the magnitudes of the force on them are also similar}

Force, F = Eq = (2.0x10

^{3}) (1.2x10^{-15}) = 2.4x10^{-12}N
(iii)

Magnitude of couple = (one) force x
perpendicular separation (of the 2 forces)

{here, the perpendicular
separation is the vertical separation between the forces [which are horizontal]
and equals [2.5x10

^{-3}]sin35 }
Magnitude of couple = (2.4x10

^{-12}) ([2.5x10^{-3}]sin35) = 3.4(4)x10^{-15}Nm
(iv)

The particle EITHER

**rotates**{in clockwise direction} to align with the field (to become horizontal) OR**oscillates**(about a position) with the positive charge nearer to the earthed plate {since the other plate is +vely charged} / clockwise
Can you explain how to get the area and why does 1.0 cm2 represent ([1.0 × 10^–2] × [2.5 × 10^3] in June 2007 Q3?

ReplyDeleteI have included some details. See if you understand now

DeleteI got it. Thank you :)

DeleteJUNE 2007 Q2(d) PAPPER 02 ? HOW TO SOLVE PLEASE?

ReplyDeleteSee solution 871 at

Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-175.html

"Without any further calculation, describe variation with distance d of acceleration of electron." Can you explain why it can't be zero at 4cm? The mark scheme rejects that. Isn't gradient proportional to E which is proportional to F hence a?

ReplyDeleteForce is proportional to E, not change in E.

DeleteGradient = change in E / change in d

O/N 2007 Q3)(b) paper 4, any help will be appreciated.

ReplyDeleteGo to

Deletehttp://physics-ref.blogspot.com/2014/09/9702-november-2007-paper-4-worked.html

Q27 c) why cant electric force be largest at 5cm? Is the gradient steepest there?

ReplyDeleteIt can also be at 5cm but since the question says that the charge is moved from x = 5.0cm. So, it is better to choose 27cm.

Delete