Wednesday, November 5, 2014

9702 June 2009 Paper 22 Worked Solutions | A-Level Physics

  • 9702 June 2009 Paper 22 Worked Solutions | A-Level Physics

Paper 22

Question 1
2 of SI base quantities and their units are mass (kg) and length (m).
3 other SI base quantities and units:
Examples: time (s), current (A), temperature (K), amount of substance (mol), luminous intensity (cdl)

Pressure p due to liquid of density ρ is related to depth h by expression      p = ρgh, where g is acceleration of free fall. Determine derived units of pressure:
Density ρ = mass / volume
Unit of density ρ: kg m-3
Unit of acceleration g: m s-2
Unit of pressure P: kgm-3 ms-2 m = kgm-1s-2

Question 2
{Detailed explanations for this question is available as Solution 1015 at Physics 9702 Doubts | Help Page 211 -}

Question 3
{Detailed explanations for this question is available as Solution 1091 at Physics 9702 Doubts | Help Page 232 -}

Question 4
Spring having spring constant k hangs vertically from fixed point. Load of weight L, when hung from spring, causes extension e. Elastic limit of spring is not exceeded.
An elastic deformation is the change of shape / size / length / dimension when a (deforming) force is removed, returns to original shape / size

Relation between k, L and e:
L = ke

Some identical springs, each with spring constant k, are arranged as shown. Load on each of the arrangements is L. For each arrangement, complete table by determining (i) total extension in terms of e, (ii) spring constant in terms of k.

{Force, L = (spring constant, k) x( extension, e). L is constant in all cases. e can be determined from the diagrams. E.g. 2 springs in parallel causes extension to be halved while 2 springs in series causes extension to be doubled. Spring constant, k is given by k = F / extension. In this question, F is constant – so, k is the reciprocal of the extension} 

Arrangement 1:
Extension = 2e
Spring constant = ½ k

Arrangement 2:
Extension = ½ e
Spring constant = 2k

Arrangement 3:
Extension = (3/2) e
Spring constant = (2/3) k

Question 5
Double-slit interference experiment is set up using coherent red light as illustrated.  Separation of slits is 0.86 mm. Distance of screen from double slit is 2.4 m. A series of light and dark fringes is observed on screen.
Coherent light is light having a constant phase difference

Estimate separation of dark fringes on screen:
Wavelength estimate 750nm to 550nm
Separation = λD / x = (650x10-9)(2.4) / (0.86x10-3) = 1.8mm

Initially, light passing through each slit has same intensity. Intensity of light passing through one slit now reduced. Explain effect, if any, on dark fringes observed on screen:
There is no longer complete destructive interference / the amplitudes no longer completely cancel. So, the dark fringes are lighter.

Question 6
2 vertical parallel metal plates are situated 2.50 cm apart in vacuum. Potential difference between plates is 350 V, as shown. An electron is initially at rest close to negative plate and in uniform electric field between plates.
Magnitude of electric field between plates:
E = V / d = 350 / (2.5x10-2) = 1.4x104NC-1

Show that force on electron due to electric field is 2.24x10-15N:
Force = Eq = (1.4x104) (1.6x10-19) = 2.24x10-15N

Electron accelerates horizontally across space between plates.
Horizontal acceleration of electron:
Force, F = ma
a = (F/m =) (2.24x10-15) / (9.1x10-31) = 2.46x1015ms-2

Time to travel horizontal distance 2.50cm between plates:
s = ½at2
2.5x10-2 = ½ (2.46x1015) t2  
Time, t = 4.5x10-9s

Why gravitational effects on electron need not be taken into consideration in calculation in (b):
EITHER The gravitational force is normal to the electric force OR The electric force is horizontal while the gravitational force is vertical.

Question 7
Network of resistors, each of resistance R, shown. The switches S1 and S2 may be ‘open’ or ‘closed’. Resistance, in terms of R, between points X and Y for switches in positions shown:
(R + R =) 2R
([1/2R + 1/2R]-1 =) R

2 cells of e.m.f. E1 and E2 and negligible internal resistance connected into network of resistors. Currents in network are as indicated. Use Kirchoff’s laws to state relation between:
I1, I2, I3 and I4:
I1 + I3 = I2 + I4

E1, E2, R and I3 in loop NKLMN:
E2 – E1 = I3R

E2, R, I3 and I4 in loop NKQN:
E2 = I3R + 2I4R

Question 8
Spontaneous and random decay of radioactive substance involves emission of either α-radiation or β-radiation and/or γ-radiation.
For a spontaneous decay, the rate of decay / activity / decay (of nucleus) is not affected by external factors / environment / surroundings

Type of emission that:
Not affected by electric and magnetic fields:
Gamma / γ

Produces greatest density of ionization in medium:
Alpha / α

Does not directly result in change in proton number of nucleus:
Gamma / γ

Has range of energies, rather than discrete values:
Beta / β


  1. for question 2.c)i) why is the velocity 4.2 and not 4?

    1. The solution has been updated.

      Both are accepted, but 4.2 is the correct value

  2. along with the mark scheme answers, can you please explain how to get to the answers too, please...

  3. q7 part a how is the resistance infinity? cant it be zero?

    1. When the switches are open, no current would flow. This means that the resistance is infinite.

      If the resistance was zero, an infinite amount of current would flow.


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