Monday, October 27, 2014

9702 June 2014 Paper 23 Worked Solutions | A-Level Physics

  • 9702 June 2014 Paper 23 Worked Solutions | A-Level Physics

Paper 23

Question 1
Underline all base quantities in following list::
ampere             charge              current             mass    second             temperature     weight

Potential energy EP stored in stretched wire is given by EP = ½Cσ2V where C is a constant, σ is strain, V is volume of wire. SІ base units of C:
Base units of σ: no units
Base units of V: m3
Base units of Ep: kgm2s-2
Base units of C: kgm2s-2 x m-3 = kgm-1s-2

Question 2
{Detailed explanations for this question is available as Solution 1016 at Physics 9702 Doubts | Help Page 212 -}

Question 3
{Detailed explanations for this question is available as Solution 608 at Physics 9702 Doubts | Help Page 120 -}

Question 4
Spring hangs vertically from point P, as shown. Mass M is attached to lower end of spring. Reading x from metre rule is taken, as shown. Fig shows relationship between x and M.
How apparatus in Fig may be used to determine load on spring at elastic limit:
Small masses can be added to cause the extension. Then, the masses hare removed to see if the spring returns to its original length. This is repeated for larger masses and the maximum mass for which, when the load is removed, the spring does return to its original length, is noted.

Explain whether Fig suggests that spring obeys Hooke’s law:
Hooke’s law requires the force to be proportional to the extension. The graph shows a straight line, hence the spring obeys Hooke’s law.

Use Fig to determine spring constant, in Nm–1, of spring:
Spring constant, k (= gradient) = force / extension
k = (0.42x9.81) / [(30 – 21.2)x10-2] = 47 (46.8) Nm-1

Question 5
Why terminal potential difference (p.d.) of cell with internal resistance may be less than electromotive force (e.m.f.) of cell:
There are lost volts / energy used within the cell / internal resistance when the cell supplies a current

Battery of e.m.f. 4.5 V and internal resistance r is connected in series with resistor of resistance 6.0 Ω, as shown. Current I in circuit is 0.65 A.
Internal resistance r of battery:
e.m.f. E = I (R + r)
4.5 = 0.65 (6.0 + r)
Internal resistance r = 0.92Ω

Terminal p.d. of battery:
V = IR and current I = 0.65A
Terminal p.d. = 0.65 x 6 = 3.9V

Power dissipated in resistor:
Power dissipated, P = V2/R    or P = I2R        or P = IV
P = (3.9)2/6 = 2.5W

Efficiency of battery:
Efficiency of battery = power out / power in
Efficiency of battery = I2R / I2(R + r) = R / (R + r) = 6.0 / (6.0 + 0.92) = 0.87

Second resistor of resistance 20 Ω connected in parallel with 6.0 Ω resistor in Fig. Explain qualitatively change in heating effect within battery:
The (circuit) resistance decreases. So, the current increases, causing more heating effect.

Question 6
{Detailed explanations for this question is available as Solution 533 at Physics 9702 Doubts | Help Page 104 -}


  1. Can you explain Q6d(ii). Why are we multiplying wavelength with 5/4

  2. in 3(b2) isn't the loss in PE = KE - PE because the whole PE doesn't get converted to KE in 2.5s please help!

    1. Details added.

      From the conservation of energy, KE cannot be greater than PE here. + there is a resistive force. Work needs to be done against it. and even if there was no resistive force, the KE would still not be greater than PE.

      The equation you gave gives a negative energy, which is wrong.

      See the details added

  3. could you please explain why, in question 2(b) (i), 22 is taken as the vertical velocity and not the resultant velocity

  4. Could you please explain why the Phase difference between P and Q is zero? question 6(c)

    1. This is from the properties of points within a loop of a stationary wave. At the next instant, both points would be moving down. They won’t move right or left since the wave is stationary. So the 2 points are in phase (phase difference = 0).


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