Suggested Answers to 2014 A Level H2 Physics 9646 PAPER 1 and 2

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H2 BIOLOGY PRACTICAL

1) C 2) A 3) A 4) D 5) D

6) C 7) C 8) D 9) C 10) D

11) B 12) B 13) B 14) C 15) A

16) B 17) A 18) C 19) A 20) D

21) D 22) C 23) B 24) D 25) A

26) D 27) D 28) B 29) A 30) D

31) D 32) C 33) D 34) B 35) D

36) C 37) A 38) D 39) A 40) C

1 ai) Y = 60.0 N

ii) θ = 13°

b) For S to be in equilibrium, Rx = Ry = 0. If weight of S and angle of rope A is maintained, then Rx = X cos 30° > 0

2 a) P-type doping introduces more holes to the semiconductor, increasing the number of charge carriers. This increases the semiconductor’s ability to conduct electric current.

b) When forward biased, the holes in the p-type will be attracted to the negative terminal of the battery while electrons of the n-type will be attracted towards the positive terminal. A large current flows through and depletion region is decreased.

When battery is reversed, the holes in p-type are attracted to negative terminaal and elecrtons of the n-type are attracted to the positive terminal. Width of depletion region increases, junction voltage increases. No current flow.

3 a) As R increases from 0, V of qx decreases, V of xq increases. Since the terminal pd is V(wq) = V(wx) + V(xq) = 5.0 + V(xq) , V(wq) increases with R.

bi) 4.84V

ii) 96.8%

ci) 1.2V

ii) 0.531m

iii) When J is moved towards Q, L will be more than 0.531m. Hence the V in pj will be greater than in pk. Therefore the pd across K and J is greater than 0, which means there is a current through the ammeter now.

4 ai) After time, the coil would have turned an angle about the axis PQ. The magnetic flux linkage is 120BA cos (2π/T)t.

According to Faraday’s law, the induced emf is directly proportional to the rate of change of magnetic flux linkage through the coil, hence E is a sinusoidal function. The negative sign comes from Lenz’s law which states that induced emf opposes the change producing it.

ii) 1. 0.170V

25.0 Hz

b) 6.94 x 10^-3 T

5 a) The work done per unit mass by an external agent in bringing a mass from infinity to that point.

bi) 1.19 x 10^27 kg

ii) -1.36 c 10^31 J

c) 1.41 x 10^7 m/s

d) Both Gravitational and electric field strength is related to their respective potentials by the negative of their potential gradient.
Gravitational potential is always < 0 whereas electric potential can be either positive or negative.

6 a) Rate of change of A decreases with increasing time. At the start, the system M and the disc moves very quickly. There is alot of resistive force from the fluid for the disc to overcome and thus A decreases at a high rate. As M and the disc moves slower, there is lesser resistive force and thus a slower energy loss, causing A to decrease at a lower rate.

bi) 5.00, 1.8, -4.0

ii) Plot the point ( 5.00 , -4.0 ), Draw the best fit line.

iii) -0.310 kg

iv) Equation 1 represents a straight line graph of ln A against 1/m with a negative gradient and a non-zero y-intercept.

v) 1. 0.124 kg/s

8.62 x 10^-2

c) 5.59 s

d) since t is directly proportional to m, as m increases, t also increases.

The lamp, detector and material are all placed in a black box.

After the lid is put in place, the intensity meter should read zero. The holes which the wires are should be kept as small as possible.

The lamp itself is isolated in a compartment and light can only enter via a converging lens. Distance between light source and material d is 1m. Leave a small gap of 1mm between the detector and material.

Set up the experiment.
Switch on the lamp and measure the intensity.
Switch off the lamp, put the slab in and measure I through the slab.
Use a pair of vernier calipers to measure t.
Repeat steps by using 2 slabs taped together, then 3 slabs, and so on, till a total of 6 results is obtained.
Record readings in a table.
Plot a graph of ln I vs t.