2014 H2 P2 solutions vjc

1 Nov 2014 H2 P2 solutions: 1(a)(i) If force X is zero, then force Y alone must balance the weight. Hence, magnitude of Y = 60.0 N, and direction of Y is upwards. So,  = 90o. (ii) For equilibrium, net horizontal force = 0 ∴ߠݏ݋ܻܿ= ݏ݋ܿܺ30௢ ∴ܻ௛ = 200ݏ݋ܿ30௢ = 173ܰ Also, net vertical force = 0 ∴ܻ௩ +݊݅ݏܺ30௢ −ܹ= 0 ܻ௩ + 200݊݅ݏ30௢ − 60.0 = 0 ∴ܻ௬ = −40.0ܰ ∴ܻ= ටܻ௫ ଶ +ܻ௬ ଶ = ඥ173ଶ + (−40)ଶ = ૚ૠૡ ۼ ߠ݊ܽݐ= ܻ௬ ܻ௫ = −40.0 173 ∴ߠ= −૚૜. ૙࢕ or 347o. 1(b) Force X has a non-zero horizontal component to the right, which must be balanced by non-zero horizontal component of force Y which can only exist if rope B is inclined to the left and not exactly vertical in direction.

2 2(a) p-type doping introduces Group 3 atoms such as indium or boron into the semiconductor material. This introduces new energy levels called acceptor levels, which are just above the valence band of the silicon atoms. Electrons at the top of the valence band of silicon are able to jump up into these acceptor levels, leaving behind holes in the valence band allowing valence band electrons to gain energy from an externally applied electric field and jump into them, implying that the electrons are able to move within the silicon solid. Thus, the electrical conductivity is increased. (b) In a p-n junction, an electrical p.d. exists across the depletion region with the n-side at a higher potential, and this p.d. prevents any net migration of electrons and holes across the junction. When an external p.d. is applied across the p-n junction in the forward-biased direction (see Fig. 1), this p.d. across the junction is reduced (depletion region narrows), making it easier for charges to diffuse across the junction, resulting in a net current flow in the p to n direction. When an external p.d. is applied across the p-n junction in the reverse-biased direction (see Fig. 2), this p.d. across the junction is increased (depletion region widens), making it harder for charges to diffuse across the junction, so that the current is effectively reduced to zero. In this way, the p-n junction can serve as a rectifier. P-type semiconductor N-type semiconductor + Fig. 1: Forward bias P-type semiconductor N-type semiconductor + Fig. 2: Reverse bias

3 3(a) As the resistance of R is increased from 0 to 12 , the terminal p.d. will increase. This is because, with a larger value of R, the external circuit resistance accounts for a larger fraction of the total circuit resistance, and so the p.d. across the external resistance will increase according to the potential-divider rule. (b)(i) By the potential-divider rule, Terminal p.d. = ସ.଴ାଷ.ହ ସ.଴ାଷ.ହା଴.ଶହ × 5.0 = 4.84 V. (ii) Efficiency of power transfer is given by: efficiency = power dissipated in external load total power dissipated =ܲ௘௫௧௘௥௡௔௟ ܲ௧௢௧௔௟ =ܫܸ ܫܧ where V = terminal p.d., and E = e.m.f. of cell. ⇒ efficiency = ܸ ܧ= 4.84 5.0 = 0.968 ≈ ૢૠ % (c) (i) p.d. across PJ = 1.2 V. [Explanation: For the ammeter to read zero current, the p.d. across PJ must be exactly 1.2 V with P at a higher potential than J, so as to counter the e.m.f. of cell C.] (ii) At balance point (current in ammeter = 0), p.d. across PJ = e.m.f. of cell C ܴ௉௃ ܴ௉ொ +ܴ+ ݎ× 5.0 = 1.2 ቀ݈ 1.0ቁ × 3.5 3.5 + 4.0 + 0.25 × 5.0 = 1.2 ∴݈= ૙. ૞૜૚ m (iii) When J is moved towards Q, the p.d. across PJ is now larger than the e.m.f. of cell C. This means that there is now a non-zero net voltage in the branch circuit (with P at a higher potential than J) that results in a current flowing from P to J through the cell C.

4 4(a)(i) According to Faraday's law, the magnitude of the induced e.m.f. in the coil is proportional to the rate of change of its magnetic flux linkage with time. ߃= −ߔ݀ ݐ݀ where magnetic flux linkage, ߔ= ܰܣܤcosߠ and  = angle between the magnetic field B and the normal to the plane of the coil. As the coil rotates, the angle  , and hence the magnetic flux linkage , changes with time, and so an e.m.f. is induced in the coil. The sinusoidal variation of the emf is obtained when the coil is rotated at constant angular velocity. When the normal to the coil’s plane is perpendicular to the magnetic field ( = 90o), the magnetic flux linkage  is changing at the greatest rate, as  varies as cos  , which has the greatest rate of change when  = 90o and 270o (greatest gradient) Hence, the induced e.m.f. in the coil is maximum when  = 90o and 270o [1] [Note to students: In Fig. 4.2, this happens at t = 0 and t = T/2.] When the normal to the coil’s plane is parallel to the magnetic field ( = 0o or  = 180o), the rate of change of the magnetic flux linkage  is instantaneously zero, as  varies as cos  , which has instantaneous zero rate of change when  = 0o and 180o (zero gradient). Hence, the induced e.m.f. in the coil is momentarily zero when  = 0o and 180o. [1] [Note to students: In Fig. 4.2, this happens at t = T/4 and t = (3/4)T.] (a)(ii) 1. Maximum value of induced e.m.f. = 0.050 V cm-1  3.4 cm = 0.17 V. 2. From Fig. 4.2, periodic time = T = 8.0 ms cm-1 5.0 cm = 40.0 ms = 0.0400 s. frequency = ଵ ் = ଵ ଴.଴ସ଴଴ = 25 Hz. (b) The maximum induced e.m.f. is given by (1) as: ߃଴ =߱ܰܣܤ where ߠ݊݅ݏ= 1, as ߠ= 90௢(normal to coil plane normal to field), where BAN = maximum magnetic flux linkage. Hence, using the answers to (a)(ii), ߃଴ =߱ܰܣܤ  0.17 =ܤ× 1.3 × 10ିଷ × 120 × 2ߨ× 25 

5 5(a) The gravitational potential at a point is defined as the work done by an external agent to bring unit mass from infinity to the point. (b) When r = 2.0  108 m,  = 6.4  108 J kg-1. Using: ߔ= −ܯܩ ݎ −6.4 × 10଼ = − 6.67 × 10ିଵଵ ×ܯ 2.0 × 10଼ ܯ= ૚. ૢ૛ × ૚૙૛ૠ kg (b)(ii) KE is found from: centripetal force = gravitational force ݒ݉ଶ ݎ= ݉ܯܩ ݎଶ ⇒ܧܭ= 1 2ݒ݉ଶ =݉ܯܩ 2ݎ ⇒݈ܽݐ݋ܶ ݕ݃ݎ݁݊݁= ܧܭ+ ܧܲ= ݉ܯܩ 2ݎ+ ൬−݉ܯܩ ݎ൰ ⇒݈ܽݐ݋ܶ ݕ݃ݎ݁݊݁= ܧ= −݉ܯܩ 2ݎ ⇒ܧ =− 6.67 × 10ିଵଵ × 1.92 × 10ଶ଻ × 8.93 × 10ଶଶ 2 × 4.22 × 10଼ ⇒ܧ= −૚. ૜૞ × ૚૙૜૚ J (c) By the law of conservation of energy, initial total energy = final total energy ܧܭଵ +ܧܲଵ =ܧܭଶ +ܧܲଶ 1 2݉ݒଵ ଶ +ݍܸଵ = 1 2݉ݒଶ ଶ +ݍܸଶ 1 2݉ݒଶ = 1 2݉ݒଶ ଶ +ݍ(ܸଶ −ܸଵ) where v2 = final velocity = 0. 1 2݉ݒଶ = 0 +ݍ(ܸଶ −ܸଵ)

6 ݒ= ඨ2ݍ(ܸଶ −ܸଵ) ݉ ݒ= ඨ2 × 1.6 × 10ିଵଽ(1.02 × 10଺ − 0) 1.67 × 10ିଶ଻ as ܸ= 1.02 × 10଺J C-1 when ݔ= 1.4 × 10ିଵହ m, and ܸ≈ 0 when ݔ= 1.0 × 10ିଵ଴ m, as this is a great distance away from S compared to the final distance. Hence, initial speed, ݒ= ૚. ૝૙ × ૚૙ૠm s-1. (d) Similarity: For both gravitation and electricity, the magnitude of the potentials decreases inversely as the distance from the source of the potential increases. Difference: For gravitation, the value of the potential at any distance from the source is always negative, whereas the potential due to the electric charge S is positive as the charge of object S is positive in this case.

7 6. (a) The rate of change of A is initially very high but decreases as time passes. This is because the gradient of the graph decreases with time. [Additional information for students: This is because the average speed of the mass's oscillation is initially very high, and this results in a large resistive force due to fluid drag. This large initial resistive force causes the energy of the system to be dissipated at a very high rate, resulting in a high rate of decrease in the amplitude A initially, as the system’s energy E and the amplitude A are related via the relationship: ܧ= ଵ ଶ݉߱ଶܣଶ] (b)(i) For m = 200 g, at t = 5.0 s, 1 ݉= 1 0.200 kg = ૞. ૙૙ ܓ܏ି૚ ܣ= ૚. ૡ × ૚૙ି૛ m. ݈݊(ܣm⁄ ) = −૝. ૙ (b)(ii) (iii) Using two points on the best fit line, (0.00, -2.50) and (8.90, -5.20), Gradient = (−5.20) − (−2.50) 8.90 − 0.00 = −૙. ૜૙૜ ܏ܓ

8 (iv) According to the expression, ܣ= ܣ଴݁ି௕௧ ଶ⁄ ௠ ⇒ܣ݈݊= ݈݊ܣ଴ + ൬−ݐܾ 2݉൰ ⇒ܣ݈݊= ൬−ݐܾ 2 ൰ 1 ݉+ ݈݊ܣ଴ Hence, according to the expression, a graph of ln A against ଵ ௠ would be a straight line with a negative gradient − ௕௧ ଶ an