Physical Sciences Grade 11 P1 September Test 2016 QP & Memo

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Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA " PHYSICAL SCIENCES P1 m (PHYSICS) : : COMMON TEST ‘ i SEPTEMBER 2016 : “Sam eeeemeceneweeenenaneamneeeeneunenenmenee” NATIONAL SENIOR CERTIFICATE ( (3 “MARKS: 50 This question paper consists of 6 pages and 2 data sheets. Copyright reserved Please turn over

~ Downloaded from hlayiso.com Physical Sciences P14 NSC September 2016 Common Test , 2 INSTRUCTIONS AND INFORMATION 41, Write your name in the appropriate space. on the ANSWER BOOK. 2. This question paper consists of 4 questions. Answer ALL the questions in the ANSWER BOOK. 3. You may use a non-programmable calculator. 4, You may use appropriate mathematical instruments. 5. YOU ARE ADVISED TO USE THE ATTACHED DATA SHEETS. 6. Number the answers correctly according to the numbering system used in ¢ this question paper. c 7. Write neatly and legibly. Copyright reserved

Physical Sciences P1 NSC Downloaded from hlayiso.com 3 QUESTION 1 Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A—D) next to the question number (1.1—1.3) in the ANSWER BOOK, for example 1.3 D. 1.1. A neutral metal sphere, M, is placed on an insulated stand. An identical metal sphere, N, carries a charge of +Q as shown. Le M and N are now brought into contact. What is the magnitude of the charge on M when M fnoves away? A Q B VQ Q Cc 3 D 0 1.2. Abar magnet is moved out of a coil, as shown in the diagram below. X and Y are two points on the conductor. <— x Y Which ONE of the following CORRECTLY describes the direction of the current and the polarity of the left end of the coil? September 2016 Common Test DIRECTION OF CURRENT POLARITY OF LEFT END OF COIL A X to Y¥ north B X to Y south Cc Y toX north D YtoX south Copyright reserved (2) (2)

~ Downloaded from hlayiso.com Physical Sciences P1 NSC September 2016 Common Test 1.3. The two resistors in circuit 1 below are identical. They are connected in series to a cell of emf V and negligible internal resistance. The power dissipated by each resistor is P. Circuit 1 The two resistors are now connected in parallel, as shown in circuit 2 below. Circuit 2 C hd ik R | Tensmnssnnnal R ——{- The power dissipated by each resistor in the circuit 2 is ... A 2. BoP, CBP, ( DP, (2) [6] Copyright reserved

( Physical Sciences P14 NSC Downloaded from hlayiso.com September 2016 Common Test 5 QUESTION 2 2.1 2.2 Draw the electric field pattern associated with a negative point charge. (2) Three point charges A, B and C of magnitudes -2 pC: + 4 uC and -2 uC respectively are placed in the diagram as shown. 2.2.1 State, in words, Coulomb’s Law of Electrostatics. (2) 2.2.2 Calculate the magnitude of the Coulombic Force that A exerts on B. (3) 2.2.3 Calculate the magnitude of the net Coulombic Force that charge B experiences as a result of the other two charges. (3) 2.2.4 Calculate the strength of the net electric field at B. (3) [13] { JUESTION 3 ) “A flat, rectangular coil consisting of 150 turns has the dimensions 25 cm by 30 cm. It is ina uniform, 1,20 T, magnetic field, with plane of the coil parallel to the field. In 0,2 s, itis rotated so that the plane of the coil is perpendicular to the field. 3.1 3.2 State Faraday's law of electromagnetic induction in words. (2) Calculate the: 3.2.1. change in magnetic flux through the coil during this rotation. (3) 3.2.2 emf induced in the coil during this rotation. (3) 3.2.3 current induced in the coil during this rotation if the coil is connected to a resistor of 50. (3) [11] Copyright reserved

Downloaded from hlayiso.com mon, Physical Sciences P1 NSC September 2016 Common Test 6 QUESTION 4 A battery of emf 2,4 V and negligible internal resistance is connected in a circuit as shown below. A current of 2A registers on ammeter Ay. 44 42 4.3 4.4 4.5 State Ohm’s law in words. Calculate the effective resistance of the circuit. Calculate the reading on: 4.3.1. The voltmeter. 4.3.2 The arnmeter Ao. Calculate the amount of energy (in kWh) that is transferred to the circuit in 48 hours. A p—k-|+-+ 60 4+ a) SF {V) The 3 Q resistor is now replaced by a 6 O resistor? How will this affect each of the following? (Choose from: INCREASES, DECREASES OR REMAIN THE SAME) 4.5.1. The reading on Ay Explain the answer. 4.5.2 The reading on Ag. Copyright reserved (3) (3) (3) TOTAL MARKS: 50

wt Downloaded from hlayiso.com Physical Sciences P14 NSC 7 September 2016 Common Test DATA FOR PHYSICAL SCIENCES GRADE 11 TABLE 1: PHYSICAL CONSTANTS PAPER 1 (PHYSICS) NAME SYMBOL VALUE Acceleration due to gravity g 9,8 ms” Gravitational constant G 6,67 x 107 N.m*.kg? Coulomb's constant k 9,0 x 10°N-m*-C* Speed of light ina vacuum c 3,0x 10° ms" Charge on electron e -1,6x 10°C ( jlectron mass Me 9,11 x 107" kg Radius of earth Re 6,38 x 10° m Mass of earth Me 5,98 x 10” kg TABLE 2: FORMULAE ELECTROSTATICS k F poms (k= 9,0 x 10°N-m*.C”) ERG E= 12 (k=9,0x 10°Nm2c% | y= r Q 4 JECTROMAGNETISN e=-nde2 ®=BAcos0 At CURRENT ELECTRICITY =k Row At Al41j414, Rf, lb ly Rett th tt... Copyright reserved

Physical Sciences P1 Downloaded from hlayiso.com NSC September 2016 Common Test 8 w W=eV at 4 At We=VIAt 2 P=VI W=FPRAt ; p=PR VAt = 2 w R pi R Copyright reserved

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Downloaded from hlayiso.com _ Physical Sciences P1 Nsc Septembéreggs’6 Common Test 3 224 Est Q _ 1018,23 4x10° = 254,5 x 10°N.C* QUESTION 3 [13] 3.1. The induced emf in a conductor is directly proportional to the rate of. change of magnetic flux through the circuit. Vv” 3.2.1 4, = BCos®. A = 1,20 Cos 90°. A= 0 Why 9, = BCosdé. A= 1,20 Cos 0 x (0,25 x 0,30) ¥ = 0, 09 Wb Ad = $F - $= 0,09 -0 = 0,09 Wbv MARK POSITIVELY FROM 3.2.2 x 3.22 e=-N 52415092 5 -67.50V¥ 323 T= = 8 1,35 AY a arr ae QUESTION 4 4.1 The current in a metal conductor is directly proportional to the potential difference across the ends of the conductor, provided the temperature remains constant. vv (2OR 0) Copyright reserved Please turn over @). (3) @) (3) [14] 2) aw Physical Sciences P1 Nee NSC September 2016 Common Test + 4 | 1 420 — = ate = Rp R ' 1 2 44gy Rp 63 Rp=20 j i Effective Resistance = Rp + Rs=2+ 1v =30V | @) OR | = §424v_ s0v | Req = T= 9g = 90 i 4.3.1 V=IRiv= 0,8 x 2v = 1,6 ' @) | / | OR } =IRY =0,8x1 =0,8W, therefore V=2,4-0,8= 1.6 V¥ @) oO I MARK POSITIVELY FROM 4.3.1 43.2 i av. lv o=48 v @) 1=0,27A 44 P=VW~=2,4x0,8¥= 1,92 W = 0,00192kW = Pi” = 0,00192 x 48v = 0,09216 kWhyY . (6) 4.5.1 { Decrease Vv { Resistance of circuit increases. “From Ohms Law, R inversely proportional to current.” (3) ~ 4.5.2. Increase, ¥ (1) [20) ” Copyright reserved | Please tum over.