# Physical Sciences Grade 11 P1 September Test 2016 QP & Memo

> Grade 11 Physical Sciences Paper 1 question paper and memo from the September Test 2016, 12 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/phy-p1-sept-qp-memo/5cd9cd8c-30c0-4707-b71c-0c3e6b7007d2
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2016
- Exam period: September Test
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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## About this document

Physical Sciences · Grade 11 · September Test · 2016. Question paper and memorandum, 12 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

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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

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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.

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Physical Sciences P1 NSC

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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

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(2)

(2)

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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]

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(

Physical Sciences P14 NSC

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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]

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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.

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(3)

(3)
(3)

TOTAL MARKS: 50

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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...

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Physical Sciences P1

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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

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_

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.

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