G.C.E. (A.L.) Support Seminar - 2014Physics - Paper I

Answer Guide

Question No

Answer Question No

Answer

(1) 1 (26) 4

(2) 2 (27) 1

(3) 5 (28) 5

(4) 4 (29) 3

(5) 5 (30) 3

(6) 3 (31) 2

(7) 2 (32) 1

(8) 3 (33) 2

(9) 2 (34) 4

(10) 2 (35) 3

(11) 5 (36) 1

(12) 3 (37) 5

(13) 1 (38) 4

(14) 2 (39) 1

(15) 5 (40) 2

(16) 2 (41) 4

(17) 1 (42) 4

(18) 2 (43) 5

(19) 5 (44) 5

(20) 5 (45) 2

(21) 4 (46) 4

(22) 4 (47) 4

(23) 5 (48) 3

(24) 5 (49) 3

(25) 5 (50) 5

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1. (a) (i) π + hρg + 0 = π + 12

ρv2

v = √2gh

(1 mark)

(1 mark)

(ii) Bernoulli's principle (1 mark)

(iii) Non viscous, Incompressible fluid A steady and a stream line flow (1 mark)

(b) (i) s = ut + 12

at2

(H − h) = 12

× gt2

t = √ 2 (H − h) g

Using

(1 mark)

(ii) s = ut

x = √2gh √ 2 (H− h) g

x = √h (H − h)2

Using

(1 mark)

(c) (i) h (cm) 0 2 4 10 16 18 20

x (cm) 0 12 16 20 16 12 0

For 4 correct answers(1 mark)

(ii) (ii)

0

2

4

6

8

10

12

14

16

18

20

2 4 6 8 10 12 14 16 18 20

x (cm)

h (cm)

For plotting the graph (1 mark)

Physics - Paper II Answer Guide

Part A - Structured Essay

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(iii) Value become maximum when ^h = 10 cm& or h have two values for one value of x' (1 mark)

(d)Water will not leave the container or velocity is zero. (1 mark)

2. a) (i) Heated water can splash out from the tube A. (1 mark)

(ii) To preventing build up of pressure inside the steam generator/ for the protection or To maintain a pressure equal to the atmospheric pressure inside the steam generator. (1 mark)

(iii)

D

thermometer

tip of the tube must be kept slightly above the water surface

stirrer

steamtrap

B - glass tube

bent pipe

A - glass tube

C - steamgenerator

E - calorimeter

(1 mark)

(b) Asbestos sheet/ Insulating sheet To minimise the flow of heat (by radiation) from the bunsen burner to the calorimeter (1 mark)

(c) Begin the experiment by reducing the temperature of water 5 °C below ambient temperature and continue to send steam until it's temperature is5 °C above the ambient temperature. (1 mark)

(d)(i) 1.Mass of the empty calorimeter + stirrer = m1

2.Mass of the empty calorimeter + stirrer + water = m2

3. Initial temperature of water = θ1 °C 4. Final maximum temperature of the system = θ2 °C

5.Mass of the empty calorimeter + stirrer + water +steam = m3

(Award 2 marks if all 5 or the first 4 are correct, 1 mark if the first 3 are correct) (2 marks)

(ii) m1 C (θ2 − θ1 ) + (m2− m1 ) Cw (θ2 − θ1 ) = (m3− m2 )L + (m3− m2 ) Cw (100 − θ2 ) (1 mark for each side of the equation)(2 marks)

(e) (Since Latent heat of vaporization of water is high) The mass of vapour collected is small. Hence the fractional error/ percentage error is large. (1 mark)

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3. (a) (i) Transmission is coherent, in same phase and linear For any two (1 mark)

(ii) In eye surgery

For decorations In surgery to remove bladder stones For any two applications

(1 mark)

or any practical application

(b) (i) 50°

X

A

B C

Ys

dr

For completing the path of a beam. (1 mark)

(ii) For the positions of X and Y. (1 mark)

(c) tan d = sr Any form (1 mark)

d = tan−1 sr

(d) (i) The beam incident on AC undergo total internal reflection, when it incident exceeding the critical angle. (1 mark)

(ii) If angle of incidence on the surface AC = θ "

n sin θ = 1 sin 90° (1 mark)

50 °

45°85°

A

B

C

θ

√2 sin θ = 1

sin θ = 1√2

θ = 45° (1 mark)

(iii) Angle of refraction at the surface AB = 5° (1 mark)

(iv) If the prism must rotate by an angle α, 1×sin α = √2 sin 5° sin α = √2 sin 5° (1 mark)

4. (a) τ = BINA cos α (1 mark)(b) τ

α

or for the sinusoidal wave (1 mark)

(c) (i) Use semi-cylindrical magnetic poles/ Use a radial magnetic fields. (1 mark)

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(ii) BINA = C θ I = C θ BNA (1 mark)

(iii) Factor Adjustment

N

A

C

B

Should increase

Should increase

Should decrease

Should increase for 4 or 3 correct answers (1 mark)

(d) (i) 100×10 = 900×R

900mA

100mA10Ω

R

1AG

10 9

= R

R = 1.11 Ω

Resistance that must be connected = 1.11 Ω (1 mark) It must be connected in parallel with the Galvanometer (Award the mark if shown in a diagram) (1 mark)

(ii) Variable resistor and a cell. (1 mark)

(iii) Scale is non - liner or Calibrated in opposite direction (deflection to opposite direction) (1 mark)

(iv) θ

R0

(1 mark)

* *

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Part B - Essay

5. (a) (i) Mineral oil, coal, solar energy For any two (1 mark)

(ii) Durability

For any three Economical benefits

Minimum environmental pollution (1 mark)

easy access to the source Minimum initial capital

(b)(i) π r2 v ρ (1 mark)

(ii) 12

× π r2 v ρ × v2 = 12

π r2ρ v3 (1 mark)

(iii)12

π r2ρ v3 × 80100

(1 mark)

(iv) 12

× 22 7

× (1.4)2 × 1.2 × 10

3 × 80

100

2956.8 W (1 mark)

(c) (i) 2956.8 × 10 = 1 2

I × 102

(1 mark)

I = 591.36 kg m2 (1 mark)

(ii) ω = ω0+ α t

10 = 0 + α × 10 α = 1 rad s−2 (1 mark)

(iii) τ = I α (1 mark)

= 591.36 × 1 = 591.36 N m (1 mark)

(d) (i) 591.36 − 443.52 = 591.36 × α

α = 0.25 rad s−2 (1 mark)

(ii) ω = ω0+ αt ω = 0 + 0.25 × 100 = 25rad s −1 For the final answer (1 mark)

(iii) 1

2× I × ω2 = τf ×

θ Or any other method (1 mark)

12

× 591.36 × 25 × 25 = 443.52 × θ

θ = 416.667 rad

n = 416.667 2 × π

= 66.2878 For dividing by 2π (1 mark)

66

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6. (a) T = Tension of the string m = Mass per unit length of the string or linear density

V = Tm √

(1 mark)

(b) (i) Fundamental 0.5 m

V = f λ 350 = f × 0.5 × 2 f = 350 Hz (1 mark)

1st over tone 350 = f × 0.5

2 × 2 f = 700 Hz (1 mark)

(2nd over tone cannot be 1050 Hz&

(ii) The maximum possible frequency for the 2nd over tone must be 1000 Hz. (1 mark) If the tension for that is T,

Tm √ = 1000 × × 20.5

3 1

(1 mark)

Initial situation

√ = 3500.15 m

2 (1 mark)

T 1

2

10002 × 0.15

3502 × 3

2

100735

0.136 N

=

=

=

by

(1 mark)

(c) (i) 350 √0.15

m =

350 √ 0.158000A

= (1 mark)

Density Mass

=Volume

8000 = m A × 1 (1 mark) A = Area of cross-section of the wire

= Y .

FA

0.15 A

0.49100

0.15 × 8000 × 3502

= Y × 0.15

2 × 1011

N m−2

Y = 8000 × 3502 × 100

0.49

0.49100

el

= Y . Since

By substitution (1 mark)

(1 mark)

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(ii) Using the equation V = Eρ √ (1 mark)

√ 2 × 1011

8000 =

= 5000 m s−1 (1 mark)

(iii)

>

>

= 2500 Hz

V 4l

f =

50004 × 0.5=

>

>

= 7500 Hz

3V 4l

f =

(1 mark)

(1 mark)

7. (a) (i) 2Tcos θ r 2T rρg

hρg =

h = (1 mark)

(ii)

h1

PA = π (atmospheric pressure)PB = π − 2T

rPC = π − 2T

r + h1ρg

PC = π − hρg + h1ρgPD = π

I. If h1 = h PC − PD

II. If h1 < h PD > PC

III. If h1 > hPC > PD

(r1 = Radius of the meniscus at the lower end&

plane meniscus.For the shape(1 mark)

concave meniscus.For the shape(1 mark)

convex meniscus.For the shape (1 mark)

... ... ... ... ... ... ... ...

... ... ... ... ... ... ...

... ... ... ... ... ... ... ...

A

B

CD

For developing the argument (1 mark)

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(iii)

According to the situation III above,

PC − PD = 2T r

(For the argument) (1 mark)

(h1 −h ) ρg = hρg

h1 = 2h (1 mark)

(iv) r1

r

h 2h0 h1

(For the shape) (1 mark)

(b) (i) Pressure in the liquid at Z = atm. pressure + 2T r

105 +

100280 N m−2

2 × 7 × 10−2

0.5 × 10−3 =

=

For the substitution

(1 mark)

(1 mark)

(ii) Pressure in the liquid at Y = 100280 N m−2 (1 mark)

Pressure in the liquid at X = 100280 N m−2 (1 mark)

(iii)

100280

100000

ZYX B

Pressure

Distance

For the shape (1 mark)

^Separations between points X, Y, Z, B are not necessary to be drawn to a scale'&

(iv) If the force on the piston, for the pressure at X to become 100280 is F"

100280 − 105 = F

π × 10 2 ( )× 10

− 3 2

F = 0.022 NF = 22 × 10 N− 3

For the substitution (1 mark)

(1 mark)

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8. (a) (i) FE = Eq (1 mark)

(ii) FB =BqV (1 mark)

(iii) ×

×

× ×

× ×

× ×

FB

FE

(1 mark)

(iv) FE > FB FE < FB FE = FB

(1 mark for each) (3 marks)

(b)

(1 mark)

To escape,

FE = FB

Eq = Bqv

E = Bv orv = EB

To not to escape,FE > FB

E > Bv or v < EB

FE < FB

E < Bv or v > EB

(Award this mark for any one/ both relationships) (1 mark)

(c) (i) Shows a circular motion (path). (1 mark)

(ii) Since the magnetic force is always perpendicular to the direction of motion/ velocity (1 mark)

(iii) B0qv = mv2

d/2 B0q = m × E d/2 B m = B0 B d q 2E

(1 mark) For the substitution of EB for v

(1 mark)For using the equation

(1 mark)

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(d) According to the results obtained in part (iii) above mα d'when mass increases, the distance d incident on the photographic plate too increases. (1 mark)

(e) Can be used for isotope separation, since it can identify ions of same charge but of different masses. (1 mark)

9. (A)(a)(i)

2 Ω

1.2 Ω 3 V

3 V

2 V

Y

XS

3μ F 2μ F

A

B

C1C2

CI

I

I

P

Q

4 Ω

4 Ω

By applying Kirchhoff's Law, 3 + 2 = I (4 + 2 + 4) I = 0.5 A

Current through cell A = 0 (1 mark)Current through cell B = 0.5A Current through cell C = 0.5A (1 mark)For obtaining the potential difference of 3V, between P andQ (1 mark)

(ii) For C1 Q = CV Q = 3 × 10−6 × 3 = 9 × 10−6C (1 mark) For C2 Q = CV Q = 2 × 10−6 × 3 = 6 × 10−6C (1 mark)

(iii) For obtaining the potentials at X andY are equal (1 mark) ∴ charge remains the same (1 mark)

(b) Same values as above (1 mark) (Since the potential difference between the plates of the capacitor is the same value)

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(c) (i)

2 Ω

1.2 Ω

R Ω

3 V

3 V

2 V

A

B

I1

C

I1

(I1−I2)

I24 Ω

4 Ω For indicating the direction of current (1 mark)

By applying Kirchhoff's Law, 6 = I1 1.2 + I2 6 + I1 R 01 (1 mark) 5 = I2 6 − (I1− I2 ) 4 02 (1 mark) When connected to the potentiometer potential drop for 100 cm = 2V

potential drop for 60 cm = 2100 × 60

= 1.2V ∴ I1 R = 1.2 03 (1 mark) By 01, 02 and03 I1 = 0.5A (1 mark) ∴ R = 2.4 Ω (1 mark)

(ii) No Voltmeter does not measure the correct potential difference since it has a finite resistance ∴ R will not be correct. (1 mark)

(B) (a)(i) (I) Potential at C is nearly zero. (1 mark)

(II) Rc = 10 − 220 × 10−3 = 400 Ω (1 mark)

(ii)(I) R2 < R1 (1 mark)

(II) Reason (Inorder to create the requirement for the saturation of the transistor since the resistance of the LDR in slight darkness is not very high. or to increaseVBE (1 mark)

(b)(i)

R1 = 1010 × 10−3 = 1000 Ω = 1kΩ

(12 − 2) = 10 × 10−3 × R

(1 mark)

(ii) VA = (12 − 9) = 3 V (1 mark)

(iii) (I) VA > VB

VQ = +12V For the value and the polarity (1 mark)

(II) VA < VB

VQ = −12V For the value and the polarity (1 mark)

^,l=Kq1&

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(iv) (I) VA > VB

(1 mark)

(II) The buzzer will activate since the B - E junction is then forward biased or if VA < VBthe buzzer will not activate since the B - E junction is then reverse biased. (1 mark)

(v) (I) 4.8 V (1 mark)

(II) The Zener voltage must be a value between 3V and 6V in order to obtain both situations VA > VB andVA < VB (1 mark)

(vi) (I) 0.5 = 4.8 I I = 0.1 A (1 mark)

(II) V = IR 12 − 4.8 = 0.1R For the substitution (1 mark) R = 72 Ω (1 mark)

10. (A) (a) For the Law (1 mark) For the conditions (1 mark)

(b) θ

t

62̊C

(2 marks)

(c)(i) Melting point is62̊C (1 mark)

(ii) dQdt = 0.1×1800 3.6

60 + 60 × 3.660

= 14.4 J s-1

(2 marks)

(1 mark) (iii) = 100 − 14.4 dQ

dt= 85.6 J s-1 (1 mark)

(iv)

1200 JK−1 = C

0.1 × 1800 × 3.6 = 0.1 × C × 4.8 + 60 × 4860 6060

(1 mark) (1 mark)

(v) 62°C 85.6 Wθ °C 100 W

At a certain temperature θ the rate of supply of heat equals to the rate of loss of heat

85.6 α (62 − 30)

100 α (θ − 30)

α (θ − θr)dQdt

(1 mark)

(1 mark)

(1 mark)

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85.6 = 32100 θ - 30(θ − 30) 85.6 = 320085.6 θ = 3200 + 30 × 85.6θ = 67.4

0C

(1 mark)

(B) (a)(i) P = 10 W

N = 10 × 6000 × 10−10

4 × 3 × 102 × 6.63 × 10−34 × 3 × 108

= 2.5 × 1016 m−2 s−1

Using E = hf (1 mark) Using C = fλ (1 mark) For the substitution (1 mark) For the answer (1 mark)

(ii) Number of photons incident on the entire area of the detector

A = 0.4 × 10−4 m2

ND = 2.5 × 1016 × 0.4 × 10−4

= 1 × 1012 (1 mark)

(iii) Number of photo electrons emitted during 1 second

For the substitution = 1 × 1012 × 0.9= 9 × 1011 electrons / sec

(1 mark)

(1 mark)

(iv) I = neePhoto current = 9 × 1011 × 1.6 × 10−19

= 14.4 × 10−8 A (1 mark)

(b) (i)

∴ Number of photons= 0.16.63 × 10−34 × 3 × 108

490 × 10−9 = 2.46 × 1017

= 2.5 × 1017

For the substitution

(1 mark)

(1 mark)

(ii)

==

2.5 × 1017 × × 1.6 × 10−19

8 × 10−3A 8 mA

Current = 20 100 For the substitution (1 mark)

(1 mark) (iii)

λ

hf = hf0 + evs (1 mark)

hC = hf0 + evs

6.63 × 10−34 × 3 × 108 = (6.63 × 10−34 × 5.2 × 1014) + (1.6 × 10−19 VS )

490 × 10−9 For the substitution VS = 0.38 V

(1 mark)

(1 mark)

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