Objective S2 2019 Answer.docx

ANSWER SECOND TERM STPM TRIAL EXAM 2014 PHYSICS PAPER 2 (960/2 )

SECTION A ( 15 marks) Answer all questions in this section ANSWER NO. 1 : C

EOS-EOQ E

Z

O

EOREOP

ANSWER No. 2 : C

√

√

q1 q2 ( 1.6 x 10−19 ) (2 x 1.6 x 10−19) mv 2 1 q 1 q2 = =¿> v= = r 4 π ε 0 r2 4 π ε 0 mr 4 π ( 8.85 x 10−12 ) ( 9.11 x 10−31 ) (5.6 x 10−11 ) = 3.00 x10 6 ms-1

ANSWER NO 3 : D Direction of both electric force; FSR & FST which act on S are toward right

8.85 x 10−12 ¿ 4π¿ towards T Q4: QR QT 4 x 10−5 S ANSWER NO. D F=F SR+ F ST = + = ¿ 1 1 1 4 π ε10 r SR12 r 2+1 12 ST 2 = + = + = =¿>CT = =4 μF C T C 1 C 2 +C3 6 4+ 8 12 3

(

)

Q 1=Q T =CT V =( 4 x 10−6 ) ( 9 )=3.6 x 10−5 F Since V2 = V3 and QαC ===>

−5 C2 4 ( −5 ) Q 2= ( Q )= 3.6 x 10 =1.2 x 10 F=12 μF C 2 +C3 4 +8

( (

Q 3=

) ( ) ) ( )(

C3 8 ( Q )= 3.6 x 10−5 )=2.4 x 10−5=24 μF C 2 +C3 4 +8

ANSWER NO. 5 : B L Use R= ρ A



R x ρ(2 L/ 4 t 2) = = 4/3 = 4 : 3 RY 6L ρ( ) 2 16 t

ANSWER NO. 6 : B Use

R= ρ

L V ; I = ∧I =Ane V d A R

V I R V V d= = = = Ane Ane AneR

V Ane ( ρ

L ) A

=

V 12.0 = =2.34 x 28 neρL ( 5.0 x 10 ) ( 1.6 x 10−19 ) ( 3,2 x 10−8 ) (2.0)

ANSWER NO. 7 : C Connected to Y ; VAY = VAJ = L1/100 x RAB = (200/(200+R)E ……..(1) Connected to Z ; VAZ = VAJ’ = L2/100 x RAB = E …………………………… (2) Eq.(1) ÷Eq(2) L1/L2 = 200 /(200+R) R = L2/L1 x 200 – 200 =58.8/40 x200 – 200 = 94.0 Ω

Answer No. 8 : C 1 1 1 = + so R XY =2 R R 4 R 4 R XY For -q

Y Eop +q

EOS

X O

EOQ

+2q

EOR

Z

W -2q

Answer : No. 9 A Refer Loop STTW : E1+ I1r1 –I2R = 0  E1 + (1.0)(1.0) – (2.0)(3.0) = 0  E1 = + 5.0 V Refer loop PQTW : E - Ir - I2R = 0  12 -(3)(r) – (2.0)(3.0) =0  r = 6/3 = 2.0 Ω

Answer NO 10 : D Use Fleming’s left Answer hand rule, the magnetic NO. 11 B: force will deflect the positive charge

carriers to Q whereas the side in the direction T is induced negatively therefoe the direction of electric field E isμQ(+) I to T(-)] B= 0 2r Use: for wires X and Y; Assuming: direction towards N = + and direction towards S = -

Resultant;

B T =B X + BY =−

μ o I μ o 4 I μ o I μo I + )= = = 0 . 5 μ o I toward N 2 r1 4 r1 2 r 1 2( 1)

ANSWER NO. 12 A : Use e.m.f. E = - dФ/dt = -d(NBA)/dt

E= -NA dB/dt

; dB/dt = gradient ]

Answer No 13 : C:. Use Lenz’s law ,Current flows in coil Q will be in reverse direction to coil P so that Q will produce magnetic field with direction reverse coil P ,hence created repulse force to against the coming of coil Q towards P

ANSWER : No. 14 B Refer formula; XC = 1/ωC  XC = 1/2πfC XC α 1/f and XC = V/IXC α 1/I

ANSWER NO. 15 : C Given V= 60 sin 300πt ; C= 400 µF

Use XC = 1/ωC = Vo/Io  Io = VoωC=(60)( 200π)(400x10-6) = 15.08A

SECTION B ( 15 marks) Answer all question in this section 16. In the circuit shown in figure below the three batteries have internal resistance r = 1.0Ω E= 15.0 V, r = 1.0 Ω A

F

X

R= 6.0 Ω

I= 1.0 A E1 , r1= 1.0 Ω

R1= 4.0 Ω B

E

Y

I1 I2=3.0 A

R2= 1.0 Ω C

D E2 , r2=1.0Ω

[ Answer : (a) 2.0 A (b) (i) 14V, 6V (ii) 13V, -13V, 23V ] (a) Determine the value current I1 I1 = I2 – I = 3.0 – 1.0 = 2.0 A

( 1 mark)

(b) the e.m.f. of E1 and E2

( 4 marks)

Refer Loop FABEF : - IR + E - Ir- + I1R1 – E1 + I1r1 =0 -(1.0)(6.0) + 15.0 –(1.0)(1.0) +(2.0)(4.0) – E 1 + (2.0)(1.0)=0 -6 +15-1+8-E1+2 = 0 E1 = 18.0 V

(2 marks)

Refer Loop FACDF : -IR+E-Ir- – E2 - I2r2- I2 R2 = 0 -(1.0)(6.0)+15.0 –(1.0)(1.0)-E 2 –(3.0)((1.0)-(3.0)(1.0) = 0 -6 + 15 -1 – E2 -3 - 3 = 0 E2 = 2.0 V

(2 marks)

(c) the potential difference across X and Y ?

( 2 marks)

Refer direction XABY E=15.0 VV E=15.0 X X

r=r=1.0Ω 1.0Ω I = 1.0A I = 1.0A

I1I1==2.0A 2.0A

R1=R1= 4.04.0 Ω Ω Y Y

B

Vxy = +E – Ir + I1R1 = 15.0 –(1.0)(1.0) + (2.0)(4.0) = 22.0 V OR Refer direction XFEY E1 =14.0 V I1 = 2.0A

R= 6.0 Ω I = 1.0A

r1 = 1.0Ω

X

Y

F

Vxy = + IR +E1 – I1 r1 = +(1.0)(6.0) + 18.0 – (2.0(1.0) =6.0 +18.0 – 2.0 = 22.0 V 17. Velocity selector + Fm

Positive Charged Ion

r

FE

Region X (magnetic field point into the plane)

Region Y (magnetic field point into the plane)

A beam of singly positive charged ions of mass 6.64 x 10 -27 Kg travels straight through the velocity selector of a Bainbridge mass spectrometer in region X which consists of an electric field E = 8.8 KV m-1 and a uniform magnetic field B1 = 0.04 T. After that the ions will enter a second region Y which consists of a uniform magnetic field B 2 = 0.08 T and make the ions move in a semicircle path.

(a) Label the direction of electric force FE and magnetic force Fm act on the positive charged ion in region X and then derive the expression of velocity of the positive charged ions which enter the region X in term of B and E. ( 3 marks) Fm=FE  qVB = qE  V = E/B (b) Derive the expression of the radius r for the semicircle path in region Y in terms of m, q, E, B1 dan B2 . ( 2 marks) 2 mV mv m E mE =qV B 2=¿>r = = = r q B 2 q B2 B1 q B2 B 1

( )

(c) calculate the value of the radius of semicircle path in region Y for the positive charged ions? ( 1 marks) r=

( 6 . 64 x 10−27 ) (8.8 x 103 ) =0 .114 m ( 1 . 6 x 10−19 ) ( 0 .08 ) (0 . 04)

(d) State what t is the effect on the radius of semicircle path and time to form the semicircle path in region Y if the linear velocity of the positive charged ions in region X increases. (2 marks) r will increase and time to form semicircle remains unchanged. ………………………………………………………………………………….. mV 2 mv =qV B 2=¿>r = =¿ >r α V r q B2 qB mr ω 2=qV B2=q ( rω ) B 2=¿>ω= 2 m

Note :



T=

2 π 2 πm = =¿=¿> T independent of v ω q B2

SECTION C (30 marks) Answer any two questions ANSWER NO 18 16.(a) Gauss’s Law states that the total electric flux Φ flowing through a closed surface that

Q εo encloses completely a charge Q within that surface and in free space is equal to where

εo

is the permittivity of free space.

(1 mark)

E

r

Spherical Closed surface O R

Choose a sphere of radius r centered at O as a suitable closed surface or Gaussian surface for the charged sphere (1 mark) Using Gauss‘s Law

∑Q φ

Electric flux ;

εo

=

∑Q εo

 EA =

(1 mark)

∑Q

 E(4π r ) = ε o For a point outside the charged sphere; r ≥ R 2

(i) 

∑¿ ¿

Q=Q

Q 2 ε  E(4π r ) = o (ii)

Q 2 E = 4 πr

 ; for r ≥ R For a point inside the charged sphere : r < R 

∑¿

[1 mark] [1 mark]

Q=0

¿

R

0 ε  E(4π r ) = o 2

r

 E=0

; for r < R

Conclusion

E=

Q 4 πr 2

0

;

; for r ≥ R

[1 mark ]

E= for r < R E

Q/4πr2

R

[2 marks] r

R

(b ) (i) Electric potentiall V at a point in electric field is the work done to Bering a unit positive vharge from infinity to the oint.) ( 1 mark) +0.01 μC

E

+0.02 μC

(i)

OS=OQ=√ 102 +102=10 √ 2

cm

Electric potential at poit O;

V 0=V 0 s+ V OQ +V OP +V ¿ q S qQ q P q R 1 ¿ + + + 4 π ε o r OS r OQ r ¿ r OP

(

;

20 cm

20 cm O

)

+0.01 μC

R

s

-0.02 µc

−6

¿

10 ( 0.01+ 0.02+ 0.01−0.02 )=1.272 x 10 3 V −12 −2 4 π ( 8.85 x 10 )( 10 √2 x 10 )

ES=ER=√ 202 +102=10 √ 5 cm Electric potential at E ;

V E =V ES+ V ER +V EP +V EQ ¿

qS qR qP qQ 1 + + + 4 π ε o r ES r ER r EP r EQ

¿

−6 10 0.01 (−0.02) 0.01 0.02 3 + + + =2.295 x 10 V −12 −2 10 10 10 5 10 5 √ √ 4 π ( 8.85 x 10 )( 10 )

(

)

(

)

Work done to bring charge q = + 3.0µC from point O to E ;

W OE =q ( V E−V O )=( 3.0 x 10−6 )( 2.295 x 103−1.272 x 103 ) =+3.069 X 10−3 J Since Charge q is moving from O(low potential) to E(high potential) will experience a potential rise thus work is done by external force on the electric field. Therefore the charge system will gain extra electric potential energy .Therefore the electric potential energy ofn the charge system will increase IMPORTANT CONCEPT: (i)

(ii)

If positive charge moves from point with low electric potential to point with high electric potential (potential rise), work is done by external force on the electric field, charge system will gain extra electric potential energy. If positive charge moves from point with high electric potential to point with low electric potential (potential drop), work is done by the electric field itself , charge system will lose electric potential energy.

ANSWER NO 19 19))a) Eletromagnetic induction laws (i) Faraday’s law state that the electromotive force (e.m.f) induced in a conductor is directly proportional to the the rate of magnetic flux linked across the conductor. (1 mark) (ii) Lenz’s law state that the direction of induced current in a conductor is always oppose the process of change in magnetic flux which produce the induced current . (1 mark)

(b) The magnetic flux linked when the conductor JK move a distance x (1 mark)

Ф = BA Cos 00 = B(Lx) Use Faraday’s law : ε=

−d ∅ d (BLX ) dx = =−BL dt dt dt

= - BLX

(2 mark)

Negative sign is due to Lenz’s law where induce current will flow in certain direction which will produce a magnetic force against the motion of rod JK to QR (1 mark)

OTHER METHODE J

Q

L

F R

X

P F’=BIL S

K

Work done to move the rod JK ; dw = Fds=-F’ds = - BIL ds x

Total work done

x

W =∫ Fds=∫ −BILds=−BILx 0

0

dW d (−BILX) dx = =−BIL =−BILV dt dt dt P −BILV =−BLV Induced e.m.f in the rod ; ∈= = I I Power

P=

( C) Normal line

θ0=ωt

Θo=ωt B

B

Initial position

(i)

Position after time t ,coil turning angle θ0=ωt

ω= 60 r.p.s = 60 x 2π rads-1 = 120π rads-1

(1 mark)

Magnetic flux linkage after time t , coil turning angle θ0=ωt Ф = NBA Cos θ= NBA Cos ωt = (850) (0.15)(0.08 x 0.05) cos (120π) t  Ф = 0.51cos 120πt (i)

Induced e.m.f ; ε =marks)

(ii)

d ∅ −d (0.51 cos 120 πt ) = dt dt

(1 marks) ( 1 mark)

= 192.3 sin 120πt

(2

Sinusoidal line = 1 mark Label magnitude of ε and t at y and x axis = 1

When ε = 0 ====>

sin 120πt = 0 

120πt = π, 2π, 3π, 4π, ……. = = 1/120 , 1/60, 1/40, 1/30 ,………

2 √¿

(iii)

¿ ( 25 . 0 ) ¿ Vo V 2 192 .3 I rms rms = √ = ¿ R R

( 2 marks)

ANSWER NO 20 20. (a) (i) Impedance is define as total opposition created by resistor and capacitor to against the flowing of alternating current in the circuit (1 mark) (ii) Reactance of capacitor ;

X C=

1 1 = ωC 2 πfC

(1 mark) (3 mark) Graph lines = 2

XC or R

XC R

0

ANSWER NO 20 (iii)

Label = 1

f

Complete Phasor diagram (1 mark)

VR=IR I

Ф VC= IXC

V=IZ

V 2=V R2 +V C 2

(1 mark)

( IZ)2 = (IR)2 + (IXC)2 Z2 = R2 + XC2

(1 mark)

√

Z =√ R 2+ X C 2= R2+(

1 2 ) 2 πfC

(1 mark)

Phase different between V and I For R-C Circuit l ; 1 2 π ( 50 ) ( 200 x 10−6 ) (28.0) ) 1 2 πfC ∅=tan −1 =tan−1 ¿ R

( )

tan ∅=

V C I XC XC = = VR IR R

( 1 mark)

= 29.60

( 1mark)

Current I is leading the V by Ф= 29.60 or 0.16π radian (b) (i) If a dielectric inserted into the capacitor, the capacitance C of the capacitor will increase due to formula C= formula

∈r ∈0 A where C α ε r . The impedance Z of the R-C circuit will decrease due to d

√

Z = R 2+(

1 2 , therefore the current flows in the circuit will increase ) 2 πfC

(2

marks) and the brightness of the lamp will increase (i) If another similar capacitor is connected series with capacitor in the circuit , the total capacitance CT will decrease doe to the formula circuit will increase due to the formula

1 1 1 = +  CT = ½ C. The impedance Z of the R-C CT C1 C2

√

Z = R 2+(

1 2 ) 2 πfC

, therefore the current flows in the

Circuit will decrease and the brightness of the lamp will decrease

( 2 marks