106_suggested Answer Scheme Mock Pspm 2 Week 17 Set 2.pdf

SUGGESTED ANSWER MOCK PSPM 2 SET 2 NO 1 (a) Oil droplet is stationary

ANSWER SCHEME equilibrium

MARKS

F  0

Resultant force,

K1

W  Fe Weight downwards = electric force upwards

mg  qE mg  q 3.26  1015 (9.81)  q

+ + + + + + +

V d

R1

Fe

1000 10  103

W

q  3.2  1019C

- - - - - - -

GJU1

Since the electric force is upwards towards the positive plate, the charge on the J1

oil droplet is negative. (b)(i) VD =

R1

kq1 kq2 + r2 r1

=

(9 109 ) (6 109 ) (9 109 )(3 109 ) + 0.05 0.03 GJU1

= 1260V (ii) VB =

=

kq1 kq2 + r4 r3 (9 109 ) (6 109 ) (9 109 )(3 109 ) + 0.05 0.12 GJU1

= –90V (iii) WBD = qoΔVDB

R1

= qo(VD–VB) = [(2×10–9) (1260–(–90)] GJU1

= 2.7×10–6 J (c) . Charge, Q (C)

label t1 , Qo, 0.63Qo, 0.37Qo,τ : D1 shape : D1

Qo Charging

0.63Qo

D2

Discharging

0.37Qo 0 1|Page

τ

t1

τ

Time, t (s)

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2 Graph shows, when S1 is closed, the capacitor is charged by the battery. When the S1 is opened and S2 closed, the capacitor discharges through the resistor R. (d) After the battery is disconnected, the charge on each plate would not change, Q K1

constant. After dielectric sheet is inserted, resultant electric field decreases From E 

V d

E V

K1

As E decreases, V across the capacitor decreases. From C 

Q , when V decreases, capacitance, C increases. V

From (U 

K1

1 1 1 Q2 ) or (U  QV ) or (U  CV 2 ) 2 2 2 C

Therefore, energy stored in the capacitor will decrease.

K1 TOTAL

NO ANSWER SCHEME 2 (a)(i) R13 = R1 + R3 = 2 + 4 = 6 

MARKS

G1

1 1 1   RT R2 R13



15%

R1 G1

1 1  3 6

JU1

RT = 2  (ii) VT = ITRT

R1

6 = IT(2) GJU1

IT = 3 A (iii) V2 = VT = 6 V

{* V2 is connected parallel in the circuit}

V2 = I2R2

K1 R1

6 = I2(3) I2 = 2 A

GJU1

(b) Σ I in = Σ I out

I3 = I1 + I2

RG1

Σ  =Σ IR Loop 1 (anticlockwise) : 20 = 2I1 + 4I3 Loop 2 (clockwise) : 15 = 3I2 + 4I3

RG1 G1

Solve using Mode EQN :

2|Page

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2

I1 = 3.08A

JU1

I2 = 0.38A

JU1

I3 = 3.46A

JU1 TOTAL

NO ANSWER SCHEME 3 (a) One Ampere - the current is the current when flowing through 2 infinitely long straight, parallel thin wires separated by a distance 1.0 meter apart in free space, produce a force of 2.0×10–7 N on each meter of the wires. (b) Assume the new distance as d’ From

F o I X I Y  l 2 d

J1

R1

____ (1)

G1

o 2 I (4 I ) 2 d 

____ (2)

G1

(2) ÷ (1) ;

3

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o I (3I ) 2 d

k 3k 

15%

3k  8I 2 2 d   k 2 d   3I 2

8d 3d 

8 d  d 9

J1

(c)(i) Apply Right Hand Force rule to each sides of the coil. F1 F2

I B

B I I

B

All correct :

D2

1 mistake :

D1

2 mistakes :

D0

D2

B F4

I F3

(ii)

 max  BINA (

max

 sin θ =1 )

R1

 (0.8)(0.2)(4)(0.18  0.09)  1.04 102 Nm 3|Page

GJU1

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2 (iii)

  B .If B decreases, the torque τ decreases. The deflection of the pointer will be less. The pointer show reading which is less than its actual value of the current, I

(d)(i)(ii)

Concept apply: FB=Fc 𝑚𝑣 2 𝑞𝑣𝐵 = 𝑟 𝑚𝑣 𝑟 = 𝑞𝐵  𝑟 ∝ 𝑣

(ii) (i)

B

C

× × × × × × × × × × × × × ×

Higher 𝑣 gives greater 𝑟 D3

× × × × × × × FB

A

J1

Electrons

D

All correct labels with arrows : (i) : D1 (ii) : D1 FB : D1

(iii) The electron is moving parallel to magnetic field  θ = 0 J1

F  qBv sin   qBv sin 0  0 N The electron is stationary OR not moving  v = 0

J1

F  qBv sin   qB(0)sin   0N TOTAL NO ANSWER SCHEME 4 (a)(i) When the switch is closed, the current from the battery produces a

magnetic field directed toward the right, along the axis of both coils. As the battery current is growing in magnitude, the induced current in the rightmost coil opposes the increasing rightward directed field by generating a field toward to the left along the axis. Thus, the induced current must be left to right through the resistor. (ii) Once the battery current, and the field it produces, have stabilized, the flux through the rightmost coil is constant (no change in flux) and there is no induced current. (iii) As the switch is opened, the battery current and the field it produces rapidly decrease in magnitude. To oppose this decrease in the rightward directed field, the induced current must produce a field toward the right along the axis, so the induced current is right to left through the resistor. (b)(i) Its mutual inductance.

15% MARKS

J1

J1

J1

Mutual inductance of the two coils.

M 

o N1 N 2 A

M

o N1 N 2 d 2



4

M = 8.88×10–5 H

4|Page

where

d  A  r 2     2

2

4 107 (1000)(50) (3 102 )2  4(50 102 )

R1

G1 JU1

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2 (ii) The induced e.m.f in the secondary coil, if the current flowing in the primary coil changes at the rate of 4.8 A s–1. If

dI  4.8 As 1 then the magnitude of the induced e.m.f in the secondary dt

coil,

2  M

dI1 dt RGJU1

 2  8.88 105 (4.8)

 2  4.26 104V (c)

U

1 2 LI 2

R1

and Lsolenoid 

Thus, U 

0 N 2 A

0 N 2 AI 2 2

R1

 4 10   68 7





 1.20  102  2 8.00  102

2







2

4   0.770  

2

G1

 2.44  106 J 2.44 μJ (d)

JU1

Vmax  I max X C Vmax 

I max 2.1  C  2  60  50 106





RG1

Vmax  111.4 V For purely capacitive circuit, I leads V. Thus Vc = Vmax sin ωt ; Ic = Imax cos ωt

I C  I O cos t  I O cos  0.5  2.1cos 

RG1

  76.22o VC  Vmax sin  VC  111.4sin 76.22o

RG1

VC  108.2 V

JU1 TOTAL

5|Page

15%

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2 NO 5 (a)(i)

(ii)

ANSWER SCHEME

MARKS

1 1 1 nmaterial (  1)    f nmedium  r1 r2 

R1

1 1  1  (1.5  1)    f   20 

G1

f = –40cm

JU1

1 1 1   f u v

R1 G1

1 1 1    40 40 v

JU1

v = –20cm (20 cm in front the lens). v 20   0.5  1 u 40 Virtual, upright, diminished (b)(i) 1 1 1   f1 u1 v1 (iii)

K2

M

Any 2 correct answers: K2

R1

1 1 1   10 25 v

v1 = 16.7 cm

GJU1

(ii) Lens 1

Lens 2

O u1

u2

v1

I1

u2 = 25 – 16.7 = 8.3 cm

25 cm

1 1 1   f 2 u 2 v2

1 1 1   5 8.3 v2

I1 becomes object for lens 2

Substitution of u2

K1

G1 JU1

v2 = 12.6 cm (behind the 2nd lens).

R1 v1 v2 16.7 12.6 M = M1M2 = u1 u 2 = 25 8.3 = 1.014

GJU1 TOTAL

6|Page

15%

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2 NO 6 (a)(i) 2t = (m + ½)

ANSWER SCHEME

MARKS

R1

2t = (7 + ½)(570x10-9)

Concept on m = 7: K1

t = 214 m

K1 GJU1

(ii) 2t = m

R1 Concept on m = 3: K1

2t = (3)(570x10-9) t = 0.86 m

K1 GJU1

(b) At the centre of a newton’s rings (the thickness of air, t = 0) :

Path difference = 0

K1

Single phase change occurred among two reflected rays  anti-phase coherent source (destructive interference). (c)(i) l 1 d= = 2×10–6 m–1  5 N 5 10 d sin = n

sin90º: K1

(2×10-6)(sin90o) = nmax(495×10–9) OR

2

K1 G1 R1 K1 K1

nmax = 4.04 = 4 GJU1

Then, number of maximum = 2nmax + 1 = 2(4) + 1 = 9 (ii) 1. By using light of lower wavelength.

J1

2. By using diffraction grating of lower N (or of higher d).

J1 TOTAL

NO

7|Page

ANSWER SCHEME

15%

MARKS

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2 7 (a)(i)

Light

Cathode

Anode

Position ammeter, A and voltmeter, V: D1 Position Cathode (emitter) & Anode (collector): D1

A

Position power supply (-ve & +ve terminals): D1

V

(ii) Kmax = eVs ; (Vs get when I=0) (iii)

D3

J1

I

D1

Shape & label new Vs: D1

J1 J1

V

Vs Vs new 0

When f decrease, the maximum kinetic energy of the photoelectrons ejected decreases. Thus, the value of Vs will decrease while current remains the same. (b)

Power, P 

R1

E t

Energy of 1 photons = hf. Hence energy of N photons = N(hf)

P

R1

E N (hf )  N  hc    t t  t  G1

8|Page

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2 34 8  N  (6.63  10 )(3  10 ) 60    250  10 9  t 

JU1

N 19    7.54  10 photons s-1  t  (c)(i) A phenomenon where under certain circumstances a particle exhibits wave properties and under other conditions exhibits particle properties but cannot be observed simultaneously. de Broglie’s relation,   (ii)

h p

All symbols must be defined: J1

h h  ........(1) p 2mK h h '   ........(2) p 2m(3K )



J1

J1 R1

' h  (2)  (1)   (3)2mK ' 

1 3



J1

TOTAL

NO 8 (a)(i)

ANSWER SCHEME β+ decay (Positron) Will happen when the number of protons is more than the number of neutrons in a nucleus. A Z

X  Z A1Y  10  v

MARKS

β- decay (Negatron) Will happen when the number of neutrons is more than the number of protons in a nucleus. A Z

X  Z A1Y  10  v

(ii) Alpha < Beta < Gamma (penetrating power abilities) Alpha - can be blocked by a sheet of paper Beta - can be blocked by a few millimetres aluminium Gamma- can be blocked by a few centimetres in lead (b)(i) Decay constant is the ratio of a radioactive material disintegrating per unit time to the number of atoms. OR



dN dt N

15%

J1 J1 J1 J1 J1 J1 J1

All symbols must be defined: J1

(ii) Initial percentage of radioactive substance, P0 = 100% Percentage of radioactive substance left after 4.0days, P = 20%

N  N 0 e  t

0.2 No  Noe (4) 0.2  e  ( 4) ln 0.2   (4)

R1 G1

JU1 9|Page

PHYSICS MUST A

SUGGESTED ANSWER MOCK PSPM 2 SET 2

   0.402 day1 @ 4.657×10–6 s–1 (c)(i)

Activity is the number of decays (or disintegrations) per second by radioactive nucleus. (ii) From equation A  A e t 0

J1 R1

ln both side, ln A  ln( A0 e t )

ln A  ln A0  ln e t

ln A  ln A0  t Rearrange and compare to y = mx + c [ ln A  t  ln A0 ] [ y = mx+ c ] gradient = –λ

K1 GJUI

06 =0.024 min–1 250  0 ln 2 ln 2 T1    28.9 min 2  0.024  

RGJU1 TOTAL

10 | P a g e

15%

PHYSICS MUST A