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RJC JC2 H2 Physics Common Test 2 (2008)Section B Solutions

21 (a) Define gravitational field strength.

Gravitational field strength at a point is defined as the gravitational force per unit massexperienced by a mass placed at that point.

(b) A spherical star of mass 302.0 10 kg undergoes gravitational collapse to an extremelydense neutron star of radius 10 km as shown in Fig. 21.1.

Fig. 21.1

(i) Calculate the weight of a 60-kg man at the surface of the neutron star.

( ) ( )

( )

( ) ( )

-11 30

23

12 -1

12

13

6.67 10 2.0 10

10 101.334 10 N kg

60 1.334 10

8.0 10 N

g

W mg

=

=

=

=

=

Comments: Some students did not convert the radius to metres when substituting thevalue into the equation. Some did not square the radius.

(ii) If the star spins about a vertical axis XY through its centre, the man feels lighter atpoint Z as compared to point X. Explain.

At point Z, the man is undergoing circular motion due to the spin of the star but notat point X. At X, the reaction force on the man is equal to his weight. At Z, thisreaction force is less than his weight as there is a resultant force/centripetal forcedirected towards the centre of the star. Hence he will feel lighter at Z as comparedto X.

Comments: Some students fail to realize that the gravitational force (weight)acting on the man at both X and Z are the same. The different feeling is due to thedifference in reaction force/contact force acting on the man at X and Z.

X

Y

Z10 km

Neutron star


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(c) A ring of particles is formed 900 km from the surface of the star. Calculate the speed ofthe particles.

( ) ( )

( )

2

2

11 30

3

7 -1

6.67 10 2.0 10

910 10

1.21 10 m s

GMm mv

rr

v

=

=

=

Comments: Some students used energy equations in this question but they are irrelevantas there are no energy changes here. The particles are just moving in uniform circularmotion.

22 (a) Since oil drop is moving at constant speed,

Fnet = mg 2kv = 0

k=mg

2v

(b) Since oil drop is moving at constant speed,

qE1 = mg+ 2kv

qE1 = mg+ 2v mg

2v= 2mg

q=2mg

E1

(c) For case 3,Drag force is kv= mgSince oil drop is moving at constant speed, the electric forcemust be upwards due to an electric field that is upwards as well

qE2 = mg kv =1

2mg

E2 = 14

E1

The direction of E2is upwards.

If E2 is assumed to be downward, qE2 + mg= kv E2 = 1

4E1

The negative here does not mean that E2 is in a direction opposite to E1 but it meansthat the assumed direction is wrong and should be upwards instead!Since the question asks for the magnitude, the negative sign is not required but theinterpretation is important.

mg

2kv

mg

2kv

qE1

mg

kv

qE2


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(d) As the oil-drop rises, it losses electric potential energy but gains gravitational potentialenergy. It is also losing energy to the surrounding air as it moves through the air. Whenthe loss in electrical potential energy is equal to the gain in gravitational potentialenergy and the loss due to drag, the kinetic energy of the oil-drop will be constant.

This question asks for energy change and hence must be answered accordingly.Abbreviations such as GPE, EPE and KE should be written in full.

23 (a) Equivalent resistance of 4 lamps in parallel, Req= 30 / 4 = 7.5

Current in circuit,total

EI

R

6 00 75

0 50 7 5= = =

+

..

. .A

Total power dissipated in the 4 lamps,20.75 (7.5) 4.21875 4.22 W= = =2

eqP=I R

(b) - 6.0 0.75(0.50) 5.625 V= = =TV E Ir

OR

T eqV IR 0.75(7.5) 5.625 V= = =

5.09.0 7.03125 V

5.0 1.4

= =

+

XYV

5.625(1.000) 0.800 m

7.03125

=

= = =

XP XP

XY XY

TXP XY

XY

L V

L V

VL L

V

(c)XY XY XY XY

XY XY

V R V R

E V R E R Ror

-= =

+ (1)

= =XY XYXY 2L L

R

A r

(2)

For larger values of potential difference across the wire XY1. change the material of wire XY to one with higher resistance RXY. (from (1))OR2. change the material of wire XY to one with higher resistivity (to increase RXY).

(from (2))3. decrease the cross-sectional area of the wire (to increase RXY). (from (2))OR4. decrease the radius / diameter of the wire (to increase RXY). (from (2))5. increase length of wire XY (to increase RXY). (from (2))6. change the driver cell to one with higher e.m.f. (from (1))7. decrease the resistance R. (from (1))

NOTE: If students state (1) or (2) and(3) or (4), (1) or (2) has to be specifically tochange the materialof the wire and (3) or (4) to change the dimensions of the wire of

the same material.CommentsGeneral: Students should show proper and neat substitution and working by using thevalues exactly as stated in the questions e.g. 6.0 V and not 6 V. Units of quantities found inintermediate steps should also be stated clearly. This will help students ensure that they givethe final answers to the correct number of significant figures and units.


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(b) Students should give their answers to 2 or 3 significant figures and not 1. Potentialdifference across length XP is the terminal p.d. of the cell in Fig. 23.1 and not the e.m.f of thecell. The potential drop across the internal resistance has to be accounted for.(c) Suggested ways should involve only the potentiometer and not the circuit in Fig. 23.1 asstated in the question. Students who wrote more than 2 ways, only the first two wereconsidered. Suggested ways should be specific and should include words like increase,decrease, larger, smaller etc. and not just state use a variable resistor. Students shouldrefer to the electromotive force (e.m.f) of a cell and not its voltage as a cell is a source of

energy.

24 (a) Force per unit length on a straight conductor carrying unit current placedperpendicularly to the field.

(b) (i)

(ii) Vertically,

o

o

0.050 9.81cos40 mg 0.6403 N

cos40

= = =

m mF F

Applying N2L horizontally,

oo 20.6403 sin40sin40 ma a 8.2 or 8.23ms

0.050

= = =

mF

(c) (i)=

m E

(V-(-V))F F Bev=Ee= e

d

2Vv=

Bd

(ii) Electron travelling path Y has greater speed on entry than electron on path X.Hence, the magnetic force on electron on path Y is greater than the electric forceon it which is independent of speed.The upward resultant force causes the electron to curve upwards (towards Vplate).

Comments:(a) Some students thought the equation for magnetic flux is the defining equation for B.

Many state as... per unit charge...(b) Incorrect angle of Fm wrt to horizontal (many labelled it 60 or 40 degree) though Fm

marked perpendicular to wire, thus leading to incorrect equations in (ii)For (i), many did not label Fmis perpendicular to wireIncorrect resolution of Fm(even though the angle was labelled correctly)

(c) (i) A few thought that q and e are different : qE = BevMany did not realise the p.d. is 2V

(c)(ii) Many thought the curved path was parabolic or circularA few thought there was only one magnetic force acting

Fm

W


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25 (a) There is an increase in magnetic flux linkage (= NBA) through loop ACYX as the rodslides downward due to a increase in area A of loop ACYX. According to Faradays lawof electromagnetic induction, the rate of change of magnetic flux linkage in a circuit is

proportional to the e.m.f. induced in the circuit. Hence there will be an e.m.f. E =d

dt

induced in the rod.

(b) 5.0

A CMagnetic

Force Induced current

X Y

Weight

Magnetic field out of the paper

B D

(c) At terminal velocity, v of the rod, resultant force on the rod = 0.

Weight of rod, Mg = Magnetic force acting on rod, BIL

I=Mg

BL

I =BLv

R

Mg

BL=

BLv

R

Terminal velocity v = 2 2MgRB L

=2 2

(0.020)(9.81)(5.0)

(0.50) (1.2)

= 2.73 m s-1

(d) Gravitational potential energy is converted to electrical energy as a induced currentflows in the rod.Electrical energy is then converted to heat energy as the current gives rise to a heatingeffect.

Comments: Faradays Law incorrectly stated.Forces not labelled.


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26 (a) Find the r.m.s current in terms of I0[3]

(b) The secondary voltage of a transformer in a furnace is 10.0 kV. The primary coil operates at anr.m.s voltage of 120 V and has a resistance of 24.0 . The transformer is 90.0% efficient.

(i) Calculate the turn ratio (s

p

N

N) of the transformer. [2]

=

= = =10000 250

83.3120 3

s s

p p

s

p

N V

N V

N

N

(ii) What is the current in the secondary coil? [3]

( )

=

=

=

120 24

5.00

p p

p

p

V RI

I

I A

( )

( )

( )( )( ) ( )

=

=

=

=

=

0.9

0.9

0.9

0.9 5 120 10000

0.054

s

p

p s

p p s s

s

s

P

P

P P

I V I V

I

I A

(iii) Suggest two reasons why the transformer is not 100 % efficient. [2]

Resistance of the wire is not equal to zero.Magnetic flux linkage is not linked completely to the secondary coil.Power loss due to eddy current.

t/s0

42

I0/2

I0

4 6 8

I/A


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27 (a) KEmax = q Vs= (1.60x10-19) (1.00) M1= 1.60x10-19J (3 sf) A1

(b) E = hc/= (6.63x10-34)(3.00x108) / (380x10-9) M1= 5.23x10-19J (3 sf) A1

(c)The graph (Vs=

1hc

e e

) is a curve or non-linear, B1

and two points are not sufficient to plot the graph accurately to determine o. B1

NB: If mentioned that graph does not cut horizontal-axis [0].

(d) Plotted both points correctly: (7.89, 1.000) & (6.00, 0.225). B2Straight-line graph joining both points. B1

(e) It is the threshold frequency of the metal, B1which is the minimum frequency of electromagnetic radiation that is required to remove

a free electron from the metal surface (or to cause photoelectric effect). B1

NB: Allow answers which state that photoelectric effect is (not) possible (below) abovefo.

(f) (i) Gradient = h/e

h = e x gradient= (1.60x10-19) [(1.000-(-0.225))/(7.89-6.00)x1014] M2= 6.56x10-34J s (a range of answers will be accepted) A1

NB: If prefix 1014missing, deduct [2]. Unit not stated, deduct [1]

= h x (f-intercept)

= (6.56x10-34) (5.50x1014) M2= 3.61x10-19J (allow ecf) A1

(or use y = mx + c to find )

NB: Must use hfrom (f)(i), else deduct [1].

(g) Parallel andhigher focompared to graph drawn in part (d). A1

(h) The work function energy of lithium is the minimum amount of energy required toremove a free (or delocalised) electron from the surface of the metal (which isdifferent from an isolated atom). B2

NB: Will not accept definition of ionisation energy because it is not stated in question.

Comments:Carelessness with sf is rampant. If unsure, it is better to leave final numerical answers to 3 sf(unless questions state otherwise).Careless in reading and plotting graphs. Students also tend to ignore axis labels, leavinganswers wrong by a factor of 1014.Did not read questions carefully and hence provide irrelevant answers that gain no marks.Poor mathematical manipulations skills are seen in many parts, especially (f). Many gothrough long procedures when simple steps suffice.

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