2010 sajc h2 prelim paper 1

1

SAJC 2010 Prelims/9646/01 [Turn Over

ST. ANDREW’S JUNIOR COLLEGE JC2 2010

Preliminary Examinations PHYSICS, Higher 2 9646/01 Paper 1 22nd September 2010 1 hour 15 minutes (1400 Hrs – 1515 Hrs) Additional Materials: Optical Mark Sheet (OMS)

READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, Civic Group and index number on the separate Optical Mark Sheet (OMS). There are forty questions in this paper. Answer all the questions. For each question there are four possible answers A, B, C, D. Choose the one you consider correct and record your choice in soft pencil on the separate Optical Mark Sheet (OMS). Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet.

Instructions for using the Optical Mark Sheet (OMS)

1. Fill in your class number (e.g. 09S03 = "03", 09S22 = “22”) in the first two rows. 2. Fill in your class register number in the next two rows. (e.g. register number 1 is filled

in as "01"). 3. Write your class and register numbers into the column on the left (ie. 2201)

For Student’s Use

Paper 1 / 40 Paper 2 / 72 Paper 3 / 80

Total / 192 Percentage / 100 Grade

This Question Paper consists of 18 printed pages

Class no.

Register no.

Write your

numbers here

2


Data speed of light in free space , c = 3.00 x 108 m s-1

permeability of free space, µo = 4 π x 10-7 H m-1

permittivity of free space, εo = 8.85 x 10-12 F m-1

(1 / (36π)) x 10-9 F m-1

elementary charge, e = 1.60 x 10-19 C the Planck constant, h = 6.63 x 10-34 J s unified atomic mass constant, u = 1.66 x 10-27 kg rest mass of electron, me = 9.11 x 10-31 kg rest mass of proton, mp = 1.67 x 10-27 kg molar gas constant, R = 8.31 J K-1 mol-1 the Avogadro constant, NA = 6.02 x 1023 mol-1 the Boltzmann constant, k = 1.38 x 10-23 J K-1 gravitational constant, G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall, g = 9.81 m s-2

3


Formulae uniformly accelerated motion, s = ut + ½at2 v2 = u2 + 2as

work done on/by a gas, W = p∆V

hydrostatic pressure, p = ρgh

gravitational potential, φ = - Gm

r

displacement of particle in s.h.m., x = x0 sinωt velocity of particle in s.h.m., v = v0 cosωt

= ± ω )xx( 22

0 −

mean kinetic energy of a molecule of an ideal gas,

E = 32 kT

resistors in series, R = R1 + R2 +… resistors in parallel, 1/R = 1/R1 + 1/R2 +…

electric potential, V = r4

Q

0πε

alternating current/voltage, x = x0 sinωt transmission coefficient, T α exp(-2kd)

where k = 2

2

h

)EU(m8 −π

radioactive decay, x = x0 exp(-λt)

decay constant, λ =

2

1t

693.0

4


1 Which of the following provides the most accurate estimate? A Mrs Quek’s black Subaru Forester 2.0-litre car has a mass of 3000 kg. B The floor area of the SAJC Cultural Centre ground floor is 700 m2. C The temperature of a yellow bunsen flame is 600 K. D The upthrust acting on a totally submerged adult round-tube float is 50 N. 2 A Fomula One car travels a distance of (100 ± 1) m from rest. If the

acceleration of the car is (23.1 ± 0.5) m s-2, what would be its final velocity at the end of the distance covered?

A (68 ± 1) m s-1 B (67.9 ± 1.0) m s-1 C (68.0 ± 1.1) m s-1 D (67.97 ± 1.08) m s-1

3 Which of the following is a random error?

A Error as a result of using g = 10 m s–2, instead of g = 9.81 m s-2 B Error due to the timing of the experimenter C Error due to a stopwatch running too fast D Zero error of a measuring instrument

4 Frankie throws a small rubber ball vertically downwards at a speed of

3.0 m s-1. It hits the ground and rebounds vertically. The graph below shows the velocity-time graph for the first 1.9 s of the motion of the rubber ball.

What is the displacement of the ball between the point at which it was first thrown and the highest point of the motion?

A zero B 1.8 m C 3.6 m D 7.2 m

time / s

v / ms-1

0.6

- 3

- 9

8

0 1.5 1.9

5


5 A device launches two identical balls, x and y, simultaneously in a horizontal direction from the same height, as shown in the diagram below. The results are as indicated.

Fig. 5.1 Which statement correctly describes what happens? A x hits the ground before y as it is closer to the launch site. B y hits the ground before x as it has a higher launch velocity. C x and y hit the ground simultaneously with the same velocity. D x and y hit the ground simultaneously with different velocities. 6 A ball of mass m is released from rest from point Y which is at a height of h

above point X and slides down a frictionless slope. The ball passes point X with a velocity v as shown below.

A second ball of mass 0.5m is released from rest from point Z which is at a

height of 2h above X. The velocity with which the second object passes point X in terms of v is

A 2v B 1.41v C v D 0.5v

Y

X

Z

2h h v

6


7 A man drives along a straight road with a constant speed from right to left and tosses a coin vertically upwards. If effects of air resistance are significant, which diagram best represents the trajectory of the coin seen by a stationary observer?

A B

C D

8 Three charges +2q, -q and +q are placed at the corners W, X and Y of a

square WXYZ respectively as shown below. A fourth charge Q is placed at Z, after which the charge X experiences a net

electrostatic force indicated by the arrow in the above diagram. What is the value of Q?

A -2.8q B -1.4q C 1.4q D 2.8q

resultant electrostatic force on X

W +2q

X -q

Y +q

Z Q

7


9 A roller coaster is traveling in a circular path in a vertical plane. The top and bottom of the track have the same radius of curvature R. The acceleration due to gravity is g. If the roller coaster has a speed v, the passenger is most likely to feel weightless

A at the top of the path when v > gR

B at the top of the path when v < gR

C at the bottom of the path when v > gR

D at the bottom of the path when v < gR

10 In a ride at an entertainment park, a person sits in a cage which moves in a

vertical circle at a constant speed.

At the instant shown, what is the direction of the force exerted by the cage on the person?

Bottom

Top

R

R

D

C

B

A

8


11 A car of mass 1000 kg travels at a constant speed of 8 m s-1 up a slope inclined at 20° to the horizontal. Given that the constant frictional force against the motion is 500 N, the engine power of the car is

A 22.8 kW B 30.8 kW C 69.7 kW D 77.7 kW 12 Which diagram shows the variation of gravitational force F on a point mass,

and of gravitational potential energy U of the mass, with its distance r from another point mass?

13 Figure shows two points X and Y at distances L and 2L from the centre of the

earth. The gravitational potential at X is - 8 kJ kg-1.

When a 1 kg mass is taken from X to Y the work done on the mass is

A - 4 kJ B - 2 kJ C 2 kJ D 4 kJ 14 Which quantity is not necessarily the same for satellites that are in

geostationary orbits around the Earth?

A angular velocity B centripetal acceleration C kinetic energy D orbital period

X

Earth

Y

2L

L

9


15 A fixed mass of an ideal gas is heated at constant volume. Which one of the following graphs best shows the variation with Celsius temperature t of pressure p of the gas?

16 In the figure below, the curve is an isotherm (a curve which joins up all the points having the same temperature) for a fixed mass of ideal gas.

Which of the following statements can be deduced for the process from A to B? A Positive work is done on the gas and heat is supplied to the gas. B Positive work is done by the gas without any heat supplied to the gas. C The internal energy of the gas decreases as heat is released by the gas. D Heat is supplied to the gas and the internal energy of the gas remains

unchanged.

A

B

pressure

volume

10


17 An ideal gas undergoes the cycle of pressure and volume changes

W→X→Y→Z as shown in the diagram.

What is the work done by the gas in the process Y→Z and the net work done by the gas as it undergoes a complete cycle of pressure and volume change?

Work done by the gas in

process Y→→→→Z Net work done by the gas

A -300 J 600 J

B 300 J -600 J

C -400 J 1200 J

D 400 J -1200 J

18 The graph shows the shape at a particular instant of part of a transverse wave travelling along a string.

Which statement about the motion of elements of the string is correct? A The speed of the element at P is a maximum. B The displacement of the element at Q is always zero. C The energy of the element at R is entirely kinetic. D The acceleration of the element at S is a maximum.

11


19 Which one of the following could be an effect of critical damping?

A A toilet door takes a long time to close after a student enters. B A rubber ball drops to ground and stops bouncing almost immediately. C A voltmeter fluctuates several times before registering a steady reading. D A passenger in a car hardly notices that the car has just crossed a

hump. 20 Light of wavelength 550 nm is incident normally on a diffraction grating having

400 lines per millimetre. What is maximum number of bright fringes that can be observed?

A 4 B 5 C 9 D 11 21 Two wave generators S1 and S2 produce water waves of wavelength 0.5 m. A

detector is placed at position X, 3 m from S1 and 2 m from S2 as shown in the diagram below. Each generator produces a wave of amplitude A at X when operated alone. The generators are operating together and producing waves which have a constant phase difference of π radians. What is the resultant amplitude at X?

A 0 B 0.5A C A D 2A

S1

S2

X

3 m 2 m

12


22 A closed insulated wire with a circular kink is placed in a uniform magnetic field with a flux density of 2.5 x 10-2 T, directed into the page. The diameter of the kink is 1.5 cm. The wire is quickly pulled taut and the kink is straightened

in a time of 0.050 s. If the wire resistance is 1.6 Ω, what is the power generated in the wire?

A 3.3 x 10-10 W B 4.9 x 10-9 W C 7.7 x 10-8 W D 8.8 x 10-7 W 23 An insulated wire is bent into a circular coil and placed above a straight portion as shown. The terminals of the wire are connected to an alternating voltage. What is the direction of the force acting at point P, the centre of the circle, on the wire?

A It is oscillating in the X and Y direction B It is pointing in X-direction C It is pointing in Y-direction D It is oscillating in and out of the page

a.c.

P

Y

X

kink

Pull Pull

13


24 An electron moves in a circular orbit in a uniform magnetic field. Which of the following statements is correct?

A The period of the orbit is independent of the speed of the electron. B The momentum of the electron is dependent on its charge. C The radius of the orbit is directly proportional to its charge. D The magnetic force on the electron is dependent on the mass of the electron. 25 Which one of the following statements about the electric potential and electric field at a point is correct? A The potential at the point is always zero when the electric field at that

point is zero. B The electric potential is given by the rate of change of electric field

intensity with distance. C The electric field at a point is zero when the potential around the point

is constant. D The potential at a point is zero when the electric field around the point

is constant. 26 Two long straight wires, X and Y, are placed perpendicularly to each other at a small distance apart. The current in wire X is flowing into the page and the current in wire Y is flowing to the right.

What is the direction of the force acting on wire Y at point P due to the magnetic field produced by wire X?

A out of the page B into the page C upwards D downward 27 The force of attraction between two unlike charges is 1.5 N. If the distance between the charges is doubled, the force of attraction between them is A 0.75 N B 0.38 N C 3.0 N D 6.0 N

x

Wire X

Wire Y P

14


28 Charges of 2 µC and -2 µC situated at points P and Q respectively as shown. X lies midway between P and Q while Y is at a mid-line. Which of the following correctly describes the directions of the electric fields and electric potentials at points X and Y?

At point X At point Y

Electric Field Potential Field Electric Field Potential Field

A right zero right zero B upwards positive upwards negative C right zero left zero D downwards negative downwards positive

29 The diagram shows a circuit consisting of five 1-Ω resistors. A multimeter is used to measure the resistance across different terminals. Which of the following statements is false?

A The resistance measured between B and C is 0.5 Ω. B The resistance measured between B and C will be smaller if an

additional resistor is connected in parallel across AC.

C The resistance measured between A and D is 1 Ω. D The resistance measured between A and D will be smaller if a zero

resistance wire is connected across BC.

A

B

C

D

P Q

X

Y

2 µC - 2 µC

15


30 A potential difference of 2.0 V is connected to a uniform resistance wire of length 3.0 m and cross-sectional area of 8.0 x 10-9 m2. A current of 0.1 A flows in the wire. What is the current flowing in the wire if the length and diameter of the wire are doubled with the same potential difference connected to it?

A 0.10 A B 0.20 A C 0.30 A D 0.40 A

31 In an ideal transformer, the most important function of the soft-iron core is A to reduce eddy-currents. B to improve the flux-linkage between the primary and secondary coils. C to dissipate the heat generated by the two coils. D to produce a uniform magnetic field in the two coils 32 A steady current I dissipates a certain power in a variable resistor. When a sinusoidal alternating current is used, the variable resistor has to be reduced to one quarter of its initial value to obtain the same power. What is the peak value of the alternating current?

A I2 B I2

C I22

D I24 33 Which of the following statements, referring to photoelectric emission, is always true? A No emission of electrons occurs for very low intensity illumination. B For a given metal there is a minimum frequency of radiation below which no emission occurs. C The velocity of the emitted electrons is proportional to the intensity of the incident radiation. D The number of electrons emitted per second is independent of the intensity of incident radiation.

16


34 The table gives relative values for three situations for the barrier tunnelling of an electron through a potential barrier. Rank the situations according to the probability of the electron tunnelling barrier, greatest first.

Electron Energy Barrier Height Barrier Thickness X E 5E L Y E 17E 2L Z E 2E 3L

A XYZ B XZY C ZXY D YZX 35 The diagram shows the energy levels of a gas atom.

____________________ 0 J ____________________ -2.4 x 10-19 J ____________________ -5.4 x 10-19 J ____________________ -21.8 x 10-19 J

A free electron of kinetic energy of 20.0 x 10-19 J collides with the cool gas atoms. What is the kinetic energy of the free electron after the collision?

A 1.8 x 10-19 J B 3.6 x 10-19 J C 5.4 x 10-19 J D 16.4 x 10-19 J

17


36 A ruby laser is a solid-state laser and emits at a wavelength of 694 nm. A CO2 laser is a gas laser and emits at a wavelength of 10.6 µm. Why are CO2 lasers (instead of ruby lasers) used in the metalworking industry to cut steel?

A CO2 lasers do not possess a metastable state B The energy of a single photon emitted by the CO2 laser is larger than

that of the energy of a single photon emitted by the ruby laser C CO2 lasers do not emit coherent photons D CO2 lasers emits laser light in the infrared region 37 Once the active medium in a LASER is excited, the first photons of light are

produced by which physical process? A Planck’s oscillation B blackbody radiation C spontaneous emission D synchrotron radiation

38 The most important property of a p-n junction is that it rectifies an alternating

current. Which of the following statements is false?

A During reverse bias condition of a p-n junction, the p-type semi-conductor becomes less negative.

B During reverse bias condition of a p-n junction, the width of the depletion region becomes larger as the externally applied p.d. adds to the junction potential.

C During forward bias condition of a p-n junction, if the applied p.d. overcomes the junction potential, electrons will cross steadily from the n-type side to the p-type side while the holes will cross steadily in the opposite direction.

D Under increasingly high reverse biased p.d., current can increase sharply through the p-n junction

39 Uranium-235 undergoes fission as shown in the equation below.

U23592 + n1

0 Cs14355 + Rb90

37 + 3 n10

195 MeV of energy was released in the reaction. Given that the binding energy per nucleon for uranium-235 is 7.6 MeV, and those for caesium and rubidium are approximately X MeV, determine the value of X.

A 8.5 MeV

B 9.7 MeV C 11.2 MeV D 13.3 MeV

18


40 The number of undecayed radioactive nuclei, N, of cobalt 6027 Co at three

different timings are as follows:

t / years N 0 500000 12 M 16 31250

What is the value of M? A 25000 B 62500 C 100000 D 125000

End of Paper

19


2010 H2 Physics Prelims Paper 1 Solutions

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10

B A B A D B D A A A 35.1% 52.0% 91.3% 45.3% 70.4% 68.5% 37.7% 32.6% 54.2% 32.4%

Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20

B C D C A D A D D C 62.0% 45.8% 71.3% 80.1% 82.2% 51.1% 74.1% 69.8% 50.5% 54.8%

Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30

A B C A C B B A D B 58.9% 62.2% 32.4% 23.4% 51.3% 62.9% 88.5% 89.7% 36.1% 79.6%

Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40

B C B B B D C A A B 80.1% 61.3% 94.1% 40.3% 55.2% 44.2% 87.2% 41.4% 68.8% 83.2%

Average score = 24/40, figures in % refers to % of cohort who got answer correct.

1 B A typical saloon car, like a Nissan Sunny, is about 1500 kg. Mrs Quek’s car has an unladen mass of 1410 kg. A Mercedes Benz weighs about 2250 kg.

An old SAF military truck used to weigh 3 tonnes (which is why it was called a 3-tonner but these days the new version is 5 tonnes (ie. 5000 kg) The cultural centre has a ground floor area of 709.5 m2. A yellow bunsen flame is typically above 1000 ºC and a blue flame can go as high as 1500 ºC. The upthrust is ρgV. Density of water, ρ = 1000 kg m-3. g is 9.81 and the volume of a tube float (totally submerged) is estimated by taking the diameter of its cross-section to be 15 cm and the diameter of the whole tube (top-view) to be 70 cm. This will give a volume of 0.0389 m3 and an upthrust of 381 N.

2 A Identify the formula => v2 = u2 + 2as, u = 0, hence, v2 = 2as

Calculating v => v = √(2as) = √(2)(23.1)(100) = 67.97

Fractional error of v => ∆vv

= ½∆aa

+ ½∆ss

Uncertainty of v, ∆v => 1.075 ∆v in one sig. fig => 1 v in same dec. pl => 68 Hence, (v ± ∆v) => (68 ± 1) m s-1

20


3 B Option A => g will always be overstated (error in one direction) Option B => The experimenter’s timing could be below or above the true

value, thus, its error can be in either direction. This constitutes random error.

Option C => timing will always be understated (error in one direction) Option D => measure will either be always overstated or always understated

(error in one direction) 4 A

The ball rebounds at 0.6 s. Area under trapezium in first 0.6 s where the ball drops to the ground = 3.6 m. The ball reaches a maximum height at 1.5 s. Hence, area of triangle from 0.6 s to 1.5 s is the upward trajectory of the ball before it changes direction, and it is also 3.6 m. Therefore, the vertical displacement between the starting point and highest point is zero.

5 D

The equation that links the height and time is sy = uyt + ½gt2

In this case, for both x and y, they have the same sy (height) and uy (both zero) and g (constant at 9.81 m s-2), Hence, the time will be same for x and y.

6 B For the ball travelling to X from either Y or Z, there is no friction or other resistive forces. Hence, the speed at X comes purely from the loss in GPE from Y or Z.

Loss in GPE = Gain in KE mgh = ½mv2

Hence, v α √(h) => v'v

= √(2hh

) => v’ = √(2) v = 1.41v

7 D

Note that answer cannot be C because the air resistance cannot cause an object to change direction horizontally. This is because when an objected afflicted by air resistance slows down to zero speed, air resistance will then cease to act on the object, as air resistance is proportional to velocity (or square of velocity). Therefore, the air resistance acting horizontally will drop to 0 when the horizontal component of the velocity is 0.

8 A

If the resultant force acting on charge X is horizontal to the left, the force that charge at Z acts on X (ie FZX) must be repulsive, hence, Q must be a negative charge (since X has a negative charge). Also, for horizontal equilibrium =>

FXY = vertical component of FZX

( )2

04 r

qq

πε =

( )2

0 24 r

Qq

πεsin45° ,where r = side of the square

Q = 2.8q

21


9 A When the passenger feels weightless, there is no normal reaction force on him (ie. N = 0). To do this question, form an equation for if the vehicle is at the top of the track (ie. options A or B) and when the vehicle is at the bottom of the track (ie. options C or D). The option where N would have been negative is the answer.

At the top => mg – N = mv2

R => N = mg -

mv2

R

In option A, v > √(gR), and N would have been negative. Hence, option A is the answer.

10 A

Since person is moving in a circle, the resultant force acting on the person must be pointing towards centre of circular motion path (ie. to the right). For that to be possible, there must be a frictional force (upwards) to counter the weight. Hence, the forces exerted by the cage are normal reaction and friction and the resultant of these two will be arrow A. The weight is not included as it is exerted by the Earth and not the cage.

11 B

The slope is at an incline of 20°. Since car is travelling at a constant speed, the resultant force = 0 (because there is no acceleration). T – mgsin20° - 500 = 0 T = mgsin20° + 500 = 3855.22 N Power, P = Tv = (3855.22)(8) = 30842 W

12 C

Force is a vector and depending on the direction declared, the force can be either positive or negative with respect to the declared direction. However, potential energy U is a scalar and gravitational potential energy is always negative, hence, answer is between options B and C.

Since F α 1r2 and U α

1r ,

Graph C gives the more accurate relative shapes for U and F.

thrust, T

weight, mg Friction, 500 N

friction

normal reaction

weight

22


13 D If the gravitational potential at X, ΦX, = -8 kJ kg-1, then ΦY would be -4 kJ kg-1,

according to the formula, Φ = - GM

r .

Work done = m∆Φ = (1)[ΦY – ΦX] = (1)[(-4) – (-8)] = 4 kJ Also note that the work done has to be positive because it is by an external agent and energy is required to bring an object against the gravitational field.

14 C

Kinetic energy depends also on mass (ie. KE = ½mv2) so if the mass is different, the KE would be different even if the velocity is the same.

15 A

Since pV = nRT, at constant volume and no change in gas quantity, both V and n as well as R are constants, and p α T. Hence, the graphs that shows a direct relationship between p and T is A or B. However, the temperature scale is in Celsius, and at t = 0°C or 273 K, the pressure could not possibly be zero, thus, the answer is option A.

16 D With reference to the 1st Law of Thermodynamics, ∆U = ∆Q + ∆W,

the temperature at both A and B are similar, meaning there is no change in internal energy from A to B and ∆U = 0. Also, there is expansion from A to B so ∆W is negative, which means ∆Q must be positive, indicating that heat is supplied to the gas.

17 A Work done by gas refers to an expansion. Work done on gas refers to a compression.

From Y to Z, it is a compression, hence, the work done by gas should be negative, and the area under YZ is 300 J. Thus, work done by gas from Y to Z = -300 J. The net work done by the gas is the area of the rectangle WXYZ and it is 600 J. It is positive because the work done through expansion process from W to X is larger than the work done through compression from Y to Z.

18 D Option A => speed at P is zero as it is a turning point for particle

Option B => displacement at Q is zero only at that instant, not always. Option C => at R where the speed is zero, the KE is zero. Option D => S is a point of maximum displacement, hence, its acceleration is

maximum, afterall, a = -ω2x and a is max. when x is max. (Note that each particle vibrates vertically in SHM)

19 D

Critical damping requires the object to return to equilibrium immediately and in the case of option D, the effect took place so fast that the passenger hardly noticed it, hence, critical damping.

23


20 C Let the maximum number of bright fringe on either side be nmax. dsinθ = nλ, but for nmax, sinθ = 1

nmax = dλ =

0.001400

550 x 10-9 = 4.545.

n must be an integer so nmax = 4, the maximum on either side. But the total number of bright fringes that can be observed includes the central bright fringe and all the fringes on the other side, so the total number is 4 + 1 + 4 = 9

21 A The path difference between S1X and S2X is 1 m or 2λ.

If the two sources are in phase, the path difference is in the form of nλ so it would be constructive interference. In this case, the two sources have a phase difference of π radians, meaning they are in anti-phase, hence, the outcome is destructive interference.

22 B The diameter of the kink is 1.5 cm or 0.015 m, hence, the radius is 0.0075 m.

ε = -∆NBA∆t

= (1)(2.5 x 10-2)(π0.00752)

0.050 = 8.8369 x 10-5

P = ε2

R = 4.8806 x 10-9 W

23 C

To do this question, trace on the diagram the path of the current in one cycle and then using Fleming’s LHR, figure out that the magnetic force acting on P is downwards. Now trace the current round the circuit in the other direction and you will realise that the force is also acting downwards.

24 A

Bev = mv2

r => Be =

mvr

=> Be = mrω

r =>

Bem

= ω = 2πT

Hence, T = 2πmBe

and independent of velocity.

25 C

Electric field at a point is its potential gradient at that point, or dVdx

.

Hence, if the potential around the point is constant, its gradient there would be zero and the electric field there would be zero.

24


26 B

Bx is parallel to the direction of the current on wire Y, so there is no force on P due to Bx. The magnetic force on P is contributed by By (downwards) and IY (to the right). Hence, by Fleming’s LHR, the force is into the page.

27 B

F = Q1Q2

4πεor2 => F α

1r2 =>

F'F

= (rr' )2 => F = (

12 )2(1.5) = 0.375 N

28 A

Electric field always points from higher to lower potential, hence, EPX and EXQ are both to the right.

Also, we can see that the vertical components of the electric fields at Y would cancel out, leaving only the horizontal components pointing to the right. The magnitudes of EQY and EPY would be the same since Y is equidistant from P and Q.

Potential at a point due to P = VP = 2µ

4πεor . Also, VQ =

-2µ4πεor

VTotal = VP + VQ is always zero because rPX = rQX and rPY = rQY. 29 D

Option A => RBC is indeed 0.5 Ω. Option B => If another resistor, say 1 Ω, is connected in parallel to AC, the

resultant resistance is 0.462 Ω. Option C => RAD is indeed 1 Ω. Note that resistor between B and C serves no

purpose as no current will go through there because the potential at B and C are equal. Since VCD = 0, IBC = 0.

Option D => This statement contradicts the factual statement in option C.

P Q

X

Y

2 µC - 2 µC

EQY

EPY

P Bx

By B

B = B-field due to wire X acting on point P Bx = horizontal component of B-field By = vertical component of B-field

25


30 B

R = ρLA

= ρL

πd2

4 ; Hence, R α

Ld2

Since L and d are both doubled, resistance is halved, and current is doubled. 31 B Flux linkage is the main function of the soft-iron core. 32 C For steady (dc) current : P = I2R

For alternating current : P = (Io√2

)2R4

Equating them : Io = √8 I = 2√2 I 33 B This is one of the four observations of the Photoelectric Effect Experiement. 34 B

The probability of the electron tunnelling the barrier depends on the transmission coefficient, T = exp(-2kd), where d = barrier thickness, L

and k = 2

2

h

)EU(m8 −π

Hence, for X=> T α exp(-√(5E – E) L) = exp(-2EL) for Y=> T α exp(-√(17E – E) 2L) = exp(-8EL) for Z=> T α exp(-√(2E – E) 3L) = exp(-3EL)

Thus, X has the greatest T value (transmission coefficient) followed by Z and then Y.

35 B It is an electron (not photon) incident on the gas atom, so the ‘all-or-nothing- rule does not apply. 20.0 Xx 10-19 J of mechanical energy can raise the atom to the -5.4 x 10-19 J level or the -2.4 x 10-19 J level. If it is raised to the -2.4 x 10-19 J level, the remaining energy which becomes the kinetic energy is 0.6 x 10-19 J, which is not one of the four answer options given. Hence, consider that the atom is raised to the -5.4 x 10-19 J level, which means that -5.4 x 10-19 J – (-21.8 x 10-19 J) = 16.4 x 10-19 J is being absorbed by the atom. Hence, the ke transferred to the free electron is 20.0 x 10-19 J - 16.4 x 10-19 J = 3.6 x 10-19 J

36 D

It is a fact that heating and cutting is usually takes place at the energy level of the infrared region.

37 C This is a reiteration of this fact which many students got wrong in BT2.

26


38 C During reverse bias condition of a p-n junction, the p-type semiconductor becomes more (rather than less) negative. The depletion region widens, as the holes in the p-type semi-conductor migrate to the positive terminal of the battery. Hence, statement A is false.

39 A

Equation : U23592 + n1

0 Cs14355 + Rb90

37 + 3 n10

Binding Energy : 235(7.6) + 0 + 195 = 143X + 90X + 0 X = 8.502 MeV

40 B

After 16 years,

NNo

= (12 )n

31250500000

= (12 )n

116

= (12 )n

(12 )4 = (

12 )n

n = 4

Therefore, 16 years is 4 half-lives. t1/2 = 4 yrs

After 12 years, NNo

= (12 )n

M

500000 = (

12 )12/4

M

500000 = (

12 )3

M = 62500

t / years N 0 500000 4 250000 8 125000 12 62500 16 31250

End of Solutions

2010 sajc h2 prelim paper 1

Documents