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Federal Board HSSC-II Examination Physics Model Question Paper

Roll No: Answer Sheet No: ___________

Signature of Candidate: ___________

Signature of Invigilator: ___________

USECTION – A

Time allowed: 25 minutes Marks: 17

Note: Section-A is compulsory and comprises pages 1-5. All parts of this section are to be answered on the question paper itself. It should be completed in the first 25 minutes and handed over to the Centre Superintendent. Deleting/overwriting is not allowed. Do not use lead pencil.

Q.1 Insert the correct option i.e. A/B/C/D in the empty box opposite each part. Each part carries one mark.

i. Which of the following is the unit of electric intensity?

A. Nm B. NCC. NCP

-1

D. NSP

-1

ii. Two point charges -10µc and +10µc are placed 10cm apart. What is the potential at the centre of the line joining the two charges?

A. -2VB. -1VC. ZeroD. 2V

iii. For which of the following AC can NOT be used?

A. HeatingB. LightingC. Transforming voltageD. Electroplating

1

DO NOT WRITE ANYTHING HERE

iii. A square sheet of side ‘a’ is held perpendicular to a uniform electric field of strength E. What is the electric flux linked with

the surface?

A. ZeroB. EaC. EaP

2

D. 4Ea

v. A cylindrical metal wire of length ‘l’ and cross-sectional area ‘A’ has resistance ‘R’, conductance ‘G’, resistivity ‘ρ’ and

conductivity ‘σ’. Which of the following expressions for ‘σ’ is valid?

A. GR/ρB. Rl/AC. ρR/GD. GA/l

vi. If a charge of 1µC experiences a force of 10P

-6PN at a point,

what will be the electric intensity at that point?

A. 10P

-12PNCP

-1

B. 10P

-6PNCP

-1

C. 1NCP

-1

D. 10P

6PNCP

-1

2

vii. What will be the magnitude of gain of an inverting op-amp having resistances RR1R= 5kΩ and RR2R = 20kΩ?

A. -5B. -4C. 4D. 5

viii. A number of capacitors each of 2µF are connected as shown in the figure given below:

What is the net capacitance between A & B?

A. 2µFB. 4µFC. 6µFD. 10µF

ix. Two copper wires A and B of lengths 1m and 9m respectively

are found to have same resistance. What is the ratio B

A

d

d

between their diameters?

A. 1:9B. 1:3C. 3:1D. 9:1

x. What is the rest mass energy of an object of mass 0.1g?

A. 3 x 10P

8PJ

B. 3 x 10P

13PJ

C. 9 x 10P

13PJ

D. 9 x 10P

16PJ

3

xi. A current flows in a wire of circular cross-section with the free electrons traveling with a mean drift velocity ‘v’. If an equal current flows in a wire of the same material but of twice the radius, what is the new drift velocity?

A. v/4B. v/2C. vD. 2v

xii. Three resistors are connected as shown in the diagram using connecting wires of negligible resistance.

What is the resistance between points P and Q?

A. 1.0 ΩB. 1.6 ΩC. 3.7 ΩD. 11 Ω

xiii. The half life of a radioactive element is such that 7/8 of a given quantity decays in 12 days. What fraction remains

un-decayed after 24 days?

A. 1/128B. 1/64C. 1/16D. 1/8

xiv. Which one of the following bulbs has the least resistance?

A. 100 wB. 200 wC. 300 wD. 400 w

4

3 6

xv. The peak value of an AC is 2 2 A. What will be its RMS value?

A. zeroB. 2 AC. 2AD. 2 2 A

xvi. In the figure given below, what is the potential drop across the resistor RR3R?

A. 3VB. 4V C. 9V D. 12V

xvii. In a transformer, laminated sheets with insulation in between are used to minimize:

A. Hysteresis lossB. Voltage lossC. Eddy currentsD. Magnetic flux

____________________For Examiner’s use only

Q. No.1: Total Marks:

Marks Obtained:

5

17

6

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Federal Board HSSC-II Examination Physics Model Question Paper

Time allowed: 2.35 hours Total Marks: 68

Note: Sections ‘B’ and ‘C’ comprise pages 1-5 and questions therein are to be answered on the separately provided answer book. Answer all the questions from section ‘B’ and section ‘C’. Use supplementary answer sheet i.e., sheet B if required. Write your answers neatly and legibly.

USECTION – B (42 marks)

NOTE: Attempt ALL the questions. The answer to each question should not exceed 3 to 4 lines.

Q.2 Apply KVL, what are the equations for voltage changes in the two loops as shown in the figure below? (2)

Q.3 Why is soft iron preferred as the core material for making transformers? (2)

Q.4 How does the frequency response of a capacitor differ from that of an inductor when subjected to a source of AC voltage? (2)

Q.5 Draw the impedance diagram for an RLC series circuit at resonance and show why its power factor is equal to one.

(2)

7

Page 1 of 5 Turn Over Q.6 Write Boolean expression for XNOR gate. What will be its output when

both inputs are made zero? (2)

Q.7 Why is Vcc made high in comparison to VRBBR as shown in the figure. (2)

Q.8 Are de-Broglie waves associated with all moving objects? Why is it not significant for macroscopic objects? (2)

Q.9 Why should the target material in the production of x-ray unit have high melting point? (2)

Q.10 Why is the mass of a nucleus always less than the total mass of all the protons and neutrons making up the nucleus? (2)

Q.11 Why do charged particles follow circular paths when projected at a right angle to the magnetic field? (2)

Q.12 What is the origin of γ-rays compared to the origin of x-rays? (2)

Q.13 What is the absolute potential at a distance of 20cm from a point charge of -4µC? (2)

Q.14 A conductor with a cross-sectional area 10P

-4P mP

2P carries an electric current of

1.2 A. If the number of free electrons be 5 x 10P

28P mP

-3P, calculate the electron

drift velocity. (3)

(OR)

An electron and a proton possessing equal momenta are injected into a region at right angles to a uniform magnetic filed. What is the ratio of their radii of curvature while moving inside the magnetic field?

(3)

8

Page 2 of 5 Turn Over Q.15 A circuit has a resistance of 100Ω. What should be the value of another

resistor to be connected to it so as to reduce the total circuit resistance to 60 Ω. Also show by circuit diagram. (3)

(OR)

A proton is moving under the influence of a perpendicular magnetic field (B) and possesses energy E. What will be the energy of the proton if the magnetic filed is increased to 4B while it is compelled to move in a circular path of same radius? (3)

Q.16 Three equal resistors connected in series across a source of e.m.f together dissipate a power of 10w. What should be the power dissipation if the same resistors are connected in parallel across the same source of e.m.f? (3)

(OR)

The eye can detect as little as 1 x 10P

-18PJ of electromagnetic energy. How

many photons of orange light whose wavelength is 600 nm does this energy represent? (3)

Q.17 A load of 20 Ω with a power rate of 5.0 watt is to be connected to a 24v battery. What is the minimum resistance of the series resistor that will prevent the power rate from increasing? (3)

(OR)

Derive an expression for the inductance of a solenoid of length ‘l’, cross-sectional area ‘A’ and having number of turns ‘N’. (3)

Q.18 Circuit diagram shows a network of resistors each of resistance 2Ω. (3)

What is the effective resistance between the points P & Q?

(OR)

9

Two wires ‘X’ and ‘Y’ each of the same length and the same material are connected in parallel to a battery. The diameter of ‘X’ is half that of ‘Y’. What fraction of the total current passes through ‘X’? (3)

Page 3 of 5 Turn Over Q.19 How much charge is stored in a 3.0µF capacitor and a 6.0µF capacitor

when joined in series with an 18V battery? (3)

(OR)

Calculate the wavelength of electrons that have been accelerated from rest through a P.D of 100V. What kind of electromagnetic radiation has wavelength similar to this value? (3)

USECTION – C(Marks: 26)

Note: Attempt ALL the questions.

Q.20 What is the need of using a transformer in an electrical network? Explain its working principle. How can this transformer be made to step up and to step down? Discuss power losses in a transformer, how can they be minimized? (8)

Q.21 a. State and explain Gauss’s law, also derive ∧

= rE02ε

σ, for an electric

field E due to an infinite sheet of charge. Where ‘σ’ is the surface charge density, εRoR is permittivity of free space and ∧

r is a unit vector. (1+2+2)

b. Two point charges q = -1.0×10P

-6PC and qR2R = +4.0×10P

-6PC, are

separated by a distance of 3.0m. Find and justify zero field location. (3)

Q.22 a. What is a galvanometer? Explain its construction and basic principle of working. (5)

b. What is meant by sensitivity of a galvanometer? How can a galvanometer be made more sensitive? (3)

c. Show by diagrams how a galvanometer can be transformed into an ammeter and a voltmeter? (2)

(OR)

a. Explain the experiment photoelectric effect, discuss its important results. (3+3)

b. A sodium surface is illuminated with light of wavelength 300nm. The work function of sodium metal is 2.46ev. Calculate:

10

(i) The max KE of the ejected electron. (2)(ii) The cut-off wavelength for sodium. (2)

____________________

Page 4 of 5 Turn Over

11

DATA

Page 5 of 5

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Federal Board HSSC-II Examination Physics Practical Model Question Paper

Time allowed: 3 hours Marks: 15

NOTE: Perform ONE experiment, as allotted by the examiner. Plot a graph where necessary.

1. Find the resistance of a voltmeter by drawing a graph between R and I/V. (10)

2. Find the characteristics of a semi-conductor diode and calculate its forward and reserve current resistance. (10)

3. Determine the e/m of an electron by deflection method (teltron tube). (10)

Practical Notebook (2)

Viva voce (3)

____________________

Page 1 of 1

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Federal Board HSSC – II Examination Physics – Mark Scheme

USECTION A

Q.1i. C ii. C iii. Div. C v. A vi. Cvii. C viii. B ix. Bx. C xi. A xii. Axiii. B xiv. D xv. Cxvi. A xvii. C

(17 × 1=17)

0BUSECTION B

Q.2 (2)Circuit

(1 mark)–IR1RRR1R–ER2R–(IR1R–IR2R)RR2R = 0

(OR)–IR1RRR1R+ER2R–(IR1R–IR2R)RR2R = 0and –(IR2R–IR1R)RR2R–IR2RRR3R+ER1R = 0

(OR)–(IR2R–IR1R)RR2R–IR2RRR3R–ER1R = 0 (1 mark)

Q.3 (2)i. Soft iron can be magnetized and demagnetized easily. (1 mark)

14

ii. Hysteresis loss for soft iron is small. . (1 mark)Q.4 (2)

XRcR = fcc πω 2

11 =

⇒ XRcR f

1∝

and XRLR = Lω = fLπ2

⇒ XRLR f∝ (1 mark)With the increase of frequency, reactance of a capacitor decreases whereas reactance of an inductor increases and vice versa (1 mark)

Q.5 (2)

⇒

(1 mark)At resonance the impedance of the circuit is resistive. Therefore current and voltage are in phase i.e. θ = oP

0P. The power factor

(Cosθ = cosoP

0P) is 1. (1 mark)

Q.6 (2)X = BABA + (1 mark)

When A = 0 & B = 0

Then X = 0×1 + 1×0 = 1 (1 mark)

Q.7 (2)In a transistor E – B Junction is always forward biased, so, small VRBBR is required (1 mark)and B – C Junction is reverse biased, so VRCCR is taken high. (1 mark)

Q.8 (2)Equation for de-Broglie wavelength

mv

h

P

h ==λ (1 mark)

For macroscopic objects ‘λ’ is v.v. small which cannot be observed.(1 mark)

15

Q.9 (2)In the production of x-rays, electrons are incident on the target material, which gives a large amount of KE to the target and target material will become very hot which may melt, so we use a target of high M.P. (2 marks)

Q.10 (2)Mass defect

∆M = MRpR + MRNR – M (nucleus) (1 mark)According to Einstein’s equation

E = ∆MCP

2

This additional mass changes into B.E of the atom. (1 mark)

Q.11 (2)Magnetic force on a charged particle when projected at right angles into a magnetic field is

)( BVqF ×= (i.e. θ = 90°)or F = q V B sin90°

F = q V B (1 mark)‘F’ provides necessary centripetal force to the charged particles. Therefore they follow a circular path. (1 mark)

Q.12 (2)γ-rays are coming out from the nucleus of an unstable atom. (1 mark)X-rays are obtained by the inner shell transition of electrons. (1 mark)

Q.13 (2)

As V = r

q×04

1

πε (1 mark)

q = –4µC = –4×10P

-6PC

r = 20cm = 0.2m

04

1

πε = 9×10P

9PNmP

2PCP

-2

V = Voltm

CCNm 52

6229

109)2.0(

)104(109 ×−=×−×× −−

(1

mark)Q.14 (3)

I = 1.2A A = 10P

-4PmP

2

n = 5×10P

28PmP

-3

q = e= 1.6×10P

-19PC

υ = ?

16

I = nqAυ (1 mark)

υ = 1641928

105.110106.1105

2.1 −−−− ×=

××××= ms

nqA

I(2 marks)

(OR)

Since the electron and proton possess same momentum mReRυReR = mRpRυRpR

e

p

m

m=

p

e

υυ

(1)

Force (FRBR) due to magnetic field provides Fc i.e. FRcR = FRBR

r

m 2υ = Beυ

υ = m

Ber(1 mark)

Then

p

pe

e

ee

p

e

m

rBm

rB

=υυ

e

p

p

e

p

e

m

m

r

r×=

υυ

⇒p

e

p

e

p

e

m

m

r

r×=

υυ

p

e

r

r=1 using equation (1) (2 marks)

Q.15 (3)To get a resultant of 60Ω, which is less than 100Ω, a resistor RR2R is connected in parallel to RR1R = 100ΩThen

1221

111111

RRRRRR−=⇒+=

150

1

300

2

300

35

100

1

60

11

2

==−=−=R

RR2R = 150Ω (2 marks)

(1 mark)(OR)

17

F = qυB = Beυ (θ = 90°)

υυBe

r

m =2

υ = m

Ber(1 mark)

ύ = υ4)4( =m

erB ( BB 4=′ )

As E = 2

2

1 υm

É = 22 )4(2

1)(

2

1 υmvm =′

É = 16

2

2

1 υm

É = 16E (2 marks)

Q.16 (3)

P = eR

V 2

P = R

V

3

2

= 10W ( eR = 3R) (1

mark)When connected in parallel

eR = 3

R

P′ = 3

2

RV

= PPR

V

R

V9

393

22

=′⇒

=×

P′ = 9×10W ( P=10W)P′ =90W (2 marks)

(OR)

Energy of light = # of photons × Energy of one photon E = nhf (1 mark)

n = hfE

= hc

Eλ(1

mark)

n = 834

918

1031063.6

10600101

××××××

−

−−

= 3 (1

mark)

18

Q.17 (3)

Since P = IP

2PRR1

IP

2P =

11 R

PI

R

P =⇒ =20

5 =0.5A (1 mark)

As R = Ω== 485.0

24

A

V

I

V

R = RR1R + RR2R ⇒ RR2R = R – RR1R RR2R = Ω−Ω 2048

RR2R = Ω24 (2 marks)

(OR)

For a solenoid of length ‘ l ’, cross-sectional area ‘A’ & # of turns ‘N’

t

BAN

tN

∆∆−=

∆∆Φ−= )(ε

t

BNA

∆∆−=ε (1 mark)

B = µRoRnI = µRoR

l

NI (for a solenoid)

t

Il

N

NAo

∆

∆−=

)(µε

t

I

l

NA o

∆∆×

−=

µε (1)

Using

t

IL

∆∆−=ε (2) (1 mark)

Comparing equation (1) and (2)

t

IL

∆∆− =

t

I

l

ANo∆∆− 2µ

nNAlAnAl

NL ooo µµµ === 2

2

(1 mark)

Which is the required expression Q.18 (3)

19

For RR1R & RR2R let their resultant be Ω=′ 4R

Resultant of 3& RR′ connected in parallel is Ω=+×=′′

3

4

24

24R (1 mark)

R ′′ & RR4R are in series their resultant is Ω=

+=′′′

3

102

3

4R

Resultant of R ′′′ & RR5R connected in parallel is R = Ω===+

×25.1

16

20

316

320

23

103

102

(2 marks)

(OR)

R = 2

4

d

L

A

L

πρρ =

V = IR = I 2

4

d

L

πρ (1 mark)

For the wire ‘x’ VRxR = IRxR 2

4

xd

L

πρ

For the wire ‘y’ VRyR = IRyR2

4

yd

L

πρ

VRxR = VRyR (as the wires are connected in parallel)

IRxR. 2

4

xd

L

πρ =IRyR 2

4

yd

L

πρ×

22y

y

x

x

d

I

d

I=

22

4

y

y

y

x

d

I

d

I= as dRxR =

2yd

xy II 4=⇒

Fraction of current which passes

through ‘x’ = 20.05

1

5===

+=

x

x

yx

xx

I

I

II

I

currenttotal

I(2 marks)

Q.19 (3)The combined capacitance ‘C’ is given by

20

C = FFFCC

CC 6

21

21 100.20.20.60.3

0.60.3 −×==

+×=

+× µµ (1 mark)

Net charge stored Q = CV = 2.0×10P

-6P×18

Q = 36×10P

-6PC (1 mark)

For capacitors connected in series, charge is same. (1 mark)(OR)

(KE)RmaxR = Ve

Vemv =2

2

1

mVemv 2=

NowmVe

h

mv

h

P

h

2===λ (1 mark)

1931

34

106.11001011.92

1063.6−−

−

×××××

×=λ

m101022.1 −×=λ (1 mark)

In the electromagnetic spectrum this would be X–Radiation. (1 mark)

1BUSECTION C

Q.20 (8)- Need of transformer (1 mark)- Explanation and working principle + figure (2 marks)- Transformer equation + step up + step down (3 marks)- Power losses and their remedies (2 marks)

Q.21 (8)a. Statement of Gauss’s law (1 mark)

Explanation of Gauss’s law using spherical body + figure (2 marks)

Derivation for electric field E = ∧r

oεσ

2 + Figure (2 marks)

b.(1 mark)

Obviously, the zero field location will be at pt. P ER1R = ER2R

21

22

21

)3(4

1

4

1

+×=x

q

x

q

oo π επ ε (1 mark)

2

66

)3(

104101

+×=× −−

xxυ(x+3)P

2P = 4xP

2

x+3 = 2xx = 3m (1 mark)

Q.22 (10)a. Definition (1 mark)

Construction + figure + working (1+1+2 = 4 marks)b. Sensitivity factor + methods to increase the sensitivity

(1+2 = 3 marks)c. Diagram for ammeter (1 mark)

Diagram for voltmeter (1 mark)

(OR)

a. Definition (1 mark)Explanation + figure (2 marks)Results (1 mark for each result) (3 marks)

b. λ= 300nm = 300× 10P

-9Pm

φ = 2.46ev = 2.46 × 1.6 × 10P

-19PJ

(KE)RmaxR = hf – φλ

φ −= hc

(KE)RmaxR = 199

834

106.146.210300

1031063.6 −−

−

××−×

×××

(KE)RmaxR = (6.63×10P

-19P – 3.936×10P

-19P)J

(KE)RmaxR = 2.693×10P

-19PJ

(KE)RmaxR = 1.68ev (2 marks)

φ = hfRoR = o

hc

λ

19

834

106.146.2

1031063.6−

−

×××××==

φλ hco

mo71005.5 −×=λ

nmo 505=λ (2 marks)

____________________

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