MCQ Physics
MCQ Physics
MCQ Physics
Roll No:
Federal Board HSSC-II Examination
Answer Sheet No: ___________
Physics Model Question Paper
Signature of Candidate: ___________
Signature of Invigilator: ___________
U SECTION – 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.
D. NS -1
P
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. -2V
B. -1V
C. Zero
D. 2V
1
DO NOT WRITE ANYTHING HERE
A. Zero
B. Ea
C. Ea 2 P
D. 4Ea
A. GR/ρ
B. Rl/A
C. ρR/G
D. GA/l
A. 10 -12NC -1
P P
P
B. 10 -6NC -1
P P
P
C. 1NC -1 P
D. 10 6NC -1
P P P
2
vii. What will be the magnitude of gain of an inverting op-amp having
resistances R 1 = 5kΩ and R 2 = 20kΩ?
R R R R
A. -5
B. -4
C. 4
D. 5
A. 2µF
B. 4µF
C. 6µF
D. 10µF
A. 1:9
B. 1:3
C. 3:1
D. 9:1
A. 3 x 10 8J
P P
B. 3 x 10 13J
P P
C. 9 x 10 13J
P P
D. 9 x 10 16J
P P
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/4
B. v/2
C. v
D. 2v
3 6
A. 1.0 Ω
B. 1.6 Ω
C. 3.7 Ω
D. 11 Ω
A. 1/128
B. 1/64
C. 1/16
D. 1/8
xiv. Which one of the following bulbs has the least resistance?
A. 100 w
B. 200 w
C. 300 w
D. 400 w
4
xv. The peak value of an AC is 2 2 A. What will be its RMS value?
A. zero
B. 2A
C. 2A
D. 2 2A
xvi. In the figure given below, what is the potential drop across the
resistor R 3?
R R
A. 3V
B. 4V
C. 9V
D. 12V
A. Hysteresis loss
B. Voltage loss
C. Eddy currents
D. Magnetic flux
____________________
For Examiner’s use only
Q. No.1: Total Marks: 17
Marks Obtained:
5
6
FBISE
WE WORK FOR EXCELL ENCE
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.
SECTION – B
U
(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.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)
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 10 -18J of electromagnetic energy. How
P P
Q.18 Circuit diagram shows a network of resistors each of resistance 2Ω. (3)
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)
SECTION – C
U
(Marks: 26)
(1+2+2)
b. Two point charges q = -1.0 × 10 -6C and q 2 = +4.0 × 10 -6C, are
P P
R R P P
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
12
FBISE
WE WOR K FOR EXCELLENCE
____________________
Page 1 of 1
13
FBISE
WE WORK F OR EXCELL ENCE
Q.1
i. C ii. C iii. D
iv. C v. A vi. C
vii. C viii. B ix. B
x. C xi. A xii. A
xiii. B xiv. D xv. C
xvi. A xvii. C
(17 ×
1=17)
0BU SECTION B
Q.2 (2)
Circuit
(1 mark)
–I 1R 1–E 2–(I 1 –I 2 )R 2 = 0
R R R R R R R R R R R R
(OR)
–I 1R 1+E 2–(I 1 –I 2 )R 2 = 0
R R R R R R R R R R R R
and
–(I 2 –I 1 )R 2–I 2R 3+E 1 = 0
R R R R R R R R R R R R
(OR)
–(I 2 –I 1 )R 2–I 2R 3–E 1 = 0
R R R R R R R R R R R R (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)
1 1
X c = ωc = 2πfc
R R
1
⇒ Xc ∝
f R R
and X L = ωL = 2πfL
R R
⇒ XL R R
∝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 = o 0. The power factor
P P
(Cosθ = coso 0) is 1. P P
(1 mark)
Q.6 (2)
X= A B + AB (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 V BB R R
is required (1 mark)
and B – C Junction is reverse biased, so V CC is taken high. R (1 mark)
R
Q.8 (2)
Equation for de-Broglie wavelength
h h
λ= = (1 mark)
P mv
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 = M p + M N – M (nucleus) R R R R (1 mark)
According to Einstein’s equation
E = ∆MC 2 P
Q.11 (2)
Magnetic force on a charged particle when projected at right angles
into a magnetic field is
F = q (V ×B ) (i.e. θ = 90°)
or F = q V B sin90°
F=qVB (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)
1 q
As V= × (1 mark)
4πε0 r
q = –4µC = –4 × 10 -6C P P
r = 20cm = 0.2m
1
4πε0 = 9 × 10 9Nm 2C -2 P P P P P
q = e= 1.6 × 10 -19C
P P P
P P
υ=?
16
I = nqAυ (1 mark)
I 1. 2
υ = = = 1.5 ×10 −6 ms −1 (2 marks)
nqA 5 ×10 ×1.6 ×10 −19 ×10 −4
28
(OR)
Since the electron and proton possess same momentum
m eυ e = m pυ p
R R R R R R R R
υe m p
= (1)
υp me
Force (F B) due to magnetic field provides Fc i.e. F c = F B
R R R R R R
mυ 2
= Beυ
r
Ber
υ= m
(1 mark)
Be re
υ me
Then υ =
e
Be r p
p
mp
υe r mp
= e ×
υ p rp me
r υ m
⇒ e = e× e
rp υ p m p
re
rp =1 using equation (1) (2 marks)
Q.15 (3)
To get a resultant of 60Ω, which is less than 100Ω, a resistor R 2 is R R
Then
1 1 1 1 1 1
= + ⇒ = −
R R1 R2 R2 R R1
1 1 1 5 −3 2 1
= − = = =
R2 60 100 300 300 150
R 2 = 150Ω
R R (2 marks)
(1 mark)
(OR)
17
F = qυB = Beυ (θ = 90°)
mυ 2
= Beυ
r
Ber
υ= m (1 mark)
( 4 B )er
ύ = m = 4υ ( B ′ = 4 B )
1
As E = 2 mυ 2
1 1
É = 2 m(v ′) 2 = 2 m(4υ) 2
1 2
É = 16 2 mυ
É = 16E (2 marks)
Q.16 (3)
V2
P= Re
V2
P= = 10W ( R = 3R)
e (1
3R
mark)
When connected in parallel
R
Re =
3
V2 V2 V 2
P′ = R = 3×
R
= 9 ⇒ P ′ =9 P
3R
3
P ′ = 9 × 10W ( P=10W)
P ′ =90W (2 marks)
(OR)
Energy of light = # of photons × Energy of one photon
E = nhf (1 mark)
E Eλ
n= =
hf (1
hc
mark)
1 ×10 −18 × 600 ×10 −9
n= =3 (1
6.63 ×10 −34 × 3 ×10 8
mark)
18
Q.17 (3)
Since
P = I 2R 1 P P
R
P P 5
I2 = ⇒I = = =0.5A (1 mark)
R1 R1 20
P P
V 24V
As R= = = 48 Ω
I 0.5 A
R = R1 + R2 ⇒ R2 = R – R1
R R R R R R R R
R 2 = 48 Ω − 20 Ω
R R
R 2 = 24 Ω
R R (2 marks)
(OR)
For a solenoid of length ‘ l ’, cross-sectional area ‘A’ & # of turns ‘N’
∆Φ ∆( BA )
ε = −N = −N
∆t ∆t
∆B
ε = −NA (1 mark)
∆t
N
B = µ o nI = µ o I (for a solenoid)
l
R R R R
N
∆( µo I)
ε = −NA l
∆t
− NA µ o ∆I
ε= × (1)
l ∆t
Using
∆I
ε = −L (2) (1 mark)
∆t
Comparing equation (1) and (2)
∆I − µo N 2 A ∆I
−L =
∆t l ∆t
N2
L = µo A = µo n 2 lA = µ o nNA (1 mark)
l
Which is the required expression
Q.18 (3)
19
For R 1 & R 2 let their resultant be R ′ = 4Ω
R R R R
4×2 4
Resultant of R ′ & R3 connected in parallel is 4+2 3
= Ω (1 mark) R ′′ =
4 10
R ′′ & R 4 are in series their resultant is R ′′′ = 3 + 2 = 3 Ω
R R
10
2× 20
3 = 3 = 20 = 1.25 Ω
Resultant of R ′′′ & R 5 connected in parallel is R = 10
R R
16 16
+2 3
3
(2
marks)
(OR)
L 4L
R= ρ =ρ 2
A πd
4L
V = IR = I ρ 2 (1 mark)
πd
4L
For the wire ‘x’ V x = I x ρ πd 2
R R R R
x
4L
For the wire ‘y’ Vy = R R I y ρ πd 2
R R
4L 4L
I x. ρ πd 2 =I y × ρ πd 2
R R R R
x y
Ix Iy
2
= 2
dx dy
4I x Iy dy
2
= 2 as d x =
dy dy 2
R R
⇒ I y = 4I x
Fraction of current which passes
Ix Ix I 1
through ‘x’ = = = x = = 0.20 (2 marks)
total current I x + I y 5I x 5
Q.19 (3)
The combined capacitance ‘C’ is given by
20
C1 × C 2 3.0 × 6.0
C = C + C = 3.0 + 6.0 µF = 2.0µF = 2.0 ×10 F
−6
(1 mark)
1 2
Net charge stored
Q = CV = 2.0 × 10 -6 × 18 P
Q = 36 × 10 -6C
P
P P (1 mark)
For capacitors connected in series, charge is same. (1 mark)
(OR)
(KE) max = Ve
R R
1
mv 2 = Ve
2
mv = 2mVe
h h h
Now λ= = = (1 mark)
P mv 2mVe
6.63 ×10 −34
λ=
2 × 9.11 ×10 −31 ×100 ×1.6 ×10 −19
λ = 1.22 ×10 −10 m (1
mark)
In the electromagnetic spectrum this would be X–Radiation. (1 mark)
1BU SECTION 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)
σ ∧
b.
(1 mark)
21
1 q1 1 q2
= × (1 mark)
4π εo x 2
4π εo ( x + 3) 2
1 ×10 −6 4 ×10 −6
=
xυ ( x + 3) 2
(x+3) 2 = 4x 2
P P
P
x+3 = 2x
x = 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 × 10 -9m
φ = 2.46ev = 2.46 × 1.6 × 10 -19J
P P
P P
hc
(KE) max = hf – φ = λ − φ
R R
R R P P
____________________
22