PRACTICE EXAMINATION - 08

FOR G.C.E A/L STUDENTS – 2020

CONDUCTED BY TAMIL STUDENTS OF FACULTY OF ENGINEERING

UNIVERSITY OF MORATUWA

PHYSICS

MARKING SCHEME WITH M.C.Q ELABORATIONS

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Answers For multiple Choice Questions

Q Ans Q Ans Q Ans Q Ans Q Ans

01. 4 02. 1 03. 3 04. 4 05. 2

06. 5 07. 2 08. 2 09. 4 10. 4

Marks Distribution

Part 𝐼 = 10 × 2 = 20 Marks

Part 𝐼𝐼 = 50 Marks

Part𝐼𝐼𝐴 (Structued Essay Question= 𝟏 × (𝟐 × 𝟏𝟎) = 20 Marks

Part𝐼𝐼𝐵(Essay Questions)= 𝟏 × (𝟐 × 𝟏𝟓) = 30Marks

Total = Part I +Part II

= 20 + 50 = 70 Marks

= 70

70× 100 = 100 Marks

M.C.Q ELABORATIONS (1)

By considering equilibrium, By considering equilibrium,

2T sin Ɵ = Mg To = T cos Ɵ = 𝑀𝑔 cos Ɵ

2 𝑠𝑖𝑛 𝜃 =

𝑀𝑔

2 𝑡𝑎𝑛 𝜃

T = 𝑀𝑔

2 𝑠𝑖𝑛 𝜃

Answer _ 4

(2)

• By Fleming’s right-hand rule, when the rod is dragged towards the left and

released, there is an induced emf from Q to P. Therefore, Q’s potential is

negative relative to P. Therefore, initially the potential of Q increases

negatively until it reaches the center of oscillation. When it passes the center

of oscillation, the magnitude o velocity of the rod decreases. So, the potential

of Q also decreases negatively.

• When the rod starts moving to the left again, potential at Q r to P increases

positively (by Fleming’s right-hand rule). P attains a maximum positive value

when passing the centre of oscillation and this value attains zero again.

• Variation of potential isn’t linear as velocity doesn’t vary linearly throughout

the motion.

Answer_ 1 (3) (0.4 - I)

V50Ω = 2V (as there is no deflection in the

galvanometer).

V = IR (50 Ω)

2 = I × 50 => I = 0.04 A

V40Ω = VR

0.04 × 40 = ( 0.4 – 0.04) × R

R = 4.4 Ω Answer_ 3

(4) g 𝛼 1

𝑟

a) True – The distance of poles is lesser from the earth’s center than the distance

between the earth’s center and equator. b) False – If the earth’s rotation rate is decreased, you decrease the acceleration due to

rotation, but the acceleration due to gravity, g, remains unaffected, so only the net acceleration increases.

c) False – g = 𝐺𝑀

𝑟2 , M = d × volume = d × 4

3𝜋𝑟3 => g =

𝐺 ⅆ4

3𝜋𝑟3

𝑟2 = 4

3𝐺𝑑𝑟

g 𝛼 r When r , g

Answer_ 4

(5)

(6)

• Field lines are directed from left to right.

Therefore, potentials at B and C are more negative relative to A. (1,4 cannot

be the answers)

• Field intensity is greater towards the right.

So, VAB < VBC.

The possible potentials at B and C are, -200 V and -450 V respectively.

Answer_ 5

(acceleration ↑ = 0)

(7) F = ma

F - 2T cos Ɵ = m (0)

F = 2T cos Ɵ

T = F / 2 cos Ɵ

F = ma (m)

T sin Ɵ = ma

a = (F tan Ɵ) / 2m

= (F.x) / (2m.√𝑎2 − 𝑥2)

Answer_ 2

(8) n, P => constant

𝐶𝑝

𝐶𝑣= 𝑟 ⇒ 𝑐𝑣 =

𝑐𝑝

𝑟

Cp – Cv = R ⇒ 𝐶𝑃 −𝐶𝑃

𝑟= 𝑅

𝑅 =𝐶𝑃(𝑟−1)

𝑟

Applying ideal gas equation, ΔQ = ΔU + ΔW

PV = nRT n Cp (ΔƟ) = ΔU + P (ΔV)

PV = (1). 𝐶𝑃(𝑟−1)

𝑟. T 1. Cp (T) = ΔU + PV

T = 𝑃𝑉𝑟

𝑐𝑝(𝑟−1) ΔU = CpT – PV = PV (

𝑟

𝑟−1− 1) =

𝑃𝑉

(𝑟−1)

Answer_ 2

(9) Weight of liquid column of height 2x = 2xadg

F = ma

2xadg = ma

a = (2𝑎ⅆ𝑔

𝑚) . x ≡ a = -w2.x

w2 = 2𝑎ⅆ𝑔

𝑚 Therefore, w = √

2𝑎ⅆ𝑔

𝑚

T = 2𝜋

𝑤 = 2𝛱√

𝑚

2𝑎ⅆ𝑔 Answer_ 4

(10)

• When the switch K is closed, there is current flowing the resistors.

V = IR

10 = I × 10 × 10−3

I = 10−3

• 10 V = Vab + Vbe

10 = (10−3 × 4 kΩ) + 𝑄

2×10−6

Q = 12 μC

Vcd = 0. Therefore, the other capacitor doesn’t get charged.

Total charges stored = 12 μC

• When the switch is open; there is no current flowing through the resistors,

Vbe = Vcd = 10 V

Total charge stored in capacitors = 2 × (CV) = 2 × 2μF × 10 = 40 μC

• Total additional charge given = 40 μC – 12 μC = 28 μC

Answer_ 4

Answers For Structured Essay Questions

(1)

i.

ii.

iii.

………..(3) marks For complete answer

iv.

………….. (2) marks

v. Angle of minimum deviation.

…………………….. (0.5) mark

vii. Ɵ1 + Ɵ2 = 90o

Ɵ2 = 90o – Ɵ1 = 55o

sin Cg = 1

𝑛=

3

4= 0.75

∴ Cg = 48.6o Ɵ2 > Cg ∴ The ray undergoes total internal reflection. The above experiment cannot be carried out.

…………………….. (1) mark

viii.

…………………….. (1) mark

ix. The direct reading when the telescope is in line with the collimator without the

prism on the prism table. …………………….. (0.5) mark

The scale reading when the image of the slit is at the position where it just turns to the opposite side when rotating the prism table. …………………….. (0.5) mark

x. Spectrometer method. …………………….. (0.5) mark

Least count of the spectrometer is much smaller. Therefore, readings are taken with the spectrometer will have a smaller percentage error. …………………….. (1) mark

(10 marks)

Answers For Essay Questions

a)

i. Mineral oil, coal, solar energy …………………….. (1) mark

ii. Durability, economic benefits, minimum environmental pollution, easy access

to source, minimum initial capital. …………………….. (1.5) mark

b)

i. πr2vρ …………………….. (1) mark

ii. 1

2 πr2vρ.v2 =

1

2 πr2ρ.v3 …………………….. (1) mark

iii. 1

2 πr2ρ.v3 ×

80

100 …………………….. (1) mark

iv. 1

2 ×

22

7 × (1.42) × 1.2 × 103 ×

80

100 …………………….. (1) mark

= 2956.8 𝑊 …………………….. (1) mark c)

i. 2956.8 × 10 = 1

2 × I × 102 …………………….. (0.5) mark

I = 591.36 kg m2 …………………….. (1) mark

ii. w = wo + αt

10 = 0 + α × 10 α = 1 rad s-2 …………………….. (1) mark τ = Iα = 591.36 × 1 = 591.36 Nm. …………………….. (1) mark

d)

i. 591.36 – 443.52 = 519.36 × α

α = 0.25 rad s-2 …………………….. (1) mark ii. w = wo + αt …………………….. (0.5) mark

w = 0 + 0.25 × 100 = 25 rad s-1 …………………….. (1) mark

iii. 1

2 Iw2 = τƟ

= 1

2 × 591.36 × 25 × 25

= 443.52Ɵ => Ɵ = 416.67 rad …………………….. (0.5) mark

No of cycles = 416.67 rad

2𝜋 rad

=416.67 rad

2×22

7 rad

=416.67 ×7

44

= 66.289

≈ 66 cycles …………………….. (1) mark

(𝟏𝟓 𝐌𝐚𝐫𝐤𝐬)

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