Posted on: 02:12 Wed, 04 Dec 2019




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Exam Time:- Thursday 5th Dec. 2019
Physics Paper III & II (Objective & Essay) – 2.00pm – 5.00pm
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PHYSICS OBJ:-
1-10: CACABDBECA
11-20: CCDBBDEECA
21-30: ECACEDACDB
31-40: CADBBDCBBC
41-50: EBCADACDAE
51-60: CDCBCEDCCB
=•=•=•=•=•=•=•=•=•=•=

PHYSICS ANSWERS:
(1a)
Electrolysis of water uses small current for a long time while electrolysis of salt water uses a large current for a short time. Electrolysis of water in its decomposition gives hydrogen and oxygen gases due to the passage of an electric current.

2Hâ‚‚O(l) + Electric energy(heat energy) --> Oâ‚‚ + Hâ‚‚O

In the electrolysis of salt water like sea water, the primary aim is to produce hydrogen at the cathod.

(1b)
x/20 = 60/80
80x = 60 x 20
80x = 1200
x = 1200/80
x = 15cm
.:. The distance between the lenses = 60cm - 15cm
= 45cm

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(2a)
(i) Acceleration
(ii) Velocity

(2b)
MSR = 8.5mm
VSR = 15mm
TSR = MSR + VSR
= 8.5 + 15
= 23.5mm

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(3i)
Draw the velocity time graph



(3ii)
magnitude of u if the distance covered during deceleration is 75ms-¹
Area of BCD = 1/2bh
=>1/2×15/1×u/1 = 75
15u = 150
u = 150/15
u = 10ms-¹

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(4a)
(i) Nuclear fission
(ii) Discharge of fossil fuels such as oil, gas and coal to the atmosphere

(4b)
h = 1.6m
V = ?
g = 10m/s²
V² = u²+2gh
V² = 2×10×1.8
V² = 36m/s
V = √36
= 6m/s

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(5)
U1 = 4m/s
U2 = 0m/s
M2 = 60g = 0.06kg
V = 2.5m/s
M = ?
M1U1 + M2U2 = (M1+M2)V
M1(4) = (M1 + 0.06)2.5
4M1 = 2.5M1 + 1.5
4M1 - 2.5M1 = 1.5
1.5M1 = 1.5
M = 1.5/1.5
=1kg

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(6a)


(6b)
(i) It is used in torches, searchlights and headlights of vehicles to get powerful parallel beams of light.
(ii) They are used as shaving mirrors to see a larger image of the face.

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(10a)
Surface tension is the attractive force exerted upon the surface molecules of a liquid by the molecules beneath that tends to draw the surface molecules into the bulk of the liquid and makes the liquid assume the shape having the least surface area.

(10b)
{Pick any 2}
(i) It is in violation of the Heisenberg Uncertainty Principle. The Bohr Model considers electrons to have both a known radius and orbit, which is impossible according to Heisenberg.
(ii) The Bohr Model is very limited in terms of size. Poor spectral predictions are obtained when larger atoms are in question.
It cannot predict the relative intensities of spectral lines.
(iii) It does not explain the Zeeman Effect, when the spectral line is split into several components in the presence of a magnetic field.
(iv) The Bohr Model does not account for the fact that accelerating electrons do not emit electromagnetic radiation.

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(11a)
(i) Negligible mass = Gamma radiation
(ii) Large Ionization Power = Alpha Particle

(11b)
λ=0.693/T½
λ=0.693/18000
λ=0.0000385 s-¹

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(12ai)
The volume of a stone is measured using a measuring cylinder by firstly measuring the volume of water in cylinder (v2) , then the stone is dropped into the cylinder (v2)
Let v1 = volume of water without stone
V2 = volume of water and stone.
Therefore the volume of stone = v2 - v1

(12aii)
(i) Nature of the liquid.
(ii) Volume of the liquid.
(iii) Shape of the container of the liquid.

(12bi)
A body is said to be on dynamic equilibrium if the body is moving continually with a constant acceleration.

(12bii)


Fy = 0(at equilibrium)
Fa + Fb - W1 - W2 = 0
Fb = W1+W2-Fa ....(1)
Taking moment about point A
W1 × 0.02 + W2×0.05 - Fb×0.07 = 0
0.07Fb = 0.02W1 + 0.05W2
Fb = 2W1+5W2/7 ...(2)

(12ci)


u = 7m/s
Time of flight = 10s

Time of flight (T) = 2usinΦ/g
10 = 2×70sinΦ/10
140sinΦ = 100
sinΦ = 100/140 = 0.7143
Φ = sin-¹(0.7143)
Φ = 45.58°
Φ = 46°

(12cii)
Maximum height = u²sin²Φ/g
H = 70²sin²45.58/2×10
H = 4900×0.7143²/20
H = 125m
Maximum height = 125m

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(14ai)
(i) Eclipse
(ii) Shadows

(14aii)
There is a shade under a tree on a sunny day because a shadow is produced by the obstruction of light by the tree(opaque object). The shade under a tree or a canopy, on a bright sunny day is a shadow produced by the opaque tree or canopy which obstructs light rays from the sun and prevents them from getting to the area of the shade.

(14b)
The walls and the ceiling of a shadow are covered with soft perforable boards because which can minimize the reflection of sound waves by absorbing them gently. Other methods of controlling reverberation are by changing curtains round the hall and by having more Springs in the walls.

(14ci)
It is preferred to plane mirror on the construction of periscope because it act to re-invert the image formed by the objective and present an erect image of the object to the eye piece which subsequently magnifies it.

(14cii)
It means that the angle of incidence in the denser medium is 42° when the angle of refraction is the less dense medium is 90°

(14di)
Data
M = 0.8g
L = 1.5m
F0 = 180H3
(i) F0 = 1/2L√T/M
F0² = T/4ML²
T = F0²×4ML²
T = (180)²×4×0.0008×1.5²
T = 233.28N

(14dii)
Speed = λF₀
V = L×F₀
V = 1.5×180
V = 270ms-¹

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(15a)
The charge on a body can be identified using a gold leaf electroscope as follows:
By touching the body to the brass disc: Let the electroscope be charged with positive charge. The leaves will be in diverged position. The body to be tested is touched with the brass disc of the electroscope and the divergence of leaves is observed.

Case (i): If the leaves diverge more, the body is positively uncharged. When the body touches the electroscope, its charges are transferred to the leaves. There is an increase in repulsion which happens due to increase in the same kind of charge, and hence, this indicates that the body is positively charged.

Case (ii): If the divergence of the leaves decreases, the body is negatively charged. When the body touches the electroscope, its charges are transferred to the leaves. There is a decrease in repulsion which happens due to decrease in the same kind of charge or increase in opposite kind of charge, and hence, this indicates that the body is negatively charged.

(15b)
F = Gm₁m₂/r²
F = mg

If m₁ = m₂
Gm₁m₂/r² = mg
.:. g = Gm/r

(15c)
R1 = 2ohms
R2 = 3ohms
First connected in parallel
1/RT = 1/2 + 1/3
1/RT = 3+2/6
1/RT = 5/6
5RT = 6
RT = 6/5
RT = 6/5ohms

To a 120v
V = IR
I = 120/6/5
I = 120÷6/5
I = 120 × 5/6
I = 100A

But when connected in series
RT = R1 + R2
RT = 2 + 3
RT = 5ohms

To a 120v
V = IR
I = V/R
I = 120/5
I = 24A
So the difference between the two arrangement in current
[100 - 24]A
= 76A