# 2019 NECO GCE PHYSICS OBJ AND ESSAY ANSWER

**Physics OBJ
1-10: CACABDBECA
11-20: CCDBBDEECA
21-30: ECACEDACDB
31-40: CADBBDCBBC
41-50: EBCADACDAE
51-60: CDCBCEDCCB
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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**

**(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)

Draw the diagram

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)

Draw the diagram

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

(2a)

(i) Acceleration

(ii) Velocity

(2b) MSR = 8.5mm

VSR = 15mm

TSR = MSR + VSR

= 8.5 + 15

= 23.5mm

(3i)

draw the velocity diagram

(3ii)

Total distance coverd = 75m

Total distance = Area of shape water graph

75m = 1/2(45) * U

75 = 22.5U

U = 75/22.5

U = 3.33m/s

(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

(5)

U1 = 4m/s

U2 = 0m/s

M2 = 60g = 0.6kg

M1 = ?

V = 2.5m/s

Using :

M1U1 + M2U2 = ( M1 + M2)V

**( M1*4) + ( 0.6*0) = ( M1 + 0.6) * 2.5
4M1 = 2.5M1 + 1.5
4M1 – 2.5M1 = 1.5
1.5M1 = 1.5
M1 = 1.5/1.5
M1 = 1kg
**

**(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**

**(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)
(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.
**

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