(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

(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