NECO 2019 Physics Practical Answer – June/July Expo «

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*NECO-PHYSICS-PRACTICALS-ANSWERS*

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(1)
Centre of gravity C G = 50.0cm
Tabulate
Y(cm). |10|15|20|25|30|
K(cm). |31.2|33.5|36.0|38.2|40.6|
X(cm). |21.2|18.5|16.0|13.2|10.6|
K¹(cm). |27.0|29.7|33.0|35.5|38.4|
X¹(cm). |17.0|14.7|13.0|10.5|8.4|

(1x)
Slope = increase in y-axis /increase in x-axis

From the graph, 
Slope = (35-10)cm/(16-6)cm
Slope (s) = 35/10=2.5

(1xi)
K =M1-Sm/S-1 
M1= 100g , Sm =50m , S =2.5

K = 100-50/2.5-1 = 50/1.5
= 500/15=100/3
= 33.3g/cm

Use any one

(1ix)
GRAPH1: CLICK HERE FOR THE GRAPh1
(1ix)


Graph2:
CLICK HERE FOR THE GRAPH2

(1xii)
(i)i avoided drought by closing the window
(ii)i avoided error due to parallax when reading the meter rule 

(1bi)
couple is a pair of forces, equal in magnitude, oppositely directed, and displaced by perpendicular distance or moment. 

(1bii)
CLICK HERE FOR THE IMAGE
Let the mass of beam = m
Taking moment about point C 5cm clockwise = sum of anitclockwise i;e moment
; 50*0.5=mx0.2
m=50*0.5/02=250/2=125g
Mass of the beam = 125g
==========


(3)
Tabulate
K,M,F 
E=3.0v
L(cm).|20|30|40|50|60|70|
I(A). |1.30|1.20|1.10|1.00|0.90|0.95|
X=E/I(V/A).
|2.31|2.50|2.73|3.00|3.33|3.16|

(3)
CLICK HERE FOR THE GRAPH

(3ix)
Slope = ΔL/Δx = 66-12/7-4 = 54/3=18cm/Ω
Intercept ,c (couldn't be shown on graph) 

(3x)
Evaluating K =4.9 10^-7 S
K =4.9 × 10^-7 × 18
K=8.82 × 10^-6cm/Ω
(3xi)
- I ensured tight connections 
- I avoided error due to parallax on the ammeter 

(3bi) 
- Keep the top or terminals clean 
- Do not tighten or removed cables while charging or discharging 
(3bii) Using ohms law, R = V/I=3.5/0.25
=14Ω
=========
Completed
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