(1ix)
Note: Use any of the graph
GRAPH1: CLICK HERE FOR THE GRAPH
GRAPH2: CLICK HERE FOR THE GRAPH

(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

(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
;- 500.5=mx0.2
m=500.5/02=250/2=125g

Mass of the beam = 125g

(3a)
CLICK HERE FOR THE IMAGE – COMPLETE SOLUTION

(v) E=3.0v
(vii) Tabulate
Under L(cm): 20.0; 30.0; 40.0; 50.0; 60.0; 70.0;
Under I(A): 0.70; 0.60; 0.55; 0.50; 0.45; 0.40;
Under X=E/I(Ohms): 4.2857; 5.0000; 5.4545; 6.0000; 6.6667; 7.5000;

(3) GRAPH:
CLICK HERE FOR THE GRAPH

(3ix)
Slope = ΔL/Δx = 66-12/7-4 = 54/3=18cm/ohms
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/ohms
(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
=14ohms

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COMPLETED!!
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Please Be Patient as this is practical, essay/main subject will come more earlier.