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    NECO-PHYSICS-PRACTICALS-ANSWERS
    (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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    Master Solution June 11, 2019 Categories: NECO 592


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