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2020 WAEC Physics Practical (a) You are provided with a metre rule, a knife-edge, set of masses, inextensible string,...

Physics
WAEC 2020

(a)
Open photo

You are provided with a metre rule, a knife-edge, set of masses, inextensible string, retort support and other necessary apparatus.

i. Place the metre rule on the knife edge. Read and record the point G where the metre rule balances horizontally, as shown in Fig (a).

ii. Suspend the metre rule at G with the aid of the string provided and attach the string to the retort support as shown in Fig 1(b). Keep the string attached to this point throughout the experiment.

iii. Attach the mass M\(_{0}\) at the 80cm mark of the metre rule. Determine the distance of y from G. Keep M\(_{0}\) at this
position throughout the experiment.

iv. Suspend a mass M= 40g on the side AG and adjust its position until the metre rule balances horizontally.

v. Measure and record the distance of x of M from G. Evaluate x\(^{-1}\)

vi. Repeat the procedure for four other values of M= 60g, 80g, 100g and 120g. Measure and record x and evaluate x\(^{-1}\) in each case.

vii. Tabulate the readings.

viii. Plot a graph of M on the vertical axis and x on the horizontal axis, starting both axes from the origin (0,0).

ix. Determine the slope s of the graph:

x. Given that s = yM\(_{0}\), determine M\(_{0}\).

xi. State two precautions taken to obtain accurate results.

(b) i. Define the moment of a force about a point.

ii. A uniform metre rule is suspended by an inextensible string at its centre of gravity. If a mass of 60g is placed at the 25cm mark, what mass should be placed at the 80cm mark of the metre rule to balance it horizontally?

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Explanation

a(i) Note: G = 50.0cm, (ii) y = 30.0cm
 (iii)

S/N M(g) x(cm) x\(^{-1}\)(cm\(^{-1}\)0 x\(^{-1}\)(x10\(^{-3}\))
1
2
3
4
6
40.0
60.0
80.0
100.0
120.0
37.50
25.0
18.75
15.00
12.50
0.027
0.040
0.055
0.067
0.08
27.00
40.00
53.00
67.00
80.00

(iv)

Open photo

 

(ix). Slope = \(\frac{\bigtriangleup M}{\bigtriangleup X^{-1}}\) = \(\frac{M_2}{X_2^{-1}} - \frac{M_1}{X_1^{-1}}\)

= \(\frac{120.0 - 40}{80.0 - 27}\) = \(\frac{80}{53}\)

= 1.509 x 10\({^3}\) = 1509

x. S = yM\(_{0}\)
1509 = 30M\(_{0}\)

M\(_{0}\) = \(\frac{1509}{30}\)

M\(_{0}\) = 50.3

xi. Precautions:
- Avoided parallax error in reading metre rule
- Noted/corrected/Avoided zero error on metre rule
- Avoided draught
- Ensured that support was rigid/firm
- Repeated readings shown on the table
- Ensured mass does not touch or rest on the table.

(b) i. Definition of the moment of a force about a point: It is the product of the force and the perpendicular distance between the line of action of the force and the point.

Open photo

Sum of clockwise moment  = sum of anticlockwise moment

M x 30 = 60 x 25
M = \(\frac{60 \times 25}{30}\) = 50


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