WAEC GCE 2019 - PHYSICS PRACTICAL ANSWER

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ANSWERS:

(1)
(i)|Xraw|Yraw| Xreal |Yreal
1 |4.9.| 5.3| 9.8| 10.6
2 |6.7 | 5.9|13.4|11.8
3 |8.3 | 9.7|16.6|19.4
4 |10.0|11.7|20.0| 23.4
5 |11.3|13.4|22.6|26.8
6 |12.7|15.0|25.4|30.0

(1aiv)
Plot your graph
[https://i.imgur.com/4h1LRRn.jpg]

(1av)
Slope(s) => change in Y/Change in X => Y2 - Y1/X2 - X1
=>26.8 - 10.6/22.6 - 9.8 = 16.2/12.8
S = 1.266

(1avi)
If S = 1.266
Then K = 1/1.266 = 0.790

(1avii)
(i) I would ensure that the reading was taking from the meniscus level before taking my reading.
(ii) I would have ensured that the tubes are vertically upright before carrying out the experiment for accuracy.

(1bi)
Pressure: It is the force applied per unit area. It is measured in N/m². Pressure = Force/Area

(1bii)
Relative density of E = Y/X = 15/10 = 1.5



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(2)




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(3)
(i)|Ri |OP(cm)|xi(cm)|X-¹|R-¹
1 |2.0| 3.70|18.50|0.054|0.500
2 |4.0| 4.30|21.50|0.047|0.250
3 |6.0| 5.00|25.00|0.040|0.167
4 |8.0| 5.40|27.00|0.037|0.125
5 |10.0|5.70|28.50|0.035|0.100

(3avi)
Slope s = ΔX-¹/ΔR-¹
=(50 - 30)×10-³/0.29 - 0.04
=20 ×10-³/0.25 = 0.02/0.25
= 0.08

Intercept, C = 26.5×10-³ = 0.0265

(3avii)
k1 = 1/C = 1/0.0265 = 37.74
K2 = 5/C = 0.08/0.0265 = 3.02

(3aviii)
(i) I would ensure right connections.
(ii) I would ensure clean terminals.

(3bi)
(i) Higher accuracy.
(ii) No errors in reading associated with the use of a Pointer in a voltmeter.

(3bii)
E1 = E2
L1 = L2
E2L1 = E1L2

Where
E1 = 1v
E2 = ?
L1 = 50cm
L2 = 72cm
50E2 = 72
E2 = 72/50
E2 = 1.44v






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