# WAEC GCE 2019: Physics Alternative to practical Questions And Solutions.

WAEC GCE 2019 PHYSICS PRACTICAL SOLUTIONS.

(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)

(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) TABULATE
(i) | Mi(kg) | li(cm) |Li(m) | T=m(g)/π² | √T

1|1.20 |2.10 |0.21 |2.52 |1.587
2|4.50 |4.30 |0.43 |19.35 |4.399
3|7.20 |5.30 |0.53 |38.16 |6.177
4|10.50 |6.30 |0.63 |66.15 |8.133
5|18.20 |8.30 |0.83 |151.06 |12.291
6|29.00 |10.40 |1.04 |301.60 |17.367

2vii) slope,s= ∆√T
——
∆L

= 10-2
——
0.72-0.32

= 8

0.4

=20//

2viii)
i) I would avoid error due to parralax when reading the meter rule
ii) I would avoid zero error on the meter rule

2bi)

2bii) resonance is a phenomenon that only occurs when the frequency at which a force is periodically applied is equal or nearly equal to one of the natural frequencies of the system on which it acts.

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(3)
TABULATE
i:1,2,3,4,5
Ri(ohm):2.0,4.0,6.0,8.0,10.0
OP(l)(cm):3.70,4.30,5.00,5.40,5.70
xi(cm):18.50,21.50,25.00,27.00,28.50
x^-1:0.054,0.047,0.040,0.037,0.035
R^-1:0.500,0.250,0.167,0.125,0.100
OR THIS TABLE
(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
DRAW THE GRAPH of x^-1 against R^-1
(3bii)
E1=E2
L1=L2
E2L1=E1L2
where E1=IV
E2=?
L1=50cm
L2=72cm
50E2=72
E2=72/50
E2=1.44v

(3iv) 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

(3vi) K2 =1/c = 1/0.0265 =37.74
K2 = 5/c = 0.08/0.0265
= 3.02

(3viii)
– I would ensure tight connection
– I would ensure clean terminals

(3bi)
– Hyper accuracy
– No errors in readings associated with the use of a pointer in a voltmeter

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