2023 Neco Physics Practical Questions and answers

Instructions: Follow the value of your teacher in school so you can get a perfect mark well if you choose to use my table you can draw the graph yourself and get your slope

(1axviii)
{CHOOSE ANY BEST 2}
(i) I avoided error due to parallax when taking readings on the meter rule.
(ii) I ensured that the support is rigid.
(iii) I avoided the zero error on the stop clock.

(1bi)
(i) Turning Fork.
(ii) Vibrating Guitar String.
(iii) Pendulum.

(1bii)

(3xi)
{CHOOSE ANY BEST 2}
(i) Key were opened when readings are not taken.
(ii) I ensured that connection was properly tightened.
(iii) I avoided parallax error when reading the ammeter.
(1a)
Lo = 13.00cm

L | Li/cm | e= L-Lo/cm | ti/s | T =t/n/S | T²/S²
1 | 21.00 | 8.00 | 7.34 | 0.734 | 0.5388
2 | 26.20 | 13.20 | 8.86 | 0.886 | 0.7850
3 | 30.50 | 17.50 | 9.53 | 0.953 | 0.9082
4 | 35.50 | 22.50 | 10.20 | 1.020 | 1.0404
5 | 40.20 | 27.20 | 10.85 | 1.085 | 1.1772

(1axvi)
Slope, S = (∆e/cm)/(∆T²/S²)

S = ([25.00-5.00]cm)/([1.20-0.40]/S²) = (20.00/cm)/(0.8/S²)
Slope, S = 25.00cm/S²

(1axvii)
K = 4π²S = 4×(22/7)²×25 = 4×(484/49)×25
K = 987.76cm/S²

(1axvii)
(i) I will avoid parallax error when taking reading on the metre rule
(ii) I will check zero error of the stop watch

(1bi)
(i) Pendulum Bob
(ii) A loaded tube suspended in a water
(iii) The vibration of the motion of a plucked guitar string

(1bii)
Mean time = 34.80+36+33.75/5 = 104.55/3
Mean time = 34.85s

Mean period= 34.85/15 = 2.323seconds
(3avii)

i | R/Ω | R-¹/Ω-¹ | IA
1 | 1.00 | 1.000 | 0.32
2 | 2.00 | 0.500 | 0.32
3 | 3.00 | 0.333 | 0.28
4 | 4.00 | 0.250 | 0.26
5 | 5.00 | 0.200 | 0.24

(3aix)
Slope, S = (∆I/A)/(∆R-¹/Ω-¹) = ([0.32-0.25]/A)/([1.00-0.160]/Ω-¹)
= (0.7/A)/(0.84/Ω-¹) = 0.083AΩ

Intercept, C = 0.24A

(3ax)
K = S/C = 0.083AΩ/0.24A = 0.35Ω

(3axi)
(i) I ensured tight connections to avoid current fluctuation
(ii) I will ensure that the key is opened when the circuit is not in used to avoid the cell being used up

(3bi)
The end of a cell is greater than it’s terminal voltage because there is some potential drop across the cell due to its small internal resistance

(3bii)
I = E/R+r
I = 1.5/6+0.5 = 1.5/6.5
I = 0.23A