2024 WAEC GCE SECOND SERIES PHYSICS: (4181)

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(1)

(1av)

Xo = 2.0cm


(1avi)

TABULATE

i: 1| 2| 3| 4| 5|

Xi(cm): 4.0| 6.0| 8.0| 10.0| 12.0|

ei(Xi-Xo)cm: 2.0| 4.0| 6.0| 8.0| 10.0|

ti(A): 3.20| 3.80| 5.20| 6.30| 7.40|

Ti=t/10(s): 0.320| 0.380| 0.520| 0.630| 0.740|

Ti²: 0.1024| 0.1444| 0.2704| 0.3969| 0.5476|


(1aix)

G = 4π²/s = 4*3.14*3.14/0.065

= 606.745cm/s²


(1ax)

(i) I avoided error due to parallax while reading the pointer position

(ii) I ensured the helical spring is not deformed


(1bi)

Hooke's law states that the extension provided in an elastic material is proportional to the force or load applied, provided it's elastic limit is not exceeded


(1bii)

Given: F = 400g*10 = 0.4kg*10 = 4N

e = 5cm = 5*10-²m

Force constant, k = F/e =4/5*10-² = 400/5

= 80N/m



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(3aiv)

TABULATE

S/N: 1| 2| 3| 4| 5| 6|

Li(cm): 2.50| 3.50| 4.50| 5.50| 6.50| 7.30|

L(cm): 25.0| 35.0| 45.0| 55.0| 65.0| 73.0|

Ii(A): 1.55| 1.25| 1.10| 0.95| 0.85| 0.75|

I-¹(A-¹): 0.6452| 0.8000| 0.909| 1.0526| 1.1765| 1.3333|


(3avii)

(i) I would ensure tight connection

(ii) I would avoid error due to to parallax when reading the ammeter


(3bi)

(i) Temperature of wire

(ii) Cross sectional area of wire


(3bii)

Reactance, R = V/I =24/2 = 12Ω



(3)

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