Home Jamb Waec Neco Nabteb
Welccome To Examloaded.Com>>>>  Home of JAMB, WAEC and NECO Qestions and Answers Expo Site

PAYMENTS 4 WAEC & JAMB 2024 EXPO ENDS ON 1ST MARCH

This is Our Only Contact | Click On The Phone Number

+2348162563540

To Chat Us on WhatsApp

# 2019/2020 Nabteb SSCE Physics Expo Questions and Answers Examination

### NABTEB PHYSICS PRACTICAL SOLUTIONS

(1a) Initial position of the pointer on the metre rule Xo = 37.20cm

Table of values In a tabular form

S/N: 1.0, 2.0, 3.0, 4.0, 5.0 M(g): 70.00, 90.00, 110.00, 130.00, 150.00 X1(cm): 39.80, 40.00, 42.00, 43.10, 44.20 e = X1 – X0(cm): 2.60, 3.70, 4.80, 5.90, 7.00

(1aviii) (i) I ensured that I avoided parallax error when taking readings on the metre rule. (ii) I ensured firm suspension of the scale pan before and after leading the masses.

(1bi) Hooke’s law states that the extension experienced by an elastic material is directly proportional to the force  applied, provided the elastic limit is not exceeded

(1bii) Pls deduce from the graph.

(1biii) 1/2mv² = Fe M = 5g = 0.005kg V = ? F = 70N e = 7cm = 0.07m 1/2 × 0.005 × v² =  70 ×0.07 V² = 70×0.07×2/0.005 = 1960 V = √1960 = 44.272 = 44.27m

Instantaneous velocity of the stone when released = 44.27m

===============================================

(2i) Tabulate S/N 1|2|3|4|5|

y(cm) 10.0|20.0|30.0|40.0|50.0|

X(cm) -30.0|60.0|30.0|24.0|21.4|

X^-1(cm^-1) -0.033|0.016|0.033|0.0416|0.046|

y^-1(cm^-1) 0.100|0.050|0.030|0.025|0.020|

(2avi) I1 = 0.068cm^-1 , I2=0.066cm^-1 2vaii) m=1/2(I1+I2)=1/2(0.068+0.066) =1/2*0.134=0.067 M=0.067cm^-1 , K=M^-1=1/0.067 =14.92 K=15cm

(2aviii) – I ensured I avoided parallax error when taking readings of the meter rule by placing eye vertically above the scale – I ensured the intensity of source of light was maintained

(2bi) K=15cm means the focal length of the concave mirror is equal to 15cm

(2bii) – To provide magnified image of the face – For applying make-up or shaving – It is used in motor vehicle head lights

(2biii) – The image is formed on the screen – The image is larger than the object – Further away than the object position – It is real and inverted =====================================

(3i) EMF=2.0V Tabulate K(Ω) | I(A) | I^-1(A^-1) 1.0 | 0.44 | 2.573 2.0 | 0.34 | 2.941 3.0 | 0.28 | 3.571 4.0 | 0.24 | 4.167 5.0 | 0.22 | 4.545

(3avii) Slope (s) =ΔI^-1(A^-1)/ΔR(Ω) =4.85-2.60/4.90-1.50 =2.25/3.40 =0.662A^-1Ω^-1 Intercept (c) =2.65A^-1 3aviii) K=E*C=2*2.65=5.3A^-1V

(3aix) – I ensured that I tightened the connection – I avoided parallax error in reading ammeter/voltmeter

(3bi) The value of K gives the overall resistance R(Ω)

(3bii) Terminal p.d is the potential difference a cell in a circuit unit. When no current is flowing the terminal p.d is equal to the e.m.f of the cell. If a current is flowing then the terminal p.d of the cell will be lower than it’s EMF as some loss of energy takes place due to its internal resistance ====================================== _COMPLETED!._ ===================================