2024 WAEC PHYSICS: 2024 WAEC GCE 2nd series Physics Practical Answers (9997)

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Welcome to official 2024 Physics WAEC answer page. We provide 2024 Physics WAEC Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct WAEC Phy Expo. WAEC Physics Questions and Answers 2024. WAEC Phy Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2024 WAEC GCE 2nd series Physics Practical Answers . 2024 WAEC EXAM Physics Questions and Answers

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(1)
Tabulate
S/n ; 1|2|3|4|5
Xi/Cm; |6.50|6.00|5.50|5.00|4.50

Xi/cm; |32.50|30.00|27.50|25.00|22.50

Mg/I; |100.00|120.00|150.00|170.00|220.00|

Xi^-¹/cm^-¹; |0.031|0.033|0.036|0.04|0.044|

Graph.
2024 WAEC GCE 2nd series Physics Practical Answers	 2024 WAEC PHYSICS: 2024 WAEC GCE 2nd series Physics Practical Answers	 (9997) Welcome to official 2024 Physics WAEC answer page. We provide 2024 Physics WAEC Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct WAEC Phy Expo

(1ix)
Slope S = DXi^-¹/cm^-¹/Δm/g = 0.044-0.031/220-100 =1.083*10^-⁴

(1x)
S=1.083*10^-⁴ =1/45(50+Mr)
50 + Mr = 1/45*1.083*10^-⁴
=205.13
Mr=205.13-50=155.13gcm

(1xi)
(i)I close the window to avoid air resistance during balancing of the meter rule
(ii)I took my reading at eye level to avoid error due to parallax

(1bi)
(i)wide base
(ii)low center of gravity

(1bii)
(i)the applied force about O point
(ii)the radius of rotation at a point

 

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(3)
Tabulate
Mc= 0.0725kg , θr= 29°c

S/n; 12345

Vi(cm); |50|70|90|110|130

θI(°c); |38|36|35|34|33

Mw=d*v(kg); 0.05|0.07|0.09|0.11|0.13

T=(θ-θr)°c; 9|7|6|5|4

T¹(/°c); 0.11|0.14|0.17|0.20|0.25

(3)
Graph

2024 WAEC GCE 2nd series Physics Practical Answers	 2024 WAEC PHYSICS: 2024 WAEC GCE 2nd series Physics Practical Answers	 (9997) Welcome to official 2024 Physics WAEC answer page. We provide 2024 Physics WAEC Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct WAEC Phy Expo

 

(3vi)
Slope S= Dm/DT^-¹ = 0.11-0.05/0.20-011=0.06/0.09=0.67kgk

(3vii)
Interprets C = - 0.022kg

(3viii)
C=McK/4200
K=4200c/mc= 4200*0.022/0.0725
= -1,274.5

(3ix)
(i)it was ensure that the mixture is heated and allowed to boil for 3 minutes
(ii)error due to parallax when reading the thermometer was avoided

(3bi)
The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by 1°C. The units for specific heat can either be joules per gram per degree (J/g°C) or calories per gram per degree (cal/g°C). This text will use J/g°C for specific heat.

(3bii)
Power =Hot=6300*(50-30)*12*60
=9,072,000J=9.072MJ

 

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