# 2021 WAEC GCE 2021nd series Physics Practical Answers : 2020 WAEC GCE 2nd series Physics Practical Answers Night before Exam

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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.

(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

2021 WAEC GCE 2021nd series Physics Practical Answers (ESSAY, OBJ, THEORY, PRACTICALS ETC)

2021 WAEC GCE 2021nd series Physics Practical Answers (ESSAY, OBJ, THEORY, PRACTICALS ETC)

(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

(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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Updated At: April 12th 2021, (07:04:47)