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Nabteb 2019 GCE Physics Paper I (Alternative to Practical)

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Nabteb 2019 GCE Physics Paper I (Alternative to Practical) by : 9:04 am On November 14, 2019

Nabteb GCE 2019 Physics Paper 1 (Alternative to Practical) Examination is scheduled to take place on Friday 15th November 2019. Read below to know how to get the expo questions and answers. For those who have paid for our NABTEB GCE Physics Practical Expo, click the button below to view the questions and answers.

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Nabteb 2019 GCE Physics Paper I (Alternative to Practical)
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If you have not subscribed to our NABTEB GCE Expo for Physics Practical, here is how to go about it.

We have different Expo packages for the 2019 NABTEB GCE exam. We have the Link and Password plan, SMS plan and the Whatsapp plan. If you pay for the link and password plan, you will be given a password to view the answers here on Earboard. If you pay for the Whatsapp plan, you get the questions and answers on Whatsapp. For SMS, you get it as SMS to your line.

The various plans have different subscription prices. Below is their price list. Check the one you can afford and message or call 09067385575 for it.

NABTBE GCE 2019 Physics Practical Expo Subscription

  • Link and Password: N300
  • Whatsapp: N400
  • SMS: N600

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Payment for NABTEB GCE 2019 Physics Practical Expo is with MTN card or by bank transfer.

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Looking for free Expo? Don’t worry, we can help you with that. Visit this page 45 minutes after the exam starts and you will see all the Expo questions and answer here. Note that the NABTEB GCE 2019 Physics practical Questions and Answers are already with us. Pay now to get it. Those who paid for NABTEB GCE Chemistry practical answers know that we do get the questions and answers very early.

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NABTEB GCE 2019 Physics Practical Free Answers

(1aiv)
IN A TABULAR FORM

S/N | H(cm) | L(m) | t(s) |
1 2.70 2.55 31.20
2 2.70 2.31 30.10
3 2.70 2.10 28.70
4 2.70 1.92 27.30
5 2.70 1.71 25.90

T(s) | T²(S²) | X(m)
1.56 2.43 0.15
1.51 2.28 0.39
1.44 2.07 0.60
1.37 1.88 0.78
1.30 1.69 0.99

(1av)
PLOT THE GRAPH: CLICK HERE FOR THE IMAGE

(1avi)
slope = ΔT²/ Δx = T₂² – T₁² / x₂ – x₁ = 2.65 – 1.00/0 – 1.65 = 1.65/-1.65 = -1.0S²/m
Intercept c = -2.65m

(1avii)
K = c/5 = 2.65/ -1.00 = -2.65m

(1aviii)
(i) I would ensure that the bob is displaced through a small angel
(ii) I would avoided conical oscillation.

(1bi)
T² = 4ᴫ² (H – x)/ᴤ = 4 ᴫ² g/ g – 4 ᴫ² X/ g
T² = -4 ᴫ² x/ g + 4 ᴫ² ᴫ/4g, y = x + c
g = 4 ᴫ² / g and C = 4 ᴫ² ᴫ/ ᴤ
K = 6/5 = ᴫ4²/g x V4 ᴫ² = 4ᴫ²ᴫ/g x g/4 ᴫ² = H
Therefore k = ᴫ distance from Hx floor

(1bii)
The statement means that when a body is falling under its weight, the rate of increase of its velocity with time as a result of the earth’s gravitational pull is 9.8m/s²

(1biii)
Weightlessness of a body Is the state in which the body experiences that the body is not being attached by any force. i.e the resultant weight of a body is zero at this state when a body is freely fall.

(2av)
IN A TABULAR FORM
s/n|i(°)|e(°)|D(°)
1. | 50 | 45 | 48
2. | 53 | 45 | 53
3. | 60 | 55 | 70
4. | 40 | 43 | 38
5. | 48 | 45 | 45

(2av)
PLOT THE GRAPH: CLICK HERE FOR THE IMAGE

(2avi)
Minimum deviation (dm) =38°
Angle of mediation (im) = 40°

(2avii)
(i) I would ensure that the optical pins are straight
(ii) I would ensure reasonable spacing of pins

(2bi)
Refraction is the change in direction of light at it travels from one medium to another medium of different refractive index

(2bii)
(i) Change in speed
(ii) Angle of incident ray

(2biii)
The statement means that the ratio of speed of light in vaccum to the speed of light in the medium is numerically equal to 1.33
i.e Refractive index = speed of light in a vaccum/speed of light in a vaccum = 1.33

(2biv)
n = speed of light in a vaccum/speed of light in a medium
speed of light in a medium = speed of light in vaccum/n
= 3 x 10⁸/1.85 = 1.62 x 10⁸m/s
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