2020 NABTEB SSCE PHYSICS PRACTICAL ANSWERS

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**WE SOLVED NUMBER 1⃣, 2⃣ & 3⃣**

**TAKE NOTE:**

**^ Means raise to the power**

**Î Means ohm’s**

**^â€“Â¹ Means Raise to the power of minus 1 or inverse**

**We are always with you all forever**

(1a)

Balance point G=50.00cm

TABULATE

S/N | m(g) | P(cm) | x(cm) | y(cm) | x^â€“Â¹ (cm^-1

UNDER S/N

1 2 3 4 5

UNDER m(g)

20.00 | 30.00 | 40.00 | 50.00 | 60.00

UNDER P(cm)

86.50 | 78.50 | 69.00 | 65.00 | 62.55

UNDER x(cm)

36.50 | 28.50 | 19.00 | 15.00 | 12.55

UNDER y(cm)

35.00 | 35.00 | 35.00 | 35.00 | 35.00

UNDER x^-1 (cm^-1)

0.027 | 0.035 | 0.053 | 0.067 | 0.080

(1av)

Click on the link to view the Graph

[img]https://i.imgur.com/Q6nWdDZ.jpg[/img]

(1avi)

Slope = âˆ†m/âˆ†x^â€“Â¹

=60 â€“ 20 / 0.080 â€“ 0.027

= 40/0.053

Slope, s = 754.72gcm

And

S/y = 754.72/35

=21.56g

(1avii)

Precautions

(i) I avoided the parallax error when reading the meter rule.

(ii) I ensued that the suspended masses did not touch the table

(1bi)

The moment of a force about a point is the product of the force and distance about that point i.e moment = Force Ã— Distance

The S.I unit of moment is Nm

(1bii)

The centre of gravity of a body is the point at which the resultant weight of the body appears to act.

The centre of gravity of a body is related to it’s stability because of

(i) The position of the centre of gravity:- The lower the position of Câ€¢G, the more stable it is and the more the position increases in height when slightly tilted, the more is the magnitude of it’s potential energy

(ii) The moment of the centre of gravity about a given axis when slightly tilted

(1biii)

(view image above)

Where m is the mass of the meter rule

Clockwise moment = Anti – clockwise moment

200 Ã— 20 = 300 Ã— m

4000=30m

m=4000/30

m=133.33g

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(3a)

Eo=3.0V

TABULATE

S/N | R(ohms) | V(V) | V^â€“1 (V^â€“1) | R^â€“1 (ohms^â€“1)

UNDER S/N

1. 2. 3. 4. 5.

UNDER R(ohms)

1.00 2.00 3.00 4.00 5.00

UNDER V(V)

1.80 2.00 2.20 2.50 2.70

UNDER V^â€“1 (V^â€“1)

0.56 0.50 0.45 0.40 0.37

UNDER R^â€“1 (ohms^â€“1)

1.00 0.50 0.33 0.25 0.20

(3av)

Click on the link to view the Graph

[img]https://i.imgur.com/gzh5i1O.jpg[/img]

Slope, s = âˆ†V^â€“Â¹ / âˆ†R^â€“Â¹

= 0.67 â€“ 0.35 / 1.0 â€“ 0.11

= 0.32/0.89

Slope, s = 0.36ohmsv^â€“Â¹

=0.36A^â€“Â¹

Intercept k = 0.31v^â€“Â¹

(3avi)

T=S/K

=0.36/0.31

=1.16ohms

(3avii)

Precautions

(i) I ensured tight connection of connecting wires to the terminals

(ii) I ensured that the key was open after each reading so that the battery did not run down quickly

(3bi)

Electromotive force of a cell (Emf) is the potential difference across the terminal of the cell when it is not delivering any current to the circuit

(3bii)

Ohmic conductors are;

(i) Coppers

(ii) Aluminium

(iii) Silver

Non ohmic conductor are;

(i) Diode

(ii) Transistor

The unit of conductivity is (Î m)^â€“Â¹

(3biii)

E=3v

R=2 ohms

I=0.5A

r=?

I=E/R + r

0.5=3/2 + r

Cross multiply

3=0.5 (2 + r)

3=1 + 0.5r

Collect like terms

3â€“1 = 0.5r

2=0.5r

Divide both sides by 0.5

r=2/0.5

r=4 ohm’s

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NABTEB Physics Practical answer Completed

**GOODLUCK!!!**

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