2017 NECO PHYSICS PRACTICAL QUESTIONS AND ANSWER

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2017/PHYSICS PRACTICAL NECO ANSWER 

     Question 1

....Tables for value 

S/n. h(cm) t(sec) T=t/n  T^2

1.     40.0. | 15.00. |0.75. | 0.563

2.    50.0.  | 17.00.| 0.85. | 0.723

3.    60.0.   |19.0 | 0.95| 0.90

4. |   70.0.  | 21.0. | 1.05 | 1.10

5. |   80.0 | 23.0 | 1.15 | 1.32

   Graph (1) 

Slope = ∆T^2 over ∆h 

         = 1.32 - 0.72 over 80-52

         =0.6 /28 

          = 0.02

Evaluate K = S/Q where 

      Q = 2/25d^2

      Q = 2/25× 80^2 

      Q= 0.0000125

: . K = S/Q 

    K = 0.02/0.0000125

    K = 154.  

Xii) Precaution to ensured 

      Accurate Results 

1. I ensured accurate measurement of the length of thread .

2. I avoided zero error in the reading of stop watch

phy-pract-An

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

Complet no1

The internal diameter is 6.90cm

Lo=125ml

1)

S/N - |1|2|3|4|5|

M(g) - |50.0|100.0|150.0|200.0|250.0|

Lo(cm) - |50.0|50.0|50.0|50.0|50.0|

L1(cm) - |60.0|70.0|80.0|90.0|100.0|

y=L1-Lv(cm^3) - |10.00|20.00|30.00|40.00|50.00|

V=0.785d^2y - |371.30|742.60|1113.90|1485.20|1856.50|

NO1 graph

1aix)

i)i avoided error due to parallel

ii)i ensured that the mass is gently dropped into the liquid

1bi)

Archimedes principle states that when a body is wholly or partly immersed in a liquid the weight of the body is equal to the mass of the fluid displaced

1bii)

volume of the wood = 2.0 × 10^-5m^3

mass = 5.0 × 10^-3kg

g = 10.0ms^-2

density of water = 1.0 × 10^3kgm^-3

volume of object = mass/density

density = mass/volume

density of the object = mass of the object/volume

= 5.0×10^-3/2.0×10^-5

= 2.5×10^2kgm^3

upthrust = volume 0f the object × density of liquid × g

=2.0×10^-5 × 1.5×10^2×10

= 0.3N

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

NO3 GRAPH

Coming soon

3)

adjusted.

TABULATE READINGS

R(N) - |2.00| |4.00| |6.00| |8.00| |10.00|

1(A) - |2.85| |2.21| |2.00| |1.75| |1.65|

R^-1 - |0.5| |0.25| |0.17| |0.13| |0.10|

PRECAUTIONS

i)i ensured clean terminals

ii)i avioved zero error on the ammeter used

iii)i ensured tight connection of all wires.  

3b)

ohm's law states that the current flowing through a metallic conductor is proportional to its portential difference provided that physical condition is constant i.e V x I

3c)

slope(s) = DR^-1/DI = 0.25-0.10/2.21-1.65

=0.15/0.56

slope(s)=0.27

intercept c = -0.35

evaluation

1/5=1/0.27=3.70

evalution

1/C = 1/-0.35

=-2.8

2017 NECO PHYSICS PRACTICAL  ANSWER 

    (2.)

You are provided with Ray box, ascreen, a concave mirror, mirror holder and meter rule...

(2a.)

F = 15cm 

 (V.)

(A)

= 60.00cm, b =20.00cm 

Hence=L=a over b = 60.00 over 20.00

 L= 3

(Vi)

Table for reading 

S/n. |b(cm).     |a(cm)|L=a over b

1      |   20.00. |60.00|  3.00

2      |  25.00.  |37.00|  1.50

3     |  30.00.    |30.00|  1.00

4     |  35.00.    |26.00|  0.75

5      | 24.00     |24.00|  0.60

(Vii)

 Slope =s = ∆l over ∆a  =(3-0.25 over 60-18.6)

S= 2.75 over 41.4  

S = 0.0066425cm^-1

phy-pract-Ans

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

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

Complet no1

The internal diameter is 6.90cm

Lo=125ml

1)

S/N - |1|2|3|4|5|

M(g) - |50.0|100.0|150.0|200.0|250.0|

Lo(cm) - |50.0|50.0|50.0|50.0|50.0|

L1(cm) - |60.0|70.0|80.0|90.0|100.0|

y=L1-Lv(cm^3) - |10.00|20.00|30.00|40.00|50.00|

V=0.785d^2y - |371.30|742.60|1113.90|1485.20|1856.50|

NO1 graph

1aix)

i)i avoided error due to parallel

ii)i ensured that the mass is gently dropped into the liquid

1bi)

Archimedes principle states that when a body is wholly or partly immersed in a liquid the weight of the body is equal to the mass of the fluid displaced

1bii)

volume of the wood = 2.0 × 10^-5m^3

mass = 5.0 × 10^-3kg

g = 10.0ms^-2

density of water = 1.0 × 10^3kgm^-3

volume of object = mass/density

density = mass/volume

density of the object = mass of the object/volume

= 5.0×10^-3/2.0×10^-5

= 2.5×10^2kgm^3

upthrust = volume 0f the object × density of liquid × g

=2.0×10^-5 × 1.5×10^2×10

= 0.3N

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

 (2.)

You are provided with Ray box, ascreen, a concave mirror, mirror holder and meter rule...

(2a.)

F = 15cm 

 (V.)

(A)

= 60.00cm, b =20.00cm 

Hence=L=a over b = 60.00 over 20.00

 L= 3

(Vi)

Table for reading 

S/n. |b(cm).     |a(cm)|L=a over b

1      |   20.00. |60.00|  3.00

2      |  25.00.  |37.00|  1.50

3     |  30.00.    |30.00|  1.00

4     |  35.00.    |26.00|  0.75

5      | 24.00     |24.00|  0.60

(Vii)

 Slope =s = ∆l over ∆a  =(3-0.25 over 60-18.6)

S= 2.75 over 41.4  

S = 0.0066425cm^-1

3)

adjusted.

TABULATE READINGS

R(N) - |2.00| |4.00| |6.00| |8.00| |10.00|

1(A) - |2.85| |2.21| |2.00| |1.75| |1.65|

R^-1 - |0.5| |0.25| |0.17| |0.13| |0.10|

PRECAUTIONS

i)i ensured clean terminals

ii)i avioved zero error on the ammeter used

iii)i ensured tight connection of all wires

3b)

ohm's law states that the current flowing through a metallic conductor is proportional to its portential difference provided that physical condition is constant i.e V x I

3c)

slope(s) = DR^-1/DI = 0.25-0.10/2.21-1.65

=0.15/0.56

slope(s)=0.27

intercept c = -0.35

evaluation

1/5=1/0.27=3.70

evaluti[truncated by WhatsApp]

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