### PHYSICS PRACTICALS AREA OF CONCENTRATION

PHYSICS PRACTICALS AREA OF CONCENTRATIONπ―π―

π¬πͺοΈπͺοΈ

NUMBER 2

Make sure you use almost that of your teacher value rapport with him or her to find out .

Note yourππ

**Observation W = ????
**IΒ°,d(cm),L(cm),e/Β°,Ξ¦/Β°,Ξ¦ = I + e/Β° any one asked π¬πͺοΈπͺοΈπ―
**Scale eg. Let 2cm represent 0.1unit on cos Ξ¦ axis
Let 4cm represent 0.1unit on sin I axis
Depending on what you are ask to place on the graph ππ
**Slope
**Intercept on the vertical/horizontal axis if asked.π―
*Make sure your graph, scale, and, intercept are well drawn and taken

#PRECAUTIONS:

I. I ensured neat traces by using sharp pencil.
II. I ensure that the pins are vertically erect.
III. I ensure that the pins are well spaced which is 4cm apart.
IV. I avoided parallax errors when reading protractor.
V. I avoided error due to parallax when reading the metre rule.

Study this incase if asked ..ππ π¬πͺοΈπͺοΈ

A. State the conditions necessary for total internal reflection of light to occur.

B. The Critical angle for a transparent substance is 39Β°. Calculate the refractive index of the substance.
C. State snell’s law
D. Calculator the critical angle for a water – air interface [ refractive index of water = 4/3 ]
WORKINGS βπΏβπΏ

A. I. Light must be traveling from a denser medium to a less dense medium.
II. The angle of incidence in the dense medium must be greater than the critical angle.

π¬πͺοΈπͺοΈπͺοΈ

B. Refractive index π = 1/sinC = 1/Sin39Β° = 1/0.6293 Therefore, n = 1.59

C. Snell’s law state that the ratio of the angle of refraction is a constant for a given pair of media.

π¬πͺοΈπͺοΈ

D. 1/SinC = a ^N w Therefore, SinC = 3/4 = 0.75 Hence C = Sin inverse 0.75 = 48.6

π¬πͺοΈπͺοΈ

MT π―π―

Study It area of concentration no. 2

π¬πͺοΈπͺοΈ

π¬πͺοΈπͺοΈ

NUMBER 3. STUDY WELL π―π―π― OJCAβπΏβπΏβπΏ

Same as num 2 rapport with your teacher to get almost the value of his π―π―

## Make sure your readings, graph, scale and intercept are well drawn and taken

π¬πͺοΈπͺοΈπͺοΈ

Precautions _OJCA

I. I ensured Tight connections.
II. I noted the zero error on the ammeter and voltmeter.
III. I avoided error due to parallax when reading the ammeter and voltmeter.
IV. I ensured that I remove the key when readings are not been taken to prevent the cells from running down.

# Study this incase if askedππ

A. Two method by which electric current can be produced

B. Explain why E. M. F. of a cell is greater than the PD. across the cell when it is supplying current though an external resistance

C. Define potential difference between two points in an electric circuit.

π¬πͺοΈπͺοΈ

#WORKINGS βπΏβπΏ

A. Electric current can be provided using:
I. Chemical Cells
II. Generator
III. Solar Cells
IV. Gas Turbines
V. Hydro Power
VI. Wind Mills
Vii. Geothermal
Viii. Nuclear Power

π¬πͺοΈπͺοΈ

B. E. M. F. (E) is the Work done across external resistance (R) and internal resistance of the cell (r) While (PD) [v] is the work done across external resistance (R) only, hence, emf of a cell is greater than the PD across the cell.
Therefore, E = V + Ir

π¬πͺοΈπͺοΈ

C. The potential difference between two points in an electric circuit is the Work done in Joules (J) in moving a charge of one Coulomb (c) from one point to the other..

Make sure you study thisπ―π―ππ

Goodluck

#
Share via