2019 NECO physics Practical Answer & Question

PHYSICS PRACTICAL:
1a)
Centre of gravity C G = 50.0cm

(1aviii)
Tabulate:
Under Y(cm): 10; 15; 20; 25; 30;
Under K(cm): 31.2; 33.5; 36.0; 38.2; 40.6;
Under X(cm): 21.2; 18.5; 16.0; 13.2; 10.6;
Under K¹(cm): 27.0; 29.7; 33.0; 35.5; 38.4;
Under X¹(cm): 17.0; 14.7; 13.0; 10.5; 8.4;

(1axii)
(i)i avoided drought by closing the window
(ii)i avoided error due to parallax when reading the meter rule

(1bi)
Couple is a pair of forces, equal in magnitude, oppositely directed, and displaced by perpendicular distance or moment.

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3av)
E=3.0v

(3avii)
Tabulate

Under L(cm): 20.0; 30.0; 40.0; 50.0; 60.0; 70.0;
Under I(A): 0.70; 0.60; 0.55; 0.50; 0.45; 0.40;
Under X=E/I(Ohms): 4.2857; 5.0000; 5.4545; 6.0000; 6.6667; 7.5000;

(3aix)
slope = ΔL/ΔX = 66 – 12/7 – 4
= 54/3 = 18cm

Intercept, C

(3ax)
K = 4.9×10^-7
K = 4.9×10^-7 × 18
K = 8.82×10^-6cm/ohm

(3axi)
-I ensured tight connections.
-I avoided error due to parallax on the ammeter.

(3bi)
keep the top or terminals clean.
Do not tighten or remove cables while charging or discharging.

(3bii)
Using ohm’s law
R = V/I = 3.5/0.25 = 14ohm

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*Question 1**

PRECAUTIONS*

I Avoided parallax error in reading spring balance-

i Tapped the wood/load slightly to avoid sticking to the table- Repeated readings shown on table.

(2) In a tabular form:

Under tita°:75, 65, 55, 45, 35

Under MO(cm):1.1, 2.0, 2.5, 3.4, 3.9

Under NO(cm):6.2, 6.4, 6.6, 6.8, 7.2

Umder H=MO/NO:0.177, 0.313, 0.379, 0.500, .542

Under Costita:0.2588, 0.4226, 0.5736, 0.7071, 0.8192

(2axiii)From the graphSlope,

S =Δcostita/^ΔH= 0.75-0.45/0.5-0.3= 0.3/0.2=1.5

(2xii)

(2axiv)

(i) I ensured both the object and the pins were in straight lines so as to avoid error due to parallax.

(ii) I made sure there was no air interference

(2bi)Snell’s law of refraction states that the ratio of the sine of angle of incidence to the sine of the angle of refraction is a constant for a given pair of media.I.e Sini/sinr = ΠWhere Π is known as refractive index.

(2bii)Given refractive index of glass = 1.5i.e aΠg = 1.5(from air to glass)SinC/sin90 = gΠaSinC/Sinn90 = 1/aΠgSinC/1 = 1/1.5SinC= 0.6667C = sin^-1(0.6667)Critical angle for glass C = 42°

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(3a)(ii) Vo = 2.00v

(3av)In a tabular formUnder S/N1, 2, 3, 4, 5

Under R(ohms):2, 3, 4, 5, 6

Under V(v):2.10, 2.30, 2.40, 2.50, 2.60

Under R^-1:0.500, 0.333, 0.250, 0.200, 0.167

Under V^-1(v^-1):0.476, 0.435, 0.417, 0.400, 0.3853vii Slope, = Δv^-1/ΔR-1= 0.5 – 0.355/0.6 – 0= 0.145/0.6S = 0.242Intercept, C = 0.355v^-1(3aviii)K = S/CK = 0.242/0.355K = 0.68

(3ix)(i) I ensured tight connections.

(ii) I ensured clean terminals.(3bi)(i) Temperature of wire(ii) Cross sectional area of wire

(iii) Length of wire(iv) Nature of wire

(3bii)Draw the diagramEffective E.m.f = 2v(parallel connection)Effective internal resistance= r * r/r + r = r/2ohmsCurrent, I , 0.8AExternal resistance R = 2AUsing,E = I(R+r)2 = 0.8(2+r/2)2.5 = (2+r/2)2.5 – 2 = r/20.5 = r/2r = 0.5*2 = 1ohms…Use your school value

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