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2020 NABTEB PHYSICS OBJ AND ESSAY QUESTIONS AND ANSWERS

 

PHYSICS+Obj
1BBBBBCCDCA
11ABAADBADDC
21BCAABBBBBA
31BCCBBABBCD
41DBDBCBCBDC

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

Theory:
INSTRUCTIONS: You are to Answer four (4) questions in all
=====================================

(1a)

(i) Acceleration=vt/t1
(ii) Retardation=0 – vt / t3 – t2
= -vt/t3 – t2
(iii) Total Distance covered=1/2(t2 – t1)(t3)(vt)

(1b)
a=v – u/t —————–(i)
S=(v + u/2)t ————-(ii)

from equ (i)
v= u + at ————————(iii)
Substitute equ (iii) in equ (ii)
S=(u + at + u/2)t
=(2u + at/2)t
=(2ut + at^2/2)
=2ut/2 + at^2/2
:. S=ut + 1/2 at^2

(1ci)

(i) x=utAB + 1/2 at^2AB
and
x=(40 + 80/2)tAB
=(120/2)tAB
x=60tAB
but a=40 – 80 / tAB
a= -40/tAB
:. x=80tAB – 1/2 x 40/tAB x (tAB)^2
x=80tAB – 20tAB
=60tAB

FOR BC
100=(40 + 0/2)tBC
100=20tBC
:. tBC=100/20
tBC=5m/s

100 + x = (80 + 0/2)(tAB + tBC)
100 + x =40(tAB + 5)
100 + x = 40(x/60 + 5/1)
100 + x = 40x/60 + 200/1
collect like terms
x – 40x/60 = 200 – 100
60x – 40x/60 = 100
Cros multiply
60x – 40x = 60 x 100
20x = 6000
x=6000/20
x=300m

(1cii)
tAB=x/60
tAB=300/60
tAB=5sec

(1ciii)
tAC=tAB + tBC
tAC=5 + 5
tAC=10sec
=====================================

(2a)
Boiling point is the temperature at which the vapour pressure is equal to the atmospheric pressure.

Procedures to determine boiling point
(i) Take about 25ml of distilled water in a boiling tube and add 2 – 3 small pieces pumice stone.
(ii) Close the mouth of the boiling tube with a rubber cork that has two bores and clamp it with the stand.
(iii) Introduces a thermometer(temperature range -10 to 110C) in one bore of the cork of the boiling tube, keep the tulb if the thermometer about 3-5cm above the surface of the water.
(iv) Then introduce one end of a delivery tube in the second bore of the cork.
(v) Place a 250ml beaker below the second end of the delivery tube to collect a condensed water.
(vi) Heat the boiling tube gently, preferably by rotating the flame.
(vii) Note the temperature(t1) when the water starts boiling.
(viii) Continue to heat the water till the temperature becomes constant and the water remains boiling. Note the constant temperature (t2).
(ix) Record the observation.

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(2c)
Mc=300g =0.3kg
tc=950C
tw=25C
t=100C
Mw=250g =0.25kg
Heat loss by copper = heat gained by water
McCc(Tc – t) = MwCw(t – tw) + ml
0.3 x 4 x 10^2(950 – 100) = 0.25 x 4.2 x 10^3 (100 -25) + 2.26 x 10^6m
102000=78750 + 2.26 x 10^6m
2.26 x 10^6m=102000 – 78750
2.26 x 10^6m=23250
Divide both sides by 2.26 x 10^6
m=23250/2.26 x 10^6
m=0.0103kg
m=10.3g

Effect of Heat
(i) change in state.
(ii) Increase in temperature.
(iii) Chemical action
(iv) Change in physical properties
=====================================

(3ai)
Resonance means the production of forced vibration of maximum amplitude of a vibrating body.

(3aii)
Two identical turning forks and sounding boxes are placed next to one another. Striking one turning fork will cause the other to resonate at the same frequency.
when a weight is attached to one turning fork. They are no longer identical. Thus, one will not cause the other to resonate, when two different turning forks are struck at the same time, the interference of their particles produce beats.

(3aiii)
The sound is louder when a struck turning fork is held against a table because the surface of the table is set into vibration. Thus, the result in more air molecules vibrating which moves the sound louder.

(3bi)

(3bii)

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

(4a)

(4b)
(i) Eddy current:- Reduced by leminating the core
(ii) Hysteresis loss:- Reduced by using a low loss nickel iron alloys in the core.
(iii) I^2R or Heat loss:- Reduced by using thick wires.

(4ci)
Vp=4400v
Ps=600w
Vs=220v

k=Ip/Is
=Vs/Vp
=Ns/Np
:. Ns/Np=220/4400
=1:20

(4cii)
95%=Ps/Pp x 100%
95%=60/Ip x 4400 x 100%
Ip=60 x 100/95 x 4400
Ip=0.014A.
=====================================

(5ai)
Nuclear fussion: It is a nuclear process in which two or more light nuclei combine or fuse to form a heavier nucleus with the release of a huge amount of energy.

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(5aii)
Nuclear fission: It is the splitting up of the nucleus of a heavy element into two approximate equal parts with the release of a huge amount of energy and neutrons.

(5aiii)
When the body is exposed to radioactive substances for long time, the radiations penetrate the body and can destroy cells in the tissues and upset natural chemical reactions. It can bring about genetic changes or mutations, causing cancer and undesirable hereditary effects. The ionising property of ∝ – particles can damage the skin. Radiation from radioactive substances can kill a person, cause sterility and cause blood abnormalities.

(5b)
Fusion is more easily achieved with lightest elements such as hydrogen. The raw materials required for fusion are more cheaply and readily available.
The reason fusion is not normally used for the generation of power is that to bring the two light nuclei together in a fusion process, very high temperature of the order of 10^6 – 10^8 degrees Celsius are required to overcome the coulomb repulsive forces between the two nuclei.
This poses severe technological problem due to the fact that the materials to withstand this high temperature are difficult to come by. It is for this reason that the fusion have not been harnessed to a nuclear power.

(5c)
2πft = 100π
2π = 100
(i) F = 100/2 = 50Hz
(ii) Io = 30A
(iii) Irms = Io/√2 = 30/√2 = 21.2A

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