2018/2019 NECO EXPO RUNZ: PHYSICS ( ESSAY AND OBJ) QUESTIONS AND ANSWERS




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*PHYSICS SOLUTIONS*

PHYSICS OBJ:
1-10: EBDBEAABCC
11-20: CCBCBACADC
21-30: EEDECACDDA
31-40: EDADCDDBBD
41-50: EBCDBEEECD
51-60: CEEDDDABED
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(1a)
water passes through a dam, and into a river below. The more water that passes through a
dam, the more energy is produced. Once a dam is
built, an artificial man-made lake is created behind the dam. Electricity is produced by a device called a turbine.
Turbines contain metal coils surrounded by magnets. When the magnets spin over the metal coils, electricity is produced. Turbines are located inside dams. The falling water spins the magnets.

(1b)
Given:
M=10
U=?
V=mu
F=+5cm
V=10u(virtual image)
using the real is positive convention
1/f=1/u+1/v
1/5=1/u-1/10u
1/5=10-1/10u
1/5=9/10u
10u=45
u=45/10
u=4.5cm
The object must be placed 4.5cm from the microscope.

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(2)
The distance covered S by a body with initail speed U and final speed V in time t is
distance = average speed x time
S = (u+v/2)t
first equation
acceleration = change in velocity/time
a=v-u/t
v-u=at
v=u+at
putting the first equation into S = (u+v/2)t we have
S = (u+u+at/2)t
S = (2u+at/2)t
S = (u+at/2/2)t
S = ut+1/2at^2
This is the second equation

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(3a)
(i) contact forces
(ii) field forces

(3b)
Time taken for ball to fall is the same as when the ball is dropped from that height
Hence:
u=omls
h=20m
g=10mls^2
t=?
Using:
h=ut + 1/29t^2
20=0(t)+1/2(10)t^2
20=5t^2
5t^2=20
t^2=20/5
t^2=4
t= √4
t=2secs
time taken for ball to fall to the ground = 2secs

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(6a)
The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane

(6b)
(i) Diverging lens
(ii) The diverge light rays entering the eye from a distance object and make them appear to be coming from the person far point

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(7a)
(i) It is used in car headlamps
(ii) It is used as a shaving mirror

(7b)
Given
object distance U =13cm
image distance V = -4cm
focal length F = ?
Using the real is positive convention
1/f=1/u+1/v
1/f=1/13-1/4
1/f=4-13/52
1/f= -9/52
f= -52/9= -5.8cm

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(9a)
(i) Cross sectional area of plates
(ii) Distanace between plates

(9b)
Given:
M=6kg
D=3m
G=6.67*10^-11 Nm^2kg^-2
using:
F= Gm^2/d^2
F= 6.667*10^-11*6^2/3^2
F= 2.668*10^-10N

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(10a)
(i) Mercury dropped on glass from droplets because the cohesion of mercury is stronger than it adhesion
(ii) Water rise in capillary tube because the force of adhesion at water molecules with glass is greater than their cohesion to each other.

(10b)
Surface tension, the ability of water to behave an elastic skin make it possible for an insect to crawl on water

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(12ai)
A trajectory is the path traced by a projectile

(12aii)
(i) the fired bullet
(ii) thrown javelin
(iii) thrown bomb

(12b)
(i)A scalar quantity has no direction while vector quantity has direction
(ii) Scalars can be added and subtracted easily by using ordinary algebra while vector quantity however have to be added or subtracted by vector diagrams and law like parallelogram,triangle law and resolution of vectors.

(12ci)
DIAGRAM

(12cii)
(i) From the diagram
slope Bc = CD/BD
14=(V-20)/20
V-20=280
V=280+20
Final velocity V=300mls

(ii) Total distance= Area of rectangle OABFO+Area Of Trapezium FBCEF
=(OA*OF) + [1/2(BF+CE)FE]
=(20*40) + [1/2(20+300)20]
=800+3200
=400metres

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(13a)
Specific heat capacity can be defined as the quantity of heat required to raise a unit mass of a material by one kelvin(k) or degrees Celsius(‘C)

(13aii)
(i) Change in temperature of phase/State
(ii) Change in resistance and in colour.

(13aiii)
(i) Linear expansivity results to expansion of metals or concrete bridges.
(ii) It as well causes cracking of glass cup when hot water is poured into the glass cup.

(13b)
(i) Heat is nothing but the amount of energy in a body. As against this, temperature is something that measures the intensity of heat.
(ii) Heat measures both kinetic and potential energy contained by molecules in an object. On the other hand, temperature measures average kinetic energy of molecules in substance.
(iii) The main feature of heat is that it travels from hotter region to cooler region. Unlike temperature, which rises when heated and falls when cooled.

(13ci)
When an inflated ballon is pierced and eleased, it obeys newton’s third law of motion. Air rushes out from the pierced hole and the balloon moves in the opposite direction

(13cii)
Jet

(13d)
using MxCx∆θ =MyCy∆θy
Given ∆θx =∆θy=(25-20)=5°c
Cx=½CY
then Mx[½CY]=MYCY
Mx/My=Cy/½Cy
Mx/My=2
Ratio of mass Of X to Y is 2:1

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(14a)
The law of reflection state that the angle which the incident ray makes with the normal is equal to the angle which the reflected ray makes to the same normal.

(14b)
(i)Effect of density
(ii)Effect of temperature:- In a gas.

(14c)
DIAGRAM

(14d)
using the lens formula
1/f=1/u+1/v
1/5=1/6+1/v
1/5-1/6=1/v
6-5/30=1/v
1/30=1/v
V=30cm (image distance from lens)
Total distance between image of len And image of mirror is
=30+12+16
=48cm

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(15ai)
Force field is a region around a body in which it experiences an effect (force) due to the presence of another body

(15aii)
(i)gravitational field
(ii)magnetic field

(15b)
Show that the escape velocity(Ve) of an object from the earth surface is √2Gme/r
To prove the escape velocity (Ve) of an object from the earth surface we can say let the kinetic energy of the object of mass M is
Ek=1/2MVe^2
When an object is at any given point P which is at a distance X from the centre of the earth the force of unity between the object and earth is
F=Gmm/X^2
Work done is taken the body against gravitational attraction DW=Fdx=Gmm/X^2 DX
Where W=work done in taking the body against gravitational attraction
W={R DW= {R Gmm/X^2 DM
=Gmm {R X^2 DM = Gmm [X^2/Y {R
= -Gmm [1/X]R = – Gmm [1/~ – 1/R]
Or, W=Gmm/R
For the object to escape from the earth surface
Ek=workdone,W
Ek=W
1/2MVe=Gmm/R
Ve=√2Gm/R(proved)

(15ci)
(i)Power supply smoothing applications
(ii)RF coupling or filter capacitor applications

(15cii)
Data given
d =4mm=4*10^-3 , 0.004m
A= 0.2m^2, V=300V
[Eo=8.85*10^-12fm^-1]

(i) Capacitance of the capacitor
using the approprate formula
C=EoAv/d
C=8.85*10^-12*0.2*200/0.004
C=8.85*10^-12*2*10^-1*2*10^2/4*10^3
C= 35.4*10^-11/4*10^-3
C= 8.85*10^-10^-11+3
C=8.85*10^-8F
the capacitance of the capacitor is 8.85*10^-8F

(ii) Electric Field intensity between the plates
E=F/q =K.q.d/d^2/q
E=kQ/d^2
E=8.85*10^-12*Q/d^2
but Q=CV
Q=8.85*10^-8*2*10^2
Q=17.7*10^-6 C,1.77*10^-7C
E=8.85*10^-12*1.77*10^-7/(4*10^-3)^2
E=15.7*10^-19/16*10^-6
E=1.57*10^-20/1.6*10^-5
E=0.98*10^-15
E=9.8*10^-1*10^-15
E=9.8*10^-16N|C or E=9.8*10^-16NC
The electric field intensity have the plates is 9.8*10^-16 NC

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(17a)
(i) Stay on top of large machinery operator training
(ii) Keep large machinery clean, and maintain a clean environment

(17bi)
Compound Microsoft:
For viewing samples at high magnification (40 – 1000x), which is achieved by the combined effect of two sets of lenses – the ocular lens (in the eyepiece) and the objective lenses

(17bii)
Film projector:
an optical device that projects an image (or moving images) onto a surface, commonly a projection screen.

(17c)
Artificial satellite orbits the earth and observes it. A rocket is designed to go to different planets, either taking people with it, or taking a space prob to take pictures of that planet.

(17d)
Data given
Rg=5n,Ig=10mA Vg=40V Rx=?
Ig=10mA =10*10^-3A
Ig=10^-2A
For Conversion Of galvanometer into voltmeter.
Vg=Ig(Rg+Rx)
Vg=Ig Rg + Ig Rx
Vg-Ig Rg =Ig Rx
Rx=Vg-IgRg/Ig
Rx=Vg/Ig-Rg
=40/10^-2 – 5/1
Rx=40/0.01 – 5/1
Rx=40-0.05/0.01
Rx=39.95/0.01
Rx=3995
the resistance of resistor is 3995


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