SaintHephzibah:
thanks *SENIORS* i'll try my best while God will do the rest

You could however reach me on whatsapp... @08166337736 , I will be available to answer all your questions... It will kinda keep me busy simultanously as well.



Anxiously waiting fr resumpion !!!!

PrinceCEE:


If he wouldn't mind though

Enry54

Mind ke! Abeg Please Holla @Whatsapp if you were admitted for computer science

VORACIOUS:
PLS WHAT IS THE EVIDENCE OF BEEN CLEARED? THE LAST PERSON TO CLEAR ME DURING MY CLEARANCE DIDNT CHECK MY ORIGINAL CREDENTIALS BUT HE GAVE ME(AUTHORITY TO PAY (LETTER)),TOLD ME TO FILL ONE FORM AND I WAS TOLD TO GO

Boss, That's not the end of clearance, I can say if one is found wanting in any requirement and escapes being screened out. Such person will eventually b fished out, like d case of lady preparing for her final year exam @ui but was later advised to withdraw by the management due to inefficiency to meet up with admission requirement with which she was admitted 4 years back.. And if I may add... Make LOTS of photocopy of the "Authority to Pay" and any other documents which will be given to you, Cause copies will be requested from you, whenever you want to pay any fee, and receipts issued to you should also be kept fr reference.

thasquirrie:
Any newly admitted computer science students in 'ere? Holla

Thanks @mathefaro, Boss I'm here oOo

Mesopotamia:

adeniran Fiyinfoluwa Samuel......aka mesopotamia, aka oluwasinzu....admitted philosophy fresher...add me 08032518198

You have been successfully added!

horpeyemmi66:

Make we wash am over Shawarma and a bottle of Night Train.

Sharp Guy! Me just Rocking Killepitsch here since Yesterday, Care to Join?? No tell your Parents Ooo!

LIST OF REQUIRED DOCUMENTS (Unofficial)
Courtesy: 08166337736

1• Letter of Invitation for the Post-UTME Screening Exercise

2• Original and 4 Photocopies of Jamb Admission Letter [if available]

3• Admission Acceptance form (downloaded from the site)

4• Original & 4 Photocopies of The UTME result Slip (imprint with photograph)

5• Original & 4 Photocopies of O'Level Results - WAEC, NECO, GCE or Cambridge Result

6• Original & 4 Photocopies Of Birth Certificate or Sworn Declaration of Age

7• Two Letter of Reference from Two Reputable Person - on Letter Headed Paper (preferably middle Level Government Employees or ClergyMen)

8• Ten Recent Passport size Photographs showing frontal Face view, (5.2 * 5.2) Names of Candidates should be written at the back.

9• Original & 4 photocopies of Bank Teller for Payment of 3000 & 20000 Admission Processing & Acceptance Fees Respectively.

10• A File With Tag

11• Candidates should go online to download and fill the following forms
•Medical Form
•Undertaking Form
•Attestation Form
•Admission Acceptance Form
•JAMB form 02 and 03
•Also fill the Biodata Form Online


Courtesy: 08166337736

Congrats To All Successful Aspirants!



A WhatsApp group. Has been Created fr you All,
So as to ensure a stress-free Registration,....



Your Presence is Mostly Anticipated... Kindly Holla



@ 08166337736 With Your Name!


However, Current students Of UI, Will be highly welcomed!




MeanWhile, How We Go Wash am ?

@Cassyrooy
@Horpeyemmi66

horpeyemmi66:

Quite laudable sir...enjoy yourself. We kinda deserve it.

To God Be the Glory!

keVeeeN:
I fail that question for the post jamb sha

Me, I learnt it after the exam ni sha ...


I just created a What'sApp Group For us to interact,
Kindly Holla @ 08166337736

Or, Better still, Reply With Your Number to be Added!

CC. All Successful Aspirants!

horpeyemmi66:

Thanks Henry, and where have you been?

Have always been around, But I usually Watch Behind Bars, I'm not the loquacious type ...
How you doing..!



Ps: Kudos to those that have kept the Group alive, Ranging from the likes of Cassyrooy, GetOvaMe, to 11michael, Olatunji,

I hope to see you all @ UI

@SegunFlames, @LexxyBoy: @All Them Boss, Kindly link me with any Computer Science Student you might Know, Thanks!

Also To Those that Made the Cut-Off...
Kindly Holla @ Me Via BBM 27C2D0E8
I'd Like to meet Y'all.... Thanks in Advance!

My 1st Official Comment Since August!!...

To God be The Glory, Great Things He Hath Done, Greater Things He Will Do...

Congrats To All Successful Aspirants....

@horpeyemmi66, Congrats, and Happy Resumption in Advance, In case you might be wondering who the heck am I? Kindly check my previous post, We used to solve Physics here!

@mathefaro, Boss of Life, I hail ooo! Congrats to You also... Greater things ahead!

And to those that Made the Cut-Off.. Congratulations! A truck of coke awaits you all!

However, To those that dint make the cut-off, keep the faith alive! All will Be Well!

The Admission Unit of the University of Lagos has released the  Cut-off Points for the 2015/16 UTME Admission exercise. Below is the list of the cut-off points:

Source: http://unilag.edu.ng/newsdetails.php?NewsID=2286

Holla I bring joy to thee... UI has started its postutme screening

morgang:
Hi all,

UI past questions and answers is available for free.... No fee charges or whatsoever! just drop your mails, and it'll be dully forwarded to you.

As for the folks who thought that we're on strike, we ain't on strike yet...probably, after first semester exam!

Incase of any piece of information, u can still reach on this platform...although not too frequent on this forum.... I presume some of u have my contact for Whatapps...

Goodluck fellas..

[email]henryogunrinde@yahoo.com [/email]

@horpeyemmi66.. I thought you were about to prove me wrong in my previous question.... I hope you now understand the way I did it? Pls note that I will neva test you guys with a question I don't understand! In case I have a question I might not understand, I will indicate it in such question.

horpeyemmi66:
*PHYSICS | LATENT HEAT 001

1. Which of the following lowers the melting point of substances?

A. Density

B. Pressure

C. Energy density

D. Specific latent heat of fusion

2. Heat is supplied to a test tube containing 100g of ice at its melting point. The ice melts completely in 1min. What is the power rating of the source of heat(latent heat of fusion of ice=336J/g)

3. The statements below were made by some students describing what happened during the determination of the melting point of solids
1. The temperature of the solid was constant until melting started
2. The temperature of the solid rose until melting started
3. During melting, the temperature was rising
4. During melting, the temperature was constant
5. The temperature continued to rise after all the solid had melted
6. The temperature stopped rising after all the solid had melted.

Which of the following gives correct statements in the right order?
A. 2, 4 & 5

B. 2, 3 & 6

C. 1, 3 & 6

D. 1, 3 & 5

E. 1, 4 & 5

4. What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100°C to 80°C?( Specific latent heat of steam is 2,260,000J/kg, Specific heat capacity of water is 4200J/kg/k)

5. 80% of the heat supplied to a 30g block of ice at 0°C. Calculate the total heat energy supplied. (Specific latent heat of Ice= 336J/g)

6. Which of the following best explains why a person suffers a more severe burn when his skin is exposed to steam than when boiling water is poured unto his skin?

A. Steam is at a higher temperature than boiling water

B. Steam possess greater heat energy per unit mass than boiling water

C. Steam spreads easily over a wider area of the skin than boiling water

D. Steam penetrates more deeply into the skin than boiling water

7. Calculate the time to heat 2kg of water from 50°C to 100°C in an electric kettle taking 5A from a 210V supply (S.H.C. of water =4200J/kg/k)

8. How many grams of water, 17°C must be added to a 42g of Ice at 0°C to melt the ice completely? (S.H.C of Water=4200J/kg/k, Specific latent heat of fusion of Ice = 3.4 x 10^5 J/kg/k).
------------------------------------------------
Humanities, where are you guys, this is not a game of monopoly, we need you guys to do your thing a symphony is what we want.

1• B
2• Heat Supplied = Heat Required For Melting Ice
H = mL = 100 x 336 = 33600J

Power Rating = Energy Used Up/ Time
Power Rating = 33600/60
= 560J/s or 560w
3• The correct option is A as temp. Rises before and after melting...
Temp. Remains constant during melting...

4• Cooling steam from 100^oC to 80^oC Comprises of
(i) 100^oC of steam (gas) - 100^oC of water i.e change of state
(ii) 100^oC of water - 80^oC of water i.e change of temperature hence, Heat given out = mL + mcθ
Cooling water from 100^oC to 80^oC, heat given out is mcθ

Difference = mL + mcθ - mcθ =mL
= 4 x 2260000
=9040000J
5• Quant. Of Heat = 30 x 336
= 10080J
Thus. 80% of Total Heat = 10080J
Total Heat = (10080 x 100)/80
= 12600
6• (B) Steam possesses greater heat energy per Unit Mass.

7• Quant. Of Heat = mc(θ₂ - θ₁)
= 2 x 4200 x (50)
= 420kJ
Thus. Quant of Heat = IVt
t= 420000/(5x210)
t= 400s

8•Mass of Water (kg) 0.042kg
Heat Required For Melting = Heat Lost By Water
0.042 x 3.4 x 10⁵ = m x 4200 x (17-0)
71400m = 14280
m=0.2kg or 200g

OGOJNR:
hen nw,which means g is equal to 0.25 nd mass is 40kg nw

g' (acceleration due to gravity at height h ) is not 0.25....

Put g= Which is 10m/s²
R= which is the radius of the earth 6.4x10⁶
h= which is the height above the earth surface 6.4x10⁶
In the equation...
g' (acceleration due to gravity at height h) = g • R²/(R + h)²
You will get 2.5 as your answer dude!

OGOJNR:
to me ooo...the answer is 10N.using g=GM/R square
g=0.25M/S2
W=Mg
mass is constant nw
w=40*0.25=10N

Guy! g=GM/R²
Thus g' at height h from the earth crust is

g'= GM/(R + h)²

Divide by g

Thus, g'/g = R²/(R + h)²

Therefore g' (acceleration due to gravity at height h) = g • R²/(R + h)²
These things are in our textbooks! You don't just consider them

horpeyemmi66:

This doesn't go down well with me sir viz:

"F2= 400 x (0.5)2= 400 x (0.25) = 100N"

F=mg, isn't it?, but look what you've done...
F2= W x g
F2= mg x g
F2= 40 x 10 x 0.25(emphasis mine)

You should have gotten the mass as 40 before comparing it with the new gravitation acting on the body at that height.

You took the weight of the body on the Earth and multiplied with the gravitation acting on the body at that height, which for the want of a better word is illogical.

Mass is constant anywhere, but gravitational force g, varies being less at the equator and more pronounced at poles.

Don't confuse people for the love of God!

Would soon vet the rest you've done...

Cc mathefaro

The equation you used to find the acceleration due to gravity at a height h above the earth crust is wrong... Check you textbook, let say the acceleration due to gravity above height h is g' with R as the radius of the earth

g=GM/R²
Thus g' at height h from the earth crust is

g'= GM/(R + h)²

Divide by g

Thus, g'/g = R²/(R + h)²

Therefore g' (acceleration due to gravity at height h) = g • R²/(R + h)²

By the way... Check how I got 0.25 I dint find g first... Take a look at the way I solved it once more...

mathefaro:
If I hear! I can say it with all authority that jamb will never set such question and neither will any school for PUTME. Because like I said earlier, it's beyond the scope of UTME even if as you put it, it's in the syllabus. For the fact that the topic is part of the syllabus doesn't mean they have to go deep to this level. For instance, when you check the math syllabus, you have calculus in it but there is a limit to which they can go in calculus questions (scope). Nevertheless, I believe horpeyemmi66 has done a good job up there. That should be the correct answer if I'm not mistaken.

Good morning to y'all

3• This, I can say it is sorta advanced but, they are all in our textbooks...

For revolution, MV²/r = Mg

Mw²r=mg

g=(2π/T)²r
g=4π²r/T²
T²=4π²r/g
T= 2π√(r/g)

Hence the mass is not dependent on the period.

Note: All questions answered are from O'level related textbooks and not A'level

mathefaro:
If I hear! I can say it with all authority that jamb will never set such question and neither will any school for PUTME. Because like I said earlier, it's beyond the scope of UTME even if as you put it, it's in the syllabus. For the fact that the topic is part of the syllabus doesn't mean they have to go deep to this level. For instance, when you check the math syllabus, you have calculus in it but there is a limit to which they can go in calculus questions (scope). Nevertheless, I believe horpeyemmi66 has done a good job up there. That should be the correct answer if I'm not mistaken.

Good morning to y'all

2• Let M be the Mass of Earth
Let r be radius of the earth
Let p be the density of earth

g=GM/r² = G(pV)/r²

V(volume)= 4/3 • πr³

Thus g= Gp • 4/3 • πr

g= (6.7x10^-11 x 5500 x 4 x π x 6400 x 10³)/ 3

Pls cross check with calc.

g= 9.87 m/s²

mathefaro:
If I hear! I can say it with all authority that jamb will never set such question and neither will any school for PUTME. Because like I said earlier, it's beyond the scope of UTME even if as you put it, it's in the syllabus. For the fact that the topic is part of the syllabus doesn't mean they have to go deep to this level. For instance, when you check the math syllabus, you have calculus in it but there is a limit to which they can go in calculus questions (scope). Nevertheless, I believe horpeyemmi66 has done a good job up there. That should be the correct answer if I'm not mistaken.

Good morning to y'all

.
1• It is a logical question and requires a logical reasoning...
The first question is all about the inverse square law of universal gravitation, I believe you know that...

Let the mass of earth be M
The mass of missile be m
While the distance is d
Thus. F=(GMm)/d²

Mere looking at the situation, GMm is constant anywhere... Right ?
So let F x d² = constant

F₁ x d₁² = F₂ x d₂²

F₂ = F₁ (d₁)²/ (d₂)²

F₂ = 400 x (6.4 x 10⁶)²/(12.8 x 10⁶)²

F₂ = 400 x (0.5)² = 400 (0.25) = 100N

mathefaro:
For candidates preparing for post utme? Way off the scope IMO

Na Jamb Questions! Based on the syllabus

PHYSICS Que. Please Don't Ignore!!!

1. A missile weighing 400N on the earth's surface is shot into the atmosphere to an altitude of 6.40 x 10^6m, taking the earth as a sphere of radius 6.40 x 10^6m. What would be the weight of the missile at that altitude?

2. Assuming that the mean density of the earth is 5500kg/m^3, that the constant of gravitation is 6.7 x 10^-11 and that the radius of the earth is 6400km. Find a value for the acceleration due to gravity at the earth's surface.

3. What would happen to the earth's period of revolution if its mass was three times as great as it is?

Odilop:
H=0.5.a ((T3)^2-(T2)^2)
=0.5. 6.( 9-4)
= 3*5 = 15m

v. Good

A body starts from rest and moves with uniform acceleration of 6m/s^2. What distance does it cover in the third second?

mathefaro:
Not possible! For the fact that it's moving with a constant speed means it's speed is neither increasing nor decreasing. Hence no change in speed and acceleration us zero

What about the case of circular motion?

OGOJNR:
finally,i got u...that questn dnt nid any calculation since u're lukin for d acceleration before cumin to rest...wen a body undergoes uniform velocity or constant speed the acceleration is always zero...the ans is 0

If a body moves with a constant speed and at the same time undergoes an acceleration, What can you say about that ?

horpeyemmi66:
*PHYSICS
1. Two Submarines The knight and The cascade were submerged gradually into a river at a depth of 120m and 180m respectively below the water surface. If the atmospheric pressure is 15m of Water, find the ratio of the pressure of The Cascade to The knight.

2. A pilot records the atmospheric pressure outside his aircraft as 83cmHg while a ground observer has a reading of 94cmHg for the atmospheric pressure on the ground. If the density of the atmosphere is constant, calculate the height of the aircraft above the ground[Relative density of Mercury=15 and Air=0.0015]

3. If the normal atmospheric pressure in a chemistry Laboratory supports a column of Mercury 0.85m high and the relative density of Mercury is 18.5, then the height of the water column which the atmospheric pressure will support in the same Chemistry laboratory at the same time is?

4. A man walks 8km north and then 5km in a direction 60° east of north. Find the distance from his starting point.

5. A boy travels 12Km eastwards to a point B and then 5km southwards to another point C. Cakculate the difference between the magnitude of the displacement of the boy and the distance travelled by him.

6. The distance travelled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resultant graph is linear.
The slope of this graph is a measure of

A. Initial displacement

B. Initial velocity

C. Acceleration

D. Half the acceleration

7. Which of the following is a set of Vectors?

A. Force, mass and moment

B. Acceleration, velocity and moment

C. Mass, weight and density

D. Mass, Volume and density

8. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2m/s and rebounds with the same speed. Determine the impulse experienced by the wall.

A. 20.0kgm/s

B. 10.0kgm/s

C. 5.0kgm/s

D. 1.3kgm/s

9. Two particles X and Y starting from rest cover the same distance. The acceleration of X is twice that of Y, the ratio of the time taken by X to that taken by Y is?

A. 1/2

B. 2

C. 1/√2

D. √2

10. The driver in a motor car of which the total mass is 800kg and which is travelling at 20m/s, suddenly observes a stationary dog in his path 50m ahead. If the car brakes exert a force of 2000N, what will most likely happen?

A. The car will be able to stop immediately the driver notices the dog
B. The car will stop 30m after hitting the dog
C. The car will stop 20m in front of the dog
D. The driver will quickly reverse the car
E. The driver will have to find a banking angle to ensure traction.


cc Polymath, Pr0ton, Dr. Host, Dr. Sage, Mikechibuzor, thaotech, AleXis0r, francistony, MedwhiteO, Bhuumhite, Funkysilver etc


*******UNA GOOD MORNING*******

1• P = A•P + Pwater
Pk = 15 + 120 = 135m of Water
Pc = 15 + 180 = 195m of Water
Thus: Pc/Pk = 1.44

2• A•P Read By Pilot = 83cm of Hg= 0.83m of Hg
A•P = 0.83 x 15000 (Conv. R•D of mecury i.e 15 to Density in Kg/m3) x 10= 124500N/m2

Thus: Distance between the Top Atmospheric Level and Aircraft
1.5(R•D of Air in Kg/m3) x 10 x h=124500

H = 8300m


A•P Read by Ground Observer = 94cm of Hg = 0.94m of Hg
A•P = 0.94 x 15000 x 10 = 141000N/m2

Thus: Distance between the Top Atmospheric Level and Ground Observer
1.5(R•D of Air in Kg/m3) x 10 x h=141000

H= 9400m

Height of The Aircraft Above Ground Level = (9400 - 8300)m
=1100m

3• A•P = 0.85m of Hg = 0.85 x 18.5 x 1000 ( R•D of Hg in Kg/m3) x 10
= 157250N/m2

Thus: Height of Water Column =
1000 x 10 x h = 157250

H=15.725m

4• If you really meant Distance = 13km but if it is displacement

D^2=25+64-2(5)(8 )Cos120
=25+64-2(5)(8 )(-Cos60)
=25+64+40
=128
D = 11.3 km

5• Magnitude Of Displacement = 13km
Distance Travelled = 17km
Their Diff. = 4km

6• C
7• B
8•A. 9• C
10• B