₦airaland Forum

MOG30:
Pls my brother has a problem, he went to register at a cybercafe and his passport was mistakely placed where his signature was meant to be. pls can he make the correction even after submitting

Yes.

DrBESTJC:
Hello UI Bosses here....

A friend of mine told me that UI does not allot equal marks to all the questions.

According to him,English Language, calculations in Chemistry and physics always carries higher marks.

Please, Is this True ?

And if yes,
Is there any other good English textbook I can buy to be more proficient in Use of English(and smash UI PostUTME)....apart from Inviscible Teacher ?

Very Serious and Urgent Please !

Cc:
Pes13
TheHistorian
Nltaliban
Condomsir
Babablogger
Rursh
Ajet001
Oldfoolnigger
Samuelizz
Bobola07
ChomaMee
Sirgalahad26
Eben331....
...and others.....

You shouldn't have tag me among the UI bosses because I am not a boss. These guys are the ones that supposed to receive the accolades of bosses...each question carries equal mark...I think Invisible Teacher should be sufficient for English but someone recommend dat I should also buy Communication skills in English by Helen Bodunde.

rasaquadri:
can someone help me send ui post utme past questions to this email
rasaquadri@gmail.com

It is on your portal, what you just have to do is to download your four jamb subject combinations.

A Physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10m/s. The instant before the stone hits the ground below, it is travelling at a speed of 30m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10m/s, what is the magnitude of the stone just before it hits the ground? (a) 10m/s (b) 20m/s (c) 30m/s (d) 40m/s. PLS SHOW UR WORKINGS!!

TeflonBlixx:
What about the mass of the vessel?

Since what we were given was heat capacity of the vessel and not the specific heat capacity of the vessel, so the mass of the vessel is not needed...from the formula of heat capacity: H=C(T2-T1) >>> C=H/(T2-T1).

TeflonBlixx:
Differential function
x = 4t - 2t^2
v = dx/dt = 4 - 4t
At time = 2.5s
4 - 4 x2.5
4 - 10
= -6m/s

Thanks bro.

Gracy3535:
A body moves over one fourth of a circular arc in a circle of radius r. The magnitude of distance travelled and displacement will be ?

1 circle = 2pie = teta...where teta=angular displacement. >>> 1/4 of circle = 1/4*2pie = pie/2 >>> therefore teta = pie/2 >>> angular displacement = pie/2...distance s = teta*radius >>> distance s = pie/2*r = pie*r/2.

Ebencoula14:
Hi,a little help needed here,the website for registration of the exam is not through,what do i do?

It may be due to many candidates trying to register. Keep trying it or you can try doing it at midnight.

Dkfpaul:
is it possible to cross from education and political science to political science in 200 level

need ur reply ASAP palz

Art students are in a better position to answer this question.

Iambaj:
My younger sister is having problem with her registration. The Subjects aren't loading also no place to upload o level any solution would be highly appreciated. Thanks.

I will advise that she should go to cybercafe. They will help her out there.

BiafranDel:
A heating coil of resistance 30ohm is connected to a 240volt supply for 300seconds to boil a quantity of water in a vessel of heat capacity 200j/k. If the initial temperature of the water is 20degree celsius and its specific heat capacity is 4200j/kg/k. Calculate the mass of water in the vessel.

R=300 ohms, V=240v, t=300s, initial temperature T1=20^oC, final temperature T2=100^oC (since it is at boiling point), heat capacity of vessel C=200J/K, specific heat capacity of water c=4200J/kg/K, mass of water in the vessel m=? >>> Note that both the water and vessel have the same initial and final temperature >>> Heat gained by water + heat gained by vessel = Heat lost by coil >>> mc(T2-T1) + C(T2-T1) = V^2t/R >>> m*4200(100-20) + 200(100-20) = (240^2*300)/30 >>> m*4200*80 + 200*80 = 57600*10 >>> m*336000 + 16000 = 576000 >>> m*336000 = 576000-16000 >>> m*336000 = 560000 >>> m=560000/336000 >>> m=1.667kg >>> m=1.67kg (approx)

BiafranDel:
2. A transformer is required to supply 0.2ampere at 100volt from the secondary using a primary voltage of 250volt. if the primary has 120turns, calculate the number of secondary turns and the primary current assuming the transformer is 100% efficient.

What I get are Ns=48 turns and Ip=0.08A

micskales:
pls in what format should i upload my o level results on UI portal, when i checked my result i didnt print it, instead i saved it as pdf, so pls is it advisable to use .pdf format? PLS I NEED ANS

I will advise that you should go to cybercafe. They will help you there.

TeflonBlixx:
Mass of Alcohol -- 21g
Density -- 0.7g/cm^3

Volume -- Mass/Density

-- 21/0.7
-- 30cm^3

Mass of Water -- 10g
Density -- 1g/cm^3

Volume--Mass/Density

-- 10cm^3

Density of the mixture -- Total mass of the mixture/Total volume of the mixture

21+10/30 + 10

-- 0.775g/cm^3
--- 0.78g/cm^3

2. B

pls show the workings to the second question.

Adolak:
This is wrong. The question is to find mass. Not mass conc.
Then you used the H conc instead of OH concentration as it is NaOH.

I got 0.0202g though. Don't know if I'm correct

can u pls show the workings??

TeflonBlixx:
Physics

4. A solid object weighs 24g in air, 16g in water and 20g in oil. The ratio of the density of the solid to that of oil is??
Show your workings..

mass in air=24g, mass in water=16g, mass in oil=20g, density of water=1g/cm^3...r.d of solid=mass in air/mass of equal volume of water=24/24-16=24/8=3 >>> density of solid=r.d of solid*density of water =3*1=3g/cm^3 >>> r.d of oil=mass in air-mass in oil/mass in air-mass in water=24-20/24-16=4/8=0.5 >>> density of oil=r.d of oil*density of water=0.5*1*0.5g/cm...therefore ratio of density of solid to density of oil=density of solid/density of oil=3/0.5=6/1 >>> 6:1

(1) If 21g of alcohol of density 0.7g/cm^3 is mixed with 10g of water, what would be the density of the resulting mixture? (a) 780g/cm^3 (b) 0.78g/cm^3 (c) 30g/cm^3 (d) 10g/cm^3 (2) For a particle having an x coordinate varies in time according to the expression x=4t-2t^2, the instantaneous velocity of the particle at t=2.5secs is (a) 12m/s (b) 6m/s (c) 0m/s (d) 10m/s PLS SHOW UR WORKINGS!!

The correct answers are marked with *

Eben331:
See me see English o (1) Is the load __ heavy that you can't carry it alone? (a) much (b) so * (c) very (d) too (2) The boys look alike that one can't know __ (a) who is whom * (b) who is who (c) which is which (d) whom is whom (3) I will soon attend __ you (a) to * (b) on (c) for (d) at (4) As __ now, Bakayoko is the best player (a) of * (b) at (c) in (d) on (5) You should be __ the alert (a) at (b) on * (c) in (d) by PLS EXPLANATIONS WILL BE NEEDED!

I have confirmed the correct answers to these questions.

Orezy5:
*CHEMISTRY*


4. Calculate the mass of sodium hydroxide that must be added to 800cm³ of water to give a solution of pH 10.5

[Na = 23, O = 16, H =1, MV =22.4dm³, NA = 6.02 x 10²³ atoms]

...

Someone should pls solve this questions.

Damayogenius18:
....u can use three formula for elastic collision.....but I will use two here.....M1V1 = M2V2....2nd formula....
U1 + V1 = U2 + V2.....so..3× 8 + 0 = 3V1 + 2V2 for 1...for 2...8 + V1 = 0 + V2..So....8 + V1 = V2.....substitute DAT into equation 1....
3 × 8 = 3V1 + 2(8+V1)
24 = 3V1 + 16 + 2V1
24 - 16 = 5V1
V1 = 8/5 ..V1 = 1.6....
V2 = 8 + V1
V2 = 8 + 1.6
V2 = 9.6ms....V1 = 1.6ms-1, V2 =9.6

Boss, can you pls show how you derive the second formula and also show the third formula.

akintosin780:
thanks,science(statistics)

You should try to help us here.

TeflonBlixx:
Mind posting Physics Utme based questions?? I'm quite rusty with physics

Rusty?? You must be joking...No, I don't mind...The ones I know I will solve them and the ones I don't know, others will solve them. This is the reason why this thread is created.

akintosin780:
Hi,am new here am tosin by name

You are highly welcome. Which department are you??

ChiomaMee:
Before I contribute my little quota, let me sigh deeply....

Ohhh...what a journey it was. Try and see to the fact that you're very familiar with all the practice questions. Like my GES tutor would say, "read like prayer won't save you and pray like reading won't help you". By God's grace, I had 68 which is the result of the lack of seriousness on my side of the whole thing.

In all, never forget God in your study. Never underestimate God's power on your exam and admission.

Pleeeaaasee...after your post utme, try as much as you can to go through some of your first semester courses. I thought first semester economics would be our secondary school economics but it's way far from it. The standard has been raised. Again, never forget God's ability in you. It definitely will help you in life. Shalom!

Thanks for this piece.

The Bosses and the Ogas- Pes13, Nltaliban, sirgalahad26, Ajet001, ChiomaMee, you guys should also help us here, and pls tell TheHistorian, Oldfoolnigger, Babablogger that their attentions are needed here for English. You all should give us tips on how we can pass the forthcoming UI PUTME excellently.

DRZAY:
Eben331 pes13 please who is able to login after C.O.I because I did mine on 17 of this month.

I didn't do change of institution.

So we don't have Art students and English gurus on this thread that can help me with the English questions I posted. That is too bad!! The questions have been there for some days now without solutions to them. Pls let help one another on this thread by answering and posting questions and, showing corrections to questions.

AASOO:
Has anyone been able to complete U.I post utme registration?
i am unable to go to the next page after uploading my passport and signature..

Are you sure that the passport and signature are in JPEG and do not exceed 50kb?? If yes, then logout and login back. You can also change your web browser to either Google Chrome or Mozilla Firefox.

BiafranDel:
How much sodium hydroxide is required to make 25cm3 of a 0.1 molar solution? ( na=23, O=16, H=1) A4g B2g C1g D5g

C=0.1mol/dm^3, V=250cm^3, m.m of NaOH =23+16+1=40g/mol Using n=CV ( n=m/m.m) >>> m/m.m=CV >>> m/40=0.1*250/1000 >>> m=1g there is a mistake in d question, the volume supposed to be 250cm^3 and nt 25cm^3 according to the options.

Damayogenius18:
....u tried bro...not correct sha

chai!!! pls show the correction.

esivue007:
no its ficsher's projection

ok, noted...thanks. You can as well help me with the English questions.

esivue007:
i think the motion of conduction is vibratory not oscillatory.

It's vibratory I choose not oscillatory.