Common Questions Asked in Post UTME Physics Exams. Day3

Common Questions Asked in Post UTME Physics Exams. Day3

Physics

Chima Sunday (Tutor)
15-06-2016 13:57:00 +0000

Try your best to solve these questions. Show working on paper when necessary and include a picture of your hand written Solution in your response. Wish you the best!

1. Which of the following serves as a nuclear moderator in a nuclear reactor
A. Cadmium rod
B. Boron rods
C. Aluminium rods
D. Heavy water
(uniben)

Updated Solution
A nuclear reactor is a device that utilizes nuclear reactions via fusion or fission reaction to generate electricity.
Moderating a nuclear reactor is a process whereby a fast moving neutron usually used in striking the fuel is slowed down, in which when there is head on collision with the neutron, extract some of the kinetic energy possessed by the neutron and consequently slowed its velocity. Examples of such moderators are Heavy water(they are water formed with deuterium), graphite rods etc

2. The half life of a radioisotope is 92 seconds, The N/N_o left after 2 hours is
A 54.22
B. 0.368
C. 650.59
D. 3.0×10^-24
(lautech)

Updated Solution
Here they are actually asking us to find the fraction of the left over(i.e N/N_o) after 2 hours, if the radioisotope has a half life of 92 seconds. there are actually two ways of doing this, but i am going to use one such ways and state the second one for you to try and use. Although the first method is a consequence of the second
methodI
recall that N/N_o = (1/2)^t/T
where T = half life = 92sec. and t = 2 * 60 * 60 = 2 * 3600sec
so N/N_o = (1/2)^2*3600/92
this can be approximated to be = (0.5)⁷⁸ = 3.3*10^-24
So correct Option is D
methodII
is by using the formular i.e N/N_o = e^-λt
where λ = 0.693/T

2 8 0

Shegx01@gmail.com Shegx01

question 1 is d(heavy water) or by using graphite to show down nuclear reaction

2 15-06-2016 14:57:00 +0000

Shegx01@gmail.com Shegx01

question 2 not clear enough sir or I can't comprehend

2 15-06-2016 14:59:00 +0000

armsy mohd

question 2. is 92sec x 60secs x 2hour / 69hrs/secs = 54.22

2 15-06-2016 15:37:00 +0000

Shegx01@gmail.com Shegx01

uniben is now live they are doing post utme CBT

2 15-06-2016 15:38:00 +0000

Chima Sunday (Tutor)

Thanks to all of that attempted to solve this question, that was a huge bold step keep it up. Also do know that this post has been updated with their detailed solutions, so try and check it up.

2 17-06-2016 15:25:00 +0000

OAA Oke

pls sir use the other method to solve number 2 que it looks confusing

2 17-06-2016 16:43:00 +0000

armsy mohd

thank you for the soln after being tried. we count on Ur support in this tutorials. God bless u.

2 17-06-2016 17:25:00 +0000

Chima Sunday (Tutor)

@OAA, recall that lin2 = log2 base e = 0.693, and recall from log that e^log a base e = a
so i think if you use this hint correctly you will fall back to the first method we applied. Thanks for your interest

2 27-06-2016 11:43:00 +0000

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