PRACTICAL PHYSICS | PART 1 | PHY107 #physics #practical #engrkingsley #exam #education

Added: 2026-05-12

[00:00:01]Okay, in today's episode on learn with engineering, we are going to be doing revision on practical phys.

[00:00:14]Let's get ready for some exam tips.

[00:00:17]Okay, we are going to be solving some massive questions, calculations and theory questions that is involved in this particular course.

[00:00:30]because credit unit is one and the the exam standard. Okay, in this course we are going to be looking at some necessary calculations from 1 to 5 in this part one. In part two we are going to take series of examples too in preparation of this exam. Just stay with us on this YouTube channel and make sure you always like, comment and share to your departmental groups and your friends. They need this.

[00:01:08]Okay, let's go back to the first question. I said, what is the relative density of a liquid if an object weighs 20 Newton in air?

[00:01:20]17 when totally immerased in water and 18 Newton when totally immerased in liquid. Okay, the first thing is to list out your parameters. You have object weighing uh 20 Newton in air. Weight Weight in air in air is 20 Newton.

[00:01:54]Weight in uh water weight in water is 17 Newton and weight in liquid what is given weight in liquid is 18 newton. Okay. The first thing we are going to do is to calculate uh the relative density. The relative density uh of liquid is up thrust.

[00:02:42]What is the meaning of up thrust? Up thrust is just simply talking about weight loss. Our intros in liquid upross in water.

[00:03:02]Okay. Liquid first is liquid up thrusting.

[00:03:08]all over up thrust water.

[00:03:22]Okay. So, what are we having here?

[00:03:27]This weight in air is generally referred to as W1.

[00:03:36]Weight in water is regarded as W2 and weight in liquid is regarded as W3.

[00:03:47]So the relative density of liquid is now equal to W1 - W3 all over W1 - W2.

[00:04:01]So we can substitute our parameters. W1 is 20 Newton minus W3 is 18 Newton all over W1 is 20 Newton minus W2 is 17 Newton.

[00:04:23]So moving on 20 - 18 will give us 2 all over 20 - 17 will give us 3.

[00:04:33]So 2 / 3 will give us 0 0.666666 do 7. Okay. Approximately you have 0.67.

[00:04:53]Okay. Let's check options that corresponds to this solution. Option B is correct. 0.67.

[00:05:02]Okay. That is for number one. Number two, if you look at number two, they said 0.2 g per cm cube is in SI unit is dash. Okay.

[00:05:23]The standard SI unit is kilogram seconds.

[00:05:32]and some others.

[00:05:34]But if you want to change gram to kilogram and want to change cm to me, what are the procedures we are going to change?

[00:05:48]Look up.

[00:05:50]If you want to change gram 1 g to kilogram, you have one which is the g / 1,000.

[00:06:02]It will be equivalent to 1 all over 10 to the^ 3. And this is still equivalent to uh 10 to the power - 3.

[00:06:21]Okay.

[00:06:23]So that is it in kilogram in kilogram. Okay. If you want to change one cm to me, you'll be having the one in cm divi time 1 all over 100. You want to change C to me, you divide by 100. Or similarly, you have 1 * 1 / 10 2 + 100 is 10 * 10 that is 10 2 which is still equivalent to 1 * 10^ - 2. Okay, moving on. You have remember that this thing is in meters. But if you observe in the question we are not talking about this one is just g that's why it's in kilogram there but this one we are talking about cm to the power 3. So when we uh cube the meter conversion we'll be having 1 * 10^ - 2 which is still 10 * 10^ - 2 all cube m the difference between this and this is that I have introduced cube to what we have finally you now see comparting that young man there will be 1 * 10 power - 2 * 3 will give us - 6 m cube.

[00:08:01]Okay, let's look at the question. The question we are converting is 0.2 g all over cm cube. Okay, if you want to do this is very simple. Just write out the figure you have 0.2. How you do it in your exam? It's very very short.

[00:08:17]multiply it with the conversion I have given you.

[00:08:20]For g you are multiplying it with * 10 raised to ^ -3 that is to convert g to kilog you multiply with 10^ - 3 but provided that that cm is cubed and you are converting it to me cube instead of being 10^ minus uh 2 because of the cube it multiply it to be 10 ra - 6.

[00:08:53]Okay, this one is in kilog and this one is in me cube.

[00:08:59]Okay, moving on. When you multiply, you have 0.2 * 10 raised to the power. According to the law of indices, this is negative. When it goes up, it change to positive. It will be -3 + 6 instead of -6 kilogram all over me cube.

[00:09:28]Okay. So we now have 0.2 2 * 10 power - 3 + 6 is 3 kilog m cube.

[00:09:48]So 0.2 * 10 ^ 3 is the answer which corresponds with option number A. Okay.

[00:09:58]The number three question they said an object has a space heat a specific heat capacity of 200 jou per kilogram per kelvin. A specific heat capacity C to be 200 jou per kilogram per kelvin if 1 * 10^ 5 of energy. So the heat energy is 1 * 10 ^ 5 j uh to change it temperature from 20° to 70°C.

[00:10:44]So we are having change change change in theta to be equal to theta 2 minus theta 1.

[00:10:56]Okay, theta_2 here is the final temperature which is 70° C. So 70° C is the same thing with degree Kelvin. And it says that even if you add 3 273 to to change it to Kelvin and add to the first one when you subtract it still give you 50. So 70° Kelvin - 20° Kelvin if you subtract it it will still give you 50° Kelvin okay we use Kelvin in calculating we don't use degrees okay so the formula of calculating heat energy is uh g uh M change MC change in theta. Okay, I think that is correct. So our aim in the question they say to determine its mass.

[00:12:00]So our aim is to make mass the subject.

[00:12:03]We divide both side by the coefficient of M which is C change in theta. We have H being divided by C change in theta.

[00:12:15]So mass which is what we are looking for is heat energy is 1 * 10 ^ 5 j all over specific heat capacity is given to us as 200 jou per kilogram per kilogram per kel and the temperature is being given to us as that is the change in theta to be 50°.

[00:12:47]Okay, moving on. We have our mass which is what we are looking for should be you plug that young man in your calculator.

[00:12:56]You plug it in, you have 1 * 10 to the power 5 all over 200 * 50 10 kilogram that is option number D.

[00:13:23]Okay, moving on to the next question which is the fourth one.

[00:13:32]Okay, this question that we just finished. Now you see that you can find heat, you can find specific heat capacity, you can find the temperature change and you can find the mass with that same formula depending on the you want to make the subject. So number four they said the standard error in the slope of a graph is 3 [clears throat] three while it range and number of of obser region are 0 62 and six respectively. The vertical width scatter is dash.

[00:14:08]So the standard error in slope is equal to 4 W all over NR. Here W is vertical W scatter which is what we are looking for in this question. Uh N is number of plotting points which is what is given to us as six. N is given to us as six. W is what we are looking for known. The vertical range they said the range is 0 0 62. So the standard error in slope is still given to us. Standard error in slope is given to us as three.

[00:14:54]Okay 3.0. So moving on we can make the vertical width scatter the subject but let's subute the parameters we have the standard error in slope to the three equal to 4 * vertical scatter unknown all over the number of plotting points is 6 * the vertical range is 0.62 62.

[00:15:22]Okay, moving on. We cross multiply. We have this time this. It will be 4 W equal to this time this will give us 6 * 0.62 * 3 that is here. Okay. By cross multiplying remember this is over 1. 1 * 4 W give us 4 W. So moving on we have 4 W to be equal to 6 * 3 is 18 6 * 3 is 18 * 0.62 62 will give us 11 116 okay making w the subject we divide both side by four so I have divided by four and divided by 4 have that four will go with four w will be equal to 11.16 / 4 will give us 2 2.7 79. So is 2.79 the option yes option A. That is all for this question is you remember that the standard error in slope is given to be 4 W all over N R.

[00:16:44]Okay. So the last question for this particular video.

[00:16:51]So please always share, subscribe, turn up your notification bell so that whenever we drop a very sweet and wonderful solution on things that is somehow difficult to you, you can engage it. Share to your friends, you subscribe very very important. So they said in an experiment the vena caliper was used to measure the breadth to be bread is given to us as 2 4 and the thickness thickness t is given to us as 0 45 cm.

[00:17:38]of the meter row. So we are talking about the middle row. They said find the volume of the meter row. But you know that volume V is length time width time breadth.

[00:17:53]Width is the same thing with thickness.

[00:17:56]Breth is the same thing with height.

[00:17:59]Width and thickness is the same. Breth and height is the same. So the standard h length for me row is the standard length for me row is 100 cm. So when you plug it in here you have 100 cm time and weight is given to us as uh 0 45 cm and breadth is given to us as 2 4 cm. So moving on you have that volume is equal to 100 * 0.45 45 * 2.40 will give us 108.

[00:18:50]CM * CM some calculation question on five to 10.

[00:19:04]Then we'll take time and digest both the theoretical aspect and the calculation aspect. Just make sure you stay with us from the beginning to the end and put your notification bell so that any time you post we don't have to come and remind you that we have dropped video you get the notification. Thank you for watching this video.