bro isn't the least count 0.11 cm Explanation: 10 VSD= 11 MSD 10 VSD = 11 * 1MSD 10VSD = 11* 0.1cm 1 VSD = 0.11 cm so least count = 0.11 cm if i am wrong reply me
@@Lama_60 Bro, 1 Vernier Scale Divison(V.S.D) = 0.11cm but Least Count(L.C) = M.S.D - V.S.D = 0.1 - 0.11 Least Count(L.C) = -0.01 Here the negative answer means that the "first marking in vernier scale" will be 0.01 ahead of "0.1 marking in main scale" So, if we just need the value, the least count will be 0.01
@@Lama_60bro I also made this mistake but figured out instantly that what you have calculated is the vernier spacing from an origin basically one vernier small division but least count doesn't mean that . Least count means the amount of distance the vernier scale is shifted from msd so you have to subtract it from msd which is 0.11-0.1= 0.01.
Man how can you be so good at teaching there is no way today in class i was so perplexed but after watching this vid i realize this is so natural and logical THANKS MAN!!
I think in the formula, we get the second decimal digit by counting the place that coincided mod 1 but because the zero error had negative shift, it would be 1- the coinciding mark so everyone got the answer as 0.6 when it was actually 0.4. This makes so much more sense thanks to you!!!
I am just regreting my whole 11th because last year I've seen one videos and it remains in my temporary memory but before 6 months of jee it came like magic really i am smiling alone after understanding this concept I don't understand why it is so underrated thanks sir 😁👀
In the last question, 1VSD>1MSD because 10VSD=11MSD (as given) Now, 1MSD=0.1cm thus 1VSD=(1.1)(0.1cm) = 0.11cm which is obv greater than 1MSD. Now the least count is 1VSD-1MSD=0.01cm. If we'll use LC=1MSD-1VSD it'll be obv negative because this takes the assumption that 1VSD is always less than 1MSD which is not the case here. Now, to the reading it'll be 2.83cm which I calculated as per usual.
I want more videos like this especially for class 11. 👍 Actually i think this is the best way of teaching as you are teaching us to reinvent something which is most important for our youth!😀
first assumption....the 0s of both the scale coincides....now, reading=2.8+x,and 8msd=7vsd+x (8*0.1=7*0.11+x) and therefore x=0.03 and hence reading=2.8+0.3=2.83... now least count is the least distance that you can measure between msd and vsd and least distance between vsd and msd(vsd-msd) will be only in that division which lies *just after* coinciding divisions(the coinciding divisions are 7th division of vsd and 8th divsion of msd) and the division which lies just after that are 8th(of vsd) and 7th(of msd) so as stated above the least distance you can measure is the distance between these division therefore 1vsd-1msd=0.01 cm.....now I know the question here can be how can i subtract msd from vsd(vsd-msd),well I have assumed that the main scale can move...because this type of question is not used in real life(why would you make a scale whose vsd is greater than msd)...so clearly that's why I am allowed to defy this rule :)( at least that's what I think)
I used to explain by saying that the number of coincident vernier division shifts by 10 if the gap is 1 mm. so for each coincident vernier division, gap increases by 0.1mm or 0.01 cm. so when coincident vernier division is 6, we need to add 0.06 cm to main scale reading in cm. your explanation however is more complete!
The reading was simple to calculate... 2.83 main scale divisions But the least count took me a while. I got it 1.11(bar).Here's how: I imagined coinciding the 10th division of VS with the 10thb division on the MS. So there is gap in between the first divisions of both the scale which is equal to 10MSD -10VSD. This 10VSD is nothing but (10/11)*10 = 9.99bar. So answer is 10-9.999 =1.111. Please confirm if it is correct... Thank you sir
I'm kinda confused at the part where you measured the divisions of vernier scale 0.09 cms each. Isn't the divisons of vernier scale the least count? which is calculated as LC= 1main scale division - 1 vernier scale division. like, in that question, it was given that 10 VSD = 9MSD ie, 1 VSD = 0.9 MSD (smallest division of main scale is 0.1 cm) then 1 VSD= 0.9(0.1cm) which is, 0.09 cms BUT, when you calculate least count then, you gotta do , LC= MSD -VSD ie, LC= 0.1 cm - 0.09cm where you get , least count = 0.01 cm. then why did we calculate the divisions as 0.09 x 6 instead of 0.01 x 6?
I think he is in line with the ideals of Khan Academy. "Free World Class Education For Anyone Anywhere in The World". Every other platform in India is paid and dont even provide the level of Khan Academy. Just try comparing Byjus app and Khan Academy. They are no match. Sure Byjus has more content. But you can't compare qualoty to quantity especially in the case of Education.
Hlo sir! This is Yuvraj this side.. I have a doubt and I came across this when I was learning what thermal velocity and drift velocity are. I'll start in 3..2..1.. My first question is when we say that the average thermal velocity of all the electrons is 0. Do we calculate this at an instant? If yes then, is my way of calculating it correct? since the motion of electrons is random inside a conductor. Thus on an average the number of electrons moving in a particular direction is equal to the number of electrons moving in opposite direction at an instant. Thus, the vector sum of the thermal velocities of all the electrons at an instant is 0..(since for every vector there is some other vector nullifying this.) Now the vector sum of the velocities of all the electrons divided by the number of electrons is the average thermal velocity .. Thus average thermal velocity is also 0.. Since 0 divided the number of electrons is 0.. If we calculate it at an instant then is it calculated the same way I have calculated? One more question if we don't consider a particular instant of time. Then the average thermal velocity of a single electron should also be zero.. Since it goes and comes back to the same spot when there is no electric field?
Love how you are comparing averaging over time for one electron vs averaging over electrons for one instant! Beautiful. Since things are random, by definition, it should be time independent. Therefore, considering properties of a single electron at multiple instants should be the same as considering multiple electrons at a particular instant. So you should get the same answer either ways! Secondly, when we say zero, what we really mean is negligible! It can’t be exactly zero. Kind of like when you toss a billion coins. You get about half a billion heads. It doesn’t mean it would be exact 0.5 billion. There may be a few 100 thousands here and there. But they are negligible. That’s the beauty of statistical processes.
Sir Please bring 10 more concepts which are asked in jee advanced and can be solved logically for next 10 days every day please This is last request please
Sir I study in class 8th I solve Pearson IIT Foundation physics. (My answer is, Reading = 2.83cm and L.C. =0.01cm because least count is smallest measurement made by a vernier callipers it doesn't matters it should be 1M.S.D - 1V.S.D. is is difference of scales of bigger and smaller units) I took about 5 minutes to solve it.
![Why less energy means more stability? [Pure Logic]](http://i.ytimg.com/vi/76n7BmqAeOs/mqdefault.jpg)
2 years of confusion, all over in just 20 mins. thanks a lot. A video explaining screw gauge too, please?
You are the one who inspire me to think and create logics behind any problem.
One day you are going to become the best teacher of India.
he IS the best teacher of india. Don't you think?
He is*
fr
For the final question:
Reading = 2.83
LC = 0.01 (which is the absolute difference in lengths of MSD and VSD)
bro isn't the least count 0.11 cm
Explanation:
10 VSD= 11 MSD
10 VSD = 11 * 1MSD
10VSD = 11* 0.1cm
1 VSD = 0.11 cm
so least count = 0.11 cm
if i am wrong reply me
@@Lama_60 Bro, 1 Vernier Scale Divison(V.S.D) = 0.11cm but Least Count(L.C) = M.S.D - V.S.D
= 0.1 - 0.11
Least Count(L.C) = -0.01
Here the negative answer means that the "first marking in vernier scale" will be 0.01 ahead of "0.1 marking in main scale" So, if we just need the value, the least count will be 0.01
Ya correct
@@Lama_60 1 VSD isnt the least count the difference of 1vsd-1msd is what matters to be the least count
@@Lama_60bro I also made this mistake but figured out instantly that what you have calculated is the vernier spacing from an origin basically one vernier small division but least count doesn't mean that . Least count means the amount of distance the vernier scale is shifted from msd so you have to subtract it from msd which is 0.11-0.1= 0.01.
Man how can you be so good at teaching there is no way today in class i was so perplexed but after watching this vid i realize this is so natural and logical THANKS MAN!!
i finally understood it after 2 years i felt emotional . i am getting ans to last question as 2.83cm and least count 0.01cm
I think in the formula, we get the second decimal digit by counting the place that coincided mod 1 but because the zero error had negative shift, it would be 1- the coinciding mark so everyone got the answer as 0.6 when it was actually 0.4. This makes so much more sense thanks to you!!!
4:20 There is a video by Sal Khan in which he explains how decimal multiplication and divisions work. 😁
Is this suggestion for Mahesh? Sorry Mahesh kidding😅
I know what to do this weekend, thanks! :D
Too bad he doesn't have a video for how to turn on autocorrect and spell it multiplication :P
dude you solved my childhood mystery
2.83 cm and 0.01 thank you for making and giving logic for stuff which nobody givess
LC = 0.01 [because it the minimum length which can be moved in the vernier calliper to take the reading]
& Reading = 2.83
I am just regreting my whole 11th because last year I've seen one videos and it remains in my temporary memory but before 6 months of jee it came like magic really i am smiling alone after understanding this concept I don't understand why it is so underrated thanks sir 😁👀
In the last question, 1VSD>1MSD because 10VSD=11MSD (as given)
Now, 1MSD=0.1cm thus 1VSD=(1.1)(0.1cm) = 0.11cm which is obv greater than 1MSD. Now the least count is 1VSD-1MSD=0.01cm.
If we'll use LC=1MSD-1VSD it'll be obv negative because this takes the assumption that 1VSD is always less than 1MSD which is not the case here.
Now, to the reading it'll be 2.83cm which I calculated as per usual.
I want more videos like this especially for class 11. 👍 Actually i think this is the best way of teaching as you are teaching us to reinvent something which is most important for our youth!😀
thanks for the awesome explanation
these are the answers i got for the last question
1) 2.83 cm
2) 0.01 cm
you clearly made it easy for me, the formula always kept me in confusion, now i am sure i can do any q , now i know logic
first assumption....the 0s of both the scale coincides....now, reading=2.8+x,and 8msd=7vsd+x (8*0.1=7*0.11+x) and therefore x=0.03 and hence reading=2.8+0.3=2.83... now least count is the least distance that you can measure between msd and vsd and least distance between vsd and msd(vsd-msd) will be only in that division which lies *just after* coinciding divisions(the coinciding divisions are 7th division of vsd and 8th divsion of msd) and the division which lies just after that are 8th(of vsd) and 7th(of msd) so as stated above the least distance you can measure is the distance between these division therefore 1vsd-1msd=0.01 cm.....now I know the question here can be how can i subtract msd from vsd(vsd-msd),well I have assumed that the main scale can move...because this type of question is not used in real life(why would you make a scale whose vsd is greater than msd)...so clearly that's why I am allowed to defy this rule :)( at least that's what I think)
Hey bro, your calculation is wrong. 2.8+0.3 is 3.1 and not 2.83..
@@udaypatel9596 he meant 0.03 ig
I used to explain by saying that the number of coincident vernier division shifts by 10 if the gap is 1 mm. so for each coincident vernier division, gap increases by 0.1mm or 0.01 cm. so when coincident vernier division is 6, we need to add 0.06 cm to main scale reading in cm. your explanation however is more complete!
Great perspective :)
your voice is unique and amazing
and the way you teach is awesome
We have 1 hr or 45 mins videos but he completed in less time and more efficiently
Mahesh sir please make a video on surface tension
Finally got what I wanted, the logic, not the formula 👍🏻
Thank u for the wonderful explanation, answers are 2.83 and 0.01
You're doin' a great job
The reading was simple to calculate... 2.83 main scale divisions
But the least count took me a while.
I got it 1.11(bar).Here's how:
I imagined coinciding the 10th division of VS with the 10thb division on the MS. So there is gap in between the first divisions of both the scale which is equal to 10MSD -10VSD. This 10VSD is nothing but (10/11)*10 = 9.99bar. So answer is 10-9.999 =1.111.
Please confirm if it is correct...
Thank you sir
nope, least count is 0.01 cm
please make same kind of video for screw gauge!
Great explanation.❤❤❤❤❤
I LOVE YOU MAHESH SHENOY
bruh my teacher made it soo complicated
My physics teacher taught us the same method and,I forgot it. and had to watch your lecture 😅😂
Now this is satisfying
sir also do this for screw gauge plz
This was so nice sir....
i wonder how this vid is not popular even after mahesh has put jee clickbait.😁😂
You are relatable :)
Great to know that!
I'm kinda confused at the part where you measured the divisions of vernier scale 0.09 cms each.
Isn't the divisons of vernier scale the least count?
which is calculated as LC= 1main scale division - 1 vernier scale division.
like,
in that question, it was given that
10 VSD = 9MSD
ie, 1 VSD = 0.9 MSD
(smallest division of main scale is 0.1 cm)
then 1 VSD= 0.9(0.1cm)
which is, 0.09 cms
BUT, when you calculate least count then,
you gotta do ,
LC= MSD -VSD
ie, LC= 0.1 cm - 0.09cm
where you get , least count = 0.01 cm.
then why did we calculate the divisions as 0.09 x 6 instead of 0.01 x 6?
@AnimeSketcheshere yeah but that's the thing right? that is what I said. the answers coming wrong with that one.
@AnimeSketcheshere nah.
Reading:2.83
LC :0.11
@FloatHeadPhysics
sir why did you choose Khan academy as a large number of platform are present in India🙄
Just
I am curious to know this
I think he is in line with the ideals of Khan Academy. "Free World Class Education For Anyone Anywhere in The World".
Every other platform in India is paid and dont even provide the level of Khan Academy. Just try comparing Byjus app and Khan Academy. They are no match. Sure Byjus has more content. But you can't compare qualoty to quantity especially in the case of Education.
Many things, but most importantly the idea of free education and the work culture!
1)2.83
2)0.01
Sir,Can you please tell me that how do you make your video? and way to make this type of animation?
Yes, will make a few videos on that!
When will second video come ??🥺
Coming out tomorrow!
When he said This is the JEE ART over here 😆😆
Thank you so much sir
Most welcome!!!
Reading 2.83cm
Least count is still 0.01cm
Hlo sir!
This is Yuvraj this side..
I have a doubt and I came across this when I was learning what thermal velocity and drift velocity are.
I'll start in 3..2..1..
My first question is when we say that the average thermal velocity of all the electrons is 0. Do we calculate this at an instant?
If yes then, is my way of calculating it correct?
since the motion of electrons is random inside a conductor. Thus on an average the number of electrons moving in a particular direction is equal to the number of electrons moving in opposite direction at an instant. Thus, the vector sum of the thermal velocities of all the electrons at an instant is 0..(since for every vector there is some other vector nullifying this.) Now the vector sum of the velocities of all the electrons divided by the number of electrons is the average thermal velocity .. Thus average thermal velocity is also 0.. Since 0 divided the number of electrons is 0..
If we calculate it at an instant then is it calculated the same way I have calculated?
One more question if we don't consider a particular instant of time. Then the average thermal velocity of a single electron should also be zero.. Since it goes and comes back to the same spot when there is no electric field?
Love how you are comparing averaging over time for one electron vs averaging over electrons for one instant! Beautiful.
Since things are random, by definition, it should be time independent. Therefore, considering properties of a single electron at multiple instants should be the same as considering multiple electrons at a particular instant.
So you should get the same answer either ways!
Secondly, when we say zero, what we really mean is negligible! It can’t be exactly zero.
Kind of like when you toss a billion coins. You get about half a billion heads. It doesn’t mean it would be exact 0.5 billion. There may be a few 100 thousands here and there. But they are negligible.
That’s the beauty of statistical processes.
Which Device u use for teaching
Answer is 2.93
Sir Please bring 10 more concepts which are asked in jee advanced and can be solved logically for next 10 days every day please
This is last request please
I have to renew my passport :(
@@Mahesh_Shenoy 😅
@@Mahesh_Shenoy sir why did you choose Khan academy as a large number of platform are present in India🙄
Just
I am curious to know this
@@jeeniuspandeyji
First go and read about sal khan ...
Why not screw gauge
Why it's not viral yet?? Best explaination❤
made my life a millionx easier
Gem video
first method is what i used in class but it took time and tr would make fun of me.
Sir I study in class 8th I solve Pearson IIT Foundation physics.
(My answer is, Reading = 2.83cm and L.C. =0.01cm because least count is smallest measurement made by a vernier callipers it doesn't matters it should be 1M.S.D - 1V.S.D. is is difference of scales of bigger and smaller units)
I took about 5 minutes to solve it.
3.15cm
Only 6k view 😢
L.C = 0.03 cm
Reading = 2.83 cm
❤
♥️♥️♥️♥️♥️♥️♥️♥️♦️
Please either you or sal Khan at least apply for Sandeep maheswari show.
Pls make video on secrew gauge also