Boyle Law

Boyle Law

Articles, Blog , , , , , , , , 45 Comments


Have you ever traveled by road to a hilly area- that is to a higher altitude place?
Did you carry chips packets with you? Have you observed that if you go to higher altitudes, the chips packets start to expand and
at times they even burst? Well, why does that happen? Think about it. So we have a person. He’s running.
So let’s count the number of times the person hits the boundaries in one minute. One, two, three, four, five, six, seven, eight. So in one minute the postman touches the
boundaries eight times. Now if the person runs at the
same speed- that is the speed of the person remains
constant, and he has to cover double the distance in
the same time- that is in one minute only the person
now has to cover double the distance and he’s running at the same speed.
How many times will he touch the boundaries? Let’s count. One, two, three four. So now if he has to cover double the
distance, he touches the boundaries half number of times.
So let’s compare this. The person in both the cases is running
at the same speed. In one minute he touches the boundaries eight number of times and if he has to cover
double the distance, the number of times he touches the boundaries
is reduced to half. So by the time he covers one distance
he has already covered double the distance. So you see that the distance covered is related to
the number of hits per unit time. If the distance covered is less-
that is if he had to cover lesser distance the number of hits per unit time
increases and if the distance covered, that he had
to cover is more the number of hits per unit time decreases
if the speed of the person remains constant. Similarly we have distance covered is
inversely proportional to the number of hits per unit time. This means greater is the number of hits per unit time,
lesser is the distance covered. So this also is the case when the speed remains constant. In physics you must have studied that
pressure at the same height remains the same. So if an experiment is
performed- in this we have different shaped tubes. Now if a liquid is poured in these
tubes, all the tubes are different in shape, so it is observed that they take the same
heights irrespective of their shapes. The height in all the tubes is the same,
so by this we know that the pressure at the same height remains the same.
This concept was used by a scientist named Robert Boyle. He performed an experiment which was known as Boyle’s experiment.
He performed the experiment in 1662. Let’s see what he performed.
He took a J-shaped tube. A gas is enclosed in this tube and this is the scale that he used to measure the
volume of the gas enclosed. Now he added mercury in the tube. So let’s see what happens. As he added mercury, the volume of the gas enclosed reduces. So now, the volume enclosed is 60cm³. This is the pressure exerted by gas and this is the atmospheric pressure.
We know that pressure at same heights remains the same, so at this point this
is the pressure of gas, at this point the pressure exerted is
the atmospheric pressure, which is 760mm of mercury. This remaining same, he added more mercury so as to make the volume of the
gas enclosed half. So when he added more mercury he saw that the
volume reduced to half. Now it occupies 30cm³. This is the pressure exerted by gas
and when he calculated the pressure on the other side, he observed that the pressure was the
atmospheric pressure, plus 760mm of mercury. So now the
pressure was two times 760mm of mercury.
Now this set-up remaining the same, he added more mercury so as to make the
volume of the gas enclosed one by third. So when he added more mercury, the volume
becomes one-third, that is 20cm³. This is the
exerted by gas. Pressure at same height remains the same,
so on the other side when he calculated- the pressure was the pressure of
atmosphere plus 1520mm of mercury. So the
overall pressure is 3 times 760mm of mercury.
So he observed as the pressure becomes three times,
the volume becomes one-third. So he concluded that if the pressure was
doubled at constant temperature- all the
calculations the entire experiment which Boyle had performed was at a constant temperature, and he was able to observe that when the pressure was doubled the volume
of the gas became half- half the original volume, and as the pressure was tripled the volume became one-third the original
volume. So as the pressure was doubled volume became half, as the pressure was tripled volume became one-third. So based on this, as we’ve seen already from this
experiment- that is a gas enclosed in a container. In nine seconds there are nine hits.
So there’s one hit per second. If the volume is reduced to half, so in the same time that is in nine seconds
only there are 18 hits. As the volume is halved the pressure doubles. So based on this experiment, Boyle gave the law. His law states that the volume is inversely proportional to pressure,
temperature remaining constant. This is something that we have seen
through the person running. So there we had the
distance covered is inversely proportional to the number
of hits per unit time, the speed remaining constant.
So, in Boyle’s law the distance covered is the volume
occupied by the gas, the number of hits per unit time is
the pressure of the gas and speed is the temperature.
So according to Boyle’s law, the volume of the gas is inversely proportional to pressure
at temperature remaining constant. So the Boyle’s law states that the volume of a
given mass of gas, so the law holds for a particular gas,
so for a given mass of gas the volume is inversely proportional,
so you have to keep in mind that the volume is inversely proportional to pressure at
constant temperature. So the law holds only at a constant
temperature for a particular gas. So from the law we have that the volume is
inversely proportional to pressure temperature remaining constant for a
particular gas. So to remove the proportionality sign we
introduce a constant, so we get volume is equal to
constant divided by pressure. Now if we take pressure on the other side,
we get pressure into volume is equal to constant. So the product of pressure and volume is
a constant value and this is the Boyle’s law.
So for a particular gas the product of pressure and volume at a constant temperature is constant. So let’s see a small experiment. So a gas is enclosed in a container. Now, we will increase the pressure which
can also be seen through this pressure gauge. We increase the pressure from top. So as the pressure is increased the volume decreases. So you see the values- as the pressure is increasing the volume which is enclosed in the container is reducing. And you see that the product of pressure
and volume is constant. So as the pressure increases volume decreases but the product of pressure and volume is always constant. So from this we can also plot the values. So if you draw a graph between the pressure and
volume of a gas we get a curve which shows that as volume increases the pressure decreases.
We know that according to Boyle’s law pressure is inversely related to volume, or we can also say that volume is inversely related to pressure. So as the volume increases the pressure decreases. Now let’s say if this 1/V is x so we get that pressure is directly proportional to 1/V
which is x. So we get pressure directly proportional to x and over here x is 1/V. So pressure is inversely proportional to volume and pressure is directly proportional to 1/V. So if we plot a graph between 1/V and pressure we get a straight line. So you see
that as one by volume increases, the pressure increases. So pressure is directly proportional
to one by volume. Since these graphs are plotted at constant temperature, so these are also known as isotherms. ‘Iso’ means same, ‘therm’ means for temperature. Since Boyle’s laws is valid for a gas at constant temperature so
these graphs are known as isotherms. That means the temperature remains constant. The pressure of a gas increases with the
increase in volume. Is it true or false? So from Boyle’s law we know that pressure is inversely related
to volume. So as volume increases, pressure decreases. So with the increase in volume, pressure decreases. So the pressure of a gas decreases with the increase in volume.
So the statement that we have is false. It does not increase, it decreases with the increase in volume. From Boyle’s law we saw that PV- that is the pressure into volume
of a gas is constant. Let us denote this constant by k
and we know this is valid for a particular gas at a constant temperature. So if we have the initial conditions- say at
some initial pressure and at initial volume, the product comes out to be k.
The final pressure of the gas is P? and the final volume is V?.

190
00:12:20,560 –>00:12:23,560
Again the product is constant which is k. So we get that P? into V? is
equal to P? into V?. So initial pressure into initial volume is equal to
final pressure into final volume which is k and which is a constant.
So from this we get that these two quantities are same. So P?V? is equal to P?V?.
That is the initial pressure of a gas, for a
particular gas and the initial volume of a particular
gas is equal to the final pressure of a particular gas into the final volume of that gas and temperature remains constant. Now let’s explain the Boyle’s law in terms of kinetic theory. So we know, we’ve seen that pressure
and volume are inversely related. So what is happening here is,
as external pressure increases the internal pressure has to increase.
This means the number of hits per unit time has to
increase and this is possible only when the volume is decreased. As the volume is decreased the number of
hits per unit time increases and the pressure increases.
Let’s recall- as outside pressure increases to
balance it the inside pressure has to increase. This means the number of hits per unit time has to increase. For the
same number of particles, this is possible only when the volume
decreases. When the volume decreases the number of
hits per unit increases, so we get as the pressure increases the volume decreases. Similarly the vice-versa case- if we decrease the external pressure, to
balance it the internal pressure has to decrease.
This means number of hits per unit time has to decrease. Now again we have the same number of molecules so to reduce the number of hits per unit time
they should travel a great distance. This is possible only when the
volume increases as the number of particles remain the same
so to get number of hits per unit time lesser, they should travel a great distance. This means the volume increases. So as pressure decreases the volume
increases. So now, did you find an answer to why the chips
packets burst when they are taken to higher altitudes? Well okay, so this follows the Boyle’s law.
When we go to a higher altitude, the outside pressure is less so to balance it the pressure inside the chips packet
should also be reduced. So now the inside pressure has to be less. How is that possible? By Boyle’s law- if the inside pressure is less
the volume increases, and if the volume increases- the
chips packet expand in size and at times they expand to such an
extent that they burst. So why do the chips packets burst
when taken to higher altitudes? At higher altitudes, the outer- that is the outside pressure is less so to balance it the
inside pressure- that is the pressure inside the chips packets that has to decrease. So by Boyle’s law
if the inside pressure that is the pressure inside the
chips packets is less the volume has to increase. As the volume increases the chips
packets expand in size and at times they expand to such an extent
that they burst. So this is the reason why the chips packets
burst when they are taken to higher altitudes. You use Boyle’s law even when you are breathing.
So while breathing there is a relationship between the volume and the pressure. While breathing in we lower the diaphragm. When we lower the diaphragm the volume
inside increases. So by Boyle’s law as the volume inside increases the pressure decreases. So the inside pressure is less and so outside air flows in. So this is how we breathe in. So we lower the diaphragm, the volume
increase, pressure decreases so to balance it the outside air from outside it moves in. This is how we breathe in. So, while breathing out what happens? We push the diaphragm up. As push the diaphragm up, the volume inside decreases, so
by Boyle’s law as the volume decreases the pressure increases, so to balance it the air inside flows outside. Since the inside pressure is more so air from inside flows outside and
this is how we breathe out. So we even use Boyle’s law during breathing. Now let’s try to solve a question.
At a constant temperature, the volume of a gas was found to be 400 cm³ at a pressure of 760 mm of mercury. If the pressure of the gas is
increased by 25 percent what should be the new volume? Okay, now let’s write down the data that we’re provided. So we have the initial pressure which is 760mm of mercury and the final pressure we know that the pressure is
increased by 25 percent this means that if this is the initial pressure then we increase the pressure by 25 percent. So what do we get? We get 125 P? by 100. So if we divide this by 25 we get 5 P? 25 fives are 125 and 25 fours are 100. So this is the final pressure. We need not write this for 760 mm of mercury because this will make the calculations
lengthier. Now we are given that the initial
volume is 400 cm³ and we have to find out the final volume. So now we know that the Boyle’s law says that the initial pressure of the gas into initial volume is equal to final pressure into final volume. So if we solve this- so let’s try to solve this. Initial pressure is P?, initial volume is 400. Final pressure- let’s write it in terms of P?. We have 5P? by 4 and we have to find out the final volume,
which is V?. So if we calculate V? from this, we can cancel out P? from both the sides. So we get V? is equal to 400 into four divided by five. This we can cancel out again. So we get 320. 8 fours are 32, so we get 320 cm³. Had we substituted the 760 instead of P? we would have just increased the calculations so we let the values remain in terms of P?. So what do you observe here?
As the pressure increases so the final pressure has increased-
so the final volume decreases.

45 thoughts on “Boyle Law

  • Amudhu students counceler Post author

    easy to understand

  • Niranjan Basu Post author

    excellent explanation

  • MD N I Sumon Post author

    You are best

  • Sourav Kumar Post author

    Wow …great idea of learning

  • Sourav Kumar Post author

    Great idea of teaching ….. thanks

  • Sagar Kumar saf Post author

    Wonderful way of teaching

  • Ntsh pwr Post author

    Very Good explanation…..

  • Sar Gam Post author

    aromatic hydrocarbons PR video banana plzzzz

  • mangalam bahavathsingh Post author

    Excellent

  • deepak kumar Post author

    osm…way of teaching

  • PRAVEEN RAWAT Post author

    Mam thank u so much for making it easy for. Me….

  • Mohammad Almasrouhy Post author

    What a talented girl I had never seen such befor I hope a marvellous future for her ; in fact super teacher & briliant scientistπŸŒ…πŸ’πŸŽ“

  • TAJAMUL ISLAM Post author

    brilliant video

  • Mahmoud Teaalab Post author

    i need this power point presentation . plz
    [email protected]

  • Girish Mahajan Post author

    Very Very Interesting…

  • Abhishek Kumar Post author

    Mst

  • Jagadeesh Jagadeesh Post author

    super madam

  • Satish Birajadar Post author

    mam thank you so much

  • Atray expert Post author

    Nice way of teaching mam

  • Rakhi madhukar Post author

    Incredible way of teaching
    Thnku so much mamπŸ€—

  • Rosy T Post author

    Nice video thq so much

  • rajashree Anand Post author

    I am just seventh standard
    i am very curious in learning more than my subjects. it's very easy. thanks for your seminar.

  • kamini mishra Post author

    I like very much

  • ANCHAL RATHORE Post author

    Please dharmodynemic pe video bnana please so easy smjh aa jaye jaise aapne ye smjhaya.. please

  • Rupali Warke Post author

    Ur just amazing

  • suneetkamaljeet randhawa Post author

    Oooo waooooo πŸ™‹ I understood…. very easily… thnku so very much….

  • Siddharth Kumar Post author

    Best explanation ever !!!!!!!

  • Anju Chakrawarti Post author

    Very helpful video

  • laa Ee Post author

    Awesome 😎

  • Sunagar Pushpa Post author

    very good teaching
    please upload low temperature in real and rare gages

  • Sunagar Pushpa Post author

    in physics plz

  • Prince Yadav Post author

    Wow ma'am best explanation…
    First of all thank you for that and your voice sweet and your voice modulation is also very nice……

  • Mubarak Beri Post author

    It's very useful lesson and your explanation is pretty clear!

  • raman k Post author

    Oh my madam tremendous unimaginable examples wonderful .it's really outstanding.give Ur mail id I vl ask some doubts in gaseous state

  • raman k Post author

    πŸ™πŸ‘πŸ€

  • Tripti Singh Tripti Singh Post author

    I hv nt taken science as a stream….nd nw preparing fr UPSC…Ur videos help me a lot to clear my concept…πŸ˜„πŸ˜„πŸ˜„

  • chandra joshna Post author

    I want πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”πŸ€”a teacher like this to our school

  • muhammad amin Post author

    Owesome explaination

  • Tushar sood Post author

    Great! well done madam…your work is osm.😊😊

  • satvik gupta Post author

    πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“

  • satvik gupta Post author

    πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“πŸŽ“v
    πŸŽ“πŸŽ“πŸŽ“πŸŽ“

  • Hemanth Kumar Singampalli Post author

    Nice explanation in this video
    I liked it

  • Uzumaki Naruto Post author

    Wow.i love this channel very very much.thanks for making lessons very easy for us.and keep on ur good worksπŸ‘πŸ‘πŸ‘

  • Manoj Gupta Post author

    In the last equation how did 25 became 125 pls explain

  • Hasina Pathan Post author

    Very well explained πŸ˜†

Leave a Reply

Your email address will not be published. Required fields are marked *