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Four identical solid spheres of mass 'm' and radius 'a' are placed as shown Find the moment of inertia of the system through A and axis parallel to DB Since the size of sphere is small it can be considered as a point object and moment of inertia about the given line can be calculated as follows. DE and BF can be calculated from respective triangles and CA and EF are the diagonals of the square. A is the mid point of EF. `I=M\timesK^2` `I=m\times 0^2 + m\times BF^2 + m\times CA^2 + m\times DE^2` =`m\timesl^2/2+m\times(sqrt(2) l)^2+m\times l^2/2` =`3ml^2`
Order of Magnitude Large and small numbers are written as power of ten in scientific notation. If the number is more than five then it's order of magnitude is power of ten plus one If the number is less than five the power of ten is it's magnitude. Example 1)Find the order of magnitude of 4.62 `\times10^3` Here the number 4.62 is less than five `\therefore` the order of magnitude is 3 2)Find the order of magnitude of 5.32 `\times10^6` Here the number 5.32 is more than five `\therefore` the order of magnitude is 6+1=7 Sai's Classes
Sai's Classes helping the students who are aiming IIT with a series of Physics problems solving through explanation in details The following video surely help those JEE preparing students . More videos will be uploaded; to get an update subscribe the You Tube channel.
Question 4 : A uniform cylinder of mass 'M' and radius 'R' is to be pulled over a step of height 'a' (a<R) by a force 'F'at it's centre 'O' perpendicular to the plane through the axis of the cylinder on the edge of the step as shown in figure. The minimum value of 'F' required is Answer: Apply the principle of moment about the edge of the step C Sum of clockwise moment=Sum of anticlockwise moment Just before rolling `F× R=Mg×d`---------------(1) (where d=CB from figure) Applying Pythagores theorem in 𝚫OBC `R^2=(R-a)^2+d^2` `d=sqrt(R^2-(R-a)^2)` substitute for d in (1) `F=(Mg/R)×R×sqrt(1-((R-a)/R)^2)` `F=Mg×sqrt(1-((R-a)/R)^2)` Watch Video with explanation
A bead of mass m stays at point P(a,b) on a wire bent in the shape of a parabola y=4Cx 2 and rotating with angular speed 𝝎.Find the value of 𝝎. Solution: Since the path of the bead is a parabola it's velocity at P can be calculated using `V=sqrt(2gh)`(in a projectile motion vertical motion is equivalent to free fall) here h=b `therefore V=sqrt(2gb)`--------------(1) since y=`4Cx^2` here x=a and y=b `therefore``b=4Ca^2` substituting in (1) V=`sqrt(2g4ca^2)` V=`2a sqrt(2gc)` 𝝎=V/r here r=a 𝝎=`2sqrt(2gc)` (Answer)
Dear students, As everyone is saying "start from basics".What is basics?. Learning new concepts from experts in subjects is the normal procedure .Very important is learning from experienced teachers who are experienced to hear the students. Expertise is important but hearing the student achieved through patience,experience and contentment counts the most Sense the issues of individual students and try to solve their problems, solving their subject problems are secondary. Teaching a mass of student with a mass of expert teachers is one method of teaching; it has some advantage also But Sai's classes have a different way! My students are still keeping a good relation even after reaching prestigious position declares that The bond between teacher and student is not bound within the classroom or till the class time It lasts for a life time.This is the basics Sai's class is believing. Regards Sai Sundar S Sai's Classes
Dear students Sai's classes started a series of lectures on Electrostatics for Plus2 students The first part is an introduction ; explaining charges and its properties. Explanations are made simple to get the feel of the actual class Click the link below to see the video lecture Electrostatics Lecture Part1
Ethylenediaminetetraacetate is a hexadentate ligand it can attach with the central atom through two Nitrogen atom and through 4 oxygen atom forming coordination compounds as shown.
The natural resources are for the future generation too. We are trying our level best to give our children good education, acquiring assets for them but how many of us are saving resources for them. A tank full of money will not be sufficient to buy a drop of water in the near future; now a small sum is enough to buy a tank full of water. If we have will we can have a well full of water with a very small investment so use the natural resources judiciously. Water crisis is growing day by day and people in the world are discussing about the issue and trying to find some solutions to it but early action is very important now we have so many methods like making rain water pits , roof top water collecting techniques (as shown in photo) But as time elapses the rain will be scanty a stitch in time is the need of this time. By water cycling and recycling alone can bring the wonder liquid back to earth So it is the time to go hand in hand to start our ...