Set B: Dimensional Analysis and Significant Figures

Transcript

0:00Introduction

Hello everyone! First, one small puzzle before we start. I make a new system of units: mass in units of one hundred grams, length in units of ten centimetres, and time in units of zero point one second. Look, every unit is ten times smaller than in S I, okay? So one newton should become a much bigger number here, right? But the answer is exactly one! Wait, how? We will crack it today, in our mixed problem class, Set B: constants in formulae, relations, significant figures and unit conversion, all JEE Main favourites. So, let's begin.

0:41Constants in a formula: work inside out

So, first tool: the dimensions of constants inside a formula. Here is a displacement: y equals small a, sine of, b t minus c x, plus d t squared, okay? We want the dimensions of a d, divided by b c. Four unknown constants. So where do you start? From the inside, yes! Inside out, always. Inside the sine, everything is a pure number. So b t has no dimensions: b is T to the power minus one. Same for c x: c is L to the power minus one. The square brackets mean, the dimensions of, okay? Now the outside. The sine is a pure number, so small a carries the displacement: a length. And d t squared is added to a displacement, so it is a length too. Divide by T squared: d is L T to the power minus two. The double arrow means, so. Clear? Now combine. Top, a into d: L squared T to the power minus two. Bottom, b into c: T to the power minus one, L to the power minus one. So the answer is L T to the power minus three, right? No, no, careful! That comes from multiplying by the bottom, not dividing. Dividing by a negative power means multiplying by a positive one, right? So L to the power two plus one, three, and T to the power minus two plus one, minus one. L cube, T to the power minus one. Nice and clean.

2:17Find a relation: match the powers

Second tool: finding a relation. The air drag on a car, F, depends on the air density rho, this Greek letter like a curly p, the speed v, and the frontal area, capital A, okay? Find the relation. So write the force as rho to the power small a, v to the power small b, and capital A to the power small c, in dimensions. Force, MLT to the power minus two. Density, M L to the power minus three. Speed, L T to the power minus one. And area, L squared. Three unknowns now. Which equation do you solve first? The easy ones, exactly! Mass appears only in density, and time only in speed. Match M: small a equals one. Match T: minus b equals minus two, so small b equals two. That's it! Now match L, with these values. Minus three, plus two, plus two c, equals one. So two c is two, and small c is one. So the drag force is proportional to rho, capital A, v squared. The open loop sign means, proportional to. Topper's habit: solve the easiest equations first. Got it?

3:34Significant figures: the weakest number decides

Third tool: significant figures, with a second puzzle. Add two point zero into ten to the power four metres, and three point one two into ten to the power two metres. The calculator says twenty thousand three hundred and twelve metres. So the answer is twenty thousand three hundred and twelve, all five digits, right? No, no, careful! This is an addition. For addition, we do not just count the digits; we look at how far each number is really known. With me? Two point zero into ten to the power four is twenty thousand, but known only to the thousands place, okay? Its hundreds, tens and units were never measured. So the sum is also known only to the thousands place: two point zero into ten to the power four metres. We added three hundred and twelve metres, and nothing changed! Clear? Alright, one more. Five point four three minus five point four zero, divided by zero point zero two five. The calculator gives one point two. Zero point zero two five has two significant figures, so keep two, right? Hmm, wait. Look at the top first. The top is zero point zero three: only one significant figure. Subtracting nearly equal numbers destroys significant figures, remember? So the answer keeps one figure: simply one. See, subtract first, then count.

5:04Example 1: 1 newton in units of 100 g, 10 cm and 0.1 s

Now, Example one, and it is our opening puzzle. In a system whose units are one hundred grams, ten centimetres and zero point one second, what is the numerical value of a force of one newton? Pause the video, and try it first. Done? Okay, let's check. The conversion formula: n two equals n one, into M one by M two to the power small a, and the same for L and T, with powers small b and small c. System one is S I, system two is the new one. Old over new, always. Force is MLT to the power minus two. So small a is one, small b is one, and small c is minus two. Now each ratio, with the same unit on top and bottom. One thousand grams over one hundred grams. One hundred centimetres over ten centimetres. And one second over zero point one second, to the power minus two. Look at that last bracket. One by zero point one is ten, and the power is minus two. So what is it? One by one hundred, yes! Zero point zero one. So, ten into ten into zero point zero one. Exactly one! Puzzle solved. The smaller mass and length units push the number up a hundred times, and the smaller time unit, with its power minus two, pulls it down a hundred times. Beautiful.

6:33Example 2: value of G in tonne, km and minute

Now a harder one, JEE level. The gravitational constant, capital G, is six point six seven into ten to the power minus eleven, in S I units. Find its value in a system with units of one tonne, one kilometre, and one minute. Pause and try. Hint: dimensions of G first. Done? Okay, let's check. The dimensions of G are M to the power minus one, L cube, T to the power minus two. So small a is minus one, small b is three, and small c is minus two. Old over new. One kilogram over one tonne, that is one thousand kilograms: ten to the power minus three. One metre over one kilometre: ten to the power minus three again. One second over one minute: one by sixty. So n two is six point six seven into ten to the power minus eleven, into each ratio, raised to its own power. Now the mass ratio is ten to the power minus three, tiny, so it makes the answer smaller, right? No, no, careful! JEE trap. The power is minus one, so the fraction flips: ten to the power three, and it multiplies. So: ten to the power three, into ten to the power minus nine from the length cubed, into sixty squared, three thousand six hundred. That is three point six into ten to the power minus three. Finally, multiply by six point six seven into ten to the power minus eleven. About two point four into ten to the power minus thirteen. The wavy equals sign means, approximately. See how fast that was?

8:18Quick revision: constants, relations, conversions, sig figs

So, quick revision. One: for constants, work inside out, and added terms have the same dimensions. Two: for relations, match the powers; for conversions, always old over new. Three: in significant figures, the least precise number decides. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!