Deriving Relations by Dimensional Analysis

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take two simple pendulums of the same length. One has a light wooden bob, the other a heavy iron bob. Pull both aside and let them swing. Which one takes longer for one swing, the light one or the heavy one? Most students guess the heavy one, right? Today we will settle it without a single experiment, using only dimensions. And look, in JEE, questions like find a, b and c, or find how one quantity depends on the others, come again and again. So, let's begin.

0:38The idea: a product of powers

So, here is the idea. Suppose a quantity, call it capital Q, depends on a few others, x, y and z, only through a product of powers. Then we write: Q equals k, times x to the power a, y to the power b, z to the power c, okay? Now tell me. What kind of thing is this k in front? Yes, exactly! It is just a number, like two, or two pi. So k is a pure number, with no dimensions at all. Remember this, it will matter in a few minutes. Clear? That leaves only the powers, a, b and c, as unknowns. And how do we find them? By matching dimensions. Remember the principle of homogeneity from this chapter? In a correct physical equation, every term has the same dimensions. So the left side and the right side must match, power by power. That one rule does all the work.

1:38The method, step by step

Now, the method. Step one: write Q equals k, x to the a, y to the b, z to the c, with k a pure number. Step two: write the dimensions of both sides. Every quantity goes into square brackets, in terms of mass, length and time. Step three: equate the powers of M, L and T, one by one. Now think. How many equations will this give us? Three! Exactly. One for mass, one for length, and one for time. And step four: solve those equations for a, b and c, then put the powers back. The relation is ready. Simple, isn't it?

2:23JEE trap: what this method cannot do

Now, the JEE trap. We have three base dimensions, so we get three equations. The double arrow on screen means, which gives. So, what if a quantity depends on four other quantities? Four unknown powers. Still easy, right? No, no, careful! Four unknowns, but only three equations. You cannot solve that, can you? With three base dimensions, we can find at most three unknown powers. Keep this in mind; it comes back in example two. And the second limit. Since k has no dimensions, it never enters these equations at all. So matching dimensions can never tell you its value. That has to come from experiment. Got it?

3:05Topper's trick: solve the easiest equation first

Now a trick that toppers use. You have three equations in three unknowns. Which one should you solve first? The easiest one, of course! So, before solving, look for a base dimension that appears in only one of the quantities, okay? Its equation has just one unknown, so solve it first. Then substitute into the other two, and they fall quickly. With me? Let's see it in action.

3:31Example 1: time period of a simple pendulum

Now, remember our pendulum puzzle? Let's finish it. The period of a simple pendulum, capital T, may depend on its length l, the mass m of the bob, and g, the acceleration due to gravity. Find the dependence. Pause the video, and try it first. Done? Okay, let's check. Step one: T equals k, l to the a, m to the b, g to the c. Three quantities, three unknown powers. Perfect. Step two, the dimensions. Careful: the period is a time, so in square brackets it becomes T, the time dimension. Length gives L to the a, mass gives M to the b, and g, an acceleration, gives L T to the minus two, all to the power c. Now, which equation should we solve first? Remember the topper's trick. Yes, mass! Only the bob has mass. On the right, the power of M is just b, and on the left there is no mass at all. So b equals zero. Next, time. On the right, only g has time in it, giving minus two c. On the left, the power is one. So minus two c equals one, and c is minus half. Last, length. The left side has no length, so a plus c equals zero, and a is plus half. Nice and smooth. Put the powers back: T equals k, times the square root of l over g. The power of mass is zero, so the period does not depend on the mass at all! See, puzzle solved: the light bob and the heavy bob take the same time per swing. And k? Dimensions cannot give it; experiment gives two pi.

5:22Example 2: Poiseuille's formula for flow through a tube

Now a harder one, JEE level. The rate of flow capital Q, that is, the volume per second, of a liquid through a narrow tube depends on the viscosity eta, the radius r, and the pressure gradient, P over l. Eta is the Greek letter that looks like an n with a long tail, okay? Find the relation. Pause and try. Hint: count your unknowns before you start. Alright, step one. Many students write eta to the a, r to the b, P to the c, and l to the power d, okay? Looks fine, right? No, no, careful! Count the unknowns: four of them, and only three equations. You will get stuck. The question itself gives P over l as one quantity, so treat it as one variable. Its dimensions are M, L to the minus two, T to the minus two. Now there are just three unknowns, a, b and c. And now it is solvable. Step two, both sides. On the left, volume per second: L cubed, T to the minus one. On the right, viscosity is M, L to the minus one, T to the minus one, to the power a, then L to the b, then M, L to the minus two, T to the minus two, to the power c. Now equate. Mass: a plus c equals zero. Time: minus a, minus two c, equals minus one. Two equations, two unknowns. Can you find a and c? Let's see. From the first, a is minus c. Put that in the second: c minus two c is minus c, and that equals minus one. So c is one, and a is minus one. Last, length: minus a, plus b, minus two c, equals three. Put in the values: one, plus b, minus two, equals three. So b is four! The flow goes as the fourth power of the radius. So Q equals k, P r to the four, over eta l. This is Poiseuille's formula, where k is pi by eight; again, not from dimensions. See? Treating P over l as one quantity turned a stuck problem into an easy one. Clear?

7:46Quick revision: product of powers, at most three unknowns

So, quick revision. Write the quantity as k times a product of powers, and remember, dimensions never give the number k. Equate the powers of mass, length and time: three equations, so at most three unknowns. And when the question gives a combination like P over l, treat it as one variable. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!