Errors in Standard Experiments

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two students do the same experiment: same voltmeter, same ammeter, same readings. They find a resistance, and then the percentage error in the heat it produces. The first student says eleven percent. The second says seven percent. Same readings, two answers? Hmm, somebody has made a mistake, right? But who? Today we learn one simple procedure for errors in standard experiments, and you will catch that mistake yourself. And look, JEE loves asking the maximum error in these experiments. So, let's begin.

0:40Four steps for any experiment

So, in JEE you will meet many standard experiments: the simple pendulum, Young's modulus, specific heat, resistance. Different apparatus, different formulas. But tell me, what question do they all end with? Yes, exactly! What is the maximum percentage error in the result? And one procedure, four steps, answers it every time. Step one. Write the formula in the quantities you actually measure: what you read on the scale, the stopwatch, the meter. Not something you calculated on the way, okay? Remember this; our puzzle is hiding right here. Step two. Take logarithms, and differentiate. Products and divisions become sums and differences, and a power comes down in front as a multiplier. Then make every term positive, because we want the maximum error; errors are never allowed to cancel. Clear? Step three. For each measured quantity, the error is the least count of the instrument you used. Or, if you have taken many readings, the mean absolute error. Step four. Add the relative errors; times one hundred gives the percentage. And if the question asks for the absolute error, multiply the relative error by the value itself. That's it, four steps. Got it?

2:02See it work: the simple pendulum

Now, let's see these steps work on the simple pendulum. g equals four pi squared l, over T squared, where l is the length of the pendulum and T is its period, okay? Now step one. Tell me, what do we actually measure in the lab? Think, which of these do you read off an instrument? Correct! The length l, with a scale, and the period T, with a stopwatch. Four pi squared is just a number; it has no error. Clear? Step two: take log on both sides. The l n you see on screen is the natural log, okay? Log g equals log of four pi squared, plus log l, minus two log T, okay? See, the square on T has come down as a two in front. Now differentiate. d g over g equals d l over l, minus two d T over T, okay? See the minus sign? So the error in T reduces the error in g, right? Hmm, think once more. No, no, careful! An error can be on either side, and we want the worst case. So every term is made positive: delta g over g equals delta l over l, plus two delta T over T, okay? This delta, the little triangle on screen, means the error. And the two in front: the error in T counts double. In the lab, you time n oscillations together, and T is t over n. Since n is just a count, T has the same relative error as the total time, right?

3:38JEE trap: a length found as a difference

Alright, now a trap that catches many students. Often you read the scale twice, at the start and at the end, and subtract: l equals x two minus x one, okay? Now tell me. Each reading has its own error. When you subtract the readings, do you subtract the errors too? Most students say yes. Let's see. No! The errors add. Each reading can be off either way, so in the worst case, both errors push the difference the same way. So delta l equals delta x one plus delta x two: the error in the difference is the sum of the two reading errors. Remember this; it comes back in Example one.

4:20Topper's trick: find the biggest term

Now let me share a trick that toppers use. After step two, look at the powers. If a quantity is raised to the power n, its relative error gets multiplied by n. Squared means double, cubed means triple. Then, after adding the terms, ask: which term is the biggest? That measurement decides the error in your result, so improve it first. Examiners love asking exactly this. In the pendulum, the period comes in squared. So if the timing gives the biggest term, how do you improve it? Yes! Time more oscillations, or use a better watch. Simple as that.

4:59Example 1: percentage error in specific heat

Let's do Example one. The specific heat of a metal is c equals Q over m delta theta. The temperature rises from twenty point zero, plus or minus zero point one degree Celsius, to thirty point zero, plus or minus zero point one degree Celsius. The mass is two hundred, plus or minus one, grams, and Q is exact. Find the percentage error in c, okay? Pause the video, and try it first. Done? Okay, let's check. We measure the mass and two thermometer readings. And careful, here delta theta is not an error. It is the temperature rise itself: thirty point zero minus twenty point zero, ten point zero degrees. Now the error in this rise. Each reading has an error of zero point one degree, so the error in the rise is zero point one degree, right? No, no! Same trap we just saw. The rise is a difference of two readings, so the two errors add: zero point one plus zero point one, zero point two degree. Relative error: zero point two over ten point zero, that is two percent. Now the mass: one gram in two hundred grams, zero point five percent. And Q is exact, so it adds nothing. Step four, add them up: two percent plus zero point five percent, two point five percent. And which term is the biggest? The temperature difference, of course. Improve that first: the topper's trick at work!

6:32Example 2: error in resistance and heat produced

Now a harder one, JEE level, and this is our puzzle from the start. R equals V over I, with V equal to one hundred, plus or minus five, volts, and I equal to ten, plus or minus zero point two, amperes. Find R, with its error. Then find the percentage error in the heat, H equals I squared R t, produced in a time t that is known exactly. Pause and try. Hint: remember step one! Done? Okay, let's check. R is one hundred over ten: ten ohms. The omega on screen is the symbol for ohm. R is V divided by I, so the relative errors add. Five over one hundred is five percent, and zero point two over ten is two percent. Total, seven percent. Step four, the absolute error: seven percent of ten ohms is zero point seven ohm. So R equals ten point zero, plus or minus zero point seven, ohms. Now the heat. H equals I squared R t, and t is exact. Easy, isn't it? I squared gives two times two percent, that is four, and R gives seven. Eleven percent, right? Hmm. That is exactly what our first student said. Wrong! Go back to step one: the quantities you actually measure. Did we measure R? No! We calculated it, as V over I, from our readings. So put V over I in place of R: H equals I squared, times V over I, times t, which is just V times I times t. Now the errors: five percent from V, two percent from I, and nothing from t, because it is exact. Seven percent! So the second student was right. Puzzle solved. And why was eleven wrong? R already carries the error of I inside it, so I squared R counts the error in I twice. Got it? Always write the result in the directly measured quantities first.

8:45Quick revision: four steps for any experiment

So, quick revision. Write the formula in what you actually measure, never in something you calculated. Take logs, differentiate, and make every term positive; a power multiplies that error. For a difference of two readings, add the two errors. Now go and try the practice problems on this page. Four steps, and you will find them easy. All the best!