Set C: Errors and Instruments

Transcript

0:00Introduction

Hello everyone! So, one small puzzle before we start. You have to measure the diameter of a thin wire, about zero point four zero millimetre, and your percentage error must stay below two percent, okay? On the table there are three instruments: vernier callipers with least count zero point one millimetre, a screw gauge with least count zero point zero one millimetre, and a finer screw gauge with least count zero point zero zero five millimetre. Which one will you pick? Most students say, the first screw gauge, it is precise enough, right? Hmm. We will check that today. This is our mixed problem class on errors and instruments: the power rule, zero errors, and the traps JEE loves to set. So, let's begin.

0:48Tool 1: in a formula, errors always add

So, first tool, a quick recall. If Z equals A to the power p, times B to the power q, divided by C to the power r, then delta Z by Z equals p times delta A by A, plus q times delta B by B, plus r times delta C by C, okay? Delta, the little triangle on screen, means the error, so delta A by A is the relative error. Each power multiplies its own error. Clear? Now, the first problem of our set. Z equals A squared, times root B, divided by C cubed. The errors are one percent in A, two percent in B, and half a percent in C, okay? Now tell me: C cubed sits in the denominator. So do we subtract its error? No, no, careful! This is the JEE trap. Errors always add, even for a quotient, because we want the maximum possible error. Write this down. So, A squared gives two times one percent. Root B is B to the power half, so half of two percent. And C cubed gives three times zero point five percent. Two, plus one, plus one point five: four point five percent. Nice and simple. One more, quickly. The speed of a body is found from its kinetic energy, with a six percent error, and its mass, with a two percent error. Speed v equals the root of two K by m, so everything inside gets the power half. Half of six plus two: four percent. And the two inside? It is an exact number, so it carries no error. Got it?

2:27Tool 2: zero error of vernier callipers

Now, second tool: zero error. Close the jaws of a vernier callipers. The vernier zero should sit exactly on the main-scale zero. If it does not, the instrument has a zero error, okay? And the rule is: corrected reading equals observed reading, minus zero error, with its sign. If the vernier zero lies to the right of the main zero, every reading is too large: a positive zero error. If vernier line number k coincides, the zero error is plus k times the least count. L C on screen means least count. With me? Now the tricky case, from our set. Callipers with twenty vernier divisions, and least count zero point zero five millimetre. With the jaws closed, the vernier zero is to the left of the main zero, and the fifteenth line coincides. So the zero error is minus fifteen times zero point zero five, right? No, no! This is where students lose marks. When the vernier zero is behind, count how far behind it is: twenty minus fifteen, that is five divisions. So the zero error is minus five times zero point zero five: minus zero point two five millimetre. Now a rod reads eighteen millimetre on the main scale, and vernier line eight coincides. What is the observed reading? Yes, eighteen plus eight times zero point zero five, that is eighteen point four zero millimetre. But this is not the answer yet, okay? True length: eighteen point four zero, minus minus zero point two five. Minus of a minus becomes plus: eighteen point six five millimetre. Makes sense, right? With a negative zero error every reading is too small, so the true length is bigger. Exactly what we expect.

4:18Example 1: choosing an instrument for a thin wire

Alright, example one, and it is our opening puzzle. The wire is about zero point four zero millimetre thick. Which instrument keeps the percentage error below two percent: the callipers of zero point one millimetre, the screw gauge of zero point zero one, or the screw gauge of zero point zero zero five? Pause the video, and try it first. Done? Okay, let's check. For a single reading, we take the least count of the instrument as the error. So the percentage error equals the least count, divided by the reading, times one hundred. Callipers first: zero point one by zero point four zero. Twenty five percent! Imagine that, a quarter of the answer is uncertain. Callipers are out. Now the screw gauge of zero point zero one millimetre. Most of you picked this one, right? Is it good enough? Zero point zero one by zero point four zero: two point five percent. More than two! So it fails, just barely. That was the trap in our puzzle. And the finer gauge: zero point zero zero five by zero point four zero, one point two five percent. Below two percent! So only the finest screw gauge does the job. See, the reading sits in the denominator, so for a thin wire you need a very small least count.

5:44Example 2: g from a simple pendulum with its error

Now a harder one, a real JEE favourite. Our pendulum has length fifty point zero centimetre, plus or minus zero point one, and makes twenty five oscillations in thirty five point five seconds. The watch has a least count of zero point one second. Find g, with its error. Pause the video, and try it. Done? Okay, let's check. Time period first. How do we get it? Exactly, thirty five point five seconds, divided by twenty five oscillations. One point four two seconds. Now, g equals four pi squared L, over T squared. The length in metres is zero point five zero zero. So four pi squared times zero point five, over one point four two squared: nine point seven nine metres per second squared. Now the error. Four pi squared is a constant, so it carries no error. Delta g by g equals delta L by L, plus two times delta T by T, because T is squared. Clear? Alright, now the time term. The watch error is zero point one second, and T is one point four two seconds. So delta T by T is zero point one by one point four two, right? No, no, careful! That would give about seven percent, far too big. We never timed one oscillation. We timed all twenty five together: thirty five point five seconds, with an error of zero point one. Dividing by twenty five, an exact count, does not change the fractional error, so delta T by T is zero point one by thirty five point five. The length gives zero point one by fifty point zero, that is zero point two zero percent, and two times zero point one by thirty five point five gives zero point five six percent. Total: zero point seven six percent. So delta g is zero point seven six percent of nine point seven nine. About zero point zero seven metres per second squared. Final answer: g equals nine point seven nine, plus or minus zero point zero seven, metres per second squared. And look, this is why we time many oscillations together: the watch error of zero point one second is shared by all twenty five. Got it?

8:08Quick revision: adding errors, zero error, least count

So, quick revision. One: errors always add, and each power multiplies its own percentage error. Two: corrected reading equals observed minus zero error, with its sign, so watch that minus sign. Three: the percentage error of one reading is the least count over the reading. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!