Zero Error and Unusual Verniers

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. You close the jaws of a vernier callipers, with nothing between them. The reading should be zero, right? But it shows zero point zero three centimetres. Now a rod reads four point two five centimetres. So do you add that zero point zero three, or subtract it? Questions on experimental skills come up often in JEE, so let us settle this today, and read some unusual verniers too. Let's begin.

0:31Zero error: the two zeros do not coincide

So, close the jaws of the callipers. The zero of the vernier scale and the zero of the main scale should coincide, okay? If they do not, the instrument has a zero error. In the left picture, the vernier zero has moved to the right of the main zero. So will this callipers read too much, or too little? Yes, exactly, it reads too much, because the vernier is already ahead before you measure anything. We call this a positive zero error. Right? Now let us find its size. Zero error equals plus, the coinciding division in brackets, times L C, the least count. The arrow shows the third division coinciding, and the least count is zero point zero one centimetres, so the error is plus zero point zero three centimetres, okay? Now the right picture, where the vernier zero lies to the left of the main zero. This is a negative zero error, and the callipers reads too little. Clear? Zero error equals minus, n minus the coinciding division in brackets, times the least count, where n is the number of vernier divisions. The seventh division coincides here, so ten minus seven gives three, and the error is minus zero point zero three centimetres. Now read the small notes under the two pictures as well: subtract zero point zero three for the positive error, add zero point zero three for the negative error. We will see why in a moment. Got it?

2:00Add or subtract? The one line that decides

Now let us go back to our puzzle. Our callipers reads too much, and that same error is sitting inside the rod's reading. So, to get the true length, should you add the error, or subtract it? Yes, exactly, you subtract it, and here is the rule for every case. Corrected reading equals observed reading, minus zero error, with its sign. Write this line down, okay? So you subtract the size of a positive error, and you add the size of a negative error, because minus of a minus becomes plus. Right? Now for our rod, four point two five, minus the plus zero point zero three in brackets, gives four point two two centimetres, and our puzzle is solved, okay? Students also mix up these words. The zero correction is equal in size to the zero error, and opposite in sign. It is what you add to the observed reading. Clear?

2:55When zero error does not matter

Zero error is a systematic error, so the same shift sits in every reading, and it cancels when two readings are subtracted. With me? The thickness of the wall of a tube is found as half the difference between its external and internal diameters, with the same callipers. Does the zero error matter here? Very good if you said no, because it cancels in the difference. But there is one condition: the two pairs of jaws must have the same zero error, okay?

3:24Other verniers: work from first principles

Now, sometimes a question gives you a differently built vernier. Do you need a new formula for each one? No, you do not, because any vernier can be worked out from first principles. Start by finding one MSD, that is one main scale division, and one VSD, that is one vernier scale division, okay? Then use one simple fact: the coinciding vernier mark lies exactly on a main scale mark. Right? Now suppose n vernier divisions equal n plus one main scale divisions, with n plus one in brackets. This is a backward vernier, and each vernier division is longer than a main scale division. Clear? The least count is still one MSD over n, okay? But the fraction you add to the main scale reading changes. It is n minus V C in brackets, times the least count, where V C is the vernier coincidence, the division that lines up. Got it?

4:20Angular vernier: same method, degrees and minutes

And one last type, the angular vernier of a spectrometer, which works the same way with degrees and minutes. Say the main scale is in half degrees, and thirty vernier divisions coincide with twenty nine main scale divisions. One main scale division is thirty minutes of arc, so the least count is thirty minutes over thirty, which is one minute of arc. The small dash after a number, called a prime, means minutes of arc. With me? Now the main scale reads forty point five degrees, the small circle meaning degrees, and the twelfth vernier division coincides. That is forty degrees and thirty minutes, plus twelve minutes, so the angle is forty degrees and forty two minutes, okay?

5:01Example 1: negative zero error, then the true length

Alright, now it is your turn. With the jaws closed, the vernier zero, with ten divisions and a least count of zero point zero one centimetres, is to the left of the main zero, and the eighth division coincides. Find the zero error. The instrument then reads three point six four centimetres, so find the true length. Pause the video and try it first. Done? Okay, let's check. So the error is minus eight times zero point zero one, which is minus zero point zero eight centimetres, right? No, no, careful, that is the first trap. The vernier zero is to the left, so we use minus, ten minus eight in brackets, times zero point zero one, which gives minus zero point zero two centimetres, okay? Now let us find the true length. Many students see the minus sign and write three point six four minus zero point zero two, which gives three point six two. Is that correct? Oh no, careful, that is the second trap, because the error already carries its own minus sign. So it is three point six four, minus, minus zero point zero two. Minus of a minus is plus, and the true length is three point six six centimetres. Clear?

6:16Example 2: reading a vernier with 10 VSD = 11 MSD

Now let us try a harder one, with the unusual vernier. In a callipers the main scale is in millimetres, and ten vernier divisions equal eleven main scale divisions. The vernier zero is just past the twelve millimetre mark, and the third vernier division coincides with a main scale mark. Find the reading. Pause the video and try it first. Done? Okay, let's check. Many students write twelve plus three times zero point one, which gives twelve point three millimetres. Is that correct? No, that does not work here, because this vernier is built differently. Ten vernier divisions equal eleven millimetres, so one VSD is one point one millimetres. Right? Now use our simple fact, with the vernier zero at capital X, okay? The third vernier mark is three times one point one further on, so capital X plus three point three must be a whole number of millimetres. With me? Write that whole number as twelve plus k, so capital X equals twelve plus k, minus three point three. And capital X must lie between twelve and thirteen, because the zero is just past the twelve millimetre mark. Which value of k does that? Yes, exactly, k equals four, because twelve plus four is sixteen, and sixteen minus three point three gives twelve point seven. So the reading is twelve point seven millimetres. Clear? Now let us see our shortcut agree. The least count is one millimetre over ten, which is zero point one millimetres. Twelve, plus ten minus three in brackets, times zero point one, gives twelve point seven millimetres again. Got it?

8:02Quick revision: corrected reading and unusual verniers

Now let us do a quick revision. Corrected reading is observed reading minus zero error, with its sign. For a positive error, the zero error is plus the coinciding division times the least count, and for a negative error, it is minus, n minus the coinciding division, times the least count. And for other verniers, find one MSD and one VSD, then use the coinciding mark. You have the rule now, so you are ready. Go and try the practice problems on this page. All the best!