Zero Error in Vernier Callipers

Transcript

0:00Introduction

Hello everyone! Before we start, here is one small puzzle for you. I take an ordinary vernier callipers: ten vernier divisions, least count zero point one millimetre. I close the jaws completely. Nothing between them. So the reading must be zero, right? But look closely. The vernier zero has slipped a little to the left of the main scale zero, and the seventh vernier line coincides with a main scale line. So what is the error? Minus zero point seven millimetre? Careful! That is what many students write, and it is wrong! By the end of this video, you will know the real answer, and how to correct every reading with it. And trust me, zero error is a favourite of JEE examiners in vernier questions. So, let's begin.

0:50What is a zero error?

So, first the basic idea. When the jaws are closed, there is nothing to measure, so the reading must be zero. That means the zero of the vernier scale should sit exactly on the zero of the main scale, okay? Now tell me. What if, with the jaws closed, the two zeros do not match? Yes, exactly! Then the callipers show a reading even when there is nothing to measure! That is called a zero error. And it can happen in two ways: the vernier zero can sit to the right of the main zero, or to the left of it. The figure shows both cases, side by side. Clear?

1:26Positive zero error: vernier zero to the right

Look at the left half of the figure. The orange box is the vernier scale. Even with the jaws closed, its zero has moved a little to the right of the main scale zero. This is called a positive zero error. Now think. The vernier zero is already ahead, before you even put the object in. So will your readings come out too large, or too small? Too large, exactly! Every reading carries this extra bit inside it. So, with a positive zero error, every reading is too large. Got it? Now, how big is the error? Same method as a normal reading. See which vernier line coincides with a main scale line. If it is line number k, the zero error is plus k times L C, the least count, okay? In the figure, the least count is zero point one millimetre, and the third line coincides: the green one. So the zero error is plus three times zero point one: plus zero point three millimetre. Simple as that.

2:27Negative zero error: vernier zero to the left

Now the right half of the figure. Here, the vernier zero sits to the left of the main zero. It is behind. This is a negative zero error. And now every reading is too small, because the vernier starts from behind the zero. With me? Alright, now back to our puzzle from the start. Ten vernier divisions, least count zero point one millimetre, and the seventh line coincides. Negative error, so just put a minus sign: minus seven times zero point one, that is minus zero point seven millimetre, right? No, no, careful! This is the JEE trap. For a negative zero error, do not simply take minus k times the least count. Count how far the vernier zero is behind the main zero. If line number k coincides, on a vernier of n divisions, the zero error is minus, n minus k, times the least count. So in the figure: minus, ten minus seven, times zero point one. Minus zero point three millimetre! And that is exactly the label written under the right half of the figure. Notice, the vernier zero is behind the main zero by only three least counts, not seven. That is why we take ten minus seven. So remember: not minus zero point seven. The answer is minus zero point three millimetre. Got it?

3:46Correcting the reading: subtract with the sign

So, now how do we use the zero error? First, take the reading exactly as you always do: the main scale reading, plus the vernier scale reading times the least count. On screen, MSR is the main scale reading, and VSR is the number of the vernier line that coincides. This is the observed reading. Then, the corrected reading equals the observed reading, minus the zero error, with its sign. One formula for both cases. Write this down, okay? Now a question. Suppose the zero error is negative. Do we add it, or subtract it? Subtract! We still subtract it, with its sign! Many students change the rule for the negative case and get confused. Don't! Keep the same formula, and let the sign do the work. Clear? Why does this work? Every reading is shifted by the same signed amount. Positive error, the readings are too large, so subtracting it brings them down. Right? Negative error, the readings are too small. And subtracting a negative number adds it back, so the reading goes up. That's all there is to it. And here is a small topper's trick. Always write the zero error with its sign first, inside brackets, and only then subtract. Then the sign never goes wrong, even in a hurry.

5:08Example 1: true length with a positive zero error

Let's do example one, an easy one. With the jaws closed, the vernier zero is to the right of the main zero, and the third vernier line coincides. The least count is zero point zero one centimetre. Then a rod gives a main scale reading of two point five centimetre, and the sixth vernier line coincides. Find the true length. Pause the video, and try it first. Done? Okay, let's check. Vernier zero to the right: positive zero error, just like the left half of our figure. The third line coincides, so the zero error is plus three times zero point zero one: plus zero point zero three centimetre. Now the rod. Two point five, plus six times zero point zero one: two point five six centimetre. That is the observed reading. Positive error, readings too large, so subtract it. Two point five six minus zero point zero three: two point five three centimetre. That is the true length. Easy, right?

6:12Example 2: sphere diameter with a negative zero error

Now a harder one, JEE pattern. And look, this is our puzzle from the start! The vernier has ten divisions, and the least count is zero point one millimetre. With the jaws closed, the vernier zero is to the left of the main zero, and the seventh vernier line coincides. Then a sphere gives a main scale reading of one point two centimetre, and the fourth vernier line coincides. Find its diameter. Pause and try. Hint: put everything in one unit first. Done? Okay, let's check. First, units. The least count is in millimetre, but the readings are in centimetre. So zero point one millimetre is zero point zero one centimetre. Clear? Now the zero error. Vernier zero on the left, seventh line coinciding. So, minus seven least counts? Careful there! Remember the trap. Count how far the vernier zero is behind the main zero. Minus, ten minus seven, times zero point zero one: minus zero point zero three centimetre. And that answers our opening puzzle: the zero error is minus zero point three millimetre, not minus zero point seven! Now the sphere. One point two, plus four times zero point zero one: one point two four centimetre. That is the observed reading. Last step: observed minus zero error, with its sign. One point two four, minus, minus zero point zero three. One point two seven centimetre! The readings were too small, so the true diameter is bigger. Makes sense, right?

7:50Quick revision: zero error sign and corrected reading

So, quick revision. Vernier zero to the right: positive zero error, plus k times the least count. Vernier zero to the left: negative zero error, minus, n minus k, times the least count, not minus k. And always, the corrected reading equals the observed reading minus the zero error, with its sign. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!