Part C: Instruments and Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You measure an object with a good vernier callipers, and you read thirty eight point three four millimetre. But the true length is thirty seven point eight eight millimetre. Almost half a millimetre has gone missing! Where did it go? Hmm, the instrument itself is hiding something, right? Today, in this problem bank, we revise the whole instrument toolkit: least count, zero error, the screw gauge, and some quick concept checks. And remember, in JEE, one small sign mistake costs you the full question. Then we solve two problems, and find that missing half millimetre. So, let's get started.

0:43Least count: the smallest step you can read

So, first tool: the least count, written L C on screen. It is the smallest measurement an instrument can read directly. For a vernier, the least count is one main scale division minus one vernier scale division. On screen, MSD means main scale division, and VSD means vernier scale division, okay? Usually, n vernier divisions are made equal to n minus one main divisions. Then the least count is simply one MSD divided by n, okay? And the double arrow you see on screen just means, therefore. Now tell me. Ordinary callipers: ten vernier divisions equal nine main divisions, and one main division is one millimetre. What is the least count? Yes, one millimetre divided by ten: zero point one millimetre. That is the least count of ordinary callipers. Clear? Now the screw gauge. In one full turn, the screw moves forward by one pitch, and its head has N circular divisions, okay? So the least count is the pitch divided by N, that's it. One turn shared equally among N divisions. Simple, right? Quick concept check. Keep the pitch the same, and increase the number of circular divisions. Does the least count go up, or down? Down, exactly! The same pitch, divided by a bigger number. And the vernier works the same way: for the same main scale division, twenty vernier divisions give a smaller least count than ten. Smaller least count, more precise instrument. Got it?

2:17Zero error: read the sign first

Second tool: the reading formula. Reading equals MSR, the main scale reading, plus VSR times the least count, minus the zero error. VSR is the number of the vernier line that coincides with a main scale line. Look, the screw gauge uses the same formula, with the circular scale reading in its place. One formula, two instruments! Now the zero error. Close the jaws, and the vernier zero should sit exactly on the main zero. Suppose it sits a little to the right. Will every reading come out too large, or too small? Too large, yes! That is a positive zero error. If vernier line number k coincides, the zero error is plus k times the least count. Now the vernier zero sits to the left: a negative zero error, and every reading is too small. So the error is just minus k times the least count, right? No, no, careful! This is the JEE trap. The vernier zero is behind the main zero, so count how far behind it is. On a vernier of n divisions, the zero error is minus, n minus k, times the least count, not minus k times the least count. Remember this one! Ten divisions, seventh line coinciding, least count zero point one millimetre. Minus, ten minus seven, times zero point one: minus zero point three millimetre, not minus zero point seven. Got it? The screw gauge has the same two cases. Close the studs gently, with the ratchet. If the zero of the circular scale goes below the index line, and division n sits on the line, the zero error is plus n times the least count. And if the zero stays above the line, with division N minus m on it, the zero error is minus m times the least count. Same idea as the vernier, okay?

4:04Concept checks: errors, backlash, spherometer

Quick concept checks now, true or false. Zero error is a random error. False! Zero error pushes every reading in the same direction, so it is a systematic error. Repeating the measurement does not reduce it. You remove it by correcting the reading, okay? That correction is our formula: corrected reading equals observed reading, minus the zero error, with its sign. Positive error, you subtract it. Negative error, minus of a minus, so you add it. That's it! Next, backlash. True or false: backlash is avoided by turning the screw in one direction? True, absolutely! When you reverse the direction, a worn screw may turn a little without moving forward. That is backlash. So always turn the screw in one direction, and use the ratchet. With me? And one more tool from this problem bank: the spherometer. It is a screw gauge standing on three legs, a distance l apart. It measures a small height h, and the radius of curvature, R, equals l squared over six h, plus h over two.

5:10Example 1: least count when 10 VSD equal 11 MSD

Now, example one, an easy one. Here, a vernier has one main scale division equal to one millimetre, and ten vernier divisions equal eleven main scale divisions. Find its least count. Pause the video, and try it first. Done? Okay, let's check. Look, something is different here. Ten vernier divisions are longer than ten main divisions, not shorter. So if you blindly write one MSD minus one VSD, you get one minus one point one: minus zero point one millimetre. Negative least count? No, no! That cannot be. Go step by step. Ten vernier divisions equal eleven millimetre, so one vernier division is one point one millimetre. Each vernier division is now a little bigger than a main division. The least count is still the difference between the two divisions: one point one minus one, zero point one millimetre. Same as ordinary callipers! Easy, right?

6:10Example 2: true length with a positive zero error

Now a harder one, JEE level. And guess what, this is our puzzle from the start! Here, a callipers has one main division equal to one millimetre, and fifty vernier divisions equal forty nine main divisions. With the jaws closed, the vernier zero is to the right of the main zero, and the twenty third line coincides. Then an object gives a main scale reading of thirty eight millimetre, and the seventeenth vernier line coincides. Find its length. Pause and try. Hint: least count first. Done? Okay, let's check. Fifty vernier divisions equal forty nine main divisions, so n is fifty, and the least count is one millimetre divided by fifty: zero point zero two millimetre. Now the zero error. The vernier zero is to the right of the main zero. Plus, or minus? Plus, that's right! Positive zero error, so every reading is too large. Twenty third line, so plus twenty three times zero point zero two: plus zero point four six millimetre. Now the object. Thirty eight, plus seventeen times zero point zero two. Seventeen times zero point zero two is zero point three four, so the observed reading is thirty eight point three four millimetre. Exactly the number from our puzzle! So the length is thirty eight point three four, right? No, no! The callipers were already ahead before you started. Corrected reading equals observed reading minus zero error. Thirty eight point three four minus zero point four six: thirty seven point eight eight millimetre. There is our missing half millimetre! It was the zero error, hiding inside the reading. Clear?

7:55Quick revision: least count, zero error and its sign

So, quick revision. Least count: one main division minus one vernier division, or for a screw gauge, the pitch divided by the number of circular divisions. Zero error: read the sign first, and for a negative one take n minus k, not k. And always, corrected equals observed minus zero error, with its sign. Now go and try the practice problems on this page: callipers, screw gauge and spherometer. All the best!