Set 2

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Suppose a screw gauge measures the diameter of a small ball, a little under seven millimetres, good to one hundredth of a millimetre. That is an error of only about zero point one five percent. So the volume of the ball is also good to about zero point one five percent, right? Hold on to your answer. This is Set two of the mixed problems on experimental skills, and questions like these come up often in JEE Main. Let's begin.

0:30Tool 1: vernier least count and zero error

So, tool one, the vernier callipers. In each picture, the upper strip is the main scale and the lower strip is the vernier. L C is the least count, and MSD and VSD are one main scale division and one vernier scale division. When n vernier divisions equal n minus one main scale divisions, the least count is one MSD minus one VSD, which is one MSD over n, okay? Now take the common laboratory callipers. Its main scale is in millimetres, and ten vernier divisions equal nine millimetres. So what is its least count? Yes, exactly, one VSD is zero point nine millimetre, so the least count is one minus zero point nine, which gives zero point one millimetre. We need this in Example one. Got it? Now close the jaws. If the two zeros do not coincide, the instrument has a zero error. In the left picture, the vernier zero lies to the right of the main scale zero, which is a positive zero error, equal to plus the coinciding division, in brackets, times the least count. The least count of these callipers is zero point zero one centimetre, and the arrow shows the third division coinciding, so the error is plus zero point zero three centimetres. Clear? In the right picture, the vernier zero lies to the left, which is a negative zero error, equal to minus, n minus the coinciding division, in brackets, times the least count, with n the number of vernier divisions. The seventh division coincides, so ten minus seven gives three, and the error is minus three times zero point zero one, which is minus zero point zero three centimetres. With me? Now the question that decides the marks. With a positive zero error, the vernier zero starts a little to the right, even before you put anything between the jaws. So does the callipers read a little too much, or a little too little? It reads too much, because that extra bit is already inside every reading, so we subtract it. Corrected reading equals observed reading minus zero error, with its sign. So a positive error is subtracted, and a negative error added, since minus of a minus is plus. That is exactly what the small notes under the two pictures say. Okay?

2:54Tool 2: reading a screw gauge

Now tool two, the screw gauge. The pitch is the distance the screw advances along the main scale, also called the pitch scale, in one complete rotation. In the figure, the main scale is on the left, and the circular scale, with fifty divisions, is on the right. Clear? The least count is the pitch over the number of circular divisions, because the circular scale cuts one full turn into equal parts. With me? This gauge has a pitch of zero point five millimetre and fifty divisions. So the least count is zero point five millimetre over fifty, which gives zero point zero one millimetre, the same as the figure shows. Got it? Now the reading. PSR, the pitch scale reading, is the last visible mark on the main scale, and CSR, the circular scale reading, is the circular division lying on the reference line. The reading equals PSR, plus CSR times the least count. Right? Now look at the figure. The red arrow marks the last visible mark, five point five millimetres, so the pitch scale reading is five point five. Division twenty eight sits on the long horizontal line, the reference line, so the circular scale reading is twenty eight. Clear? So the reading is five point five, plus twenty eight times zero point zero one, which gives five point seven eight millimetres, exactly what the figure shows on the right. Keep this gauge in mind, because Example two uses one just like it.

4:22Tool 3: errors in a product of powers

Got it? Now tool three, the key to our puzzle. Say Z equals capital A to the power small a, times capital B to the power small b, over capital C to the power small c. Delta, the Greek capital letter like a small triangle, means the error, so delta Z over Z is the relative error, okay? The relative errors add, each times its power, even for C, which divides. Right? For a single reading, the error is simply the least count of the instrument that took it. Clear? So the quantity with the highest power, and the one with the smallest value, needs the most careful measurement, because its relative error counts the most. Got it?

5:03Example 1: true length from a positive vernier zero error

Now Example one, the easier one. On a vernier callipers, ten vernier divisions equal nine millimetres. When the jaws touch, the vernier zero lies to the right of the main scale zero, and the fourth division coincides. An observed length is three point two seven centimetres. Find the true length. Pause the video and try it first. Done? Okay, let's check. We found this least count in tool one: one millimetre over ten, which is zero point one millimetre. Right? The vernier zero is to the right, so the zero error is positive: plus four times zero point one, which is plus zero point four millimetre, or plus zero point zero four centimetres. Change it to centimetres, because the reading is in centimetres. Clear? Now many students write three point two seven plus zero point zero four, which gives three point three one centimetres. Is that correct? No, it is not, and that is the trap. The callipers reads too much, so we subtract the zero error, with its sign. Three point two seven, minus zero point zero four, gives three point two three centimetres. Got it?

6:17Example 2: ball diameter and percentage error in volume

Now Example two, our opening puzzle. Here it is: a screw gauge has a pitch of zero point five millimetre and fifty circular divisions. For a ball, the main scale shows six point five millimetres and the circular scale twenty eight. Find the diameter, and the percentage error in the volume. Pause the video and try it first. Done? Okay, let's check. The least count is zero point five millimetre over fifty, which gives zero point zero one millimetre, just like the gauge in the figure. With me? The diameter, d, is the pitch scale reading plus the circular scale reading times the least count. Six point five, plus twenty eight times zero point zero one, gives six point seven eight millimetres. Right? Now the volume. This is one reading, so the error in d is one least count, zero point zero one millimetre, about zero point one five percent of six point seven eight. So is the volume, V, also good to about zero point one five percent? Not quite, and this is the heart of our puzzle. The volume goes as d cubed, and the sign on screen means proportional to. So by tool three, delta V over V is three times delta d over d, because the power three multiplies the error. Okay? So it is three times zero point zero one over six point seven eight, times one hundred percent, which gives about zero point four four percent, and our puzzle is solved. The wavy equals sign means approximately.

7:57Quick revision: zero error, screw gauge, errors with powers

Clear? So, quick revision. For a vernier, the corrected reading is the observed reading minus the zero error, with its sign. For a screw gauge, the least count is the pitch over the number of divisions, and the reading is PSR plus CSR times the least count. And relative errors add, each times its power, so a cube triples the relative error. Now try the rest of this set on this page. All the best!