IIT-JEE Questions (2003–2011): Measurement

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle from an old IITJEE paper. You measure the diameter of a ball with a screw gauge, and the error in that reading is less than half a percent. But in the density, that same error adds more than one percent. So one tiny error has somehow grown, and by the end of this video, you will see exactly why. Measurement questions like this come up often in JEE, and the good news is that only five tools sit behind these nine old questions. So, let's go.

0:32Tool 1: the vernier callipers

Tool one is the vernier callipers, with the main scale on top and the vernier scale below it. L C is the least count, MSD is one main scale division, and VSD is one vernier scale division. The least count is one MSD minus one VSD, okay? Now here is a shortcut, but it comes with a condition. If n vernier divisions equal n minus one, in brackets, main scale divisions, then the least count is one MSD over n, and only under that condition, right? Keep an eye on that condition, because it comes back to trick us in the first example. MSR is the main scale reading, the mark just before the vernier zero, and V C is the vernier coincidence, the division that lines up with a main scale mark. The reading is MSR plus V C times the least count. In the figure, the fourth division coincides, so two point three plus four times zero point zero one gives two point three four centimetres, clear?

1:28Tool 2: the screw gauge

Tool two is the screw gauge. The pitch is the distance the screw advances along the main scale in one complete rotation, and the least count is the pitch over the number of circular scale divisions. In the figure, that is zero point five millimetres over fifty, which gives zero point zero one millimetres, okay? PSR is the pitch scale reading, and CSR is the circular scale reading, so the reading is PSR plus CSR times the least count. In the figure, the last visible mark, five point five, plus twenty eight times zero point zero one gives five point seven eight millimetres, with me?

2:00Tool 3: zero error and its sign

Tool three is zero error. With the jaws closed, the two zeros should coincide. On the callipers, a vernier zero to the right of the main zero means a positive error, and a vernier zero to the left means a negative error. On the screw gauge, with the studs in contact, a circular zero below the reference line means a positive error, and above the line, a negative error, okay? In both, the corrected reading equals the observed reading minus the zero error, with its sign. So when the zero error is negative, do we add or subtract? Yes, we add, because taking away a negative is the same as adding. So a positive error is subtracted, and a negative error is added, got it? This sign is an easy place to lose marks, so be careful with it.

2:49Tool 4: errors, where powers become weights

Tool four is the one behind our puzzle. Suppose capital Z is capital A to the power small a, times capital B to the power small b, over capital C to the power small c. Capital C is in the denominator, so does its error get subtracted? No, it does not, because errors always add. The little triangle is the Greek capital delta, and delta A means the error in capital A, okay? So delta Z over Z equals small a times delta A over A, plus small b times delta B over B, plus small c times delta C over C, clear? So each power becomes a weight, and a small value makes its fraction, delta A over A, large. That is why the quantity with the highest power, and the one with the smallest value, need the most careful measurement, okay?

3:39Tool 5: significant figures in the answer

Tool five is significant figures. When you multiply, divide, or take a power or a root, the result keeps the fewest significant figures among the measured values, right? Here is a quick one. The edge of a cube is measured as one point two times ten to the power minus two metres. Cubing gives one point seven two eight times ten to the power minus six cubic metres. So tell me, how should the volume be recorded? The edge has only two significant figures, so writing all four digits would claim more than we measured. The volume keeps two, and we record one point seven times ten to the power minus six cubic metres, got it?

4:20Example 1: least count when 20 VSD match 16 MSD

Now an easy one, from IITJEE two thousand and ten. The main scale has one millimetre marks, and twenty equal vernier divisions match with sixteen main scale divisions. Find the least count. Pause the video and try it first. Have you done it? Okay, let us check. The shortcut says one millimetre over twenty, which is zero point zero five millimetres. Is that the least count? No, and this is exactly the trap, because the shortcut needs twenty divisions to match nineteen, and here they match sixteen. So we go to first principles. Twenty VSD equal sixteen MSD, which is sixteen millimetres, okay? So one VSD is sixteen over twenty millimetres, which gives zero point eight millimetres. The least count is one MSD minus one VSD, so one minus zero point eight, which equals zero point two millimetres. Very good if you got that, and when the numbers look unusual, always work from first principles, clear?

5:25Example 2: diameter of a ball from a screw gauge reading

Now a harder one, from IITJEE two thousand and eleven, and it is our puzzle. The diameter of a solid ball is measured with a screw gauge of pitch zero point five millimetres and fifty circular divisions. The main scale reads two point five millimetres, and the circular scale twenty divisions. The mass has a relative error of two percent. Find the percentage error in the density. Pause the video and try it first. Have you got something? Good, let us check it together. The least count is zero point five over fifty, which gives zero point zero one millimetres, with me? The diameter, capital D, is two point five plus twenty times zero point zero one, which gives two point seven zero millimetres. One reading is uncertain by the least count, so delta D, the error in capital D, is zero point zero one millimetres, clear? Density is mass over volume, and the volume depends on the diameter. So which power of capital D goes into the density?

6:32Example 2: the density error

Yes, exactly, it is the cube. Rho, the Greek letter on screen, is the density, so rho equals small m, the mass, over capital V, the volume. For a ball, the volume is pi over six, times D cubed, so rho equals six times m, over pi times D cubed, okay? Delta rho over rho equals delta m over m, plus three times delta D over D, and we add, even with capital D in the denominator, right? Delta m over m is given as two percent. Delta D over D is zero point zero one over two point seven zero, about zero point three seven percent. So the total is about two point four percent. Is that the answer? Oh no, careful, this is the trap, because that answer forgets the three. Three times zero point zero one over two point seven zero gives about one point one percent, okay? So two percent plus one point one percent equals three point one percent. And there is our puzzle solved. The diameter error was less than half a percent, but the diameter is cubed, so it added more than one percent, right?

7:39Quick revision: least count, zero error, adding errors

So, quick revision. The least count is one MSD minus one VSD, or the pitch over the circular divisions. The corrected reading is the observed reading minus the zero error, with its sign. And relative errors add with the powers as weights, and then we keep the right figures. You now have all five tools, so go and try the other seven questions on this page, and all the best!