AIEEE and JEE Main Questions (2008–2018): Measurement

Transcript

0:00Introduction

Hello everyone, here is a strange one. Close the two jaws of a screw gauge, with nothing between them. It should read zero, right? But in a JEE Main question from two thousand sixteen, the jaws are touching, and the forty fifth division sits on the line. So is the instrument lying to us? Today we solve AIEEE and JEE Main questions on measurement, using three ideas: least count, zero error and combining errors. So, let's begin.

0:32Screw gauge: pitch, least count, reading

So, here is a screw gauge, with its main scale and its circular scale. Pitch is the distance the screw moves in one complete rotation, that is, distance moved over number of rotations. Okay? Now, in AIEEE two thousand eight, two full turns cover one millimetre on the main scale. So what is the pitch? If you said one millimetre, careful, that is the trap. One turn covers half of it, so the pitch is zero point five millimetre. With me? The least count, the L C on screen, is the pitch over the number of circular divisions. In the figure, zero point five over fifty gives zero point zero one millimetre. Right? And the reading is the pitch scale reading, PSR, from the main scale of the gauge, plus the circular scale reading, CSR, times the least count. In the figure, the last visible mark is five point five millimetres and division twenty eight is on the line, so five point five, plus twenty eight times zero point zero one, gives five point seven eight millimetres. Clear?

1:36Vernier scales: straight and angular

Now the vernier, where n vernier scale divisions, VSD on screen, equal n minus one main scale divisions, the MSD on screen. So the least count is one main scale division over n, okay? The reading is the main scale reading, MSR, plus the vernier coincidence, V C, times the least count. In the figure, the vernier zero is just past two point three centimetres and the fourth division coincides, so two point three, plus four times zero point zero one, gives two point three four centimetres. Got it? An angular vernier works the same way. In AIEEE two thousand nine, one main scale division is half a degree, written with the little circle on screen, and thirty vernier divisions match twenty nine main divisions. What is the least count? Yes, exactly, it comes to one minute. Half a degree is thirty minutes, written with the little tick for minutes, and thirty minutes over thirty gives one minute. Right?

2:33Zero error: the sign decides

Now we come to our puzzle, the zero error. With the jaws closed, the two zeros should coincide, and if not, the instrument has a zero error. The corrected reading is observed minus zero error, with its sign. In the figure, on a vernier callipers, the positive error of zero point zero three centimetre is subtracted, and the negative one of the same size is added. Okay? On a screw gauge, if the zero of the circular scale is n divisions below the line, the gauge already reads a little, so the error is positive: plus n times the least count. Right? If it is above the line, the error is negative: minus n times the least count, and the division seen on the line is capital N minus n, where capital N is the total number of divisions. Got it? Back to AIEEE two thousand eight, where fifty circular divisions give a least count of zero point zero one millimetre. For a thin wire, the main scale reads three millimetres and the thirty fifth division is in line, so we observe three point three five millimetres. The zero error is minus zero point zero three millimetre. So do we add it or subtract it? Yes, we add it, because two minus signs make a plus. Three point three five, minus, minus zero point zero three, gives three point three eight millimetres. With me?

3:50Combining errors: they add

The third idea is combining errors. Take capital Z as A to the power a, times B to the power b, over C to the power c. Delta, the Greek triangle, means the error, so delta Z over Z is the relative error, and it equals a times delta A over A, plus b times delta B over B, plus c times delta C over C, all added. Okay? In AIEEE two thousand twelve, resistance is voltage over current, with errors of three per cent each. It is one quantity over the other, so do the errors cancel? No, they do not, and that is the trap. Relative errors add even in a quotient, so capital R, V over I, has three per cent plus three per cent, which is six per cent. Right? In JEE Main two thousand eighteen, the mass and the length of a cube have relative errors of one point five per cent and one per cent. Density is mass over length cubed, so one point five plus three times one gives a maximum error of four point five per cent. Clear?

4:52Example 1: diameter of a wire from a screw gauge reading

Time for Example one, from AIEEE two thousand eleven. Here a screw gauge gives these readings for the diameter of a wire: main scale reading zero millimetre, circular scale reading fifty two divisions. One millimetre on the main scale corresponds to one hundred divisions of the circular scale. Find the diameter. Pause the video and try it first. Done? Okay, let's check. One millimetre over one hundred gives a least count of zero point zero one millimetre. Okay? The reading is zero, plus fifty two times zero point zero one, which gives zero point five two millimetre. Now, what is that in centimetres? Ten millimetres make a centimetre, so the diameter is zero point zero five two centimetre. Be careful, because zero point five two centimetre would make a thin wire over five millimetres thick. Clear?

5:49Example 2: thickness of a sheet with negative zero error

Now Example two, our puzzle from the start, from JEE Main two thousand sixteen. The gauge has a pitch of zero point five millimetre and fifty circular divisions, and it measures a thin sheet of aluminium. With the jaws in contact, the forty fifth division coincides with the line, and the zero of the main scale is barely visible. For the sheet, the main scale reads zero point five millimetre and the twenty fifth division coincides. Find the thickness. Pause the video and try it first. Done? Okay, let's check. Zero point five over fifty gives a least count of zero point zero one millimetre. Right? Now comes the key step of this problem. Is the zero error plus forty five divisions, or minus five divisions? It is minus five divisions, and if you said that, very good. Division forty five on the line means capital N minus n is forty five, so the circular zero is five divisions above the line. The main scale zero is barely visible, so the screw is very near its zero and the error must be small. Fifty minus forty five, times zero point zero one, with a minus sign, gives minus zero point zero five millimetre. Clear? The observed reading is zero point five, plus twenty five times zero point zero one, which gives zero point seven five millimetre. With me? So what is the thickness: zero point seven zero millimetre, or zero point eight zero millimetre? Yes, exactly, it is zero point eight zero millimetre: zero point seven five, minus, minus zero point zero five. Getting zero point seven zero is the trap, because it loses the sign. So our puzzle is solved: the gauge was not lying, it was only showing us its zero error.

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

Okay? So, let us do a quick revision. On a screw gauge, the least count is pitch over divisions, and the reading is PSR plus CSR times the least count. The corrected reading is observed minus zero error, with its sign. And relative errors add, each weighted by its power. Now go and try the practice problems on this page. You can absolutely do these, so all the best!