JEE Advanced Questions (2013–2022): Measurement

Transcript

0:00Introduction

Hello everyone, here is a small puzzle. Take a screw gauge with a main scale division of half a millimetre and one hundred circular divisions. So the least count is half a millimetre over one hundred, which gives zero point zero zero five millimetres, right? In a JEE Advanced question from twenty twenty two, that quick answer would be wrong. Today we solve two JEE Advanced measurement questions, and this puzzle is one of them, so let's begin.

0:30Vernier callipers: start from 1 MSD and 1 VSD

These past questions can give unusual scales, so a memorised least count will not help. So, with this vernier callipers, find one main scale division, written MSD, and one vernier scale division, written VSD, okay? When each vernier division is a little shorter, their difference is L C, the least count. Clear? For this vernier, the reading is MSR, the main scale mark just before the vernier zero, plus V C, the number of the vernier division that coincides, times the least count. In the figure, two point three plus four times zero point zero one gives two point three four centimetres.

1:07Screw gauge: pitch first, then least count

With me? Let us move on to the screw gauge. Pitch is the distance the screw advances along the main scale in one complete rotation, so it is the distance moved over the number of rotations. Remember, it need not be one main scale division. Okay? The least count is the pitch over the number of circular divisions. In the figure, zero point five millimetres over fifty gives zero point zero one millimetres. Right? The reading is PSR, the pitch scale reading, plus CSR, the circular scale reading, times the least count. In the figure, five point five plus twenty eight times zero point zero one gives five point seven eight millimetres. Got it?

1:46Zero error, the mean, and powers

Now for the instrument's own mistake, the zero error. With the jaws closed or the studs touching, the two zeros should coincide. If they do not, the corrected reading equals the observed reading minus the zero error, with its sign, okay? Now let me ask you this, and keep that sign in mind. If the zero error is negative, do we add its size or subtract it? We add its size, because taking away a negative number is the same as adding. And a positive error is simply subtracted from the reading. Students often flip these two, so please be careful here. Clear? Now suppose we take several readings of the same thing. Then we report their mean. And the mean absolute error is the mean of the magnitudes of the deviations from that mean, rounded to the least count. Got it? Now let us look at powers. Take the area, pi times r squared, where r is the radius. Let delta r, with the little triangle called capital delta, mean the error in the radius, so delta r over r is its relative error. What number multiplies delta r over r in the relative error of the area? Very good if you said two, because relative errors add, with the powers as weights. So the power two multiplies it, and the relative error in the area is two times delta r over r. With me?

3:12Example 1: cylinder diameter by vernier, JEE Advanced 2013

Let us start with the easier question, from JEE Advanced twenty thirteen. The diameter of a cylinder is measured with a vernier callipers with no zero error. The zero of the vernier lies between five point one zero and five point one five centimetres of the main scale. The vernier scale has fifty divisions equivalent to two point four five centimetres, and its twenty fourth division coincides exactly with a main scale division. Find the diameter. Pause the video and try it first. Done? Okay, let's check. Look at the two marks the question gives, five point one zero and five point one five. So how long is one main scale division? That is correct, five point one five minus five point one zero gives zero point zero five centimetres. So please do not assume one millimetre here, that is the trap in this question. Okay? Next, one vernier division is two point four five over fifty, which gives zero point zero four nine centimetres. Right? The least count is the difference, so zero point zero five minus zero point zero four nine gives zero point zero zero one centimetres. Clear? Now let us take the reading. Of five point one zero and five point one five, which one is our main scale reading? It is five point one zero, the mark just before the vernier zero. The twenty fourth division coincides, and there is no zero error to remove. So the diameter is five point one zero plus twenty four times zero point zero zero one, which gives five point one two four centimetres. Got it?

4:51Example 2: wire area by screw gauge, JEE Advanced 2022

Now the harder one, our puzzle, from JEE Advanced twenty twenty two. The area of cross-section of a wire is measured with a screw gauge. One main scale division is zero point five millimetres, the circular scale has one hundred divisions, and one full rotation shifts the main scale by two divisions. With the arms touching, it reads zero main scale divisions and four circular divisions. With the wire, two attempts give four main scale with twenty circular, and four main scale with sixteen circular. Find the diameter and the area of cross-section. Pause the video and try it first. Done? Okay, let's check. One full rotation shifts the main scale by two divisions, and that sentence decides everything. So what is the pitch of this gauge? Yes, exactly, two times zero point five gives a pitch of one point zero millimetres. So the least count is one point zero over one hundred, which gives zero point zero one millimetres. Our puzzle is solved, because zero point zero zero five comes only from missing that sentence. Clear? With the arms touching, the gauge reads four circular divisions. So what is the zero error, with its sign? The gauge already reads a little, so the error is positive, and plus four times zero point zero one gives plus zero point zero four millimetres, okay? Now let us take the first attempt. Be careful here, how many millimetres do four main scale divisions give? They give two millimetres, because each division is still zero point five millimetres. So two point zero zero plus twenty times zero point zero one gives two point two zero. Removing the zero error, two point two zero minus zero point zero four gives two point one six millimetres. Right? The second attempt is the same, so two point zero zero plus sixteen times zero point zero one gives two point one six, and taking away zero point zero four gives two point one two millimetres. With me? The mean of these two is two point one four, and each is zero point zero two away from it, so the mean absolute error is zero point zero two. So d, the diameter, is two point one four, plus or minus zero point zero two, millimetres, where the plus or minus sign shows the error. Okay? For the area, what are the radius and its error? We take half of each, so the radius is one point zero seven with an error of zero point zero one. The area is pi times one point zero seven squared, and the relative error is two times zero point zero one over one point zero seven. One point zero seven squared is about one point one four, and one point one four times that relative error comes to about zero point zero two. So the area is pi times, in brackets, one point one four plus or minus zero point zero two, in square millimetres. If you got this one by yourself, that is really good work.

7:55Quick revision: least count, pitch, zero error, powers

So, here is a quick revision. For a vernier with shorter vernier divisions, find both divisions first, and the reading is main scale reading plus vernier coincidence times least count. For a screw gauge, the pitch is the shift in one rotation, and the least count is the pitch over the circular divisions. The corrected reading is the observed reading minus the zero error, and relative errors add with the powers as weights. Now go and try the practice problems on this page. All the best!