Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. One tuning fork gives two resonances in a tube, at twenty five point four centimetres and at seventy eight point six centimetres. The second length should be three times the first, right? But three times twenty five point four is only seventy six point two. So where do the extra two point four centimetres come from? We will answer that in the second example. This is part A of the problem bank for experimental skills, at JEE Main level. So, let's begin.

0:35Screw gauge: pitch, least count, reading

First, the screw gauge, or micrometer. In the figure, the main scale, or pitch scale, is on the left, in millimetres, the m m you see. The circular scale, with fifty divisions, is on the right. The pitch is the distance the screw advances along the pitch scale in one complete rotation. Okay? Now let me ask you this. In one full turn, the screw moves by one pitch. So if you turn it by just one circular division, how far does it move? It moves the pitch shared equally among all the divisions, yes, exactly! That tiny step is the least count, written L C on screen. So the least count equals the pitch over capital N, the number of divisions on the circular scale. In the figure, zero point five over fifty gives zero point zero one millimetres, a common value. Remember it, because it will try to trick you. Clear? Reading equals the pitch scale reading, PSR, plus the circular scale reading, CSR, times the least count; the cross is the times sign. The arrow at the top points to the last visible mark, five point five millimetres, and twenty eight sits on the reference line, the long horizontal line. So five point five plus twenty eight times zero point zero one gives five point seven eight millimetres. Right? Now, zero error. With the two studs in contact, the reading should be zero; if it is not, the gauge has a zero error. The rule is always the same: corrected reading equals the observed reading, minus the zero error. With me? Now the sign of the zero error. If the zero of the circular scale sits below the reference line, the gauge already reads a little, so the error is positive, the plus sign. If it sits above the line, the error is negative, the minus sign. So a positive error is subtracted, and a negative error is added. Got it?

2:29Resonance tube: two lengths, one speed

Now the resonance tube. The figure shows one tube at its first resonance and at its second, with water in grey closing it below. We hold a fork of frequency f over the open end, though it is not drawn. The orange curves show how much the air moves. They close to a node at the water, and open out at the upper dashed line, a little above the mouth. That extra length, between the two dashed lines, is the end correction, small e, okay? The arrows mark l one and l two, the lengths from the open end down to the water; the small one and two below the l are just labels. From a node to the next antinode is a quarter of a wavelength, written lambda, this Greek letter. So l one plus e equals lambda over four, and l two plus e equals three times lambda, over four. Right? But we do not know e, and it sits in both lines. So how would you get rid of it? Subtract the first line from the second, yes, exactly! The e cancels, and l two minus l one is lambda over two. So lambda equals two times the bracket, l two minus l one. Clear? Now small v, the speed of sound, is f times lambda. So v equals two times f, times the same bracket, and the end correction has cancelled out. The figure shows it too. With me? The same two readings give a bonus. Rearrange the two lines, and e equals l two minus three times l one, all over two. The wavy equals sign means approximately, and e is roughly zero point three times d, where d is the internal diameter of the tube. Got it? Look at what this says. The end correction is positive, so l two is a little more than three times l one, not exactly three times. Remember our puzzle? This is the clue. Okay?

4:17Example 1: screw gauge reading with 100 circular divisions

Now, example one, the easier one. Here, a screw gauge of pitch zero point five millimetres has one hundred circular divisions. The main scale reads four point five millimetres, and the circular scale reads sixty two. Find the reading. Pause the video and try it first. Done? Okay, let's check. The common gauges have a least count of zero point zero one millimetres, like the one we just read. So can we use zero point zero one here too? No, we cannot, so count the divisions first. This gauge has one hundred divisions, not fifty. So the least count is zero point five millimetres over one hundred, which gives zero point zero zero five millimetres, half the usual value. Right? The main scale reading is four point five millimetres. So the reading is four point five, plus sixty two times zero point zero zero five. Okay? Sixty two times zero point zero zero five gives zero point three one. Add four point five, and the reading is four point eight one millimetres. If you checked the least count first, very good!

5:28Example 2: speed of sound and end correction from l1, l2

Now example two, the harder one, which is our puzzle from the start. Here, a fork of three hundred and twenty hertz, the H z on screen, gives resonances at twenty five point four centimetres and seventy eight point six centimetres; c m means centimetres. Find the speed of sound and the end correction. Pause the video and try it first. Done? Okay, let's check. First, l two minus l one. Seventy eight point six minus twenty five point four equals fifty three point two centimetres. Okay? So the wavelength is fifty three point two centimetres, right? No, it is not, and this is the trap. Successive resonances are only half a wavelength apart. So lambda is two times fifty three point two, which gives one hundred and six point four centimetres. With me? For the speed, v equals two times f, times the bracket, with the bracket in metres, zero point five three two metres. Two times three hundred and twenty times zero point five three two gives about three hundred and forty metres per second, written m s to the power minus one on screen. Got it? Now the end correction, and our puzzle. Three times l one is three times twenty five point four, which gives seventy six point two centimetres. But l two is seventy eight point six, which is two point four centimetres more. Right? Now use e equals l two minus three times l one, all over two. Seventy eight point six minus seventy six point two, all over two, gives one point two centimetres. So those extra two point four centimetres were just two end corrections, and our puzzle is solved! If you got both answers, really well done.

7:13Quick revision: least count, zero error, end correction

So, quick revision. For the screw gauge, the least count is the pitch over capital N, so always count the divisions. The corrected reading is the observed reading minus the zero error, sign and all. For the resonance tube, v equals two times f, times l two minus l one, and e equals l two minus three times l one, all over two. Now attempt the rest of part A on this page as a test, before reading the solutions. All the best!