Set 3

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. In a lab at seventeen degrees Celsius, a resonance tube gives its first loud sound at sixteen centimetres. Now we use the same tube and fork on a hot day, at thirty seven degrees. Does the loud spot stay at sixteen centimetres, or does it move? Hold on to your answer. This is Set three of the mixed problems on experimental skills, with the screw gauge and the resonance tube today. Problems like these come up often in JEE Main, so let's begin.

0:33Tool 1: reading a screw gauge

So, tool one, the screw gauge. In the figure, the main scale, also called the pitch scale, is on the left, in millimetres, and the circular scale, with fifty divisions, is on the right. The pitch is how far the screw moves in one full rotation. On screen, the least count, L C, equals the pitch over capital N, the number of divisions on the circular scale. In the figure, zero point five millimetres over fifty gives zero point zero one millimetres. Okay? Now the reading. Reading equals the pitch scale reading, PSR, plus the circular scale reading, CSR, times the least count. In the figure, the orange arrow points to the last visible mark, five point five millimetres, and division twenty eight sits on the long horizontal line, the reference line. So five point five, plus twenty eight times zero point zero one, gives five point seven eight millimetres. Right? Now the zero error. Bring the two studs together, with nothing between them. The zero of the circular scale should sit on the reference line, but suppose it sits below it. In the figure, the circular scale numbers increase upwards. So tell me, is this zero error positive or negative? It is positive, yes, because the gauge already reads a little with nothing between the studs. If the zero is small n divisions below the line, the zero error is plus small n times the least count. Clear? And if the zero is small n divisions above the line, the error is negative, minus small n times the least count. With me? Corrected reading equals the observed reading minus the zero error. Take the sign along with it, so a positive error is subtracted, and a negative error is added. Got it?

2:23Tool 2: the resonance tube and temperature

Now tool two, the resonance tube. The figure shows the same tube at the first resonance and at the second, with the water in grey, and a tuning fork of frequency small f, not drawn, over the open end. The orange curves show the stationary wave: the water surface is a displacement node, and the antinode lies a little above the open end. That extra length is the end correction, small e, the gap between the two dashed lines at the top. Okay? Today we need only the first resonance, on the left, not the other formulas. The arrow marks l one, the l with a small one below, the length down to the water. Node to antinode is a quarter of a wavelength, so l one plus e equals lambda over four, with lambda, the Greek letter, the wavelength. And e is approximately equal to zero point three times d, the internal diameter of the tube; the wavy equals sign on screen means approximately. Right? Now, the wavelength lambda equals v over f, where small v is the speed of sound. So l one plus e equals v over, four times f. Clear? Here is a question for you. When the air in the tube gets warmer, does sound travel faster or slower? Yes, exactly, it travels faster. On screen, v is proportional to the square root of capital T; this sign, like a stretched alpha, means proportional to. And capital T here is the absolute temperature, in kelvin, never in degrees Celsius. Got it? Now here is a topper's trick. With the same fork, f does not change, and with the same tube, e does not change. So l one plus e is proportional to the square root of T, and a ratio is all you need. But notice, it is l one plus e that scales, not l one alone. With me?

4:16Example 1: screw gauge reading with a +0.03 mm zero error

Now the examples, the easier one first. Here it is: a screw gauge has a pitch of zero point five millimetres and fifty divisions on its circular scale. It has a zero error of plus zero point zero three millimetres. For a wire, the main scale reads two point five millimetres, and the circular scale reads seventeen. Find the diameter. Pause the video and try it first. Done? Okay, let's check. The least count is zero point five over fifty, which gives zero point zero one millimetres. Okay? Now the observed reading. Two point five, plus seventeen times zero point zero one, gives two point six seven millimetres. With me? Now the zero error is plus zero point zero three millimetres. So we add it, and the diameter is two point seven zero millimetres, right? Be careful, because that is the trap. The gauge already read zero point zero three with nothing between the studs, so every reading is too big by that much. Corrected reading equals observed minus zero error: two point six seven minus, in brackets, plus zero point zero three, which gives two point six four millimetres. Clear?

5:34Example 2: new resonance length when the room warms to 37 C

Now Example two, our opening puzzle. Here it is: a resonance tube with an end correction of zero point six centimetres gives its first resonance at sixteen point zero centimetres at seventeen degrees Celsius, the small circle and capital C on screen. At what length does the first resonance occur with the same fork at thirty seven degrees Celsius? Pause the video and try it first. Done? Okay, let's check. Many students take the square root of thirty seven over seventeen. Is that the right ratio? No, it is not, and that is the first trap. Capital T must be in kelvin, the capital K on screen. Seventeen plus two hundred and seventy three gives two hundred and ninety kelvin, and thirty seven plus two hundred and seventy three gives three hundred and ten kelvin. T one and T two, with small labels below, are the two temperatures. With me? Now the second trap: we scale l one plus e, not l one alone. Call the new length l one prime, with the small dash called a prime. So l one prime plus e equals, in brackets, sixteen point zero plus zero point six, times the square root of three hundred and ten over two hundred and ninety. Okay? Sixteen point zero plus zero point six gives sixteen point six. The square root of three hundred and ten over two hundred and ninety is about one point zero three four. So sixteen point six times one point zero three four gives about seventeen point two centimetres. Right? Last step: take away the end correction, which does not change. Seventeen point two minus zero point six gives sixteen point six centimetres. So our puzzle is solved, and the loud spot does move. Sound is faster in warmer air, so the column is longer, sixteen point six centimetres instead of sixteen point zero.

7:35Quick revision: zero error sign, end correction, T in kelvin

Clear? So, quick revision. For the screw gauge, the reading is PSR plus CSR times the least count, and the least count is the pitch over capital N, the number of divisions. Corrected reading equals observed minus zero error, with its sign. For the resonance tube, l one plus e equals v over, four times f, and v goes as the square root of T, with T in kelvin. You have now seen three classic traps, and you know how to avoid every one of them, so well done. Now try the other problems of this set on this page, and all the best!