JEE Main Questions (2019–2020): Measurement

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. You tighten a screw gauge fully, with nothing between its studs. So it should read exactly zero, right? But the zero of its circular scale sits three divisions below the line. The gauge is showing a reading for an object that does not exist! So do we add those three divisions, or subtract them? Measurement questions like this one came in JEE Main in twenty nineteen and twenty twenty, and one wrong sign changes the whole answer. So, let's begin.

0:35Screw gauge: pitch and least count

Let us start with the screw gauge. The screw moves along a scale called the main scale, or the pitch scale. The distance it moves in one complete rotation is the pitch. So the pitch is the distance moved, over the number of rotations, okay? Now, as the screw turns, its circular scale turns too, and one full rotation carries every circular division past the reference line. So how far does the screw move for just one circular division? Yes, exactly, it moves the pitch over the number of circular divisions. That is the least count, written L C on screen. With me?

1:13Screw gauge: reading the scales

Now look at this figure. On the left is the main scale, and on the right is the circular scale, with fifty divisions. The pitch here is zero point five millimetres, so the least count is zero point five over fifty, which gives zero point zero one millimetres. Clear? So let us read the gauge. The reading is the pitch scale reading, written PSR, plus the circular scale reading, written 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 the reading is five point five, plus twenty eight times zero point zero one, which gives five point seven eight millimetres. Got it?

1:52Screw gauge: the sign of the zero error

Now comes the real question, the zero error. Bring the studs into contact, with nothing between them. The zero of the circular scale should sit exactly on the reference line, right? But suppose it sits below the line. Has the gauge already read a little, or has it not even reached zero yet? Yes, exactly, it already reads a little, so the error is positive. Call the number of divisions small n, and the zero error is plus n times the least count, okay? And if the zero sits above the line, the error is negative, so the zero error is minus n times the least count. Right? Now here is a trap. When the zero sits above the line, the division you see on the line is capital N minus small n, where capital N is the total number of circular divisions. So do not take the number you see as the error, clear? Suppose you have found a positive zero error. Do we add it to the observed reading, or subtract it? We subtract it, because the corrected reading is the observed reading minus the zero error, with its sign. So a positive error is subtracted, and a negative error is added, because minus of a minus becomes plus. Write this line down. With me?

3:09Vernier callipers: the same rule

Now the vernier callipers, and the correction rule is the same. The reading itself is the main scale reading, written MSR, plus the vernier coincidence, written V C, times the least count. The vernier coincidence is the number of the vernier division that lines up with a main scale mark. Clear? Look at the left figure, with the jaws closed. The vernier zero lies to the right of the main zero, and the third division coincides. That is a positive zero error, the coinciding division times the least count. The least count here is zero point zero one centimetres, so the error is plus zero point zero three centimetres. As the figure says, subtract it from the reading. Right? On the right, the vernier zero lies to the left of the main zero, and the seventh division coincides. Here is the trap: the error is not minus seven times the least count. It is minus, in brackets n minus the coinciding division, times the least count. And this n is the total number of vernier divisions, not the n we used for the screw gauge. So that is minus, in brackets ten minus seven, times zero point zero one, giving minus zero point zero three centimetres, so we add that amount. Got it?

4:25Topper's trick: subtract two readings

Now let me share a small trick. Zero error is a systematic error. So take two readings that carry the same zero error, and subtract one from the other. What happens to the zero error? Yes, it cancels, because the same error is there in both readings. So when a question needs only such a difference, the zero error does not matter, okay?

4:49Example 1: least count from 6 rotations and 50 divisions

Alright, Example one, an easy one from JEE Main twenty twenty. If the screw of a screw gauge is given six rotations, it moves by three millimetres on the main scale. There are fifty divisions on the circular scale. Find the least count. Pause the video and try it first. Are you done? Okay, let us check it together. First we need the pitch. The question says three millimetres, so the pitch is three millimetres, right? No, careful, that is the trap, because the pitch is the distance for one rotation, and three millimetres is for six rotations. So the pitch is three over six, which gives zero point five millimetres, okay? The least count is the pitch over the number of circular divisions. Zero point five over fifty gives zero point zero one millimetres. Clear? Let us also write this in centimetres, so watch the unit. Ten millimetres make one centimetre, so the least count is zero point zero zero one centimetres.

5:55Example 2: sheet thickness with a positive zero error

Now the harder one, Example two from JEE Main twenty nineteen. The pitch of a screw gauge is zero point five millimetres, and it has one hundred circular divisions. When it is fully tightened without any object, the zero of its circular scale lies three divisions below the mean line. For a thin sheet, the main scale reads five point five millimetres, and the circular scale reads forty eight. Find the thickness of the sheet. Pause the video and try it first. Are you done? Okay, let us check it together. First we need the least count. Zero point five over one hundred gives zero point zero zero five millimetres, right? Now let us find the observed reading. The main scale gives five point five, and forty eight times zero point zero zero five gives zero point two four. That makes five point seven four millimetres, okay? But this is only what the gauge shows. Now we use the three divisions from our puzzle. The mean line is our reference line, and the zero lies below it. Is this zero error positive or negative? It is positive, because below the line means the gauge already reads a little. So the zero error is plus three times zero point zero zero five, which gives plus zero point zero one five millimetres. Clear? Now let us find the thickness. Five point seven four plus zero point zero one five gives five point seven five five millimetres. Is that correct? No, that is the trap, because a positive error is always subtracted. The thickness is five point seven four minus zero point zero one five, which equals five point seven two five millimetres. So there is our puzzle solved: those three divisions were never part of the sheet, so we took them out.

7:44Quick revision: pitch, least count, sign of the zero error

So, quick revision. First, the pitch is the distance for one rotation, and the least count is the pitch over the circular divisions. Second, a circular zero below the line gives a positive error, and above the line a negative error. Third, the corrected reading is the observed reading minus the zero error, with its sign. Now try the other solved questions on this page, before you read their solutions. The screw gauge and vernier zero error ones use exactly this method, so start with those. All the best!