Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. In the lab, we find an unknown mass with a metre scale, a knife edge and one known mass, using the principle of moments. Now take the same set to the Moon, where everything weighs much less. Do we get a different answer? Today we do concept checks from this unit's problem bank, and this Moon puzzle is one of them. Questions like these do come up in JEE, so let's begin.

0:30Check 1: random error or systematic error?

So, check one. Two errors from the lab: the zero error of a screw gauge, and the scatter in repeated timings of a pendulum. Which one is random, and which one is systematic? Take the zero error first. With the two studs in contact and nothing between them, the gauge should read zero, but it may already read a little. That small reading is the zero error. Okay? Now its sign, as the course gives it. If the zero of the circular scale is small n divisions below the reference line, the error is plus n times L C, the least count. If the zero is above the line, the error is minus n times the least count. Right? Then the corrected reading equals the observed reading, minus the zero error, with its sign. So a positive error is subtracted, and a negative error is added. Clear? Now see what it does. The zero error shifts every reading by the same amount, in the same direction. So it is systematic, as many of you guessed, and we remove it by correcting each reading. Got it? Now the pendulum. Time twenty oscillations again and again, and the numbers differ slightly each time. That scatter is random, and averaging reduces it, so we report the mean. With me?

1:49Check 2: backlash error in a screw gauge

Now check two, on the screw gauge. In the figure, the main scale, or pitch scale, is in millimetres, the m m on screen, and the circular scale has fifty divisions. The pitch, the distance moved in one full turn, is zero point five millimetres, so the least count is zero point five over fifty, which gives zero point zero one millimetres. Okay? The orange arrow marks five point five millimetres, the last visible mark, and division twenty eight is on the reference line. So the reading is five point five, plus twenty eight times zero point zero one, which gives five point seven eight millimetres. Right? Now the check itself. What is backlash error in a screw gauge, and how is it avoided? Inside, the screw turns in a nut, and there is always a little play, a little looseness, between them. That play causes backlash. Clear? Suppose you overshoot, and turn back. On reversing the direction of rotation, the head turns a little before the screw starts to move. So the circular scale changes while the screw stands still. Got it? The cure is simple. Always turn the screw in the same direction while approaching the final setting. Never go back and forth at the end. With me?

3:03Check 3: why is a null method more accurate?

Now check three, the metre bridge. The cell at the top, the long and short lines, drives the current. The orange line, from capital A to capital B, is a uniform wire one metre long. Capital R, the known resistance, is in the left gap, and capital S, the unknown, in the right gap. The circle marked capital G is the galvanometer, joined to the point D, and its arrow is the jockey, touching the wire at the point J, okay? We tap the jockey along the wire until the galvanometer shows no deflection: that is the null point, at a distance l from the end marked capital A, in centimetres, as the arrows below the wire show, with one hundred minus l on the other side. Then capital R over capital S equals l, over one hundred minus l. The formula beside it, for rho, the resistivity, is for another lesson. Right? Now the check. Why is a null method like this more accurate than reading a current on a meter? The answer is in the word null. At balance, no current flows through the galvanometer at all, so it only has to show zero. Okay? So the result does not depend on the calibration of the galvanometer, on whether its scale is marked correctly. Got it? It does not depend on the e m f of the cell, its electromotive force, or on its internal resistance either. Only the known resistance and the two lengths decide the answer, while a meter is only as good as its calibration. With me?

4:33Example 1: does a mass found by moments change on the Moon?

Now example one, the easier one, and it is our puzzle from the start. Would a mass found by the principle of moments come out differently on the Moon? Pause the video and try it first. Done? Okay, let's check. In the figure, the metre scale, marked from zero to one hundred centimetres, rests on a knife edge, the grey triangle, at its centre of gravity, so its own weight has no moment. Small m, the known mass, hangs at distance d one, the d with a small one below it, and capital M, the unknown, at distance d two. Okay? Now, on the Moon, g, the acceleration due to gravity, is much smaller, so both weights shrink. So will the balance shift, and the answer change? Let us not guess, let us write the balance. Small m times g times d one equals capital M times g times d two: the anticlockwise moment equals the clockwise moment, each a force times a perpendicular distance. Right? Now look closely. Both weights carry the same g, so g cancels from the balance equation. Capital M equals small m times d one, over d two, with no g at all. Clear? So with the same known mass and the same distances, we get the same capital M on the Moon as on the Earth. Nothing changes, and that solves our puzzle. Got it?

5:56Example 2: which error matters most in Searle's method

Now example two, a little harder. In Searle's experiment, in which of the quantities capital M, capital L, small d and small l does a given percentage error matter most, and with which instrument is it measured? Pause the video and try it first. Done? Okay, let's check. In Searle's experiment, a wire of length capital L and diameter small d carries a load of mass capital M, and small l is its extension. Young's modulus, capital Y, is stress over strain: four times capital M times g times capital L, over pi times d squared times small l. Okay? Now, the extension, small l, is so tiny that we read it with a micrometer. So its error must matter the most, right? No, and that is the trap, because the question fixes the percentage error, the same in each quantity. Delta, the Greek letter like a triangle, means the error in a quantity. Delta Y over Y equals delta M over M, plus delta L over L, plus two times delta d over d, plus delta l over small l. With me? Now look at the powers. Y varies as one over d squared, so the percentage error in d is counted twice, while every other quantity appears only once. So for the same percentage error, the diameter adds double, and it matters the most. Clear? And the diameter is measured with a screw gauge, as the last line says. So checks one and two come back: correct its zero error, and turn it one way only. That is the full answer: the diameter matters most, and a screw gauge measures it. Got it?

7:38Quick revision: zero error, backlash, null point, Searle's

So, quick revision. Zero error is systematic, so we correct it; scatter is random, so we average it. Turn a screw gauge one way only, to avoid backlash. At a null point the galvanometer's calibration drops out, and in a balance of moments, g cancels. In Searle's experiment, the diameter needs the most care. You handled these checks really well. Now try the practice problems on this page. All the best!