Past JEE Questions: Mechanics Experiments

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle from a JEE Main paper. Take a pendulum with a period of half a second, and a watch that reads only to one full second. The error of the watch is double the period itself! And yet the error in g, the acceleration due to gravity, comes out at just a few per cent. So how can that be? Let's solve that puzzle. This part of the course has eight past JEE Main questions, six on the simple pendulum and two on capillary rise. So, let's begin.

0:35Pendulum: which error counts double?

Let us start with the pendulum. For small swings, g equals four pi squared times L, over capital T squared. Pi is the Greek letter, L is the effective length, and capital T is the period. Both L and capital T carry errors. So which error counts double? Yes, exactly, the error in capital T counts double, because capital T is squared. Delta, the little triangle, means the error in a quantity. So delta g over g equals delta L over L, plus two times delta capital T over capital T, okay?

1:12Pendulum: where does the watch error sit?

Now here is where our puzzle lives. The watch times many swings together. So where does its error sit, on one period, or on the whole time? It sits on the whole time, and that is the key point. Let small t be the total time and n the number of oscillations. Then capital T is small t over n, and delta capital T is delta small t over n, so the swings share the error. Clear? Now divide one by the other, and the n cancels completely. So delta capital T over capital T equals delta small t over small t. Right? So here is the recipe. Delta g over g equals delta L over L, plus two times delta small t over small t. You divide the watch error by the total time, never by one period, okay? And delta small t is the resolution of the watch, which some questions call the least count. With me?

2:10JEE Main 2026: the graph of L against T

Now a question from JEE Main twenty twenty six, on the plot of length against period. Rearrange our formula, and L equals g over four pi squared, times capital T squared. So what shape does L against capital T make? It is tempting to draw a straight line, but capital T is squared here. So L against capital T is a parabola through the origin, curving upwards. Got it? And L against capital T squared is a straight line through the origin, okay?

2:43JEE Main 2024: capillary rise, hot or cold?

Now the second experiment, capillary rise. We will come to the question on screen in a moment, so first look at the figure. Water climbs a height h up a tube of bore two times r, where r is the radius. The rise h equals two times capital T times cos theta, over r times rho times g. Be careful, because capital T now means the surface tension, not the period. Theta is the Greek letter for the angle of contact, and rho is the Greek letter for density. Clear? The notes in the figure solve this for capital T, with a small meniscus correction, r over three, added to the height. For water and clean glass, theta is nearly zero, and that is the wavy equals sign. So, with the correction neglected, capital T is r times rho times g times h, over two. Both h and r are read with a travelling microscope. Right? Now the question on screen. The tube is dipped in hot water and in cold water, in a different order in each statement. Statement one says the rise is higher in hot water, and statement two says it is higher in cold water. So which one do you believe? The key fact is that surface tension falls as the temperature rises. So hot water, with its smaller capital T, gives a smaller rise. Got it? So statement one is false, and statement two is true.

4:09Example 1: error in g, 100 swings timed by a 1 s watch

Now let us do Example one, the puzzle we started with, from JEE Main twenty twenty two. In an experiment to find g with a simple pendulum, a time period of zero point five seconds is measured from the time of one hundred oscillations, with a watch of one second resolution. The length is ten centimetres, known to one millimetre. Find the accuracy in the determination of g. Pause the video and try it first. Are you done? Okay, let us check it together. The watch error is one second, and the period is zero point five seconds. So delta capital T over capital T is one over zero point five, which is two hundred per cent, right? No, that is the trap, and it is what made the experiment look hopeless. The watch timed all one hundred oscillations. So the total time is one hundred times zero point five, which gives fifty seconds, and the one second error sits on that. Got it? Now use our recipe. One millimetre is zero point one centimetres, so delta L over L is zero point one over ten. Then add two times one over fifty. Okay? That gives one per cent plus four per cent, which is five per cent. So the puzzle is solved. One hundred oscillations shared that one second error.

5:35Example 2: accuracy of pendulum energy from g and T errors

Now the harder one, Example two, from JEE Main twenty twenty one. The acceleration due to gravity is found to an accuracy of four per cent on a planet. The energy supplied to a simple pendulum of known mass m, to make it oscillate with time period capital T, is being estimated. The time period is measured to an accuracy of three per cent. To what accuracy is the energy known? Pause the video and try it first. Are you done? Okay, let us check it together. For a given angular amplitude, the energy, capital E, is proportional to m times g times L, and the curly symbol on screen means proportional to. Right? The mass is known, and the question says nothing about the length. So is the answer just the four per cent error in g? No, careful, that is the trap, because g and capital T fix the length. L equals g times capital T squared, over four pi squared. Put this in, and the energy is proportional to g squared times capital T squared. With me? Now use the same rule as in our recipe. So what does each per cent get multiplied by? Each per cent gets multiplied by the power, and both powers are two. Two times four per cent gives eight per cent, and two times three per cent gives six per cent. Together that makes fourteen per cent. If you got that, very good! So here is a trick worth keeping: first write the quantity using only what was measured, and then count the powers.

7:14Quick revision: error in g, L-T graphs, capillary rise

So, quick revision. First, delta g over g equals delta L over L, plus two times delta small t over small t, where small t is the total time. Second, L against capital T squared is a straight line, and L against capital T is a parabola. Third, surface tension falls as the temperature rises, so the rise is higher in cold water. Now try the other past questions on this page. All the best!