Past JEE Questions: Electricity Experiments

Transcript

0:00Introduction

Hello everyone, here is a small puzzle from an old JEE paper. In a metre bridge, the null point is at twenty centimetres. Now we make the smaller resistance four times as big. So the null point also goes four times as far, to eighty centimetres, right? That is wrong, and today you will see exactly why. This page has five past JEE questions on electricity experiments, and today we solve two of them on the metre bridge. So, let's begin.

0:31The metre bridge at balance

First, let us recall how the metre bridge balances. In the figure, the orange line from A to B is a uniform wire, one hundred centimetres long, and the battery at the top drives a current along it. In the gaps above the wire, R, the known resistance, sits on the left, and S, the unknown one, sits on the right, okay? Now the circle marked G is a galvanometer, joined at D, the point between the two resistances. The arrow from it ends on J, the jockey, which we tap gently on the wire, never slide. At one point the galvanometer shows zero deflection, so no current flows through it. That is the null point, or balance point. Clear? Its distance from the end A is small l, and the rest of the wire is a hundred minus l, as the double arrows below show. So at balance, R over S equals what? Yes, exactly, R over S equals l over a hundred minus l. The metre bridge is a Wheatstone bridge whose other two arms are the two parts of the uniform wire. The note on the right of the figure says the same. Below it, rho, the Greek letter for resistivity, uses pi, the diameter small d and the length capital L, and we do not need it today. Got it?

1:49Swap the two gaps: the mirror point

Now let us swap the two gaps, so S goes on the left and R on the right. Call the new null point l prime, which is l with the small dash on screen. Then S over R equals l prime, over a hundred minus l prime, okay? But from the old balance, S over R is just the old ratio turned upside down, which is a hundred minus l, over l. Right? Now compare these two lines. So where does the new null point land? Yes, it lands at a hundred minus l, the mirror point on the other side of the middle. Both lines have the same form, so l prime equals a hundred minus l. With me? Here is an example from the metre bridge lesson earlier in this chapter. Ten ohms are in the left gap, and capital omega, this Greek letter, is the symbol for ohm. The null point is at forty point zero centimetres, and after the swap it moves to sixty point zero centimetres. Forty and sixty add up to a hundred. Clear? That is why, in the lab, we swap the gaps and take the mean of the two results. It reduces the effect of the resistance at the ends of the wire and of any non-uniformity. Keep this mirror idea in mind for Example two. Okay?

3:06Example 1: X, Y null at 20 cm; find l for 4X and Y

Now Example one, our puzzle from the start, from A I triple E two thousand and four. In a metre bridge experiment, the null point is obtained at twenty centimetres from one end of the wire, when resistance X is balanced against another resistance Y, with X less than Y, which is the less than sign on screen. Where will the null point be, from the same end, if four times X is balanced against Y? Pause the video and try it first. Are you done? Okay, let us check it together. X becomes four times as big, so the null point also goes four times as far. Four times twenty gives eighty centimetres, right? That is exactly the trap, so be careful. The length is not proportional to X; the ratio, l over a hundred minus l, is. Since X is less than Y, X faces the shorter part of the wire, the twenty centimetres, and the other part is eighty. So X over Y is twenty over eighty, which is one quarter. Got it? Now four times X, over Y, is four times one quarter, which is exactly equal to one. Right? So l over a hundred minus l must equal one, which means l equals a hundred minus l. So l is fifty centimetres, right in the middle of the wire. Puzzle solved, the answer is fifty, not eighty. Very good if you worked with the ratio!

4:33Example 2: R1 + R2 = 1 kilo-ohm, swap shifts null by 10 cm

Now Example two, the harder one, from JEE Main twenty eighteen. On interchanging the resistances, the balance point of a metre bridge shifts to the left by ten centimetres. The resistance of their series combination is one kilo ohm, written as k with capital omega, which is one thousand ohms. How much was the resistance in the left slot before the interchange? Pause the video and try it first. Are you done? Okay, let us check it together. Let R one, which is R with a small one below it, be the resistance in the left slot, and R two the one in the right slot. Before the swap the balance point is at l, so R one over R two equals l over a hundred minus l, okay? Now we swap them, and the balance point shifts ten centimetres to the left. So where is the new balance point? Yes, at l minus ten, because the lengths are measured from the left end. The rest of the wire is a hundred minus the bracket, l minus ten, which is a hundred and ten minus l. So R two over R one equals l minus ten, over a hundred and ten minus l. Right? Now multiply the two lines. On the left, R one over R two, times R two over R one, is exactly equal to one, as the line says. So l times the bracket, l minus ten, equals the bracket, a hundred minus l, times the bracket, a hundred and ten minus l. With me? Open the brackets. The left side is l squared minus ten times l, and the right side is eleven thousand, minus two hundred and ten times l, plus l squared. The l squared cancels, so two hundred times l equals eleven thousand, and l is fifty five centimetres. Clear? Now, R one plus R two is one thousand ohms. So how do we get R one alone? From the first line, R one is l parts out of a hundred of the total, so fifty five parts. So R one equals fifty five over a hundred, times one thousand, which gives five hundred and fifty ohms, and R two is four hundred and fifty ohms. The question asks for the left one, so the answer is five hundred and fifty ohms. Got it? And here is the quicker way, with our mirror rule. After the swap, the new point must be at a hundred minus l. So l minus ten equals a hundred minus l, two times l is a hundred and ten, and l is fifty five again, in one line. If you spotted the mirror, very good!

7:12Quick revision: balance ratio, swap rule, 4X trap

So, let us do a quick revision. At balance, R over S equals l over a hundred minus l, for a uniform wire. Then, swap the gaps, and the null point moves to the mirror point, a hundred minus l. And always work with the ratio, never the length, because four times X does not give four times l. Three more past questions wait on this page, on the diode, a non-uniform wire and a galvanometer. Try them first, and all the very best to you.