Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. There is a block floating in water, with three fifths of it under the surface. Now we put the same block in oil, which is less dense than water. Will it float higher, or sink deeper? Hold on to your answer; we will settle it today. And here is the bigger picture. In JEE, questions on this unit mix the topics freely: a stretched wire, then a floating block, then a capillary, then a cooling body. This video is your map for the mixed problems of this unit. So, let's begin.

0:36How this chapter is built

So, first, what makes this chapter different? JEE mixes the topics of this unit freely. Okay? Look at the arrows on screen; each arrow means, and then. First a wire's extension, then a floating block, then a capillary, then a cooling body. Four different chapters, back to back. Right? Now, where are the real exam questions? The JEE Main questions from twenty twenty one to twenty twenty six, and the JEE Advanced questions, with the older IITJEE ones in brackets, are at the end of chapters one to six. The dashes on screen mean to. Clear? This chapter is different. Its problems are original, written in the same style as those real questions. So, fresh practice. Got it? There are three sets: set one easy, set two medium, and set three hard, each with thirteen solved problems and three practice problems. Today, example one comes from set one, and example two from set two. With me?

1:40Values to take, unless stated

Next, the values. Unless a question says otherwise, this chapter takes small g, the acceleration due to gravity, as ten metres per second squared. Ten, not nine point eight. Okay? Rho, the Greek letter, with water written below, is the density of water: one thousand kilograms per cubic metre. Right? Small s, with water written below, is the specific heat of water: four thousand two hundred joules per kilogram per kelvin, or, in brackets, one calorie per gram per degree Celsius; the small circle before the C means degree. Four thousand two hundred, and one. Can they really be the same value? Yes, they can! One calorie is four point two joules. So one calorie per gram is four point two joules per gram, which is four thousand two hundred joules per kilogram. And, as the screen says, a change of one degree Celsius is a change of one kelvin. Clear? And L f, capital L with a small f below, f for fusion, is the latent heat of melting ice: eighty calories per gram. Keep it ready for example two. Got it?

2:56Toolkit 1: a stretched wire, a floating block

Now, your toolkit: one first line for each type of question. First, a wire stretched by a load. Delta L, with the Greek capital delta, the little triangle, is the change in length. It equals F L over A Y: F is the force, capital L the length, capital A the area of cross section, and Y is Young's modulus. With me? But tell me, does this formula work for any load at all? No! Only within the elastic limit, where Hooke's law holds and stress is proportional to strain. Beyond that limit, this formula does not apply. Okay? Second, a floating block. Any floating body displaces its own weight of liquid, so V sub over V, the fraction of its volume under the surface, equals rho b over rho l. The small sub below the V is for submerged, b is for the body, and the curly l is for the liquid. Clear?

3:53Toolkit 2: a capillary, a cooling body

Third, a capillary: a thin tube dipped in a liquid. The liquid rises to a height h, equal to two T cos theta, over r rho g. Theta, the Greek letter, is the angle of contact, small r is the radius of the tube, and rho is the density of the liquid. Right? Now, capital T in this formula. Is it the temperature? No, no, careful! In the capillary formula, capital T is the surface tension, not a temperature. Remember that. Got it? Now the fourth tool, for a cooling body. Here capital T is the temperature of the body, and T nought, with a small zero below, is the temperature of the surroundings. d T by d t, the rate of change of temperature, equals minus k times, T minus T nought, with k a constant. Same letter, two meanings, in one paper! Okay? This is Newton's law of cooling, and it holds only for a small excess temperature, T minus T nought, over the surroundings. Right?

4:55Example 1: fraction of a floating block submerged in oil

Now, back to our puzzle. Example one, an easy one from set one. So, a block floats in water with three fifths of its volume submerged. What fraction is submerged in oil of density eight hundred kilograms per cubic metre? Pause the video, and try it first. Done? Okay, let's check. Look at part a of the figure, on the left. The block floats, so its weight, m g, the black arrow, equals the upthrust, the green arrow: rho l times V sub times g. Right? So the submerged fraction is rho b over rho l. In water, the block's density, rho b, is three fifths times one thousand: six hundred kilograms per cubic metre. Clear? Now the oil, at eight hundred, is less dense than water. So does more of the block go under, or less? More! Very good! Six hundred over eight hundred: three quarters of the block is under the oil. Each cubic metre of oil pushes up less, so more of the block must go under. With me? And three quarters is more than three fifths; the pointed sign means greater than. So the block sinks deeper in oil. Puzzle solved! If you said it floats higher, that was the trap. Got it?

6:17Example 2: final temperature when ice melts in warm water

Now a harder one, a medium problem from set two. Fifty grams of ice at zero degrees Celsius is added to two hundred grams of water at forty degrees Celsius. Find the final temperature. Pause the video, and try it. Hint: use the chapter's values. Done? Okay, let's check. Some students simply average: two hundred times forty, over two hundred and fifty grams, gives thirty two degrees. Is that right? No, no, careful! That forgets the melting. Look at the graph of temperature against heat supplied: at zero degrees, the line goes flat while the ice melts. Heat goes in, but the temperature does not rise. So, first the heat available. Cooling the water from forty down to zero, the arrow on screen, is a sloping part of the graph: Q equals m s delta theta, where theta, the Greek letter, is the temperature here. Two hundred grams, times one, the specific heat, times forty. Eight thousand calories. Okay? Next, the heat to melt all the ice. That is a flat part of the graph: Q equals m L, mass times latent heat. Fifty times eighty, four thousand calories. Right? Eight thousand is more than four thousand, so all the ice melts, and the answer is above zero. With ice, always do this check first. Clear? The four thousand calories left over warm all two hundred and fifty grams of water up from zero, at one calorie per gram per degree. So eight thousand minus four thousand equals two hundred and fifty times capital T, the final temperature. With me? So capital T is four thousand over two hundred and fifty: sixteen degrees Celsius. Sixteen, not thirty two. The latent heat made all the difference. That's it!

8:12Quick revision: values to take, capital T trap, ice check

So, quick revision. Unless a question says otherwise, take small g as ten metres per second squared, rho water as one thousand kilograms per cubic metre, and L f as eighty calories per gram. Capital T means surface tension in a capillary, and temperature in cooling. And with ice, check that all of it melts before you equate heats. Now go and try the three sets of this chapter. With this map, you are ready for them. All the best!