Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take a black body at twenty seven degrees Celsius. Now heat it to three hundred and twenty seven degrees Celsius. The Celsius reading went up more than twelve times. So, how many times more power does it radiate? Twelve times? Or twelve to the power four, more than twenty thousand times? We will check at the end. Today: a mixed set from the whole unit, wires, fluids, bubbles and heat. Questions like these come up often in JEE Main, each in about a minute. So, let's begin.

0:38Tool 1: stretching a wire

So, tool one: stretching a wire. Delta L, the extension, equals F L over A Y, okay? Delta, the little triangle, means change in. F is the load, L the length, capital A the cross section area, and Y is Young's modulus Y of the wire. And the bracket matters: this holds only within the elastic limit. Clear? From our set: two wires of the same material have lengths in the ratio one to two, and diameters in the ratio two to one. They carry equal loads. What is the ratio of their extensions? Let's see. Same material means the same Y, and equal loads mean the same force. So delta L is proportional to L over capital A; this curly sign means proportional to. For a round wire, the area grows as the diameter d squared, so delta L is proportional to L over d squared. With me? The first wire is half as long, and twice as thick. Half times half, so the ratio is one to four, right? No, no, careful! The diameter gets squared. Delta L one over delta L two, small one and two for the two wires, equals one over two squared, over two over one squared: one quarter over two, one eighth. So the ratio is one to eight, not one to four. Got it?

2:06Tool 2: pressure at a depth, and flow in a pipe

Now, tool two: fluids. At a depth h below the free surface, P equals P naught plus rho g h. P naught, with the small zero, is the pressure on the surface, here the atmosphere. Rho, the Greek letter, is the density, and g is the acceleration due to gravity. Okay? From our set: at what depth in a lake is the total pressure twice the atmospheric pressure, two P naught? Take P naught as ten to the power five pascal, written P a. The total pressure is two P naught. So rho g h must equal P naught: ten to the power five pascal, and one pascal is one newton per square metre. The double arrow means so. Right? For water, rho is one thousand kilograms per cubic metre, and g is ten. So one thousand times ten times h equals ten to the power five, and h is ten metres. Just ten metres of water adds one whole atmosphere! Clear? Now flow. For an ideal fluid, incompressible and non viscous, in steady flow, capital A one v one equals capital A two v two. Area goes as diameter squared, so the speed v is proportional to one over d squared. With me? Try this one. Water flows at one metre per second in a pipe of diameter four centimetres, which narrows to two centimetres. The diameter halves, so the speed doubles, two metres per second, right? Oh no, careful! Same trap as the wire. The arrow shows d going from four to two centimetres. So v equals one, times four over two, the whole squared: four metres per second, written m s to the power minus one. Four, not two! Got it?

4:01Tool 3: excess pressure in drops and bubbles

Next, tool three: surface tension. Across a curved liquid surface, the pressure inside is higher than outside. For one surface of radius r, this excess pressure, delta P, equals two T over r. Careful: here capital T is the surface tension, not temperature. That covers a drop, or an air bubble in a liquid. Okay? Now, a soap bubble in air. Look closely at its thin film. How many surfaces does it have? Two! Yes, exactly! An outer surface and an inner surface. So the excess pressure doubles: delta P equals four T over r. Right? From our set: radius two centimetres, surface tension zero point zero three newton per metre, written N m to the power minus one. Two centimetres is zero point zero two metres. Four times zero point zero three, over zero point zero two: six pascal. Used two T over r? Then you got three pascal, half the answer. Count the surfaces first. Clear?

5:12Example 1: extension of a 2 m wire under a 10 kg load

Now, Example one. Here is a wire two metres long, with cross section one square millimetre, written m m squared. Its Young's modulus Y is two times ten to the power eleven pascal, and it carries a ten kilogram load, written k g. Find its extension. Pause the video and try it. Done? Okay, let's check. The load is given in kilograms, but F must be a force. So F equals m g, mass times g: ten times ten, one hundred newtons, written capital N, taking g as ten. Okay? Now the area. One millimetre is ten to the power minus three metres. So one square millimetre is ten to the power minus three square metres, right? No, no, wait! The conversion gets squared too. Ten to the power minus three, the whole squared, is ten to the power minus six. So capital A is ten to the power minus six square metres. With me? Now put everything in. Delta L equals F L over A Y: one hundred times two, over ten to the power minus six, times two, times ten to the power eleven. Clear? The bottom is two times ten to the power five, and the top is two hundred. So delta L is ten to the power minus three metres: just one millimetre. Huge Young's modulus, stiff wire, tiny stretch. Perfect!

6:45Example 2: Stefan law, black body heated 27 C to 327 C

Now, remember our puzzle from the start? Let's finish it. Here, the small circle and C mean degrees Celsius. So, a black body is heated from twenty seven to three hundred and twenty seven degrees Celsius. By what factor does its radiated power increase? Pause the video and try it. Hint: which temperature scale? Done? Okay, let's check. This is the Stefan Boltzmann law: P equals e sigma capital A T to the power four. P is the radiated power, capital A the surface area, capital T the temperature, and sigma, the Greek letter, is the Stefan Boltzmann constant. The small e is the emissivity, and for a black body, e is one. Okay? The power goes as T to the power four. So the factor is three hundred and twenty seven over twenty seven, to the power four, more than twenty thousand, right? No, no, careful! T to the power four needs the temperature in kelvin; the capital K on screen stands for kelvin. So add two hundred and seventy three. T one, with a small one, is three hundred kelvin, and T two is six hundred kelvin. Got it? Same body, so e, sigma and capital A all cancel. P two over P one equals six hundred over three hundred, to the power four. That is two to the power four: sixteen! Not twelve, not twenty thousand: just sixteen. Puzzle solved!

8:20Quick revision: d squared traps, 4T over r, T in kelvin

So, quick revision. One: diameter given? Area goes as d squared, so delta L goes as L over d squared, and v as one over d squared. Two: a soap bubble has two surfaces, so delta P is four T over r, not two T over r. Three: units first. One square millimetre is ten to the power minus six square metres, and T to the power four needs kelvin. Now try the three practice problems on this page, about a minute each. All the best!