Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Hang a twenty newton load on a wire, and it is one hundred point two centimetres long. Hang forty newtons instead, and it is one hundred point four centimetres. So, how long is the wire with no load at all? No area given, no Young's modulus given. Tricky, right? We will crack it in example one. Today is Part A of this unit's problem bank, at JEE Main level. Three quick ideas first, then two full problems. So, let's begin.

0:34Idea 1: a floating body

Idea number one, from fluids. Here, a body floats in water with twenty percent of its volume above the surface. Find its density, okay? First, the rule for floating: the body displaces its own weight of liquid. So V sub, the volume under the liquid, over V, the whole volume, equals rho b over rho l. Rho, the Greek letter, means density; the small b is for the body, and the small l for the liquid. Clear? Twenty percent is above the surface, so the density is zero point two times one thousand: two hundred, right? No, no, careful! The rule needs the part under water, not the part above. Under water is one hundred percent minus twenty percent: eighty percent. Got it? So rho b equals zero point eight times one thousand, the density of water: eight hundred kilograms per cubic metre, written k g m to the power minus three. Less dense than water, so it floats. Perfect!

1:34Idea 2: flow through a narrow tube

Idea number two, from viscosity. By what factor does the flow rate through a tube change if we halve its radius? Same pressure difference, streamline flow. With me? The tool is Poiseuille's equation. Q, the volume flowing per second, equals pi P r to the power four, over eight eta l. P is the pressure difference, r the radius, l the length of the tube, and eta, the Greek letter, is the viscosity. And the bracket matters: it holds only for streamline flow. Okay? With P, eta and l all unchanged, Q is proportional to r to the power four; this curly sign means proportional to. Clear? Now, half the radius means a quarter of the area. So a quarter of the flow, right? Oh no, careful! Q goes as r to the power four, not r squared. The arrow means becomes: r becomes r over two, so Q becomes one half to the power four times Q, which is Q over sixteen. One sixteenth of the flow, not one quarter. Got it?

2:40Idea 3: volume expansion of a solid

Idea number three, from heat. Here, a brass sphere of one hundred cubic centimetres, written c m cubed, is heated by one hundred kelvin. Alpha, the Greek letter, is its coefficient of linear expansion: two times ten to the power minus five per kelvin, written capital K to the power minus one. Find its increase in volume. Many students put alpha straight in, and get zero point two cubic centimetres, right? No, no, wait! Alpha is for length. For volume we need gamma, the coefficient of volume expansion. For an isotropic solid, one that expands equally in every direction, gamma equals three alpha: six times ten to the power minus five per kelvin. Okay? Now delta V, the change in volume, equals V gamma delta T, where delta, the little triangle, means change in, and delta T is the rise in temperature. So, one hundred, times six times ten to the power minus five, times one hundred. So delta V is zero point six cubic centimetres, three times the wrong answer. Length uses alpha, volume uses gamma. Clear?

3:50Example 1: natural length of a wire from two loads

Now, example one: our puzzle from the start. Here, a wire is one hundred point two centimetres long under a twenty newton load, and one hundred point four centimetres under forty newtons. The c m means centimetres, and capital N means newtons. Find its natural length. Pause the video, and try it first. Done? Okay, let's check. Look at this stress strain curve. Stress is force per unit area; strain is the fractional change in length. On the straight part, O A, stress is proportional to strain: that is Hooke's law, and its slope, Y, is Young's modulus. Point A is the proportional limit, and point B the elastic limit, or yield point. Beyond B come the plastic region, D, the ultimate strength, and E, the fracture, okay? On the straight part, below the proportional limit, the extension is proportional to the load. So L equals L naught plus k F: L naught, with the small zero, is the natural length, F is the load, and k F is the extension, with k a constant. Right? Now think. The load doubled, from twenty newtons to forty. Does the length double too? Of course not! Only the extension, k F, doubles; L naught stays the same. So write this line for both readings, and eliminate k. L naught equals L one F two, minus L two F one, over F two minus F one. The small one and two mark the two readings. With me? Now the numbers: one hundred point two times forty, minus one hundred point four times twenty, over twenty, the difference of the loads. That is four thousand and eight, minus two thousand and eight: two thousand, over twenty. So L naught equals one hundred point zero centimetres! Puzzle solved, and we never needed the area or Young's modulus. Very good!

5:46Example 2: heat to turn ice at -10 C into steam

Example two, a longer one. How much heat turns ten grams of ice at minus ten degrees Celsius into steam at one hundred degrees Celsius? The small circle and C mean degrees Celsius, and g means grams. The specific heat of ice, s ice, is zero point five. L f, the latent heat of fusion, is eighty, and L v, the latent heat of vaporisation, is five hundred and forty, in calories per gram, written c a l g to the power minus one. Pause the video and try it. Hint: count the stages. Done? Okay, let's check. Many write just two terms: melt the ice, then boil the water. Is that all? No, no, careful! Look at the curve for heating ice at a steady rate: temperature up the side, heat supplied along the bottom. On the sloping parts, the heat Q equals m s delta theta: mass, times specific heat, times delta theta, the change in temperature theta. On the flat parts, it is m times L, the latent heat. First slope: ice, minus ten to zero degrees, as the arrow shows. Ten times zero point five times ten: fifty calories. Okay? Next, the first flat part: melting at zero degrees, with no change in temperature. Ten times eighty: eight hundred calories. Then the water, from zero to one hundred degrees. The specific heat of water is one calorie per gram per degree Celsius. Ten times one times one hundred: one thousand calories. Right? Then the long flat part: boiling at one hundred degrees. Ten times five hundred and forty: five thousand four hundred calories. We stop there: the last slope, heating the steam, is not needed. Got it? Total: fifty plus eight hundred plus one thousand plus five thousand four hundred. Seven thousand two hundred and fifty calories, written c a l. Boiling is the biggest part by far. Never skip a stage!

7:49Quick revision: floating, Poiseuille, Hooke, ice to steam

So, quick revision. Floating: the fraction under water equals rho b over rho l. Poiseuille: Q goes as r to the power four. For solids, gamma equals three alpha, and below the proportional limit, L equals L naught plus k F, okay? Ice to steam: the sloping parts need m s delta theta, the flat parts need m times L, and you count every stage. Now try the three practice problems on this page. You have got this. All the best!