 So, here are next set of questions, so let us take one by one. So question number 11, if a current in a conductor is 1.5 ampere, the number of free electrons are given and the area of cross section is also given, we need to find the drift velocity of the electron. So, looking at all these data immediately, the formula current is equal to NEAVD that comes in our mind, so you can directly substitute the values and get the answer. So, it is a straightforward question, the current is 1.5 that divided by number of free electrons which is 8 into 10 is power 28 to charge of an electron 1.6 into 10 is power minus 19 multiplied by area of cross section. Area of cross section is 5 mm square, so that is 5 into 10 is power minus 6, so this you can write it as 1.5 divided by 8 into 1.6 into 5, in the denominator 10 is power 3 will come, so that you can take it in the numerator, so that is 10 is power minus 3. So, here you will get 1 divided by 5 is of 40, so that into 15 by 16, so this into 10 is power minus 3 is what you will get. Now, 1 divided by 40, let us say you multiply and divide by 10, so this 10 is power minus 3 will become 10 is power minus 4, so this will come out to be 0.25 into 15 by 16 into 10 is power minus 4 meter per second, so you can say that this is 0.25 into 15 by 16, sorry 0.25 into 15 by 16 mm per second, now you can safely assume that 15 by 16 when you multiply that with 0.25 mm, the answer will come close to 0.02 mm, so that is the option number 1 is correct. So, you can see that this is a straightforward question, direct formula substitution and these kinds of questions should be solved first when you are solving any question paper. Here is the next question, so we have an arrangement which is as shown here, if a temperature difference T1 minus T2 is 120 degree Celsius, so between this point and that point, the temperature difference is 120 degree Celsius, you need to find temperature difference between A and B, so A is this and B is that, okay. Now, if you look at this structure, it is nothing but a collection of thermal resistances, right. So, you can just draw the circuit diagram like this, okay, so it will look like this, now each arm has the same, okay, it looks same, each arm looks same but they are not, okay. For example, this arm has a length of L by 2, that is L by 2, okay, and this one has length L, so let's say this is R1, this is also length L, let's say this is, this resistance will be equal to the top resistance, so let's say this is R1, let's say this is R2, that is R2 and this one R3 and this one is R3, okay, now thermal resistance is nothing but L by KA, okay, so assuming that all the conductors are made up of same material, you can assume K is same for all, okay, and area of procession also, we are assuming it to be same, okay, so if you write L R1, it will be R1, we are considering R1 to be this part, right, that is L by KA, R2 will be equal to what, L by 2 divided by KA, right, and R3 will come out to be L by 4 divided by KA, alright, so now these two temperatures are given to us, so T1 and T2, T1 and T2, so T2, sorry, T1 minus T2 is given to us, we need to find temperature difference between these two points, okay, so you can first find the equivalent, this thing, equivalent resistance you can find out, you can see that R1, R3, R1, R3 and this R3, they are in series and that is connected parallel to R1, okay, so first you should get a relation over here that R1 is equal to 4 times of R3, okay, and R2 is equal to 2 times of R3, fine, so we try to write down everything in terms of R3 resistance, okay, so R1 is 4 times of R3, so you have 4 R3, R3 and R3, so you have 6 R3 parallel with 4 R3, okay, so you can see that 6 R3 parallel with 4 R3, this will be the equivalent resistance, okay, so 2, so 1 R3 is also gone, so you will have 12 by 5 R3 as the equivalent between these two points, okay and total equivalent resistance will be equal to what, total equivalent resistance will be 2 R2 plus this resistance, fine, so R2 is 2 R3, so 2 R2 becomes 4 R3, so 4 R3 plus this is your total equivalent resistance, so that is what 5 was at 20, so this will be 22 by 5 times R3, okay, so this current, the heat current dQ by dt which is in this branch like this will be equal to temperature difference that is 120, okay, divided by this, so that is 22 by 5 R3, okay, this is your dQ by dt, okay, now this dQ by dt will split between these two parts, okay, so let us say this is dQ1 by dt and this is dQ2 by dt, alright, so this I can write it as dQ1 by dt plus dQ2 by dt, alright, now dQ1 by dt that you can write it as delta T between these two points, okay, that divided by this dQ1 by dt, okay, the upper one, so delta T divided by R1 and R1 is 4 R3, this is 4 R3, okay, plus dQ2 by dt which is this one, now total resistance in the lower branches 2 times R3 plus R1, so that is 6 R3, so delta T divided by 6 R3, alright, so this will be equal to, let me simplify this a bit, so 120 into 5 divided by 22 R3, okay, just cancel out R3 like this, okay, then can divide with 2, so 211 and this is 60, fine, so solving this equation will get the value of delta T easily, alright, so you can see that calculation is heavy but the concept wise it is again a very very simple question, so if you keep your head cool you will be able to arrive at the answer very easily, okay, so let us move to the next one, alright, so you can see this is a mechanics question you can easily make out and this will be about the collisions only because coefficient of restitution is given, so you can read, before reading itself you can get an idea about this, so three blocks M again another mass M and capital M, they are kept on a frictionless floor as shown in the figure, okay, the left most block is given a velocity v towards the right, all the collisions between the blocks are perfectly inelastic, okay, so inelastic collision after collision the velocity separation becomes zero, so it will be as if two mass move together, okay, the loss in kinetic energy after all the collision is 5-6 the initial kinetic energy, you need to find ratio of capital M and small m, okay, so loss in kinetic energy is 5-6 of the initial kinetic energy, okay, so the final kinetic energy will be what, finite kinetic energy will be equal to initial minus loss, right, so it will be 1-6 of the initial kinetic energy, alright, so it will be equal to 1-6 multiplied by half M into v square, this should be your finite kinetic energy, alright, now let us say, let us see how we can proceed further, so this M when it hits that mass and if you conserve the momentum you will get M into v is equal to 2M into v1, so with this velocity v1 equals to v by 2 this M and that M, they will move together, okay forward and then they will again collide with capital M, again that will be inelastic only, right, if you conserve the linear momentum, so it will be 2M into the velocity that should be equal to 2M plus capital M into finite velocity v2, fine, so now v2 you will get as M into v divided by 2M plus capital M, right, so this is the finite velocity with which all three blocks after collision will start moving, okay, so the finite kinetic energy should be equal to half total mass which is 2M plus capital M into velocity square, so that is MV divided by 2M plus capital M whole square, fine, so this should be equal to one-sixth of half MV square, fine, so this square I can put it over here and one of this will get cancelled, so numerator here will have just coefficient one, so this half is gone, okay, 1M is gone and v square is also gone, so only M is remaining now, so you can see that 6M will be equal to 2M plus capital M, okay, so you will get capital M to be equal to four times of small M, okay, so capital M by small M should be equal to four, all right, so hence you will get option number three to be correct over here, okay, here is another question, we have a mixture of two moles of helium and one mole of argon, all right, and we need to find vRMS of helium divided by vRMS of argon, so again it is a direct formula based question, vRMS is what, vRMS formula is what, under root 3RT by molecular mass, okay, now you can see that the vRMS does not depend on number of moles you take, okay, it only depends on the temperature and molecular mass, okay, now since it is a mixture, I can say that both the gases are at the same temperature, okay, so vRMS of helium divided by vRMS of argon should be equal to under root of molecular mass of argon divided by molecular mass of helium, okay, so all you have to do is substitute over here, right, so molecular mass of helium is substitute here four and then molecular mass of argon you substitute there, you will get the answer, okay, that is how you solve this particular question and this is again a direct formula based question, so we have seen that in 2019 there are many, many simple questions, so the cutoff will be very high and that is why you need to be even more careful, okay, because if you make a silly error in a simple question, you know, since it is a simple question many of the students will get it right, so if you do a silly error then you may be behind many, many students, okay, so be very careful in a question paper which is very simple, all right, here is another question, the light of wavelength lambda 1, lambda 2 both wavelengths are given, they are incident on metallic surface, okay, so if the ratio of speed of the electrons ejected is 2, so let us say v1 by v2 is 2, okay, we need to find the work function of the metal, now which particular wavelength will correspond to higher velocity of the electron, of course the lower wavelength, right, lower wavelength means the frequency is higher and the energy of the photon is higher, so that particular photon will eject the electron with higher kinetic energy, okay, so let us quickly solve this one, so we have Einstein's relation, right, that is k max is equal to h mu minus work function, right, so this is k max, okay, so we have two different frequencies, so h mu 1 is equal to h mu 1 minus phi is equal to k max 1 and h mu 2 minus phi is equal to k max 2, all right, now if you take a ratio, if you take a ratio, you will get a simplistic term, so h mu 1 minus phi divided by h mu 2 minus phi is equal to ratio of kinetic energy, now ratio of kinetic energy will of course be equal to ratio of the velocity squares ratio, right, because mass will be constant, right, mass of electron is same, so using this, if I square it, I will get v 1 square by v 2 square to be equal to 4, now I can multiply half times mass of electron in the numerator and denominator, so ratio of the kinetic energy comes out to be 4, this will be equal to 4, okay, so h mu 1 minus phi comes out to be 4 h mu 2 minus 4 times phi, okay, so phi will be equal to one-third of, phi will be equal to one-third of, if you take h common, it will be mu 2 minus mu 1, okay, so just substitute the values now, mu is what, c by lambda, right, so you can simplify that as well, so it will be equal to h c by 3, 1 by lambda 2 minus 1 by lambda 1, okay, so like this you get the expression, right, so you will get this, let me check it quickly, if there are any cellarors, so 4 h mu 2 minus this, so I take this side, okay, so there will be 4 times of mu 2 over here, so 4 by lambda 2 minus 1 by lambda 1, okay, so now all you have to do is substitute the values and get the answer, so this problem, although conceptually it is just, you know, evaluation of few data points and you will arrive at the expression quickly, but the calculation wise, this is little involved, so you need to be good at calculation, so that is another reason why I tell that whenever you have some expression to be calculated, don't reach out to the calculator, because towards the end of your preparation, you can't suddenly make yourself very good at, you know, hand calculation, so don't reach out to calculators when you are doing practice.