Test Papers – ICSE – Class – X
All you need to know about JEE Advanced 2016
Introduction to JEE Advanced
JEE Advanced is one of the most awaited and toughest exams occurring every year in India. JEE Advanced is guided and controlled by Joint Admission Board (JAB). The exam is conducted by the zonal IITs and hence it is often referred as IIT JEE Advanced examination. Aspirants who want to apply for JEE Advanced must clear the JEE Mains cut-off. Candidates who clear the JEE Advanced exam get the opportunity to be an esteemed part of the 16 IITs of the nation or ISM. Candidates get the choice to opt for a bachelor degree program or integrated Master/Dual degree program.
JEE Advanced 2016 Eligibility Criteria
The year 2016, brought a new pile of challenges in front of the JEE aspirants. As usual, the JEE Mains was tougher than the last year and the results of the Mains examination decided the fate of millions of aspirants. First 1, 50,000 Mains qualifiers were chosen to apply for the JEE Advanced based on the CML –
- For General Category – 50.5%
- OBC (Non-Creamy Layer) – 27%
- SC- 15%
- ST- 7.5%
Age Limit
- For General/OBC (NCL) – Aspirants must be born after Oct. 1, 1991
- For SC/ST/Pwd – Aspirants must be born after Oct. 1, 1985
Important Topics
Maths – Mathematics paper generally worries the JEE Mains & Advanced aspirants. However those who follow a smart but constant regimen and strategy during the JEE preparation, succeed for sure in the exam. The basic pillars of the JEE Advanced Mathematics are Differential & Integral Calculus and Algebra (complex numbers, quadratic equations, AP, GP and HP, Permutation & Combination, Binomial Theorem, complete matrix section, 3D analytical geometry and Vectors). These are the parts of Mathematics from where a JEE Advanced aspirant many expect numerous and tough questions.
Physics – For Mechanics section, an aspirant must go through relative velocity, projectiles, conservation of linear momentum and mechanical energy, center of mass, elastic and non-elastic collisions, Gravitational potential and field, Escape Velocity, parallel and perpendicular axes theorems, moment of inertia, Hooke’s law, Young’s modulus, Viscosity, surface tension, Stoke’s Law, Bernoulli’s theorem, Resonance & Doppler Effect.
The General Physics is totally unpredictable and hence one needs to go through what he/she can i.e; from Unit & Dimensions to Resistance.
In Thermal Physics, an aspirant must work hard on Idea Gas Laws (Cv, Cp), bulk modulus, Kirchhoff’s law, Stefan’s law, Wien’s displacement law, Coulomb’s Law, Capacitance, Ohm’s law (series & parallel), heating effect of current, Biot–Savart’s law, Ampere’s law and EMI.
For Optics, the important topics are Huygen’s principle, TIR, Combination of thin lenses and mirrors and double-slit experiment.
Chemistry – For Physical Chemistry, the vital topics are mole concept, oxidation-reduction problems, molarity, molality, normality, ideal gas equation, Kinetic theory of gasses, Bohr’s Model, De Broglie hypothesis, s, p & d orbital shapes and hybridisation, electronic configuration of elements, Aufbau’s Principle, Hund’s rule, Pauli’s exclusion principle, VSEPR model, Enthalpy, Entropy, Hess’s Law, pH and buffer solutions, Le Chatelier’s principle, Electrochemistry, Solid state, Nuclear Chemistry and Chemical Kinematics.
The Organic and Inorganic Chemistry should be prepared completely as per the Official JEE Advanced Syllabus without escaping any part. Candidates are advised to follow the Official JEE Syllabus only and prepare the maximum they can. However the primary parts have been discussed above to reduce the stress a bit.
Result & Top Colleges
JEE Advanced results have been declared on 12th June 2016. The result brought joy and disappointment at the same time for millions.
The appeared candidates can check their results on the Official JEE portal link – http://cbseresults.nic.in/jee/jee_2016.htm
They need to enter their JEE Advanced registration number along with their DOB.
The TOP 3 IIT JEE Advanced RANK holders are –
- Aman Bansal
- Bhavesh Dhingra
- Kunal Goyal
The TOP Colleges after clearing IIT JEE Advanced that one can anticipate include:
- 16 IITs & ISM Dhanbad
- IISc, Bangalore
- IIST, Thiruvananthapuram
- RGIPT, Rae Bareli
Making learning easier with the flipped class room
Imagine this scenario. A child is watching his lessons on the tab or smartphone. He is enjoying the lesson because it is filled with animations and presentations. The lesson is short so he doesn’t get bored and at the end of the lesson the child can assess himself by answering the objective type questions attached to the lesson. The software analyses the areas where the child is weak and presents the analysis to the child, who can go back to the lesson and relook at the concept to understand it thoroughly. Ultimately, he does comprehend the concept and passes the test on his device and also in school.
Many parents would go love the part where it says that the child enjoys the lesson, because that is impossible to imagine. However, it is believable once the word ‘animation’ enters the conversation. This plot is not from a sci-fi movie, but a concept that is fast catching on in the world of education. This is called the flipped class room. So called because the student learns his lessons at home leaving the class room time free for discussions, experiments, activity-based learning and tests.
The flipped class learning has been experimented upon in the west and it has been found that it increases student engagement, thus addressing the failure rate and the drop-out rate. The experiment was initially started for students who missed class lectures. The teachers were encouraged to record the lectures and email them to the students. Students without a computer at home were provided one in school to catch-up on the lectures without taking up class time.
The experiment has now grown to a concept that the lectures are presented in such a way that it includes animation, engages the students and helps in explaining concepts, rather than the teacher trying to tell examples told orally. It has been found the students who subscribed to the flipped class room learning method actually improved their grades since their conceptual understanding had increased.
MT Educare has been pioneering this innovative idea in India. MT Educare’s Robomate+ now offers lectures on Android mobile devices free of cost for the benefit of students. Students can download the free app and access the lectures they wish to see. The app was formally launched by Mr. Amitabh Bachchan, who is an active advocate of female literacy, along with being a legendary actor.
MT Educare is continuously improving the app and students are urged to take advantage of the service. It can help not only students who easily get bored of the usual classroom type, traditional learning, but also the class high rankers who can now improve their marks also.
AIIMS MBBS Exam 2016
AIIMS MBBS Exam 2016
Introduction to AIIMS MBBS Exam
All India Institute of Medical Sciences (AIIMS) is a well known and the most resourceful medical university and research institution under Indian government located in Delhi, India. The government organization conducts AIIMS MBBS Exam every year in India. Candidates who clear the AIIMS MBBS exam get the opportunity to experience advanced medical training from the expert faculties of AIIMS during their undergraduate course. Undoubtedly AIIMS MBBS is the toughest exam in the medical field in India but an undergraduate degree from AIIMS is capable of opening the routes to fortune for sure.
AIIMS 2016 MBBS Eligibility Criteria
Candidates who want to appear in the AIIMS MBBS Examination have to qualify a certain eligibility criteria in terms of age, passing percentage, academic board of education and primary subjects.
The minimum percentage requirement in the 10+2 results for the candidates to be eligible for AIIMS MBBS 2016 is
Category General SC/ST
Aggregate marks (%) 60 50
Candidates applying for AIIMS MBBS 2016 Exams must have PCB (Physics, Chemistry and Biology) and English in their 10+2 exam.
Candidates applying for AIIMS MBBS 2016 Exams must have accomplished their 10+2 from CBSE or a recognized Board.
Candidates applying for AIIMS MBBS 2016 Exams must be of 17 years of more on the date December 31, 2016.
Candidates applying for AIIMS MBBS 2016 Exams must be Indian.
Exam Pattern & Syllabus
The examination pattern for AIIMS MBBS 2016 is as follows –
| SUBJECTS | NO: OF QUESTIONS |
| PHY | 60 |
| CHEMISTRY | 60 |
| BIOLOGY | 60 |
| GENERAL KNOWLEDGE | 20 |
Candidates who applied for AIIMS MBBS 2016 Exams had the preference to choose Hindi or English language for their question papers. There was negative marking for the 200 MCQs. The primary subjects for the AIIMS MBBS aspirants are Physics, Chemistry, Zoology & Botany. The brief syllabus for each of the subjects have been mentioned below –
Result & Top Colleges
The result of the AIIMS MBBS 2016 Exam, which occurred on May 29, 2016, is out now. AIIMS MBBS result is one of the most awaited results of the year and now finally it calmed down the appeared candidates. For some, it’s a high time while for many it’s the opportunity to put some extra efforts next time. The cutoff marks of the AIIMS MBBS 2016 Exams are –
In the first counseling of AIIMS MBBS, a total of approx 24,000 students have been chosen. Deserving candidates among the selected ones will be offered the 672 seats in the seven AIIMS institutions.
Candidates can check their results on http://mbbs.aiimsexams.org
The top 3 rank holders of AIIMS MBBS 2016 Exams are –
- Sathvik Reddy Erla
- Nikhil Bajiya
- Patel Lajja Ben Jayesh Kumar
All the three fortunate candidates belong from OBC category.
After the first counseling dates i.e; on 4th, 5th, 7th and 8th July 2016, the selected candidates will get the seats in the AIIMS institutes including –
- AIIMS New Delhi
- AIIMS Bhopal
- AIIMS Bhubaneswar
- AIIMS Jodhpur
- AIIMS Patna
- AIIMS Raipur
- AIIMS Rishikesh
Turning the Class Inside out – the flipped classroom.
It is usual for most children to show promise while growing up. Most parents love to display their child’s talent of memorizing poetry and songs to relatives and friends. However, the same parents lament the loss of interest in studies as the children grow up. “He can do it,” says the parent. “If only he would take interest.” It is not rare to find a teacher scolding her student by recounting the songs her student can remember, but when it comes to studies, the teacher is disappointed at the student’s ability to remember simple theorems. Sometimes, it even happens that certain students show promise in class, but the next day they are struggling with the same lessons.
Why do some students fall behind in studies?
There are many answers to question why. Most of them are related to the interest levels of students, and that the lectures are boring. However, is there a solution to this? What if lectures were made interesting? They could include animation and presentations! It has long been accepted that children retain most information when it is presented to them in a pleasing manner.
What about the problem of students forgetting lessons once they reach home?
How about teaching the students at home, rather than in class? What? Why not? Instead of watching television, they could watch interestingly presented lectures in video format. These pre-recorded video lectures can be viewed and reviewed as many times as needed for understanding of the student.
Then what use would the class be?
The classes could be used for solving queries and questions that some students would still have after watching the videos. The class time can be better used for assessment of the student and the teacher can keep an eye on the progress of the student to see if they are ready for the exam. The time lost by teachers and students writing important notes on the board can be saved and when the teacher is explaining a concept in class she would be actually explaining a finer point rather than explaining something new.
Welcome to the flipped class room. This is the place where students and teachers interact freely while trying to assess the weaknesses of students and remedy those shortcomings. This is where technology helps students learn better because, they can now plan their studies and assess themselves through questions built into the software.
These kinds of class rooms have already been experimented upon in some countries and have shown promising results. India, a giant in the software sector can easily take the lead using this approach to take advantage of its young and growing population. This approach although not new is still in its development stage and the field is open to design new pathways and perspectives. With the help of new and improved technology the question of learning new concepts maybe easily answered.
MT Educare has taken the lead in this field and launched the Robomate+app. The app is free to download and has video lectures by expert tutors along with presentations and animation. The software has tests at the end of each lecture to assess the progress of the student. The student can plan his studies and make notes with the help of the software.
This is a good start and MT Educare’s Robomate+ is the world’s largest curriculum based study app. It can be hoped that in the coming years the situation can only improve and students can study smarter.
JEE MAIN & ADVANCE 11th PCM Maths Demo Videos
Hello, students, welcome back. In this module we will discuss the Method of Difference. Now, let us see, what does this concept mean. Method of difference, let say I am given sequence a1, a2, a3 till an and this sequence doesn’t look like A.P. or G.P. or A.G.P. and we have to find the sum of this sequence. Now, students, if we are able to find the nth term we can use summation concept. And, students, what you observe over here is, that the difference between consecutive terms is forming an A.P. or G.P. The given sequence is not an A.P. or GP but the difference between the consecutive terms is forming an A.P. or G.P. Then, students, we can use this concept to find the nth term. Students, what I am going to do is, I have written the sum of this series S as the first equation. I am going to rewrite the given sum and I am writing that as second equation. Now, students, I am going to subtract these two series. I know subtracting these two series, S will cancel out with S will obtain zero. And I am going to subtract, students, see over here. S minus S will become equal to zero. a1 the first term remains as it is, a2 minus a1, I have formed in the bracket, a3 minus a2 I have formed second bracket and so on till an minus an minus 1. The last an over here which should be over here and minus an has gone over there on the left hand side. Now, students, an is over here and now, students, we know that difference between the consecutive terms is forming A.P. or G.P. therefore a2 minus a1, a3 minus a2 and so on, can be written in form of A.P. or G.P. and hence their sum could be found out. And hence, I will be able to find out an that is the general term and now since I am able to find the general term everything comes down to this summation series and we know, how to do that.
Now, students, let us take an example and understand exactly what we have to do over here. Let’s say students, I am given the series 5, 7, 11, 17, 25 and so on. And I have to find the sum of nth term of this series. Students, note 5, 7, 11, difference between 7 and 5 is 2, 11 and 7 is 4, 17 and 11 is 6, so difference is 2, 4, 6 and so on. So, difference is forming in A.P. 2, 4, 6 and so on. Therefore students, over here to find the nth term, I will use the method of difference. Sum S, I have already written. Again S I am going to write it but shifting the terms, taking the differences, students, so when I subtract I will get tn. You can see over here, students, on the right hand side I have done this, over there on the left hand side there is zero, tn from here will go on left hand side. So, tn will be equal to 5 plus, you can see over here, students, the bracket there are n minus 1 terms. First term is 2 then 4, 6, 8 that is an A.P. that is sum of n minus 1 terms in A.P. sum of n minus 1 term in A.P. I have written the formula. It just a simple calculation, students. So, tn comes out as n square minus n plus 5. So, when I have to find this sum of series, it’s simple summation of tn and now the sum of terms would be r raised from 1 to n, summation r square minus summation r raised from 1 to n from r plus 5 times summation r raised from 1 to n again 1. Now, it’s the simple application of formula, friends, summation r square is n into n plus 1 into 2n plus 1 divided by 6. Summation r is n into n plus 1 by 2 and we know summation r raised from 1 to n, 1 is simple n, so it becomes 5n. Calculating, is simple calculation, friends, taking out the common and now taking the LCM and then now simplifying it. I obtain the final result, n by 6 into 2n square plus 28. That gives us the final answer, students.
Let us take one more example and understand it. So, let’s say, students, I have to find the sum of series 1, 5, 11, 19, 29, you have to find the sum of these n terms. Now, 1 and 5 difference is 4, 5 and 11 difference is 6, 11 and 19 difference is 8. So, students we can see 4, 6, 8, so the difference again forms an A.P. So again to find the nth term, students, I will use the method of difference. Rewriting the series you can see over here, students, taking the difference on subtracting, I again obtain this series, this equation is obtained taking tn on the left hand side tn will be equal to 1 plus sum of n minus 1 terms. First term is 4, common difference is 2. So, n by 2 into 2n minus n to t. It’s simple calculations, students, calculating it, expanding the brackets to simplify. So, when I expand tn comes out as n square plus n minus 1. Sum of terms that is Sn would be the summation r raise from 1 to n, tr and that would be equal to summation r raise from 1 to n r square plus r minus 1. Now, splitting this summation and now we obtain summation r raise from 1 to n and r square plus summation r raise from 1 to n r minus summation r raise from 1 to n, 1. Substituting the formula, friend, that is very easy, now we have remembered that formula and now its calculation. So, taking out common and then calculating it and so, friends, we calculate and we obtain the final result to it. And final result comes out as that would be equal to n into n square plus 3n minus 1 whole divided by 3.
Students, I hope you understand this example, thank you.
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JEE MAIN & ADVANCE 11th PCM Chemistry Demo Videos
Hello, students, let us continue our discussion on Mole and Equivalent Concept.
Now in this module, first we will take an example on percentage purity and then we will discuss what is the percentage yield of a reaction. Let us take the example first.
An impure 6 grams of NaCl is dissolved in water and then treated with excess of silver nitrate. The mass of AgCl precipitate is found to be 14.35 grams. What is the percentage purity of NaCl? Now, in these cases first of all we will see what is the reaction. It is a double displacementry. It is a double displacementry reaction NaCl plus silver nitrate gives you AgCl plus NaNO3 and now, we will first develop the strategy here. What should be our strategy? Means any AgCl is precipitated, it will precipitate with the actual NaCl. So, using the mass AgCl precipitate we will first calculate the mass of NaCl that is present and then using this information we find the percentage purity. We have actual and theoretical it is given purity 6 gram, we can calculate. So, we write down the reaction and then first we do stoichiometric analysis. So, 1 mole NaCl gives you 1 mole AgCl. So, using the molar mass data, now they have given the atomic rates. Silver is 108, sodium is 23 and chlorine is 35.5. So, from here we can calculate the molar mass of NaCl as 58.5 and that of AgCl as 143.5. Now, from here converting moles into mass, we can write down 58.5 grams of NaCl is producing 143.5 grams AgCl. Now, we will do reverse analysis. 143.5 grams of AgCl is going to be produced by 58.5 grams of NaCl. So, 1 gram of AgCl is going to be produced by 58.5 divided by 143.5 grams of NaCl. Now, we are getting 14.35 grams of AgCl. So, that is going to produce, rather that is going to be produced by 58.5 by 143.5 into 14.35 grams of NaCl, which comes out as 5.85 grams of NaCl. This was actual amount. Theoretically how much is given, 6 grams. So, percentage purity is simply what 5.85 divided by 6 times 100 that is equal to 97.5.
Now, let us discuss a very important term percentage yield. What is Percentage Yield? Sometimes the reaction may not get completed according to the initial amount taken. Means whatever amount you have taken, whatever you predict it doesn’t happen. What are the possible reasons? The first possible reason is establishment of equilibrium. Now, this chapter you will read later in the portion of chemistry but let me tell you, that the amount of reactant you take, after a particular time it stabilises, it doesn’t go ahead. So, that is called as equilibrium. And therefore when you expect the corresponding product after the entire reaction, it doesn’t happen. Now, for the second reason, some of the reactants are not able to react due to formation of a protective layer. Suppose, there is a solid and over it we pour some liquid that reacts with the surface and then covers it up. Now, the remaining solid cannot react. And therefore we will not get the actual amount that we are thinking. Now, the third possible case would be when we take gaseous reactants they may simply escape out of the reaction chamber and will not give you the products. And the fourth possible reason can be other reactions happening in parallel for the same reactants. Like, till now you have studied N2 and H2 can give you ammonia, NH3, but they can also combine to give you N2H4, which is called as Hydrazine. So, you may not get ammonia in total that you are thinking. Now, in such cases the efficiency of the process is measured using percentage yield. What is percentage yield? It is the product mass that you actually get divided by the product mass that you theoretically predict times 100. Now, let us take an example for this.
A on controlled oxidation gives X according to the reaction. Stoichiometric is given to us. They are saying some of the reactant A oxidizes and only 70% of maximum yield is obtained. What mass of X is produced if 200 grams of A is taken? Now, what should be your strategy? First of all, we will calculate it normally. We will think that it is 100% yield and let’s find out how much mass of A that we can produce and then using the mass of X that we have calculated, we will calculate the actual mass by using the percentage yield 70%. So, let us see how we do it? We write down the stoichiometry of the reaction. 2A plus 9B gives you 2X plus 4Y plus 5Z. But what is the interesting part for me? 2A and 2X, so we write down. 2 mole A is going to produce 2 mole X. So, 1 mole A is going to produce 1 mole X and we first convert the moles into mass. So, 128 grams of A will give you 128 grams of X based on the molecule data given to us. So, from here we can say 128 grams of A produces 148 grams of X and therefore 1 gram of A is going to produce 148 by 128 grams of X. By taking 100% yield, 200 grams of A is going to produce 148 by 128 times 200 grams of X. Now, since percentage yield is equal to product mass actual divided by product mass theoretical times 100. We will rearrange this equation because we want mass actual. We can write it down as mass actual is equal to percentage yield by 100 times mass theoretical. Now, putting in the value, yield value is 70%, so, 70 by 100 times this much mass and that is going to give us 161.875 grams. So, first we calculate the actual amount or rather the theoretical amount and then we calculate the actual amount using the yield.
Thank you.
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