Friction is a force which comes into existence when two bodies in contact have relative motion or tending to have relative with respect to each other and oppose the motion.

**Types of Friction:**

There are three types of Friction:

- Kinetic Friction
- Static Friction
- Rolling friction

**What is a Static Friction?**

The frictional force which exists between the two surfaces which are at rest or which are on the verge of relative motion with respect to each other is called the static friction.

**Laws of Static Friction:**

Limiting static friction between any pair of dry unlubricated surfaces is independent of apparent area in contact.

Limiting static friction between any pair of unlubricated surfaces depends on nature and material of the surfaces in contact.

Limiting static friction between any pair of dry unlubricated surfaces is directly proportional to the normal reaction.

If F_{s} is static friction and R (N) is a normal reaction.

F_{s} α R

∴ F_{s} = μ_{s }R

Where,

μ_{s} = Coefficient of Static Friction.

∴ \({{\mu }_{s}}\,\,=\,\,\frac{{{F}_{s}}}{R}\).

The coefficient of static friction is defined as the ratio of the limiting static friction to the normal reaction.

]]>An equation of the form \(f’\left( y \right)\frac{dy}{dx}+Pf\left( y \right)=Q\)… (i)

Where P and Q are constants or functions of x alone, can be reduced to linear form by putting f (y) = v.

\(f’\left( y \right)\frac{dy}{dx}=\frac{dv}{dx}\),

Then, Equation (i) becomes \(\frac{dv}{dx}+Pv=Q\). Which is linear in v and x.

**Example: **Find the solution of {xy³ (1 + cosx) – y} dx + x dy = 0.

**Solution:** Given that {xy³ (1 + cosx) – y} dx + x dy = 0.

The given equation can be written as \(\frac{dy}{dx}+{{y}^{3}}\left( 1+\cos x \right)-\frac{y}{x}=0\).

i.e, \(\frac{1}{{{y}^{3}}}\frac{dy}{dx}-\frac{1}{{{y}^{2}}x}=-\left( 1+\cos x \right)\).

Using the transformation \(\frac{-1}{{{y}^{2}}}=u\),

Differentiation with respect ‘u’ We get

\(\frac{2}{{{y}^{3}}}dy=du\).

The above equation reduces to \(\frac{1}{2}\frac{du}{dx}+\frac{u}{x}=-\left( 1+\cos x \right)\).

Whose \(IF={{e}^{\int{\frac{2}{x}}dx}}={{e}^{2\ln x}}={{x}^{2}}\).

Hence, the solution of the given differential equation is given by ux² = -2 ∫x² (1 + cosx) dx,

i.e, \(\frac{{{x}^{2}}}{2{{y}^{2}}}=\int{{{x}^{2}}}dx+\int{{{x}^{2}}}\cos x\,dx=\frac{{{x}^{3}}}{3}+{{x}^{2}}\sin x-\int{2x}\sin x\,dx\).

\(=\frac{{{x}^{3}}}{3}+{{x}^{2}}\sin x+2x\,cosx-2\int{\cos x}+C\).

\(=\frac{{{x}^{3}}}{3}+{{x}^{2}}\sin x+2x\,cosx-2\sin x+C\).

]]>Alliance University aspires to be among the best universities in the world by the year 2025 through a series of strategically crafted moves, precisely calibrated action plans and an unwavering commitment to the pursuit of excellence

**Eligibility Criteria:**

⇒ Passed 10+2 examination with Physics and Mathematics as compulsory subjects along with one of the Chemistry/ Biotechnology/ Biology/ Technical Vocational subjects.

⇒ Obtained at least 45% marks (40% in case of candidate belonging to reserved category) in the above subjects taken together.

⇒ JEE (Main); JEE (Advanced); Karnataka CET; COMED-K; Alliance University Engineering Entrance Test (AUEET); or any other State-level engineering entrance examination

**Admission Process:**

All short-listed applicants are required to appear for the Admission Selection Process at the Alliance College of Engineering and Design, Alliance University. The Process has two stages.

**Stage I:** Candidates are required to appear for the Alliance University Engineering Entrance Test (AUEET) conducted by the Alliance College of Engineering and Design.

**Stage II:** Candidates are required to appear for a Personal Interview.

**Mode of Application:** Online & Offline

**Application Fee:**

Indian Nationals: 1000/-

NRI/ SAARC Nationals: 2000/-

Foreign Nationals: USD$50.

**Test Pattern:**

Subjects | Questions | Marks | Duration |

Physics | 45 | 45 | 2 Hours (120 Minutes) |

Chemistry | 45 | 45 | |

Mathematics | 45 | 45 | |

Total | 135 | 135 |

**Important Dates:**

Application: Started

]]>Application started for admissions into UG & PG Courses 2019 – 2020 in Jain University: https://www.jgigroup.com/JU-UG/.

]]>All India Institute of Medical Sciences, New Delhi has proposed to Initiate Online Registration of applicants well in advance. Registrations will be a two-stage process comprising of Basic Registration and Final Registration.

From past experience it has been found that a majority of rejections are due to Improper filling of details other than educational qualifications. In the first stage applicants desirous of appearing in the AIIMS-MBBS Entrance Examination in future will be able to initially register by filling basic details and uploading images almost six month in advance (Basic Registration). These particulars and images will be screened, there would be adequate time to correct errors and deficiencies. The candidates will be informed whether their registration is accepted, and their details will be stored with an unique identification number, Thus, candidates will not be subjected to the pressures of last minute rush and the fear of rejection of applications without any time left to make amends. Basic Registration process is free of any charges and thus no payment shall be collected at this stage. Since applicants may not have completed their eligibility of examinations etc, certain details may not be collected at the time of Basic Registration.

At a specified time prior to the entrance examination for the year (AIIMS-MBBS-2019) the Basic Registration process will be closed and no further applications for that particulars year will be allowed. The Prospectus with Examination city will be uploaded before start of Final Registration. The applicants who have completed Basic Registration and this has been accepted can thus decide to either appear for the examination of this year or subsequently. In case any applicant does not wish to appear this year, he/she need not proceed to fill Final Registration but his/her Basic Registration details will be stored and will be valid for the subsequent year, In case the applicant desires to take the examination as announced, he/she can proceed for Final Registration. Please note that rejected Basic Registration applicants are not eligible to proceed for Final Registration. In Final Registration applicants can fill remaining details such as qualification details, choose the available city for examination and make the necessary payments. These applicants will be eligible to be issued Admit Cards for that examination subject to fulfilment of eligibility criteria as mentioned in the respective Prospectus.

After completion of Final Registration and uploading of Admit Cards, the Basic Registration process will then resume and continue for subsequent year. Those who have already completed Basic Registration once need not fill these details again whether they appear for AIIMS-MBBS Entrance Examination-2019 or not. However, every time the dates for Final Registration for subsequent Entrance Examinations are made, the Final Registration will have to be completed. Thus, while Basic Registration is enduring, Final Registration is examination year specific.

Thus, all prospective candidates will have the facility of ensuring their basic registration details are completed and accepted in all respects, well in advance. It would also ensure that only those applications that are complete in all respects and accepted at the Basic Registration stage are permitted to make the Final Registration including payment. Basic Registration details need not be filled, and images need not be uploaded repeatedly.

**This prospective Applicants Advanced Registration (PAAR) facility is being initiated for AIIMS-MBBS Entrance Examination 2019 onwards and will be continued in future also. For details please visit www.aiims.org from time to time.**

* The expected date for starting of Basic Registration is 2^{nd} week of November-2018 and Final Registration is 1^{st} week of February-2019.* Again, it must keep in mind that only those candidates who will fill the Basic Registration form and correct their errors and deficiencies within stipulated time as well as found their Basic Registration in order, will only be eligible to fill Final Registration for AIMS-MBBS-2019 Entrance Examination.

An equation of the form \(\frac{dy}{dx}+Py=Q{{y}^{n}}\). Where P and Q are Functions of X only and n ≠ 0, 1 is known as Bernoulli’s Equation.

It is easy to reduce the equation into linear form as below on dividing both sides by yⁿ. We get

\({{y}^{-n}}\frac{dy}{dx}+P{{y}^{1-n}}=Q\).

Put y¹⁻ⁿ = z.

\(\left( 1-n \right){{y}^{1-n}}\frac{dy}{dx}=\frac{dz}{dx}\).

∴ Given Equation becomes \(\frac{dz}{dx}+\left( 1-n \right)Pz=\left( 1-n \right)Q\).

Which is a linear differential equation in z.

Here, \(IF={{e}^{\int{\left( 1-n \right)P.dx}}}\).

Required solution is \(z\left( IF \right)=\int{\left( 1-n \right).Q.{{e}^{\int{\left( 1-n \right)P.dx}}}dx}\).

**Note:** If equation is of the form \(\frac{dx}{dy}+Px=Q{{x}^{n}}\), where n ≠ 0, 1 and P and Q are functions of y only. Then divide by xⁿ and put x¹ ⁻ ⁿ = z.

\(\left( 1-n \right){{x}^{-n}}\frac{dx}{dy}=\frac{dz}{dx}\).

**Example:** What is the solution of \(x\left( \frac{dy}{dx} \right)+y={{y}^{2}}\log x\).

**Solution:** The given equation can be written as \(\frac{dy}{dx}+\frac{1}{x}y=\frac{{{y}^{2}}}{x}\log x\) (since dividing throughout by x).

(Or) \(\frac{1}{{{y}^{2}}}.\frac{dy}{dx}+\frac{1}{x}.\frac{1}{y}=\frac{\log x}{x}\) … (i) (since dividing throughout by y²)

Now, put 1/y = v, so that \(\left( -\frac{1}{{{y}^{2}}} \right)\frac{dy}{dx}=\left( \frac{dv}{dx} \right)\).

With these substitutions Equation (i), becomes

\(-\frac{dv}{dx}+\frac{1}{x}.v=\frac{\log x}{x}\).

Or \(\frac{dv}{dx}-\frac{1}{x}.v=-\frac{\log x}{x}\).

This is linear with v as the dependent variable.

Here, P = -1/x and Q = – (log x)/ x

∴ \(IF={{e}^{\int{Pdx}}}\),

\(={{e}^{\int{\left( -\frac{1}{x} \right)dx}}}\),

= e^{-log x}

\(={{e}^{\log \left( \frac{1}{x} \right)}}\),

= 1/x.

Hence, the solution is \(v\left( \frac{1}{x} \right)=\int{\left\{ -\frac{\log x}{x} \right\}.\left( \frac{1}{x} \right)dx+C}\),

\(\frac{v}{x}=-\int{\left( \frac{1}{{{x}^{2}}} \right)\log x\,dx+C}\),

\(=-\left[ \left( -\frac{1}{x} \right)\log x-\int{\left( \frac{1}{x} \right)\left( -\frac{1}{x} \right)dx} \right]+C\) (integrating by parts 1/x² as the second function)

\(=\left( \frac{1}{x} \right)\log x-\int{\left( \frac{1}{{{x}^{2}}} \right)}dx+C\),

\(=\left( \frac{1}{x} \right)\log x+\frac{1}{x}+C\),

Or \(\frac{1}{\left( xy \right)}=\left( \frac{1}{x} \right)\left( 1+\log x \right)+C\),

Or 1 = y (1 + log x) + C xy.

]]>Friction is a force that tries to oppose motion between the two surfaces in placed contact. When we throw a ball on the floor it starting with some velocity. But once it is rolled ideally no force is acting in the direction of motion and according to Newton’s first law the ball should keep on rolling but this does not happen. The ball stops after moving certain distance, so a force must be acting on it. That force is known as Friction.

**What is a Kinetic Friction?**

The friction force act on the body when the body is moving is called the Kinetic friction. It always acts in opposite direction of the motion, not like static friction which act in the direction of motion. Kinetic friction is slightly less than the maximum static friction. Kinetic friction depends on the kinetic coefficient.

Force that resists motion of one body over another with which it is in contact, it is denoted by f_{k}. As motion starts static friction (f_{s}) vanishes and Kinetic friction (f_{k}) appears. It is independent of the area of contact and velocity of the body. It varies with Normal reaction (N).

f_{k} α N

f_{k} = Constant x N

f_{k} = μ_{k} x N

Where,

μ_{k} = Co-efficient of Kinetic Friction.

There are three cases can arise in a body’s motion.

**Case 1: **When applied force > f_{k}

F_{a} > f_{k}

(F_{a} – f_{k}) = ma

\(Acceleration(a)\,\,=\,\,\frac{\left( {{F}_{a}}-{{f}_{k}} \right)}{m}\).

**Case 2: **When applied force = f_{k}

F_{a} = f_{k}

∴ a = 0

The body moves with uniform velocity.

**Case 3: **When applied force = 0

F_{a} = 0

\(a\,\,=\,\,-\frac{{{f}_{k}}}{m}\),

No motion occurs, the body stops.

]]>SRM Institute of Science and Technology (formerly known as SRM University) is one of the top-ranking universities in India with over 38,000 students and more than 2600 faculty across all the campus, offering a wide range of undergraduate, postgraduate and doctoral programs in Engineering, Management, Medicine and Health sciences, and Science and Humanities.

**Eligibility Criteria:**

Kashmir migrants are eligible for relaxation in eligibility for admission subject to minimum eligibility requirements as per UGC.

**(a) Nationality and Age:**

Resident Indian or Non-Resident Indian (NRI), holder of PIO or OCI card issued by Government of India are eligible to apply for SRMJEEE (UG).

**Note:** Should have attained the age of 16 + on the 31^{st} July of the calendar year in which the 12^{th} Board examination is to be held (source).

**(b) Qualifying Examination:**

Minimum 50% aggregate in PCM/ PCB

(i) Passed in Higher secondary examination (10+2 pattern) or appearing in Higher Secondary examination in the current academic year with Physics, Chemistry and Mathematics or Biology or Biotechnology as major subjects in regular stream from any state board within India, CBSE, ISCE, Matriculation, or NIOS

(ii) GCE A-level or International Baccalaureate (IB) diploma or IB certificate with Physics, Chemistry and Mathematics as major subjects in any International schools within India B. Tech Biotechnology, B. Tech Biomedical Engineering and B. Tech Genetic Engineering:

**Note:** Only those Candidates who have passed the above qualifying examination in not more than **2 attempts** including app Testnce for improvement will be considered for admission

**(c) Direct Admission:**

To encourage and support students of exemplary talent, SRM Institute of Science and Technology (formerly known as SRM University) offers direct admission and scholarships to ¬rst rank students of all the central and state boards in India, top 1000 rankers in IIT JEE, top rankers in each district of Tamil Nadu and exemplary sports persons at National and International level.

**Admission Criteria:**

For majority of the programs, admission is done through National level entrance examination conducted by SRM Institute of Science and Technology (formerly known as SRM University) or authorized/ statutory boards or agencies. For few programs, admission is done directly. In addition to the above, the eligibility for admission is subject to fulfilling the requirement of minimum marks in the qualifying examination as well.

**Mode of Application:** Online & Offline

**Application Fee:**

Online Application Form: Rs.1100/-

Offline Application Form: Rs.1160/-

**Mode of Exam:** Online

**Exam Pattern:**

Parts | Subjects | No. of Questions | Marks per question | Marks | Duration |

Part 1 | Physics | 35 | 3 | 105 | 2 Hours 30 Minutes |

Part 2 | Chemistry | 35 | 3 | 105 | |

Part 3 | Mathematics/ Biology | 35 | 3 | 105 | |

Total Marks | 315 | ||||

**No Negative Marking |

**Important Dates:**

Application | Started |

Last date of Application | 31^{st} March 2019 |

Exam Dates | 15^{th} April, 2019 – 25^{th} April, 2019 |

Slot Booking | Candidates have to book their convenient test slots, subject to the availability of slots for a particular centre/ date. |

**What is Weight?**

Weight is the measure of the force of gravity acting on a body,

The formula for weight it is given by:

Weight(W) = Mass(M) x Acceleration due to gravity(g)

As weight is a force its SI unit is also the same as that of force, SI unit of weight is Newton(N). The direction of the weight acts downwards towards the centre of the earth. It is sometimes also called as gravitational force (or) gravity.

**What is Mass?**

Mass is one of the fundamental quantities in physics and the most basic property of matter. We can define mass as the measure of the amount of matter in a body. The SI unit of mass is Kilogram (kg).

Mass depends on the number and composition of atoms and molecules.

Mass(m) = Weight(W)/Gravity(g).

The mass of a body does not change at any time. Only for certain extreme cases when a huge amount of energy is given or taken from a body. For example, in a nuclear reaction tiny amount of matter is converted into a huge amount of energy, this reduces the mass of the substance.

**Mass and Weight Relation:**

Weight(W) = Mass(M) x Acceleration due to gravity(g)

Mass is directly proportional to weight.

Thus, if mass increases weight increases and vice versa, if mass decreases weight also decreases.

**Difference between Mass and Weight:**

Mass |
Weight |

Mass can never be zero. | Weight can be zero. As in space if no gravity acts upon an object, its weight becomes zero. |

Mass is commonly measured in kilograms and grams. | Weight is commonly measured in Newton’s. |

Mass doesn’t change according to location. | Weight varies according to location. |

Mass is a scalar quantity. It has magnitude. | Weight is a vector quantity. It has magnitude and is directed toward the centre of the Earth or other gravity well. |

Equations in which the variables are separable are those equations which can be expressed that the coefficient of dx is only a function of x and that of dy is only a function of y.

Thus, the general form of such an equation is f(x) dx + g(y) dy = 0. The solution of this equation is obtained by integrating f(x) and g(y) with respect to x and y respectively. i.e., solution is given by

∫f(x) dx = ∫ g(y) dy + C

**Example:** Find the solution of the differential equation sec²x. tan y dx + sec²y.tanx dy = 0

**Solution: **Given, sec²x tany dx + sec²y tanx dy = 0

On separating the variables, we get

sec²x tany dx = -sec²y tanx dy

\(\frac{{{\sec }^{2}}x}{\tan x}dx=-\frac{{{\sec }^{2}}y}{\tan y}dy\),

On integrating both sides, we get

\(\int{\frac{{{\sec }^{2}}x}{\tan x}dx}=-\int{\frac{{{\sec }^{2}}y}{\tan y}dy}\),

Let us consider

tan y = v

\({{\sec }^{2}}y=\frac{dv}{dy}\),

\(dy=\frac{dv}{{{\sec }^{2}}y}\),

\(\int{\frac{{{\sec }^{2}}x}{u}\frac{du}{{{\sec }^{2}}x}}=-\int{\frac{{{\sec }^{2}}y}{v}}\frac{dv}{{{\sec }^{2}}y}\),

\(\int{\frac{du}{u}=-\int{\frac{dv}{v}}}\),

log |u| = -log |v| + log |C|

log |tanx| = -log |tany| + log |C|

log |tanx tany| = log |C| (∵ log m + log n = logmn)

tanx. tany = C (∵ logm = logn → m = n)

Which is the required general solution.

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