What is TS EAMCET?
Telangana State Engineering, Agriculture and Medical Common Entrance (TS EAMCET) 2018 is being conducted by JNT University Hyderabad (JNTUH) on behalf of Telangana State Council of Higher Education (TSCHE). This state level examination is the prerequisite for admission into various professional courses offered in University / Private Colleges in the state of Telangana and held once in a year.
Eligibility:
a) Candidates should be of Indian Nationality or Persons of Indian Origin (PIO) / Overseas Citizen of India (OCI) Card Holders.
b) Candidates should belong to the state of Telangana / Andhra Pradesh. The candidates should satisfy local / non – local status requirements as laid down in the Telangana / Andhra Pradesh Education Institutions (Regulations of Admission) order, 1974 as subsequently amended.
c) Candidates should have passed or appeared for the final year of Intermediate Examination (10+2 pattern) with Mathematics, Physics along with Chemistry /Biotechnology / Biology as optionals or related vocational courses in the fields of Engineering and Technology, conducted by the Board of Intermediate Education, Telangana / Andhra Pradesh along with bridge course or courses conducted by it for candidates enrolled during 2000-2002 and subsequent batches, or any other examination recognized as equivalent thereto by the Board of Intermediate Education, Telangana / Andhra Pradesh, provided that candidates who have passed or appeared for the final year of Intermediate Examination (10+2 pattern) with Biology, Physics and Chemistry as optionals along with the bridge course examination in Mathematics conducted by the Board of Intermediate Education, Telangana / Andhra Pradesh shall also be eligible for the Bio-Technology course.
d) Candidate should obtain at least 45% of marks (40% in case of candidates belonging to reserved category) in the subjects specified taken together at 10+2 pattern.
Important Dates:
Commencement of Submission of Online application forms | 04-03-2018 |
Last date for submission of online applications without late fee | 04-04-2018 |
Correction of online application data already submitted by the candidate | 06-04-2018 to 09-04-2018 |
Last date for submission of online applications with additional late fee of Rs.500/- | 11-04-2018 |
Last date for submission of online applications with additional late fee of Rs.1000/- | 18-04-2018 |
Last date for receipt of online applications with additional late fee of Rs.5000/- | 24-04-2018 |
Last date for receipt of applications with additional late fee of Rs.10000/- | 28-04-2018 |
Downloading of Hall Tickets from the website eamcet.tsche.ac.in | 20-04-2018 to 01-05-2018 |
Date of TS EAMCET-2018 Examination | |
(a) Agriculture and Medical (AM)
FN: 10-00 A.M. to 1-00 P.M. AN: 3-00 P.M. to 6-00 P.M. |
2-5-2018 (FN)
2-5-2018 (AN) 3-5-2018 (FN) 3-5-2018 (AN) |
(b) Engineering (E)
FN: 10-00 A.M. to 1-00 P.M. AN : 3-00 P.M. to 6-00 P.M. |
4-5-2018 (FN)
4-5-2018 (AN) 5-5-2018 (FN) 5-5-2018(AN) 7-5-2018 (FN) 7-5-2018 (AN) |
Application Fee:
Stream |
Registration Fee |
Agriculture & Medical (AM) | Rs.400/- (for SC/ST candidates)
Rs.800/- (for others) |
Engineering (E) | |
Candidates who wish to write the test in both E & AM streams | Rs.800/- (for SC/ST candidates)
Rs.1,600/- (for others) |
If u and v be two functions of x, then the integral of product of these two functions is given by \(\int{uv}.dx=u\int{vdx-\int{\left[ \frac{du}{dx}\int{vdx} \right]}}dx\).
In applying the above rule care has to be taken in the selection of the first function (u) and the second function (v) depending on which function can be integrated easily. Normally we use the following methods for making this choice:
If both functions are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually we use the following preference order for the first function.
(Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)
This rule is called as ILATE which states that the inverse function should be assumed as the first function while performing integration. Hence, the functions are to be assumed in the order from left to right depending on the type of functions involved. Various other cases have been discussed in detail in the later sections.
Examples
1. Evaluate∫ xsec²x dx.
Solution: Let I = ∫ xsec²x dx.
= x (tanx) – ∫ tanx dx.
= x tanx – log|secx|+c
2. Evaluate∫logx/x² dx.
Solution: ∫logx/x² dx = (logx)(-1/x) + ∫1/x.1/x dx
=-1/x logx – 1/x² + c
3. ∫(1ogx)² dx
Solution: ∫(1ogx)² dx = (1ogx)²x – ∫x.2logx.1/x dx
= (1ogx)² x – 2∫logx. dx
= (1ogx)² x – 2(xlogx-∫x.1/x dx)
= (logx)² – 2x.logx + x + c
]]>Soon after Rutherford’s scattering theory had been confirmed by experiment (about 1913), the one-to-one association of an atomic number Z with each element was solidified by the work of Henry Moseley (1887- 1915). He used Bohr’s model of atomic structure to determine the energy emitted when low-level electrons change orbitals. This energy has a strong dependence on atomic number, so that by measuring the energy of the x-rays characteristic of an element, its atomic number Z can be unambiguously determined. In today’s lab you will measure the x-ray spectra of a number of elements and also identify several unknown elements by looking at their characteristic x-ray spectra.
A law that relates the frequency of the spectral lines of the characteristic X radiation of a chemical element to its atomic number. This law was experimentally established by H. Moseley in 1913. According to Moseley’s law, the square root of the frequency v of a spectral line of the characteristic radiation of an element is a linear function of its atomic number Z.
\(\sqrt{\frac{\nu }{R}}\,=\,\frac{Z\,-\,{{S}_{n}}}{n}\)
Where,
R = Rydberg constant,
S_{n} = Screening constant,
N = Principal quantum number.
On a Moseley plot, the dependence of √v on Z is a series of lines (Such as the K_{α} lines,
L_{α} lines, and M_{α} lines, which correspond to the values n = 1, 2, and 3).
Moseley’s law was incontrovertible proof of the correctness of the arrangement of the elements in D. I. Mendeleev’s periodicsystem of the elements and the law helped to clarify the physical significance of Z.
According to Moseley’s law, the characteristic X – ray spectra do not display the periodic regularities that are inherent in optical spectra.
This indicates that the inner electron shells of the atoms of all elements, which are manifested in the characteristicX-ray spectra, have an analogous structure.
Subsequent experiments revealed some deviations from a linear.
Moseley plot for the K_{α} lines, L_{α} lines, and M_{α} lines of characteristic X – radiation. The atomic number of the element Z is plotted along the axis of abscissas and the quantity √v/RC is plotted along the axis of ordinates.
Dependence for the transition groups of elements: The deviations being due to the change in the order in which the outer electron shells are filled and also for heavy atoms, in which that the velocities of the nucleus and the state of the outer electron shell. The study of these shifts makes it possible to obtain detailed information about the atom.
]]>Suppose f : R→R is a continuous periodic function with periodic T. let a ϵ R. Then prove that for any positive integer n, \(\int\limits_{a}^{a+nT}{f(x)\ dx\ =n\int\limits_{a}^{a+T}{f(x)\ dx}\ }\)
Problems:
1. Evaluate \(\int\limits_{0}^{\pi /2}{\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\ dx}\)
Solution: let us I = \(\int\limits_{0}^{\pi /2}{\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\ dx}\) … (1)
= \(\int\limits_{a}^{\pi /2}{\frac{f(\sin (\pi /2-x)\ dx}{f[\sin (\pi /2-x]+f[\cos (\pi /2-x]}}\ \),
\(\begin{align}& \\& I=\int\limits_{0}^{\pi /2}{\frac{f(\cos x)}{f(\cos x)+f(\sin x)}\ dx} \\\end{align}\) … (2)
Equations 1 + 2
2I \(=\int\limits_{0}^{\pi /2}{\frac{f(\cos x)}{f(\cos x)+f(\sin x)}\ dx}+\int\limits_{0}^{\pi /2}{\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\ dx}\).
\(=\int\limits_{0}^{\pi /2}{\frac{f(\sin x)+f(\cos x)}{f(\cos x)+f(\sin x)}\ dx}\),
2I = \(\int\limits_{0}^{\pi /2}{1.\ dx}\).
2I = \(x|_{0}^{\pi /2}\),
I = π/4.
2. Evaluate \(\int\limits_{0}^{\pi /2}{\frac{f(\tan x)}{f(\tan x)+f(\cot x)}\ dx}\).
Solution: Let us I = \(\int\limits_{0}^{\pi /2}{\frac{f(\tan x)}{f(\tan x)+f(\cot x)}\ dx}\) … (1)
I \(=\int\limits_{0}^{\pi /2}{\frac{f(\tan (\pi /2-x)\ dx}{f[\tan (\pi /2-x]+f[\cot (\pi /2-x]}}\),
I =\(\int\limits_{0}^{\pi /2}{\frac{f(\cot x)}{f(\cot x)+f(\tan x)}\ dx}\) … (2)
Equations 1+ 2
2I = \(\int\limits_{0}^{\pi /2}{\frac{f(\cot x)}{f(\cot x)+f(\tan x)}\ dx}+\int\limits_{0}^{\pi /2}{\frac{f(\tan x)}{f(\tan x)+f(\cot x)}\ dx}\),
=\(\int\limits_{0}^{\pi /2}{\frac{f(\tan x)+f(\cot x)}{f(\tan x)+f(\cot x)}\ dx}\),
2I = \(\int\limits_{0}^{\pi /2}{1.\ dx}\),
2I = \(x|_{0}^{\pi /2}\),
I = π/4.
]]>The Gauss law is a very convenient tool to find the electric field of a system of charges. It is especially useful when situations of symmetry can be easily exploited. The electric flux (\(d\phi \)) through a differential area is the dot product of the electric field and the area vector (this vector is normal to the area pointing towards the convex end; its magnitude is the area magnitude).
\(d\phi \,=\,\overrightarrow{E}.\overrightarrow{dA}\,=\,E.dA\,\cos \theta \).
Where \(\theta \) is the angle between the electric field and area vector. The net flux through a surface is simply the integral of the differential flux \(d\phi \).
\(\phi \,=\,\int{d\phi }\,=\,\int{\overrightarrow{E}.\overrightarrow{dA}}\).
For a closed surface such as that of a sphere, torus or a cube, this integral is represented as a loop integral \(\phi \,=\,\oint{\overrightarrow{E}.\overrightarrow{dA}}\).
Gauss Theorem: The net flux through a closed surface is directly proportional to the net charge in the volume enclosed by the closed surface.
\(\phi \,=\,\oint{\overrightarrow{E}}.\overrightarrow{dA}\,=\,\frac{{{q}_{n}}et}{{{\varepsilon }_{0}}}.\).
In simple words, the Gauss law relates the ‘flow’ of electric field lines (flux) to the charges within the enclosed surface. If there are no charges enclosed by a surface, then the net electric flux remains zero. This means that the number of electric field lines entering the surface is equal to the field lines leaving the surface.
The electric flux from any closed surface is only due to the sources (positive charges) and sinks (negative charges) of electric fields enclosed by the surface. Any charges outside the surface do not contribute to the electric flux. Also, only electric charges can act as sources or sinks of electric fields. Changing magnetic fields. For example, cannot act as sources or sinks of electric fields.
The net flux for the surface on the left is non – zero as it encloses a net charge. The net flux for the surface on the right is zero. Since it does not enclose any charge. Note that the Gauss Law is only a restatement of the Coulomb’s Law. If you apply Gauss Law to a point charge enclosed by a sphere, you will get back the Coulomb’s Law easily.
Application to an infinitely long line of Charge: Consider an infinitely long line of charge with the charge per unit length being λ. We can take advantage of the cylindrical symmetry of this situation. By symmetry, the electric fields all point radially away from the line of charge, there is no component parallel to the line of charge.
We can use a cylinder (with any arbitrary radius r and length l) centred on the line of charge as our Gaussian surface.
As you can see in the above diagram, the electric field is perpendicular to the curved surface (hence parallel to the area vector) of the cylinder. Thus, the angle between the electric field and area vector is zero and cos θ = 1. The top and bottom surfaces of the cylinder lie parallel to the electric field. Thus the angle between area vector and electric field is 90 degrees and cos θ = 0.
Thus the electric flux is only due to the curved surface:
\(\phi \,=\,\oint{\overrightarrow{E}}.\overrightarrow{dA}\),
\(\phi \,=\,{{\phi }_{curved}}\,+\,{{\phi }_{top}}\,+\,{{\phi }_{bottom}}\),
\(\phi \,=\,\oint{\overrightarrow{E}.\overrightarrow{dA}}\,=\,\int{E.dA.\cos \,0\,+\,\int{E.dA.\cos \,{{90}^{0}}}}+\,\int{E.dA.\cos \,{{90}^{0}}}\),
\(\phi \,=\,\int{E.dA.1}\).
Due to radial symmetry, the curved surface is equidistant from the line of charge and the electric field in the surface has a constant magnitude throughout.
\(\phi \,=\,\int{E.dA}\,=\,E\,\int{dA}\,=\,E.2\pi rl\).
The net charge enclosed by the surface is:
\({{q}_{n}}et\,=\,\lambda .l\).
Using Gauss theorem,
\(\phi \,=\,E.2\pi rl\,=\,\frac{{{q}_{n}}et}{{{\varepsilon }_{0}}}\,=\,\frac{\lambda .l}{{{\varepsilon }_{0}}}\),
\(E.2\pi rl\,=\,\frac{\lambda .l}{{{\varepsilon }_{0}}}\),
\(E\,=\,\frac{1}{2\pi {{\varepsilon }_{0}}}\times \frac{\lambda }{r}\).
]]>AP EAMCET (Engineering, Agriculture and Medical Common Entrance Test) is being conducted on behalf of APSCHE.
Eligibility: A pass in 10+2 or equivalent examination with 50 % marks aggregate.
Mode of Application: ONLINE
Important Dates:
Commencement of Application |
28.02.2018 |
Last Date of Application |
29.03.2018 |
Last Date of Application with Fine of Rs.500/- |
06.04.2018 |
Last Date of Application with Fine of Rs.1000/- |
11.04.2018 |
Last Date of Application with Fine of Rs.5000/- |
16.04.2018 |
Last Date of Application with Fine of Rs.10000/- |
21.04.2018 |
Download of Hall Ticket |
18.04.2018 |
Time Table:
Stream |
Date(s) | Timings (2 Sessions per Day) | Fee |
Engineering (E) | 22^{nd} to 25^{th} April, 2018 | 10.00 AM to 01.00 PM & 02.30 PM to 05.30 PM |
Rs.500/- |
Agriculture (A) |
25^{th} & 26^{th} April, 2018 | Rs.500/- | |
E & A (Both Streams) | 24^{th} & 25^{th} April, 2018 |
Rs.1000/- |
Mode of Exam: Computer Based Test (ONLINE)
]]>We know that there are two types of connection series and parallel. Series circuit is one in which current remains the same along each element. Now we will talk about LCR circuit. The LCR circuit analysis can be understood better in terms of phasors. A phasor is a rotating quantity.
For an inductor (L), if we consider I to be our reference axis, then voltage leads by 90° and for the capacitor the voltage lags by 90°. But the resistance, current and voltage phasors are always in phase.
Analysis of an LCR Circuit – Series Circuit: Let’s consider the following LCR circuit using the current across the circuit to be our reference phasor because it remains the same for all the components in a series LCR circuit.
As described above the overall phasor will look like below:
From the above phasor diagram we know that,
V² = (V_{R})² + (V_{L} – V_{C})² … (1)
Now Current will be equal in all the three as it is a series LCR circuit. Therefore,
V_{R} = IR … (2)
V_{L} = IX_{L} … (3)
V_{C} = IX_{C} … (4)
Using equations 1, 2, 3 and 4. We get:
\(I\,=\,\frac{V}{\sqrt{{{R}^{2}}\,+\,{{\left( {{X}_{L}}\,-\,{{X}_{C}} \right)}^{2}}}}\).
Also the angle between V and I is known phase constant,
\(\tan \,\phi \,=\,\frac{{{V}_{L}}\,-\,{{V}_{C}}}{R}\).
It can also be represented in terms of impedance,
Tan φ = X_{L} – X_{C} R
Depending upon the values of X_{L} and X_{C}.
We have three possible conditions:
1. Find the order and degree of \({{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{3}} \right)}^{\frac{6}{5}}}=6y\)
Solution: Given that \({{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{3}} \right)}^{\frac{6}{5}}}=6y\)
\({{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{3}} \right)}^{\frac{6}{5}}}^{\times \frac{5}{6}}={{\left( 6y \right)}^{\frac{5}{6}}}\)
\(\left( \frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{3}} \right)={{\left( 6y \right)}^{\frac{5}{6}}}\)
That is \(\left( \frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{3}} \right)={{\left( 6y \right)}^{\frac{5}{6}}}\)
The highest differentiation is order that power is degree, hence Order is 2, degree = 1
2. Find the order of the family of the differential equation obtained by eliminating the arbitrary constant b and c from xy = ce^{x} + be^{-x} + x²
Solution: Equation of the curve is xy = ce^{x} + be^{-x} + x²
Number of arbitrary constant in the given curve is 2
Therefore, the order of the corresponding differential equation is 2
3. Find the order of differential equation of the family of all circle with their centers at the origin
Solution: Given family of the curve x² + y² = a² … (1)
Where a is parameter
From equation differentiation with respect to the x
2x + 2yy₁ = 0
Hence differential equation is x + yy₁ = 0
Order of the differential equation is 1
4. From the differential equation from the relation xy = ax² + b/x ny eliminating the arbitrary constant a, b.
Solution: Given that xy = ax² + b/x … 1
Take LCM
yx² = ax³ + b
Differentiation with respect to the x
x²y₁ + 2xy = 3ax²
xy₁ + 2y = 3ax … 2
Again differentiation with respect to the x
xy₂ + y₁ + 2y₁ = 3ax
xy₂ + 3y₁ = 3a
From equation (2)
xy₁ + 2y = x (xy_{2} + 3y₁)
xy₁ + 2y = x²y₂ + 3xy₁
x²y₂ + 2xy₁ – 2y = 0
Differential equation x²y₂ + 2xy₁ – 2y = 0
]]>Consider a dipole with charges +q and –q forming a dipole since they are a distance d away from each other. Let it be placed in a uniform electric field of strength E such that the axis of the dipole forms an angle θ with the electric field.
The force on the charges is
\(\overrightarrow{{{F}_{+}}}\,=\,+\,q\overrightarrow{E}\)
\({{\overrightarrow{F}}_{-}}\,=\,-\,q\overrightarrow{E}\)
The components of force perpendicular to the dipole are:
\({{F}_{+}}^{\bot }\,=\,+\,qE\,\sin \theta \)
\({{F}_{-}}^{\bot }\,=\,-\,qE\,\sin \theta \).
Since the force magnitudes are equal and are separated by a distance d, the torque on the dipole is given by:
Torque (\(\tau \)) = Force \(\times \) Distance Separating Forces
\(\tau \,=\,dqE\,\sin \theta \)
Since dipole moment is given by:
\(p\,=\,qd\)
And the direction of the dipole moment is from the positive to the negative charge, it can see from the above equation that the torque is the cross product of the dipole moment and electric field. Notice that the torque is in the clockwise direction (hence negative) in the above figure. If the direction of Electric Field is positive.
Thus,
\(\tau \,=\,-\,pE\,\sin \theta \)
Or,
\(\overrightarrow{\tau }\,=\,\overrightarrow{p}\,\times \,\overrightarrow{E}\)
\(\left| \overrightarrow{\tau } \right|\,=\,\left| \overrightarrow{p}\,\times \,\overrightarrow{E} \right|\,=\,pE\,\sin \theta \)
]]>What are the Fundamental Forces in Nature?
There are four universal or fundamental forces in nature. Without these forces, all matter in the world will fall apart. Force as such is any pull or push that causes an object to alter its physical state (in terms of motion or deformity). Newton defined a force as anything that causes an object of mass ‘m’ to move with an acceleration ‘a’. A brief overview of the four fundamental forces is given below.
Gravitational Force: Why is the universe not filled with floating human beings and cows and cars and other such things? Obviously because the gravitational force of the earth holds us to the planet. Gravitational force is the force responsible for holding planets in their orbits and this is possible only because of their infinitely long range.
Gravitational force can be represented as follows:
F_{g} = Gm₁m₂/r²
Where,
F_{g} = Gravitational force.
G = Universal gravitational constant.
m_{1} and m_{2} are the masses of the objects in consideration.
r = Distance between the centres of the two objects in consideration.
When considered for massive objects, like the sun, or giant planets, gravitational force is considered to be strong as the masses of these objects are also large. On an atomic level, this force is considered weak.
Electromagnetic Force: A fundamental force in nature, the electromagnetic force acts between charged particles and is the combination of all electrical and magnetic forces. The electromagnetic force can be attractive or repulsive. It has infinite range, although its strength is inversely proportional to the inverse square of the distance (that is, doubling the distance between two charged particles decreases the force between them by a factor of four). Aside from gravity, the electromagnetic force is responsible for all forces we experience directly (the only other two known fundamental forces are the “strong force” and the “weak force,” which hold together the particles in the nucleus of an atom). For example, the electromagnetic force holds atoms together in molecules, causes friction, and attracts iron to a magnet.
Strong Nuclear Forces: Out of the four fundamental forces, nuclear forces are the strongest attractive forces. Electromagnetism holds matter together, but there was no explanation on how the nucleus is held together in the atom. If we consider only electromagnetism and gravity, then the nucleus should actually fly off in different directions. But it doesn’t, implying that there exists another force within the nucleus which is stronger than the gravitational force and electromagnetic force. This is where nuclear forces come into play. Strong nuclear forces are responsible for holding the nuclei of atoms together.
Weak Nuclear Forces: Weak nuclear forces are responsible for the radioactive decay, specifically the beta decay neutrino interactions. It has a very short range (less than 1 fm) and this force is, as the name suggests, weak in nature.
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