The post Orbital Velocity of Satellite appeared first on MyRank.

]]>Satellites are natural or artificial bodied describing orbit around a planet under its gravitational attraction. Moon is a natural satellite while INSAT – 1B is an artificial satellite of earth. Condition for establishment of artificial satellite is that the center of orbit of satellite must coincide with center or earth or satellite must move around great circle of earth. Orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the earth.

For revolution of satellite around the earth, the gravitational pull provides the required centripetal force.

\(\frac{m{{v}^{2}}}{r}=\frac{GMm}{{{r}^{2}}}\).

\(v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{g{{R}^{2}}}{R+h}}=R\sqrt{\frac{g}{R+h}}\) (∵ GM = gR²; r = R + h)

\(v=R\sqrt{\frac{g}{R+h}}\).

Orbital velocity is independent of the mass of the orbiting body and is always along the tangent of the orbit. i.e.., satellites of different masses have same orbital velocity, if they are in the same orbit.

Orbital velocity depends on the mass of central body and radius of orbit and the orbital velocity of the satellite when it revolves very close to the surface of the planet and for a given planet, greater the radius of orbit, lesser will be the orbital velocity of the satellite.

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]]>The post Equation of Line passing through Two given Points appeared first on MyRank.

]]>**Vector Form:**

From the figure, \(\overrightarrow{OP}\,=\,\overrightarrow{r},\,\overrightarrow{OA}\,=\,\overrightarrow{a}\) and \(\overrightarrow{OB}\,=\,\overrightarrow{b}\).

Since \(\overrightarrow{AP}\) is collinear with \(\overrightarrow{AB}\), \(\overrightarrow{AP}\,=\,\lambda \,\overrightarrow{AB}\) for some scalar λ, we have

\(\overrightarrow{OP}\,-\,\overrightarrow{OA}\,=\,\lambda \,\left( \overrightarrow{OB}\,-\,\overrightarrow{OA} \right)\).

or \(\overrightarrow{r}\,-\,\overrightarrow{a}\,=\,\lambda \,\left( \overrightarrow{b}\,-\,\overrightarrow{a} \right)\).

or \(\overrightarrow{r}\,=\,\overrightarrow{a}\,+\,\lambda \,\left( \overrightarrow{b}\,-\,\overrightarrow{a} \right)\) … (i)

Therefore, the equation of a straight line passing through \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is \(\overrightarrow{r}\,=\,\overrightarrow{a}\,+\,\lambda \,\left( \overrightarrow{b}\,-\,\overrightarrow{a} \right)\).

**Cartesian Form:**

We have \(\overrightarrow{r}\,=\,x\hat{i}\,+\,y\hat{j}\,+\,z\hat{k},\,\overrightarrow{a}\,=\,{{x}_{1}}\hat{i}\,+\,{{y}_{1}}\hat{j}\,+\,{{z}_{1}}\hat{k}\) and \(\,\overrightarrow{r}\,=\,{{x}_{2}}\hat{i}\,+\,{{y}_{2}}\hat{j}\,+\,{{z}_{2}}\hat{k}\).

Substituting these values in (i), we get

\(x\hat{i}\,+\,y\hat{j}\,+\,z\hat{k}\,=\,{{x}_{1}}\hat{i}\,+\,{{y}_{1}}\hat{j}\,+\,{{z}_{1}}\hat{k}\,+\,\lambda \left[ \left( {{x}_{2}}\,-\,{{x}_{1}} \right)\hat{i}\,+\,\left( {{y}_{2}}\,-\,{{y}_{1}} \right)\hat{j}\,+\,\left( {{z}_{2}}\,-\,{{z}_{1}} \right)\hat{k} \right]\).

Equating the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), we get

\(x\,=\,{{x}_{1}}\,+\,\lambda \,\left( {{x}_{2}}\,-\,{{x}_{1}} \right)\); \(y\,=\,{{y}_{1}}\,+\,\lambda \,\left( {{y}_{2}}\,-\,{{y}_{1}} \right)\); \(z\,=\,{{z}_{1}}\,+\,\lambda \,\left( {{z}_{2}}\,-\,{{z}_{1}} \right)\).

On eliminating λ, we obtain \(\frac{x\,-\,{{x}_{1}}}{{{x}_{2}}\,-\,{{x}_{1}}}\,=\,\frac{y\,-\,{{y}_{1}}}{{{y}_{2}}\,-\,{{y}_{1}}}\,=\,\frac{z\,-\,{{z}_{1}}}{{{z}_{2}}\,-\,{{z}_{1}}}\,=\,\lambda \).

which is the equation of the line in Cartesian form.

**Example:** The Cartesian equation of a line is \(\frac{x-3}{2}=\frac{y+1}{-2}=\frac{z-3}{5}\).
Find the vector equation of the line

**Solution:** Given that \(\frac{x-3}{2}=\frac{y+1}{-2}=\frac{z-3}{5}\).

Note that it passes through (3, -1, 3) and

Is parallel to the line whose direction ratios are 2, -2, and 5.

Therefore, its vector equation is \(\overrightarrow{r}=3\hat{i}-\hat{j}+3\hat{k}+\lambda \left( 2\hat{i}-2\hat{j}+5\hat{k} \right)\).

Where λ is a parameter

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]]>The post Vignan’s Scholastic Aptitude Test (VSAT) 2020 Notification Released appeared first on MyRank.

]]>- Vignan’s Foundation For Science, Technology and Research (VFSTR) has made it, as its mission to prepare globally acceptable readily deployable, industry ready skilled professional and navigate towards the socio-economic transformation of the region. It provides quality education in a diverse and intellectually stimulating environment.
- It imparts value addition training to students to make them competent and inspired engineers. The institute celebrate the power of knowledge, cultivates vision and build awareness about the self and society around.

Agricultural Engineering | Civil Engineering | Mechanical Engineering |

Automobile Engineering | Computer Science & Engg. (CSE) | Petroleum Engineering |

Biotechnology* | Electronics & Communications Engg. (ECE) | Textile & Fashion Technology |

Bioinformatics* | Electrical & Electronics Engg. (EEE) | B. Pharmacy* |

Biomedical Engineering* | Food Technology* | |

Chemical Engineering | Information Technology (IT) |

- Candidates born on or after 1
^{st}July, 1999 and a pass in Intermediate or its equivalent with minimum CGPA of 7.0 in Andhra Pradesh or above 60% aggregate marks in Telangana State are eligible to appear for the admission test. - Candidates qualified in AP or TS EAMCET/ JEE(Mains)/ JEE(Advanced) with a minimum of CGPA of 7.0 in Andhra Pradesh or above 60% aggregate marks in Telangana State in the +2/ Intermediate examination.

Sections | Subjects | No. of Questions | No. of Marks |
---|---|---|---|

Section I | Mathematics or Biology | 30 | 30 |

Section II | Physics | 30 | 30 |

Section III | Chemistry | 30 | 30 |

Section IV | English/ Aptitude | 30 | 30 |

Total | 120 | 120 | |

*Duration: 150 Minutes (2 ^{1}/_{2} Hours) |
|||

*There is No Negative Marking |

Sl. No | Events | Dates |
---|---|---|

1 | Issue of Application forms | Started |

2 | Last date for receipt of filled in applications | 05 April 2020 |

3 | On-line Test Dates | 14 – 24 April 2020 |

4 | Test Time | Slot 1: 08.00 A.M to 10.30 A.MSlot 2: 11.00 A.M to 01.30 P.MSlot 3: 02.30 P.M to 05.00 P.M |

5 | Announcement of V-SAT 2020 result | May first week 2020 |

6 | Counseling of merit listed candidates for admission | May 2020Venue: Vignan’s Deemed to be University, Vadlamudi, Guntur – 522 213 (A.P)Ph: +91 863 2344777/ 78Toll Free: 1800-425-2529 |

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]]>The post Triangle Law of Vector Addition of Two Vectors appeared first on MyRank.

]]>The physical quantities may be broadly classified into the vectors and the scalars. The quantities with magnitude and direction both are known as vector quantities, it means a vector is a physical quantity that has both magnitude and direction. If two non-zero vectors are represented by the two sides of a triangle taken in same order then the resultant is given in opposite order.

i.e.., \(\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}\) \(\because \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\)

**1) Magnitude
of Resultant Vector: **In ΔABN, \(\cos \theta =\frac{AN}{B}\) ⇒ AN = B cos θ

\(\sin \theta =\frac{BN}{B}\) ⇒ BN = B sinθ; ΔOBN, we have OB² = ON² + BN².

R² = (A + B cosθ)² + (B sinθ)²

R² = A² + B² cos²θ + 2 AB cosθ + B² sin² θ

R² = A² + B² (cos²θ + sin²θ) + 2AB cosθ = A² + B² (1) + 2 AB cosθ

\(R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\).

**2) Direction of resultant vectors: **If θ is angle between \(\overrightarrow{A}\) and \(\overrightarrow{B}\), then

\(|\overrightarrow{A}+\overrightarrow{B}|\,=\,\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\) ; If \(\overrightarrow{R}\) makes an angle α with\(\overrightarrow{A}\), then in ΔOBN,

\(\tan \alpha =\frac{BN}{ON}=\frac{BN}{OA+AN}=\frac{B\sin \theta }{A+B\cos \theta }\).

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]]>The post Vectors – Equation of Straight Line appeared first on MyRank.

]]>**Equation of Straight Line passing
through a given Point and Parallel to a given Vector**

**Vector Form: **Line passing through point \(A(\overline{a})\) and parallel to vector \(\overline{b}\).

Let A be the given point and let AP be the given line through A.

Let \(\overline{b}\) be any vector parallel to the given line.

Position vector of point A is \(\overline{a}\).

Let P be any point on line AP, and let its position vector be \(\overline{r}\).

Then, we have

\(\overline{r}=\overline{OP}=\overline{OA}+\overline{AP}\)\(\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\) … (1)

Here, \(\overline{r}\) is the position vector of any point P (x, y) on the line. So \(\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\).

In, particular, the equation of straight line through origin and parallel to \(\overline{b}\) is \(\overrightarrow{r}=\lambda \overrightarrow{b}\).

**Cartesian Form: **Let the coordinates of the given point A be (x₁, y₁, z₁) and the
direction of the line be a, b and c. Consider the coordinates of any point P be
(x, y, z). then

\(\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\),

\(\overrightarrow{a}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}\) and

\(\overrightarrow{b}=a\hat{i}+b\hat{j}+c\hat{k}\).

Substituting these values in (1) and equating the coefficients of \(\hat{i},\ \hat{j}\) and \(\hat{k}\). We get

x = x₁ + λa

y = y₁ + λb

z = z₁ + λc

These are parametric equation of the line.

Eliminating the parameter λ, we get \(\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\).

Note:

- Here any point on the line is (x, y, z) ≡ (x₁ + λa, y₁ + λb, z₁ + λc)
- Since the x, y and z – axes pass through the origin and have direction cosine (1, 0, 0), (0, 1, 0) and (0, 0, 1), their equations are

Equation of x – axis: y = 0, z = 0,

Equation of y – axis: x = 0, z = 0,

Equation of z – axis: x = 0, y = 0.

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]]>The post Cylindrical Capacitor appeared first on MyRank.

]]>A cylindrical capacitor consists of two coaxial cylinders of radii a and b, say b > a. The outer one is earthed. The cylinders are long enough so that we can neglect fringing of electric fields at the ends. The electric field at a point between the cylinders will be radial, and its magnitude will depend on the distance from the central axis.

Consider a Gaussian surface of length y and radius r such that a < r < b. Flux through the plane surface is zero because the electric field and the area vector are perpendicular to each other.

For the curved part:

\(\phi =\int{\overrightarrow{E}.\overrightarrow{ds}}=\int{Eds}=E\int{ds}=E2\pi ry\),

Consider inside the Gaussian surface is: \(q=\frac{Qy}{L}\),

From Gauss’s law: \(\phi =E2\pi ry=\frac{Qy}{L{{\varepsilon }_{0}}}\),

\(E=\frac{Q}{2\pi {{\varepsilon }_{0}}Lr}\),

Potential difference is: \({{V}_{b}}-{{V}_{a}}=-\int\limits_{a}^{b}{\overrightarrow{E}}.\overrightarrow{dr}=-\int\limits_{a}^{b}{\frac{Q}{2\pi {{\varepsilon }_{0}}Lr}dr}=-\frac{Q}{2\pi {{\varepsilon }_{0}}L}\int\limits_{a}^{b}{\frac{1}{r}dr}\),

\({{V}_{a}}=\frac{Q}{2\pi {{\varepsilon }_{0}}L}\ln \frac{b}{a}\,\,\,\,\,\,\,\,\left( \because \,\,{{V}_{b}}=0 \right)\),

Now, the capacitance is: \(C=\frac{Q}{{{V}_{a}}-{{V}_{b}}}=\frac{Q}{{{V}_{a}}}=\frac{2\pi {{\varepsilon }_{0}}L}{\ln \left( \frac{b}{a} \right)}\).

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]]>The post Homogeneous Differential Equation appeared first on MyRank.

]]>A differential equation \(\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}\) is said to be a homogeneous differential equation in x, y if both f (x, y), g (x, y) are homogeneous function of same degree in x and y.

To find the solution of the Homogeneous Differential Equation put y = v x, then \(\frac{dy}{dx}=v+x\frac{dv}{dx}\). Substituting these values in given differential equation, then it reduces to variable separable form. Then we find the solution of the differential equation.

**Example:** Solve the differential equation \(\frac{dy}{dx}=\frac{x-y}{x+y}\).

**Solution: **Given that \(\frac{dy}{dx}=\frac{x-y}{x+y}\) … (1)

Equation (1) is a Homogeneous Differential Equation

Put y = vx

\(\frac{dy}{dx}=v+x\frac{dv}{dx}\),

\(v+x\frac{dv}{dx}=\frac{x-vx}{x+vx}\),

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

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

\(x\frac{dv}{dx}=\frac{1-2v-{{v}^{2}}}{1+v}\),

\(\int{\frac{(1+v)dv}{1-2v-{{v}^{2}}}}=\int{\frac{dx}{x}}\),

\(\frac{1}{2}\int{\frac{2(1+v)dv}{1-2v-{{v}^{2}}}}=\int{\frac{dx}{x}}\),

\(\frac{-1}{2}\int{\frac{-2(1+v)dv}{1-2v-{{v}^{2}}}}=\int{\frac{dx}{x}}\) \(\left( \because \int{\frac{f'(x)}{f(x)}}=\log |f(x)|+c \right)\).

\(\frac{-1}{2}\log |(1-2v-{{v}^{2}})|=\log x+\log c\),

\(\frac{-1}{2}\log |(1-2v-{{v}^{2}})|=\log xc\),

\(\frac{-1}{2}\log |(1-\frac{2y}{x}-\frac{{{y}^{2}}}{{{x}^{2}}})|=\log xc\),

\(\frac{-1}{2}\log |\left( \frac{{{x}^{2}}-2yx-{{y}^{2}}}{{{x}^{2}}} \right)|=\log xc\),

\(\log |{{\left( \frac{{{x}^{2}}-2yx-{{y}^{2}}}{{{x}^{2}}} \right)}^{\frac{-1}{2}}}|=\log xc\),

\(\log |{{\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)}^{\frac{1}{2}}}|=\log xc\),

\({{\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)}^{\frac{1}{2}}}=xc\),

\(\left( \frac{{{x}^{2}}}{{{x}^{2}}-2yx-{{y}^{2}}} \right)={{x}^{2}}{{c}^{2}}\),

\({{x}^{2}}-2yx-{{y}^{2}}=\frac{1}{{{c}^{2}}}=k\),

\({{x}^{2}}-2yx-{{y}^{2}}=k\).

The post Homogeneous Differential Equation appeared first on MyRank.

]]>The post Sathyabama University All India Engineering Entrance Exam (SAEEE) 2020 Notifications Released appeared first on MyRank.

]]>- Sathyabama Institute of Science and Technology (A Christian Minority Institution) formerly known as Sathyabama Engineering College was established in 1987 by Dr. Jeppiaar.
- It is a premier institute imparting knowledge in the areas of engineering, science, technology and education.
- The institution’s progress and contribution to the field of technical education for over two decades resulted in granting Deemed University status on 16
^{th}July, 2001. - The University has been awarded as Category “A” University by Ministry of Human Resources Development (MHRD), Government of India and accredited by NAAC with Grade “A” in the year 2017, certified with DNV-GL ISO 9001 standard and offering exemplary education from last 29 years.

Courses | Specializations |

B.E | Aeronautical Engineering |

Automobile Engineering | |

Civil Engineering | |

Computer Science and Engineering | |

Electrical and Electronics Engineering | |

Electronics and Communication Engineering | |

Electronics and Instrumentation Engineering | |

Electronics and Telecommunication Engineering | |

Mechatronics | |

Mechanical Engineering | |

B.Tech | Chemical Engineering |

Information Technology | |

Biomedical | |

Biotechnology | |

B.Des | Bachelor of Design |

B.Arch | Bachelor of Architecture |

- A pass in the 1
^{th}class or Equivalent Examination with a minimum aggregate of 60% marks or “6.0” CGPA. - A pass in the 10+2/ HSC/ ICSE or equivalent examination with Mathematics, Physics and Chemistry with an average of 45% marks and above (in Mathematics, Physics and Chemistry).
- Candidate opting for these programmes should appear for Mathematics, Physics and Chemistry in the entrance examination.

- A pass in the 10+2/ HSC/ ICSE or equivalent examination with Mathematics, Physics and Chemistry with a minimum average of 45% marks (in Mathematics, Physics and Chemistry) and a valid NATA marks (National Aptitude Test in Architecture).
- Candidate opting for this programme should appear for Mathematics, Physics and Chemistry in the entrance examination

- Passed 10+2 examination. Obtained at least 45% marks( 40 % in case of candidates belonging to reserved category) in the qualifying examination.

Sl. No | Subjects | Questions |
---|---|---|

1 | Mathematics | 60 |

2 | Physics | 30 |

3 | Chemistry | 30 |

Sl. No | Events | Dates |
---|---|---|

1 | Issue of Entrance Examination Application Form (Online) | 18^{th} October, 2019 |

2 | Last Date for Submission of Application Form | 4^{th} April, 2020 |

3 | SAEEE 2020 Date of Examination | 14^{th} to 19^{th} April, 2020 |

4 | Result | 30^{th} April, 2020 11.30 AM. (tentative) |

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]]>The post Spherical Capacitor appeared first on MyRank.

]]>A spherical capacitor consists of two concentric spherical conducting shells of radii \(a\) and \(b\), say \(b>a\). The outer shell is earthed. Place a charge \(+Q\) on the inner shell. It will reside on the outer surface of the shell. A charge \(-Q\) will be induced on the inner surface of the outer shell. A charge \(+Q\) will flow from the outer shell to earth.

Consider a Gaussian spherical surface of radius \(r\) such that \(a<r<b\). Form Gauss’s law, the electric field at distance \(r>a\) is \(E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\).

The potential difference is: \({{V}_{b}}-{{V}_{a}}=-\int\limits_{a}^{b}{\overrightarrow{E}.\overrightarrow{dr}}=-\int\limits_{a}^{b}{\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}}dr\).

Since, \({{V}_{b}}=0\) we have:

\({{V}_{a}}=\int\limits_{a}^{b}{\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}}dr=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{a}-\frac{1}{b} \right)=\frac{Q\left( b-a \right)}{4\pi {{\varepsilon }_{0}}ab}\).

Now, the capacitance is: \(C=\frac{Q}{{{V}_{a}}-{{V}_{b}}}=\frac{Q}{{{V}_{a}}}=\frac{4\pi {{\varepsilon }_{0}}ab}{\left( b-a \right)}\).

Therefore, the capacitance of a spherical capacitor is: \(C=\frac{4\pi {{\varepsilon }_{0}}ab}{\left( b-a \right)}\).

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]]>The post Complex Numbers – Equation of a Circle appeared first on MyRank.

]]>**Equation of a Circle:** Consider a fixed complex number zₒ and let z be any complex
number which moves in such a way that its distance from zₒ is always to r. this
implies z would lie on a circle whose center is zₒ and radius is r. and its
equation would be

|z – zₒ| = r

Squaring on both sides

|z – zₒ|² = r²

\((z-{{z}_{0}})(\overline{z}-\overline{{{z}_{0}}})={{r}^{2}}\),

\(z\overline{z}-z\overline{{{z}_{0}}}-\overline{z}{{z}_{0}}+{{z}_{0}}\overline{{{z}_{0}}}-{{r}^{2}}=0\).

Let -a = zₒ and \({{z}_{0}}\overline{{{z}_{0}}}-{{r}^{2}}=b\). Then

\(z\overline{z}+a\overline{z}+\overline{a}z+b=0\).

It represents the general equation of a circle in the complex plane.

Now, let us consider a circle described on a lime segment AB \((A({{z}_{1}}),B({{z}_{2}}))\) as its diameter. Let P(z) be any point on the circle. As the angle in the semicircle is \(\frac{\pi }{2}\), So

\(\angle APB=\frac{\pi }{2}\)\(\arg \left( \frac{{{z}_{1}}-z}{{{z}_{2}}-z} \right)=\pm \frac{\pi }{2}\),

\(\frac{z-{{z}_{1}}}{z-{{z}_{2}}}\) is purely imaginary

\(\frac{z-{{z}_{1}}}{z-{{z}_{2}}}+\frac{\overline{z}-\overline{{{z}_{1}}}}{\overline{z}-{{\overline{z}}_{2}}}=0\),

\((z-{{z}_{1}})(\overline{z}-\overline{{{z}_{2}}})+(z-{{z}_{2}})(\overline{z}-\overline{{{z}_{1}}})=0\).

The post Complex Numbers – Equation of a Circle appeared first on MyRank.

]]>