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Embarking on the journey of Class 11 chemistry can be both exciting and challenging. With its intricate concepts and detailed theories, chemistry requires a solid understanding and efficient study techniques. One of the most effective ways to grasp and retain this knowledge is through the art of handwritten notes. Handwritten notes are not just a mere transcription of textbook information; they are a powerful tool that enhances learning and memory retention. By engaging multiple senses in the process of writing, you can create a personalized and dynamic resource that caters to your unique learning style. In this post, we will explore the benefits of handwritten notes for Class 11 chemistry, share tips on how to create effective notes, and provide you with strategies to make the most out of this invaluable study method. Whether you are aiming for top grades or seeking a deeper understanding of chemical principles, mastering the skill of note-taking can significantly impact your academic journey. Let’s dive in and discover how you can transform your chemistry studies with the power of handwritten notes! TRANSCRIPT: 1. Introduction to Atomic Structure Atomic structure refers to the arrangement of subatomic particles within an atom. Understanding atomic structure is fundamental to explaining chemical behavior. 2. Subatomic Particles Protons: Positively charged particles found in the nucleus. Mass = 1 atomic mass unit (amu). Neutrons: Neutral particles found in the nucleus. Mass = 1 amu. Electrons: Negatively charged particles found in electron shells outside the nucleus. Mass ≈ 1/1836 amu. 3. Atomic Models Dalton’s Atomic Theory: Atoms are indivisible particles that make up all matter. Thomson’s Plum Pudding Model: Atoms consist of electrons scattered within a positively charged "soup". Rutherford’s Nuclear Model: Atoms have a small, dense, positively charged nucleus with electrons orbiting around it. Bohr’s Model: Electrons orbit the nucleus in specific energy levels or shells. 4. Quantum Mechanical Model Wave-Particle Duality: Electrons exhibit both wave-like and particle-like properties (de Broglie hypothesis). Heisenberg Uncertainty Principle: It is impossible to simultaneously know the exact position and momentum of an electron. Schrödinger’s Equation: Describes the behavior of electrons in atoms as wave functions (ψ). 5. Quantum Numbers Quantum numbers describe the properties and locations of electrons in atoms: Principal Quantum Number (n): Indicates the main energy level or shell (n = 1, 2, 3,...). Azimuthal Quantum Number (l): Indicates the subshell or orbital shape (l = 0 to n-1). s (l=0), p (l=1), d (l=2), f (l=3). Magnetic Quantum Number (m_l): Indicates the orientation of the orbital (m_l = -l to +l). Spin Quantum Number (m_s): Indicates the spin of the electron (m_s = +1/2 or -1/2). 6. Electron Configuration Aufbau Principle: Electrons fill orbitals starting with the lowest energy level. Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers. Hund’s Rule: Electrons will fill degenerate orbitals (orbitals of the same energy) singly before pairing up. 7. Atomic Orbitals s-Orbital: Spherical shape, one per energy level. p-Orbital: Dumbbell shape, three orientations per energy level (px, py, pz). d-Orbital: Cloverleaf shape, five orientations per energy level. f-Orbital: Complex shapes, seven orientations per energy level. 8. Electromagnetic Radiation and Atomic Spectra Electromagnetic Radiation: Energy waves that travel through space at the speed of light. Wavelength (λ): Distance between two consecutive peaks of a wave. Frequency (ν): Number of wave cycles passing a point per unit time. Speed of Light (c): c = λν. Planck’s Quantum Theory: Energy is quantized and can be absorbed or emitted in discrete amounts (quanta). E = hν (where h is Planck’s constant). Photoelectric Effect: Emission of electrons from a material when light shines on it, explained by Einstein using quantum theory. Atomic Emission Spectrum: Set of frequencies of light emitted by an atom's electrons moving from higher to lower energy levels, unique to each element. Hydrogen Spectrum: Explained by Bohr’s model, showing specific lines corresponding to electron transitions between energy levels. 9. Modern Atomic Theory Modern atomic theory incorporates both the wave-like and particle-like nature of electrons, utilizing quantum mechanics to describe electron behavior in atoms. Understanding atomic structure is crucial for delving into more complex chemical concepts and reactions, providing insight into the behavior of elements and compounds at the atomic level. Happy studying!

Are you ready to take your JEE preparation to the next level? Dive into the world of handwritten notes, a powerful tool that can revolutionize the way you study for one of the most competitive exams in the world. In this comprehensive guide, we will explore the benefits of using handwritten notes for JEE preparation, how to create effective notes, and tips for maximizing their impact on your learning. Whether you are a seasoned JEE aspirant or just starting your journey, this guide will equip you with the knowledge and strategies you need to harness the full potential of handwritten notes for JEE preparation. Let's embark on this transformative journey together and unlock the key to success in your JEE endeavors. TRANSCRIPT OF NOTES: Electric Charge: Matter contains electric force. Electric charges exert forces on each other. Law of Conservation of Charge: Charge is neither created nor destroyed but is transferred from one body to another. The total charge in an isolated system remains constant. Quantization of Charge: Charge exists in discrete packets called quanta. The smallest charge is the charge on an electron. Charge is always a multiple of the fundamental charge e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}e=1.6×10−19C. Coulomb's Law: The force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically: F=kq1q2r2F = k \frac{q_1 q_2}{r^2}F=kr2q1​q2​​, where kkk is Coulomb's constant (k≈8.99×109 N m2/C2k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2k≈8.99×109N m2/C2). Electric Field: The region around a charged particle where its force is exerted on other charges. Electric field EEE at a point is defined as E=FqE = \frac{F}{q}E=qF​. The direction of the field is the direction of the force on a positive test charge. Electric Dipole: Consists of two equal and opposite charges separated by a small distance. The electric dipole moment ppp is given by p=q⋅dp = q \cdot dp=q⋅d, where ddd is the distance between charges. Electric Field of a Dipole: Along the axial line of the dipole, the electric field EEE at distance rrr is E=14πϵ02pr3E = \frac{1}{4 \pi \epsilon_0} \frac{2p}{r^3}E=4πϵ0​1​r32p​. Along the equatorial line, E=14πϵ0pr3E = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^3}E=4πϵ0​1​r3p​. Gauss's Law: The total electric flux through a closed surface is equal to 1ϵ0\frac{1}{\epsilon_0}ϵ0​1​ times the charge enclosed by the surface. Mathematically: ΦE=∮E⋅dA=Qencϵ0\Phi_E = \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0}ΦE​=∮E⋅dA=ϵ0​Qenc​​. Potential Energy in an Electric Field: The work done to move a charge qqq in an electric field EEE is given by W=qEdW = qEdW=qEd. Electric potential VVV at a point is defined as the work done per unit charge to move a test charge from infinity to that point: V=WqV = \frac{W}{q}V=qW​. Conductors and Insulators: Conductors allow free movement of electric charge, while insulators do not. In a conductor, electric field inside is zero in electrostatic equilibrium. The charge resides on the surface of the conductor. Electric Flux: The measure of the number of electric field lines passing through a given area. Mathematically: ΦE=E⋅A⋅cos⁡(θ)\Phi_E = E \cdot A \cdot \cos(\theta)ΦE​=E⋅A⋅cos(θ). Capacitance: The ability of a system to store charge per unit potential difference. Mathematically: C=QVC = \frac{Q}{V}C=VQ​, where CCC is the capacitance, QQQ is the charge, and VVV is the potential difference. For a parallel plate capacitor, C=ϵ0AdC = \frac{\epsilon_0 A}{d}C=dϵ0​A​, where AAA is the area of the plates and ddd is the separation between them. Example Problems: 1. Transfer of Charge Between Two Point Charges: Two point charges AAA and BBB with charges +Q+Q+Q and −Q-Q−Q are placed a distance apart with force FFF between them. If 25% of the charge from AAA is transferred to BBB, the new force between the charges can be calculated using Coulomb's law with the modified charges. 2. Force Between Two Ions: Two positive ions, each with charge qqq, are separated by distance ddd. The force FFF of repulsion between them can be determined using Coulomb's law. The number of electrons missing from each ion can be calculated by equating q=n⋅eq = n \cdot eq=n⋅e. 3. Electric Field Between Parallel Line Charges: Two parallel infinite line charges with linear charge densities +λ+\lambda+λ and −λ-\lambda−λ are placed at a distance 2R2R2R apart. The electric field midway between the two line charges is determined by superimposing the electric fields due to each line charge. These notes provide a concise overview of key concepts in electrostatics, including fundamental laws, principles, and example problems. For further details and handwritten notes, visit crookshanksacademy.com/jeeneet.

The question given to us is: Answer the following for 5-fold cross validation on a training data set of 45 tuples: (i) How many rounds of learning will be performed? (ii) State the size of the training and testing set for each round of learning. (iii) How would the accuracy of the classifier be calculated in 5-fold cross validation? Let us solve this step-by-step. (i) In 5-fold cross validation, the entire dataset is divided into 5 equally sized folds. Each fold is used once as a test set while the remaining 4 folds form the training set. Therefore, there will be 5 rounds of learning. (ii) Given a dataset of 45 tuples: Each fold will consist of 45/5​=9 tuples. In each round, the training set will consist of the data from 4 folds, and the testing set will consist of the data from the remaining fold. So, for each round: The training set will have 4×9=36 tuples. The testing set will have 9 tuples. (iii) The accuracy of the classifier in 5-fold cross validation is calculated as follows: Perform 5 Rounds of Learning: In each round, train the classifier on the training set (36 tuples) and test it on the testing set (9 tuples). Calculate the accuracy for each of the 5 rounds. The accuracy for a round is the proportion of correctly classified instances out of the total instances in the test set. Calculate the Average Accuracy: After obtaining the accuracy for each of the 5 rounds, compute the overall accuracy by taking the average of these 5 accuracy values. Formally, if Ai is the accuracy in the i-th round, then the overall accuracy A is given by:

This is the official question paper of Computer Graphics Paper of B.Sc. (Hons.) Computer Science Course at the University of Delhi. Question 1 to 7 are compulsory. Attempt any four questions from question 8 to question 14. Part of a question must be answered together. Some symbols may not be visible on mobile devices. Hence we recommend that you use a desktop to view the solutions to the questions. Download the Question Paper as PDF Question 1 (a): What is a Polygon Mesh? List any one polygon mesh representation. View Solution Question 1 (b): Consider a polygon with vertices ABCD with coordinates A(1,2), B(6,6), C(8,3) and D(5,10). Trace the contents of the Active Edge Table according to scan line fill algorithm. View Solution Question 2 (a): Define horizontal and vertical retracing. View Solution Question 2 (b): What are the properties of unweighted area sampling technique of anti-aliasing? View Solution Question 3 (a): Why is depth-sort algorithm for visible surface determination called painter’s algorithm? View Solution Question 3 (b): Prove that parallel lines remain parallel under 2-D Transformations. View Solution Question 4 (a): Differentiate between orthographic and oblique projections. View Solution Question 4 (b): Rotate a triangle with coordinates A(0,0), B(1,1), C(5,2) by 45° about coordinate C in clockwise direction. View Solution Question 5 (a): How to convert RGB color model to CMY color model? View Solution Question 5 (b): Draw a 3 × 3 pixel grid pattern to display 10 intensities on a bi-level system display. Show patterns for all the intensity values. View Solution Question 6 (a): What is the condition to switch from region 1 to region 2 of the first quadrant of an ellipse in mid point ellipse drawing algorithm? View Solution Question 6 (b): What is diffuse reflection? How is it different from specular reflection? View Solution Question 7 (a): Differentiate between cabinet and cavalier parallel projections. View Solution Question 7 (b): Write the 4×4 3-D transformation matrices for each of the following transformations respectively : (a) Uniform scaling to double the size of an object. (b) Translate an object 2 units in x direction and 3 units in y direction. View Solution Question 8 (a): Explain briefly raster scan display architecture. View Solution Question 8 (b): Give the steps to clip the lines PQ and RS (having coordinates P(5,12), Q(20,25), R(11, 8) and S(25,16)) against the clip rectangle ABCD (having co-ordinates A(10,20), B(20,20), C(10,0), D(20,10)) using Cohen Sutherland Line Clipping Algorithm. View Solution Question 9 (a): Consider a 3D object with coordinate points P(0,3,3), Q(3,3,6), R(3,0,1) and S(0,0,0). Perform a local scaling on the object with scaling factors of 2, 3 and 3 along X, Y and Z axes respectively, to obtain the new coordinates of the transformed object. View Solution Question 9 (b): A cubic Bezier curve segment is described by control points P0(2,2), P1(4,8), P2(8,8) and P3(9,5). Another curve segment is described by Q0(a,b), Q1(c,2), Q2(15,2) and Q3(18,2). Find the values of a, b, and 10. (ac such that the curve segments join smoothly and C1 continuity exists between them. View Solution Question 10 (a): Write steps to shade an object using Phong shading method of polygon rendering? How does it overcome the drawback of Gouraud shading method? View Solution Question 10 (b): Consider a line from (0,0) to (5,5). Rasterize the line using Bresenham's line drawing algorithm. View Solution Question 11 (a): Reflect the polygon whose vertices are A(-1,0), B(0, -2), C(1,0) and D(0,2) about the line x =2 using homogeneous co-ordinates. View Solution Question 11 (b): Clip the polygon ABCD with the vertices A(7,0), B(5,12), C(7,7) and D(6,2) against the window P (2,0), Q(10,0), R(10,10) and S(2,10) using the Sutherland-Hodgeman Polygon Clipping algorithm. Also show output vertex array at each step. View Solution Question 12 (a): Explain Hue, Saturation and Value in HSV color model. View Solution Question 12 (b): Consider a line segment AB parallel to the Z axis with end points A[3 2 2 1] and B[3 2 4 1]. Overall scale to double the size of line AB followed by two point perspective projection with COP along X-axis and Y-axis as Xc=10 and Yc=20 respectively. Also, write the corresponding vanishing points. View Solution Question 13 (a): Explain depth sort algorithm for visible surface determination. View Solution Question 13 (b): A rectangular parallelopiped is given. Its length on x-axis. y-axis and z-axis is 3, 2 and 1 respectively. Perform a rotation by an angle 90° about x axis followed by a rotation by an angle 90° about y axis. View Solution Question 14 (a): Consider a rectangular parallelepiped with the given coordinates: Apply a trimetric projection on the given position vectors by an angle of Φ=30° about the y-axis, followed by a rotation by angle θ=45° about the x-axis, followed orthographic parallel projection onto the z=0 plane. Also, find the three foreshortening ratios. View Solution Question 14 (b): Given two keyframes for an object transformation. The first keyframe contains a triangle and the second keyframe contains a quadrilateral. Convert the triangle into the quadrilateral by equalizing vertex counts. View Solution END OF PAPER

Both cabinet and cavalier projections are types of oblique projections used in technical drawing and computer graphics to represent three-dimensional objects on a two-dimensional surface. Cabinet Projection: In cabinet projection, the object is projected onto the drawing plane at an angle of 45 degrees. The depth of the object is shown at half scale compared to its actual size along the z-axis, giving it a reduced appearance. Lines parallel to the z-axis are projected at half their true length. This projection retains the true shape of objects but distorts their depth. Cabinet projection is often used in technical and architectural drawings where a clear representation of object dimensions is required, albeit with some reduction in depth perception. Cavalier Projection: In cavalier projection, the object is projected onto the drawing plane without any reduction in the depth dimension. The object is tilted at an angle of 45 degrees, similar to cabinet projection, but lines remain at their true length along the z-axis. Unlike cabinet projection, cavalier projection does not suffer from depth distortion. However, because the depth is not reduced, the resulting drawings may appear elongated along the z-axis, making them less aesthetically pleasing. Cavalier projection is less commonly used in technical drawing due to its lack of depth scaling, but it can be useful in certain artistic and illustrative contexts where a more dramatic presentation is desired. In summary, cabinet and cavalier projections are both forms of oblique projection used in technical drawing and computer graphics, with cabinet projection featuring reduced depth scaling for a more compact representation, while cavalier projection maintains full depth perception but may result in elongated object appearances.

This is a question based on halftone patterns. Halftoning is a technique used in printing and digital imaging to simulate continuous-tone images using binary or limited-color output devices, like printers or screens. The key idea is to create the illusion of different shades of gray or color by varying the density and pattern of dots.

The question is given as: A cubic Bezier curve segment is described by control points P0(2,2), P1(4,8), P2(8,8) and P3(9,5). Another curve segment is described by Q0(a,b), Q1(c,2), Q2(15,2) and Q3(18,2). Find the values of a, b, and 10. (ac such that the curve segments join smoothly and C1 continuity exists between them. Now we are given that C1 continuity exists in the two curve segments. For C1 continuity, the tangents of the curve are equal in magnitute and direction. But before that if two curves are continous, then the endpoint of the first curve is the same as the start point of the next curve. So, Q0(a,b) = P3(9,5). Therefore, a = 9 b = 5 Now we know that tangent vector of a cubic bezier curve is given by: Now either we can use this monstrous equation or we have a simple trick to solve this question. The tangent vector at the end point of the first curve segment (which is also Q0​) is given by the difference between the last two control points of the first curve: T1 = P3 - P2 = (9,5) - (8,8) = (1,-3) T2 = Q1 - Q0 = (c,2) - (9,5) = (c-9,-3) c-9 = 1 c = 10 Hence we have found the value of a, b and c. We can of course use the equation to find the value of c too. But that would be far too complex and we should try to avoid it.

The given question is as follows: Consider a line segment AB parallel to the Z axis with end points A[3 2 2 1] and B[3 2 4 1]. Overall scale to double the size of line AB followed by two point perspective projection with COP along X-axis and Y-axis as Xc=10 and Yc=20 respectively. Also, write the corresponding vanishing points. So, first of all let us double the size of the line segment. We have been given homogenous coordinates of A and B. Let us apply our scaling matrix to it. So the new coordinates are [6 4 4 1] and [6 4 8 1]. Now we can apply two point perspective projection along it. Please note that since it is happening along x axis and y axis we will take z=0 plane for this projection. The transformation matrix for two point perspective projection along z=0 plane is given by: Now let us apply transformation matrix on our endpoints of the line. As we can see that although it was two point projection it turned out to be one point since both the vanishing points lie at the same coordinates.

To morph a triangle into a quadrilateral, we need to equalize the vertex counts by converting the triangle (3 vertices) to have the same number of vertices as the quadrilateral (4 vertices). This can be done by splitting one of the edges of the triangle to create an additional vertex. Suppose the triangle is ABC and the quadrilateral is EFGH then equalizing vertex counts can be shown as:

Phong shading is a technique in computer graphics used to shade objects with smooth surfaces by interpolating surface normals across the polygons of the object and computing the color at each pixel. Here's a step-by-step explanation of how the Phong shading method is implemented, along with how it addresses the drawbacks of Gouraud shading. Steps for Phone Shading Calculate Vertex Normals: For each vertex of the polygon, calculate the normal vector. This is typically done by averaging the normals of the faces that share the vertex. Interpolation of Normals: During rasterization, interpolate the vertex normals across the surface of the polygon. Unlike Gouraud shading, which interpolates the vertex colors, Phong shading interpolates the normals. Per-Pixel Normal Calculation: For each pixel within the polygon, calculate the normal vector using the interpolated normals. Compute Illumination Model: For each pixel, use the interpolated normal to compute the color using the Phong reflection model, which includes ambient, diffuse, and specular components. Ambient Component: This is the constant illumination that affects all parts of the surface equally. Diffuse Component: This is the light that scatters evenly in all directions from the surface. Specular Component: This is the light that reflects in a specific direction, creating highlights. Combine Components: Sum the ambient, diffuse, and specular components to get the final color for the pixel. Drawbacks of Gourand Shading Gouraud shading interpolates vertex colors across the surface of the polygon and computes the illumination at the vertices only. This method has some drawbacks: Mach Banding: Gouraud shading can produce noticeable "Mach bands" or artifacts where there are abrupt changes in color between polygons. This occurs because the interpolation is linear and the illumination is computed only at the vertices, which can lead to a loss of detail in areas with high specularity. Specular Highlight Issues: Specular highlights can be missed entirely or appear inaccurately because the highlights may not align with the vertices where illumination is calculated. Overcoming the Drawbacks Phong Rendering overcomes these drawbacks by: Smooth Shading: Phong shading provides a much smoother appearance by interpolating normals instead of colors. This reduces the appearance of Mach bands and creates a more continuous shading effect. Accurate Specular Highlights: By computing the illumination at each pixel using the interpolated normal, Phong shading can accurately capture specular highlights. This results in more realistic rendering of shiny surfaces as the highlight moves smoothly across the surface, irrespective of the vertex positions. Phong shading improves upon Gouraud shading by performing per-pixel lighting calculations and interpolating normals instead of colors. This results in smoother shading and more accurate rendering of specular highlights, making it particularly effective for rendering shiny surfaces and complex lighting effects.

Let's first write transformation matrix for uniform scaling to double the size of an object. Uniform scaling means scaling equally in all directions. To double the size of an object, the scaling factor s is 2. The transformation matrix for uniform scaling in 3D is: For a scaling factor s=2" Now let's write the matrix to double to translate an object 2 units in the x direction and 3 units in the y direction. Assuming no translation in z axis,

In the midpoint ellipse drawing algorithm, the decision to switch from Region 1 to Region 2 of the first quadrant of an ellipse occurs based on the midpoint of the region's boundary. This algorithm divides the ellipse into different regions and iteratively calculates points along its boundary. To switch from Region 1 to Region 2 in the first quadrant, the algorithm typically compares the midpoint of the region with a decision parameter. The condition for this transition is usually based on the value of a function that represents the ellipse's boundary. The general condition for switching from Region 1 to Region 2 in the midpoint ellipse drawing algorithm is: This transition condition ensures that the algorithm progresses along the ellipse's boundary accurately while minimizing the number of calculations required.