computer-nec-license
  • NEC-Computer
  • 1. Concept of Basic Electrical and Electronics Engineering
    • 1.1 Basic Concepts
    • 1.2 Network Theorems
    • 1.3 Alternating Current Fundamentals
    • 1.4 Semiconductor Device
    • 1.5 Signal Generator
    • 1.6 Amplifiers
    • MCQs
      • MCQs On Basic Electrical
        • set-1
        • set-2
      • MCQs On Basic Electronics
        • set-1
        • set-2
  • 2. Digital Logic and Microprocessor
    • 2.1 Digital Logic
    • 2.2 Combinational & Arithmetic Circuit
    • 2.3 Sequential Logic Circuits
    • 2.4 Microprocessor
    • 2.5 Microprocessor System
    • 2.6 Interrupt Operations
    • MCQs
      • MCQs On Digital Logic
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
        • set-7
        • set-8
        • set-9
        • set-10
        • set-11
        • set-12
      • MCQs On Microprocessor
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
        • set-7
        • set-8
        • set-9
  • 3. Programming Language and Its Applications
    • 3.1 Introduction to C Programming
    • 3.2 Pointers, Structures, and Data Files
    • 3.3 C++ Language Constructs with Objects and Classes
    • 3.4 Features of Object-Oriented Programming
    • 3.5 Pure Virtual Functions and File Handling
    • 3.6 Generic Programming and Exception Handling
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
  • 4. Computer Organization and Embedded System
    • 4.1 Control and CPU
    • 4.2 Computer Arithmetic and Memory System
    • 4.3 I/O Organization and Multiprocessor
    • 4.4 Embedded System Design
    • 4.5 Real-Time Operating and Control Systems
    • 4.6 Hardware Description Language (VHDL) and IC Technology
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
      • set-6
      • set-7
      • set-8
      • set-9
      • set-10
      • set-11
  • 5. Concept of Computer Network and Network Security System
    • 5.1 Introduction to Computer Networks
    • 5.2 Data Link Layer
    • 5.3 Network Layer
    • 5.4 Transport Layer
    • 5.5 Application Layer
    • 5.6 Network Security
    • MCQs
      • Basic Networking
        • set-1
        • set-2
      • Advanced Networking
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
  • 6. Theory of Computation and Computer Graphics
    • 6.1 Introduction to Finite Automata
    • 6.2 Introduction to Context-Free Languages (CFL)
    • 6.3 Turing Machines (TM)
    • 6.4 Introduction to Computer Graphics
    • 6.5 Two-Dimensional Transformation
    • 6.6 Three-Dimensional Transformation
    • MCQs
      • MCQs on Theory of Computation
        • set-1
        • set-2
        • set-3
      • MCQs On Computer Graphics
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
  • 7. Data Structures and Algorithm, Database System and Operating System
    • 7.1 Introduction to Data Structures, Lists, Linked Lists, and Trees
    • 7.2 Sorting, Searching, Hashing and Graphs
    • 7.3 Introduction to Data Models, Normalization, and SQL
    • 7.4 Transaction Processing, Concurrency Control, and Crash Recovery
    • 7.5 Introduction to Operating System and Process Management
    • 7.6 Memory Management, File Systems, and System Administration
    • MCQs
      • MCQs ON DSA
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
      • MCQs On DBMS
        • set-1
        • set-2
      • MCQs On Operating System
        • set-1
        • set-2
        • set-3
        • set-4
        • set-5
        • set-6
        • set-7
        • set-8
        • set-9
        • set-10
        • set-11
        • set-12
  • 8. Software Engineering and Object-Oriented Analysis & Design
    • 8.1 Software Process and Requirements
    • 8.2 Software Design
    • 8.3 Software Testing, Cost Estimation, Quality Management, and Configuration Management
    • 8.4 Object-Oriented Fundamentals and Analysis
    • 8.5 Object-Oriented Design
    • 8.6 Object-Oriented Design Implementation
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
      • set-6
      • set-7
      • set-8
      • set-9
  • 9. Artificial Intelligence and Neural Networks
    • 9.1 Introduction to AI and Intelligent Agents
    • 9.2 Problem Solving and Searching Techniques
    • 9.3 Knowledge Representation
    • 9.4 Expert System and Natural Language Processing
    • 9.5 Machine Learning
    • 9.6 Neural Networks
    • MCQs
      • set-1
      • set-2
      • set-3
      • set-4
      • set-5
      • set-6
      • set-7
      • set-8
      • set-9
  • 10. Project Planning, Design and Implementation
    • 10.1 Engineering Drawings and Its Concepts
    • 10.2 Engineering Economics
    • 10.3 Project Planning and Scheduling
    • 10.4 Project Management
    • 10.5 Engineering Professional Practice
    • 10.6 Engineering Regulatory Body
    • MCQs
      • MCQs On Engineering Drawing
        • set-1
        • set-2
      • MCQs On Engineering Economics
      • MCQs On Project Planning & Scheduling
      • MCQs On Project Mangement
      • MCQs On Engineering Professional Practice
      • MCQs On Engineering Regulatory Body
  • Questions Sets
    • Set 1 (Chaitra, 2080)
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Set 2 (Aasadh, 2081)
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Set 3 (Asojh, 2080)
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Model Set - Computer Engineering By NEC
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
    • Model Set - Software Engineering By NEC
      • Short Questions (60*1=60 Marks)
      • Long Questions (20*2=40 Marks)
  • Tips & Tricks
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  1. 6. Theory of Computation and Computer Graphics
  2. MCQs
  3. MCQs On Computer Graphics

set-5

201. Sound pressure levels are measured in ______.

  1. Decibels.

  2. Ounce

  3. Pound

  4. Fathom

Show me the answer

Answer: 1. Decibels.

Explanation:

  • Sound pressure levels are measured in decibels (dB).

  • Decibels are a logarithmic unit used to measure the intensity of sound.

  • The formula for sound pressure level in decibels is:

    Lp=20log⁡10(pp0)L_p = 20 \log_{10}\left(\frac{p}{p_0}\right)Lp​=20log10​(p0​p​)

    where:

    • LpL_pLp​ = sound pressure level in dB,

    • ppp = measured sound pressure,

    • p0p_0p0​ = reference sound pressure (typically 20μPa20 \mu Pa20μPa).

202. MIDI stands for ______.

  1. Music instruction digital interface.

  2. Musical instrument digital interface.

  3. Musical instrumental digital interface

  4. Music instrument digit interface

Show me the answer

Answer: 2. Musical instrument digital interface.

Explanation:

  • MIDI stands for Musical Instrument Digital Interface.

  • It is a protocol used for communicating musical information between devices.

  • MIDI does not transmit audio signals but rather instructions like note-on, note-off, pitch, and velocity.

203. Digital audio data is the actual representation of a ______.

  1. Light.

  2. Music

  3. Sound

  4. Noise

Show me the answer

Answer: 3. Sound.

Explanation:

  • Digital audio data is the actual representation of sound.

  • It is created by sampling analog sound waves and converting them into digital form.

  • The process involves:

    • Sampling: Capturing the amplitude of the sound wave at regular intervals.

    • Quantization: Rounding the amplitude values to the nearest integer.

    • The sampling rate is given by:

      fs=1Tsf_s = \frac{1}{T_s}fs​=Ts​1​

      where:

      • fsf_sfs​ = sampling frequency,

      • TsT_sTs​ = sampling interval.

204. Digital audio data is also called as ______.

  1. Tracks.

  2. Probable

  3. Examples

  4. Samples

Show me the answer

Answer: 4. Samples.

Explanation:

  • Digital audio data is also called samples.

  • Each sample represents the amplitude of the sound wave at a specific point in time.

  • The number of samples per second is determined by the sampling rate, fsf_sfs​.

205. In Windows, system sounds are ______ files.

  1. .Rar.

  2. .Wav.

  3. .3gp.

  4. .Wmv.

Show me the answer

Answer: 2. .Wav.

Explanation:

  • In Windows, system sounds are stored as .wav files.

  • The WAV format is commonly used for uncompressed audio.

  • The file size of a WAV file can be calculated using:

    File Size=fs×Bit Depth×Channels×Duration\text{File Size} = f_s \times \text{Bit Depth} \times \text{Channels} \times \text{Duration}File Size=fs​×Bit Depth×Channels×Duration

    where:

    • fsf_sfs​ = sampling frequency,

    • Bit Depth = number of bits per sample,

    • Channels = number of audio channels (e.g., stereo = 2),

    • Duration = length of the audio in seconds.

206. MIDI files are ______ than CD quality digital audio files.

  1. Larger.

  2. Too large

  3. Smaller

  4. Equal

Show me the answer

Answer: 3. Smaller.

Explanation:

  • MIDI files are smaller than CD-quality digital audio files.

  • This is because MIDI files contain instructions for playing music rather than the actual audio data.

  • For example, a MIDI file might store a note as:

    Note=(Pitch,Velocity,Duration)\text{Note} = (\text{Pitch}, \text{Velocity}, \text{Duration})Note=(Pitch,Velocity,Duration)

    whereas a digital audio file stores the actual waveform.

207. The sampling frequencies often used in multimedia are ______.

  1. KHz.

  2. MHz

  3. GHz

  4. DHz

Show me the answer

Answer: 1. KHz.

Explanation:

  • The sampling frequencies often used in multimedia are in kilohertz (KHz).

  • Common sampling rates include:

    • 44.1 KHz44.1 \text{ KHz}44.1 KHz (CD quality),

    • 48 KHz48 \text{ KHz}48 KHz (DVD quality).

  • The Nyquist theorem states that the sampling rate must be at least twice the highest frequency in the signal:

    fs≥2×fmaxf_s \geq 2 \times f_{\text{max}}fs​≥2×fmax​

    where:

    • fsf_sfs​ = sampling frequency,

    • fmaxf_{\text{max}}fmax​ = highest frequency in the signal.

208. The value of each sample is rounded off to the nearest integer known as ______.

  1. Samplization.

  2. Quantification

  3. Quantization

  4. Digitations

Show me the answer

Answer: 3. Quantization.

Explanation:

  • The process of rounding off the value of each sample to the nearest integer is called quantization.

  • Quantization introduces a small error called quantization noise.

  • The signal-to-noise ratio (SNR) for quantization is given by:

    SNR=6.02×Bit Depth+1.76 dB\text{SNR} = 6.02 \times \text{Bit Depth} + 1.76 \text{ dB}SNR=6.02×Bit Depth+1.76 dB

    where:

    • Bit Depth = number of bits per sample.

209. The amount of information stored about each sample is the ______.

  1. Sample no.

  2. Sample size.

  3. Sample volume.

  4. Sample unit

Show me the answer

Answer: 2. Sample size.

Explanation:

  • The amount of information stored about each sample is called the sample size.

  • It is typically measured in bits (e.g., 8-bit, 16-bit).

  • For example, a 16-bit sample size allows for:

    216=65,5362^{16} = 65,536216=65,536

    possible amplitude values.

210. LEDs operate on ______ frequencies.

  1. Optical.

  2. Infra

  3. Infrared

  4. Electro

Show me the answer

Answer: 3. Infrared.

Explanation:

  • LEDs (Light Emitting Diodes) often operate on infrared frequencies.

  • The wavelength of infrared light is given by:

    λ=cf\lambda = \frac{c}{f}λ=fc​

    where:

    • λ\lambdaλ = wavelength,

    • ccc = speed of light (3×108 m/s3 \times 10^8 \text{ m/s}3×108 m/s),

    • fff = frequency.

211. Digitized sound is the ______ sound.

  1. Sampled.

  2. Covered

  3. Diluted

  4. Modified

Show me the answer

Answer: 1. Sampled.

Explanation:

  • Digitized sound is the sampled sound.

  • It is created by capturing discrete samples of an analog sound wave.

  • The sampling process is governed by the Nyquist theorem:

    fs≥2×fmaxf_s \geq 2 \times f_{\text{max}}fs​≥2×fmax​

    where:

    • fsf_sfs​ = sampling frequency,

    • fmaxf_{\text{max}}fmax​ = highest frequency in the signal.

212. Sample size are ______ bits.

  1. 2

  2. 4

  3. 6

  4. 8

Show me the answer

Answer: 4. 8.

Explanation:

  • Sample sizes are typically 8 bits for basic audio quality.

  • Higher-quality audio uses 16-bit or 24-bit samples.

  • The dynamic range for an 8-bit sample is:

    20log⁡10(28)≈48 dB20 \log_{10}(2^8) \approx 48 \text{ dB}20log10​(28)≈48 dB

213. A 16-bit sample provides a staggering ______ units.

  1. 16,384.

  2. 32,786.

  3. 65,536.

  4. 84,658.

Show me the answer

Answer: 3. 65,536.

Explanation:

  • A 16-bit sample provides 65,536 possible values, as:

    216=65,5362^{16} = 65,536216=65,536

214. Removing of dead air or blank space from a front of the recording is known as ______.

  1. Trimming.

  2. Cutting

  3. Editing

  4. Erasing

Show me the answer

Answer: 1. Trimming.

Explanation:

  • Removing dead air or blank space from the front of a recording is called trimming.

215. MIDI sounds are typically stored in files with the ______ extensions.

  1. .MDI.

  2. .MID.

  3. .MIDI.

  4. .MDII.

Show me the answer

Answer: 2. .MID.

Explanation:

  • MIDI sounds are typically stored in files with the .MID extension.

216. A popular effect in which one image transforms into another is known as ______.

  1. Animation.

  2. Encrypting

  3. Modifying

  4. Morphing

Show me the answer

Answer: 4. Morphing.

Explanation:

  • The effect where one image transforms into another is called morphing.

217. PAL stands for ______.

  1. Pass alternative line.

  2. Pass alternate line.

  3. Phase alternative line.

  4. Phase alternate line

Show me the answer

Answer: 4. Phase alternate line.

Explanation:

  • PAL stands for Phase Alternate Line, a color encoding system used in television broadcasting.

218. HDTV stands for ______.

  1. High defined television.

  2. Higher definition television.

  3. High definition television

  4. Higher defined television.

Show me the answer

Answer: 3. High definition television.

Explanation:

  • HDTV stands for High Definition Television, which provides higher resolution than standard-definition TV.

219. Light comes from an ______.

  1. Electron.

  2. Cathode

  3. Atom

  4. Radium

Show me the answer

Answer: 3. Atom.

Explanation:

  • Light is emitted from atoms when electrons transition between energy levels.

220. VGA stands for ______.

  1. Video graphics array.

  2. Video graph array.

  3. Visual graphics array.

  4. Visual graph array.

Show me the answer

Answer: 1. Video graphics array.

Explanation:

  • VGA stands for Video Graphics Array, a display standard for computers.

221. The area on a display device to which a window is mapped is called a ______.

  1. Window.

  2. View port

  3. Coordinate

  4. Section

Show me the answer

Answer: 2. View port.

Explanation:

  • The area on a display device to which a window is mapped is called a viewport.

222. A world coordinate area selected for display is called ______.

  1. Window.

  2. View port

  3. View point

  4. Section

Show me the answer

Answer: 1. Window.

Explanation:

  • A world coordinate area selected for display is called a window.

223. The region against which an object is to clipped is called a ______.

  1. Clipping.

  2. View port

  3. Window

  4. Clip window

Show me the answer

Answer: 4. Clip window.

Explanation:

  • The region against which an object is to be clipped is called a clip window.

224. The location of the point relative to the boundaries of the clipping rectangle is called ______code.

  1. Location.

  2. Binary

  3. Region

  4. Area

Show me the answer

Answer: 3. Region.

Explanation:

  • The location of the point relative to the boundaries of the clipping rectangle is called the region code.

225. The region code of the clipping rectangle is ______.

  1. 0001

  2. 1000

  3. 1111

Show me the answer

Answer: 1. 0000.

Explanation:

  • The region code of the clipping rectangle is 0000.

226. A region code is a ______ digit binary code.

  1. 16

  2. 8

  3. 4

  4. 2

Show me the answer

Answer: 3. 4.

Explanation:

  • A region code is a 4-digit binary code.

227. All-or none ______ strategy is used to keep all of the string inside a clip window.

  1. Word clipping.

  2. Character clipping

  3. Object clipping

  4. String clipping

Show me the answer

Answer: 2. Character clipping.

Explanation:

  • The all-or-none character clipping strategy is used to keep all of the string inside a clip window.

228. The picture parts to be saved are those that are outside the region is referred as ______clipping.

  1. Outside.

  2. Exterior

  3. External

  4. Extreme

Show me the answer

Answer: 1. Outside.

Explanation:

  • The picture parts to be saved are those that are outside the region, referred to as outside clipping.

229. ______ are used to describe interior properties by partitioning the spatial region containing an object into a set of small, non-overlapping contiguous solids.

  1. Surface rendering.

  2. Surface identifications.

  3. Space partitioning representation

  4. Boundary representations.

Show me the answer

Answer: 3. Space partitioning representation.

Explanation:

  • Space partitioning representations are used to describe interior properties by partitioning the spatial region containing an object into a set of small, non-overlapping contiguous solids.

230. ______ generally refers to any time sequence of visual changes in a scene.

  1. Computer animation.

  2. Visualization

  3. Graphics

  4. Frame work

Show me the answer

Answer: 1. Computer animation.

Explanation:

  • Computer animation generally refers to any time sequence of visual changes in a scene.

231. Analogy with two-dimensional polar co-ordinates the dimensions unit for solid angles is called the ______.

  1. Steradian.

  2. Angle

  3. Solid angle.

  4. Co-ordinates.

Show me the answer

Answer: 1. Steradian.

Explanation:

  • The unit for solid angles is called the steradian.

  • A solid angle is defined as:

    Ω=Ar2\Omega = \frac{A}{r^2}Ω=r2A​

    where:

    • Ω\OmegaΩ = solid angle in steradians,

    • AAA = surface area of the sphere,

    • rrr = radius of the sphere.

232. A ______ model is a method for explaining the properties or behavior of color within some particular control.

  1. Single color model

  2. Color

  3. Light color

  4. Spectral color

Show me the answer

Answer: 2. Color.

Explanation:

  • A color model is a method for explaining the properties or behavior of color within some particular control.

233. The dominant frequency is also called as the ______.

  1. Hue.

  2. Color

  3. Frequency

  4. Wavelength

Show me the answer

Answer: 1. Hue.

Explanation:

  • The dominant frequency is also called the hue.

  • Hue is related to the wavelength of light:

    λ=cf\lambda = \frac{c}{f}λ=fc​

    where:

    • λ\lambdaλ = wavelength,

    • ccc = speed of light,

    • fff = frequency.

234. The term ______ is used to refer collectively to two properties describing color characteristics: purity and dominant frequency.

  1. White light source.

  2. Purity

  3. Chromaticity

  4. Saturation

Show me the answer

Answer: 3. Chromaticity.

Explanation:

  • The term chromaticity refers to the combination of two properties:

    • Purity: How saturated or vivid a color is.

    • Dominant frequency: The wavelength of light that determines the color.

  • Chromaticity is often represented in a chromaticity diagram, which maps colors based on their hue and saturation.

235. ______ are the most commonly used boundary representation for a 3D graphics object.

  1. Data polygon

  2. System polygon

  3. Surface polygon

  4. Area polygon

Show me the answer

Answer: 3. Surface polygon.

Explanation:

  • Surface polygons are the most commonly used boundary representation for 3D graphics objects.

  • A polygon is a 2D shape defined by vertices and edges.

  • In 3D graphics, surfaces are often approximated using polygons (e.g., triangles or quadrilaterals).

  • The area of a polygon can be calculated using:

    A=12∣∑i=1n(xiyi+1−xi+1yi)∣A = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1} - x_{i+1} y_i) \right|A=21​​i=1∑n​(xi​yi+1​−xi+1​yi​)​

    where:

    • (xi,yi)(x_i, y_i)(xi​,yi​) are the coordinates of the polygon's vertices.

236. A three-dimensional object can also be represented using ______.

  1. Method

  2. Equation

  3. Point

  4. Line

Show me the answer

Answer: 2. Equation.

Explanation:

  • A three-dimensional object can be represented using equations.

  • For example:

    • A sphere can be represented by:

      (x−x0)2+(y−y0)2+(z−z0)2=r2(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2(x−x0​)2+(y−y0​)2+(z−z0​)2=r2

      where:

      • (x0,y0,z0)(x_0, y_0, z_0)(x0​,y0​,z0​) is the center of the sphere,

      • rrr is the radius.

    • A plane can be represented by:

      ax+by+cz+d=0ax + by + cz + d = 0ax+by+cz+d=0

      where:

      • a,b,ca, b, ca,b,c are the coefficients,

      • ddd is the constant term.

237. ______ is a simple object space algorithm that removes about half of the total polygons in an image as about half of the faces of objects are back faces.

  1. Wire frame model

  2. Constructive solid geometry methods

  3. Isometric projection

  4. Back face removal

Show me the answer

Answer: 4. Back face removal.

Explanation:

  • Back face removal is a simple object space algorithm.

  • It removes polygons that are not visible because they are facing away from the viewer.

  • This is based on the dot product of the surface normal and the view vector:

    If N⋅V>0, the face is visible.\text{If } \mathbf{N} \cdot \mathbf{V} > 0, \text{ the face is visible.}If N⋅V>0, the face is visible.

    where:

    • N\mathbf{N}N is the surface normal,

    • V\mathbf{V}V is the view vector.

238. By which, we can take a view of an object from different directions and different distances.

  1. Projection

  2. Translation

  3. Rotation

  4. Scaling

Show me the answer

Answer: 1. Projection.

Explanation:

  • Projection allows us to take a view of an object from different directions and distances.

  • There are two main types of projection:

    • Perspective projection: Objects farther away appear smaller.

    • Orthographic projection: Objects retain their size regardless of distance.

  • The perspective projection formula is:

    xp=xz,yp=yzx_p = \frac{x}{z}, \quad y_p = \frac{y}{z}xp​=zx​,yp​=zy​

    where:

    • (x,y,z)(x, y, z)(x,y,z) are the 3D coordinates,

    • (xp,yp)(x_p, y_p)(xp​,yp​) are the projected 2D coordinates.

239. The projection that can be viewed as the projection that has a center of projection at a finite distance from the plane of projection is called ______.

  1. Parallel projection

  2. Isometric projection

  3. Perspective projection

  4. Geometric projection

Show me the answer

Answer: 3. Perspective projection.

Explanation:

  • Perspective projection has a center of projection at a finite distance from the plane of projection.

  • This creates a realistic view where objects farther away appear smaller.

  • The formula for perspective projection is:

    xp=xz,yp=yzx_p = \frac{x}{z}, \quad y_p = \frac{y}{z}xp​=zx​,yp​=zy​

    where:

    • (x,y,z)(x, y, z)(x,y,z) are the 3D coordinates,

    • (xp,yp)(x_p, y_p)(xp​,yp​) are the projected 2D coordinates.

240. The surfaces that are blocked or hidden from view in a 3D scene are known as ______.

  1. Hidden surface

  2. Quad tree

  3. Frame buffer

  4. Area buffer

Show me the answer

Answer: 1. Hidden surface.

Explanation:

  • Hidden surfaces are surfaces that are blocked or hidden from view in a 3D scene.

  • Hidden surface removal algorithms are used to determine which surfaces are visible.

  • Common algorithms include:

    • Z-buffer algorithm,

    • Painter's algorithm,

    • Ray tracing.

241. ______ surface algorithm is based on perspective depth.

  1. Depth comparison

  2. Z-buffer or depth-buffer algorithm

  3. Subdivision method

  4. Back-face removal

Show me the answer

Answer: 2. Z-buffer or depth-buffer algorithm.

Explanation:

  • The Z-buffer algorithm is based on perspective depth.

  • It uses a depth buffer to store the depth (z-value) of each pixel.

  • The algorithm compares the depth of each new pixel with the depth stored in the buffer:

    If znew<zbuffer, update the pixel and depth buffer.\text{If } z_{\text{new}} < z_{\text{buffer}}, \text{ update the pixel and depth buffer.}If znew​<zbuffer​, update the pixel and depth buffer.

242. In ______ year, the Z-buffer algorithm was described.

  1. 1995

  2. 1974

  3. 1945

  4. 1981

Show me the answer

Answer: 2. 1974.

Explanation:

  • The Z-buffer algorithm was described in 1974.

  • It is one of the simplest and most widely used algorithms for hidden surface removal.

243. Z-buffer algorithm is ______.

  1. Simplest algorithm

  2. Largest algorithm

  3. Complex algorithm

  4. Poor algorithm

Show me the answer

Answer: 1. Simplest algorithm.

Explanation:

  • The Z-buffer algorithm is the simplest algorithm for hidden surface removal.

  • It is easy to implement and works well for most 3D rendering tasks.

244. The painter algorithm is based on the property of ______.

  1. Polygon buffer

  2. Depth buffer

  3. Frame buffer

  4. Area buffer

Show me the answer

Answer: 3. Frame buffer.

Explanation:

  • The painter algorithm is based on the property of the frame buffer.

  • It works by drawing objects from back to front, similar to how a painter paints a scene.

  • Objects are sorted by their depth, and the frame buffer is updated accordingly.

245. ______ type of projection does not have the projection rays parallel to each other.

  1. Axonometric projection

  2. Orthographic projection

  3. Oblique projection

  4. Perspective projection

Show me the answer

Answer: 4. Perspective projection.

Explanation:

  • Perspective projection does not have the projection rays parallel to each other.

  • Instead, the rays converge at a single point (the center of projection).

  • This creates a realistic view where objects farther away appear smaller.

246. ______ are the three principal planes in orthographic projection.

  1. Front, top, profile

  2. Top, front, right side

  3. Back, top, profile

  4. Frontal, horizontal, profile

Show me the answer

Answer: 4. Frontal, horizontal, profile.

Explanation:

  • The three principal planes in orthographic projection are:

    • Frontal plane,

    • Horizontal plane,

    • Profile plane.

  • These planes are used to create 2D views of a 3D object.

247. The painter algorithm was developed in ______.

  1. 1972 by Newell

  2. 1974 by Catmull

  3. 1972 by Evans

  4. 1976 by Evans

Show me the answer

Answer: 1. 1972 by Newell.

Explanation:

  • The painter algorithm was developed in 1972 by Newell.

  • It is one of the earliest algorithms for hidden surface removal.

248. All the hidden surface algorithms employ image space approach except ______.

  1. Back face removal

  2. Scan line method

  3. Depth buffer method

  4. Depth sort method

Show me the answer

Answer: 1. Back face removal.

Explanation:

  • All hidden surface algorithms employ an image space approach except back face removal.

  • Back face removal is an object space algorithm.

249. ______ are the two types of projections that give a pictorial view of the object without convergence.

  1. Orthographic and perspective

  2. Perspective and oblique

  3. Oblique and axonometric

  4. Isometric and orthographic

Show me the answer

Answer: 3. Oblique and axonometric.

Explanation:

  • Oblique and axonometric projections give a pictorial view of the object without convergence.

  • These projections are often used in technical drawings.

250. The name of a visible surface detection algorithm is ______.

  1. Back face detection

  2. Ray tracing

  3. Back face removal

  4. Area tracing

Show me the answer

Answer: 3. Back face removal.

Explanation:

  • The name of a visible surface detection algorithm is back face removal.

  • It removes polygons that are not visible because they are facing away from the viewer.

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