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The correct option is (b) 110 cm.

A light pen is a pointing device shaped like a pen and is connected to a monitor. The tip of the light pen contains a light-sensitive element which detects the light from the screen enabling the computer to identify the location of the pen on the screen. Light pens have the advantage ofedrawing’ directly onto the screen, but this becomes hard to use, and is also not accurate.

The correct option is (d) 225 cm.

Different types of mouse available are:

Mechanical Mouse, Optical, Laser Mouse, Air Mouse, 3D Mouse, Tactile Mouse,firgonomic Mouse and Gaming Mouse.

(c) Power on Self Test

Conversion of centimeter into millimeter

4 centimeter = …………. millimeter

∵ 4 = 1 cm. + 1 cm. + 1 cm. + 1 cm.

= 10 mm + 10 mm. + 10 mm. + 10 mm. = 40 mm.

or 4 x 10 mm. = 40 mm.

(i) 140 cm.

(ii) 170 cm.

(iii) 130 cm.

Value of 1 meter is equal to 100 cm.

The most commonly used Input devices are: 

(i) Keyboard:

Keyboard (wired / wireless, virtual) is the most common input device used today. The individual keys for letters, numbers and special characters are collectively known as character keys. This keyboard layout is derived from the keyboard of original typewriter.

The data and instructions are given as input to the computer by typing on the keyboard. Apart from alphabet and numeric keys, it also has Function keys for performing different functions. There are different set of keys available in the keyboard such as character keys, modifier keys, system and GUI keys, enter and editing keys, function keys, navigation keys, numeric keypad and lock keys.

(ii) Mouse:

Mouse (wired/wireless) is a pointing device used to control the movement of the cursor on the display screen. It can be used to select icons, menus, command buttons or activate something on a computer. Some mouse actions are move, click, double click, right click, drag and drop.

(iii) Scanner: 

Scanners are used to enter the information directly into the computer’s memory. This device works like a Xerox machine. The scanner converts any type of printed or written information including photographs into a digital format, which can be manipulated by the computer.

(iv) Barcode Readers: 

A Bar code is a pattern printed in lines of different thickness. The Bar code reader scans the information on the bar codes transmits to the Computer for further processing. The system gives fast and error free entry of information into the computer.

(v) Digital camera:

It captures images J videos directly in the digital form. It uses a CCD (Charge Coupled Device) electronic chip. When light falls on the chip through the lens, it converts light rays into digital format.

1. 100

2. distance / length

3. 100

4. 15000

5. 1

6. 8000

7. 5

1000 meter in 1 km.

Speakers produce voice output (audio). Using speaker along with speech synthesize software, the computer can provide voice output. This has become very common in places like airlines, schools, banks, railway stations, etc.

Hardware is the physical component of a computer like motherboard, memoiy devices, monitor, keyboard etc. while software is the set of programs or instructions.

(i) 9 meter

(ii) 5 meter 60 cm.

(iii) 2 meter 40 cm.

Generally, length is measured in meter – cm.

Conversion of meter into centimeter

3 meter = ……….. centimeter

∵ 3 meter = 1 meter + 1 meter + 1 meter

= 100 centimeter + 100 centimeter + 100 centimeter = 300 centimeter

or 3 x 100 centimeter = 300 cm.

Inch tape used by tailor has marks like meter scale but inch tape longer then meter scale.

Appropriate unit of measurement for long distance in km.

Registers are the high speed temporary storage locations in the CPU. Although the number of registers varies from computer to computer, there are some registers which are common to all computers

Meter scale is of 1 meter while inch tape is long in length, Meter scale cannot be folded while tape can be put by folding it.

Meter scale is authentic way of measuring length while rod is made according to our need by measuring meter scale. Meter scale have inch, cm and mm, marks while rod does not have cm and mm marks on it.

In 1 meter = 100 centimeter

Therefore 1/4 meter 

= 1/4 x 100 centimeter 

= 25 centimeter.

(i) 12 kilometer 

To convert kilometer into meter multiply by 1000.

Therefore 12 km = 12 × 1000 = 12000 meter

(ii) 10 1/2 kilometer

To convert kilometer into meter, multiply by 1000

Therefore 10 1/2 km = 10 km + 1/2 km

= 10 x 1000 meter + 1/2 x 1000 meter

= 10000 + 500 meter = 10500 meter

(iii) 25 1/4 meter

To convert meter into centimeter multiply by 100.

Therefore 25 1/4 meter

= 25 meter + 1/4 meter

= 25 x 100 cm + 1/4 x 100 cm

= 2500 cm + 25 cm

= 2525 cm.

(iv) 15 3/4 meter

To convert meter into centimeter multiply by 100

Therefore meter 15 3/4 cm. = 15 meter + 3/4 meter

= 15 x 100 cm. + 3/4 x 100 cm.

= 1500 cm. + 3 x 25 cm.

= 1500 cm. + 75 cm.

= 1575 cm.

(v) 4 1/5 cm.

To convert centimeter into millimeter multiply by 10.

Therefore 4 1/5 cm.

= 4 cm. + 1/5 cm.

= 4 x 10 mm. + 1/5 x 10 mm.

= 40 mm. + 2 mm.

= 42 mm.

(vi) 1 4/5 cm.

To convert centimeter into millimeter multiply by 10.

Therefore 1 4/5 cm. = 1 cm. + 4/5 cm

= 1 x 10 mm. + 4/5 x 10 mm.

= 10 mm. + 4 x 2 mm.

= 10 mm. + 8 mm

= 18 mm.

Robotics deals with the design, construction, operation and use of robots, as well as computer systems for their control, sensory feedback and processing.

(i) Chanchal jumped the maximum distance 1 meter 70 centimeter.

Distance = 1 meter 70 centimeter

= 100 cm. + 70 cm.

= 170 cm.

(ii) Ankita jumped shortest distance.

Distance = 1 meter 10 centimeter

= 100 centimeter + 10 centimeter

= 110 centimeter.

Speech Recognition, Natural Language Understanding and Natural Language Generation.

Speech Recognition, Natural Language Understanding and Natural Language Generation

No, the vocabulary of a corpus does not remain the same before and after text normalization. 

Reasons are:

● In normalization the text is normalized through various steps and is lowered to minimum vocabulary since the machine does not require grammatically correct statements but the essence of it. 

● In normalization Stop words, Special Characters and Numbers are removed. 

● In stemming the affixes of words are removed and the words are converted to their base form.

So, after normalization, we get the reduced vocabulary.

In Text Normalization, we undergo several steps to normalize the text to a lower level. After the removal of stop words, we convert the whole text into a similar case, preferably lower case. This ensures that the case-sensitivity of the machine does not consider same words as different just because of different cases.

Bag of Words is a Natural Language Processing model which helps in extracting features out of the text which can be helpful in machine learning algorithms. In bag of words, we get the occurrences of each word and construct the vocabulary for the corpus. 

Bag of Words just creates a set of vectors containing the count of word occurrences in the document (reviews). Bag of Words vectors are easy to interpret.

The communications made by the machines are very basic and simple. Human communication is complex. There are multiple characteristics of the human language that might be easy for a human to understand but extremely difficult for a computer to understand.

For machines it is difficult to understand our language. Let us take a look at some of them here:

Arrangement of the words and meaning - There are rules in human language. There are nouns, verbs, adverbs, adjectives. A word can be a noun at one time and an adjective some other time. This can create difficulty while processing by computers.

Analogy with programming language- Different syntax, same semantics: 2+3 = 3+2 Here the way these statements are written is different, but their meanings are the same that is 5. Different semantics, same syntax: 2/3 (Python 2.7) ≠ 2/3 (Python 3) Here the statements written have the same syntax but their meanings are different. In Python 2.7, this statement would result in 1 while in Python 3, it would give an output of 1.5.

Multiple Meanings of a word - In natural language, it is important to understand that a word can have multiple meanings and the meanings fit into the statement according to the context of it.

Perfect Syntax, no Meaning - Sometimes, a statement can have a perfectly correct syntax but it does not mean anything. In Human language, a perfect balance of syntax and semantics is important for better understanding.

These are some of the challenges we might have to face if we try to teach computers how to understand and interact in human language.

Term Frequency: Term frequency is the frequency of a word in one document. Term frequency can easily be found from the document vector table as in that table we mention the frequency of each word of the vocabulary in each document.

Inverse Document Frequency: The other half of TFIDF which is Inverse Document Frequency. For this, let us first understand what does document frequency mean. Document Frequency is the number of documents in which the word occurs irrespective of how many times it has occurred in those documents. 

The document frequency for the exemplar vocabulary would be:

Talking about inverse document frequency, we need to put the document frequency in the denominator while the total number of documents is the numerator. 

Here, the total number of documents are 3, hence inverse document frequency becomes:

The formula of TFIDF for any word W becomes:

TFIDF(W) = TF(W) * log (IDF(W))

The words having highest value are – Mumbai, Famous

Text Normalizationin Text Normalization, we undergo several steps to normalize the text to a lower level.

Sentence Segmentation - Under sentence segmentation, the whole corpus is divided into sentences. Each sentence is taken as a different data so now the whole corpus gets reduced to sentences.

Tokenisation- After segmenting the sentences, each sentence is then further divided into tokens. Tokens is a term used for any word or number or special character occurring in a sentence. Under tokenisation, every word, number and special character is considered separately and each of them is now a separate token.

Removing Stop words, Special Characters and Numbers - In this step, the tokens which are not necessary are removed from the token list.

Converting text to a common case -After the stop words removal, we convert the whole text into a similar case, preferably lower case. This ensures that the case-sensitivity of the machine does not consider same words as different just because of different cases.

Stemming In this step, the remaining words are reduced to their root words. In other words, stemming is the process in which the affixes of words are removed and the words are converted to their base form.

Lemmatization -in lemmatization, the word we get after affix removal (also known as lemma) is a meaningful one.

With this we have normalized our text to tokens which are the simplest form of words present in the corpus. Now it is time to convert the tokens into numbers. For this, we would use the Bag of Words algorithm

Time taken to cross the river

=\(\frac {Width\,of\,the\, river}{Speed\, in\, this\,direction}\)

= \(\frac{1km}{4km^{-1}}\)

= \(\frac{1}{4} \,h = 15 \,min\)

Distance along the river covered in this time 

= Speed of the river x Time taken to cross the river

= 3 km h^-1 × 15 min = 0.75 km.

If an athlete takes jump with velocity u and at an angle θ with horizontal, then

\(H = \frac{u^2sin^2\theta}{2g}\)

and \(R=\frac{u^2\,sin\,2\theta}{g}\)

∴ \(\frac{R}{H}=\frac{u^2\,sin\,2\theta}{g}\times\frac{2g}{u^2sin^2\theta}\)

 = \(\frac{2\,sin\,2\theta}{sin^2\,\theta}=\frac{2\times2\,sin\,\theta\,cos\,\theta}{sin^2\,\theta}\)

 = 4 cot θ

or R = 4H cot θ

Thus, the span of the jump depends on the height of the jump and the angle at which the athlete jumps.

Since we all know that the language of computers is Numerical, the very first step that comes to our mind is to convert our language to numbers.

This conversion takes a few steps to happen. The first step to it is Text Normalization. Since human languages are complex, we need to first of all simplify them in order to make sure that the understanding becomes possible. Text Normalization helps in cleaning up the textual data in such a way that it comes down to a level where its complexity is lower than the actual data.

Let \(\vec A\) and \(\vec B\) be the two vectors. 

\(\vec A\) =\(\vec{(A_x}\hat{i}+\vec{A_y}\hat{j}+\vec{A_z}\hat{k})\)

 and \(\vec R\) = \(\vec{(B_x}\hat{i}+\vec{B_y}\hat{j}+\vec{B_z}\hat{k})\)

\(\vec A\) × \(\vec B\) =\(\vec{(A_x}\hat{i}+\vec{B_y}\hat{j}+\vec{B_z}\hat{k})\)

A_xB_x(\(\hat{i}\times\hat{i}\)) + A_xB_y(\(\hat{i}\times\hat{j}\)) + A_xB_y(\(\hat{i}\times\hat{k}\)) + A_yB_x(\(\hat{j}\times\hat{i}\)) + A_yB_y(\(\hat{j}\times\hat{j}\)) + A_yB_z (\(\hat{j}\times\hat{k}\)) + A_zB_x(\(\hat{k}\times\hat{i}\)) + A_zB_x(\(\hat{k}\times\hat{j}\)) 

+ A_zB_y (\(\hat{k}\times\hat{k}\))

or\(\vec{A}\) × \(\vec{B}\) = (A_yB_z − A_zB_y)\(\hat{i}\)– (A_xB_y − A_zB_x)\(\hat{j}\)+ (A_xB_y − A_yB_x)\(\hat{k}\)

or \(\vec{A}\) × \(\vec{B}\) = A_xB_x(0) + A_xB_y(\(\hat{k}\)) + A_xB_z (\(-\hat{j}\)) + ̂ A_yB_x(\(-\hat{k}\)) + A_yB_y(0) + A_yB_z (\(\hat{i}\)) + A_zB_x(\(\hat{j}\)) + A_zB_y(\(-\hat{i}\)) + A_zB_z(0)

It can be written in determinant from as

 \(\vec{A}\) ×\(\vec{B}\) = \(\begin {vmatrix} \hat{i} & \hat{j} &\hat{k} \\ A_x &A_y&A_z \\ A_x&B_y&B_z \end{vmatrix}\)

Clearly the two particles are at the two ends of the diameter of a circular path. It is further clear that each particle will return to its initial position after describing one circle.

If t = required time then,

t = \(\frac{2\pi r}{v}=\frac{\pi d}{v}\)

Where,

v = speed of particles.

r = radius of the circular track

d = constant distance between particles

2r = diameter of the circular path

The passenger inside the car is in the reference frame of the car so he experiences centrifugal force

= \(\frac{mv^2}{r}\) acting on the car. So, to remain in circular path force should be larger as it is proportional to v². For an observer on the road side finds that the centripetal force \(\frac{mv^2}{r}\) acting on the car becomes inadequate with increases in v. Since again v² ∝ r is insufficient so the car skids.

The angular velocity of hour hand of a watch is greater than the angular velocity of earth around its own axis.

Explanation – We know that angular velocity (ω) of an object having time period (T) is given by ω = \(\frac{2\pi}{v}\) …(i)

T for hour hand of a watch is 12 ℎ

∴ω_h \(=\frac{2\pi}{12}=\frac{\pi}{6}\) …(ii)

T for each is 24 ℎ

∴ ω_e   \(=\frac{2\pi}{24}=\frac{\pi}{12}\) rad h^-1

Equations (ii), (iii) gives,

\(\frac{\omega_h}{\omega_e}=\frac{\frac{\pi}{6}}{\frac{\pi}{12}}=2\)

or \(\omega_h=2\omega_e\)

or \(\omega_h>\omega_e\)

Correct option is (D) None

\(\because\) The cost of 12 fruits = Rs 27

\(\therefore\) The cost of 18 fruits \(=Rs\,(\frac{27}{12}\times18)\)

\(=Rs\,(\frac{27}{2}\times3)\) = Rs (27 \(\times\) 1.5)

= Rs 40.5

Correct option is  D) None

Correct option is (B) ₹ 176

\(\because\) The cost of 5 litres of petrol = Rs 110.

\(\therefore\) The cost of 8 litres of petrol \(=Rs\,(\frac{110}5\times8)\)

\(=Rs\,(22\times8)\)

= Rs 176

Correct option is  B) ₹ 176

We use the relation \(\frac{\text{x}}{y}=\frac{\text{x}}{y}\) Here x₁ = 5, y₁ = 210 and x₂ = 2

Here, \(\frac{3}{72}=\frac{\text{x}{_1}}{120}\)

⇒ x₁ × 72 = 3 × 120

⇒ x₁ = \(\frac{3\,\times\,120}{72}=5\)

Now, \(\frac{3}{72}=\frac{\text{x}{_2}}{192}\)

⇒ x₂ × 72 = 3 × 192

⇒ x₂ = \(\frac{3\,\times\,192}{72}=8\)

And \(\frac{3}{72}=\frac{10}{y{_2}}\)

⇒ y₂ × 3 = 10 × 72

⇒ y₂ = \(\frac{10\,\times\,72}{3}=240\)

If more dolls are bought the cost will also be more. Therefore it’s directly proportional.

Let the number of dolls be x, 18/630 = x/455

630 × x = 18 × 455

630x = 8190

x = 8190/630

x = 13

∴ 13 dolls can be bought for ₹455.

Here we check the values of x × y

(i) 6 × 9 = 54, 10 × 15 = 150, 14 × 21 = 294, 16 × 24 = 384 ; unequal

(i) 5 × 18 = 90, 9 × 10 = 90, 15 × 6 = 90, 3 × 30 = 90,45 × 2 = 90; equal

(i) 9 × 4 = 36, 3 × 12 = 36, 6 × 6 = 36, 6 × 9 = 54, 36 × 1 = 36 ;unequal

The variable x is inversely proportional to y.

xy = k(constant)

Since, we know that two quantities x and y are said to be in inverse proportion, if an increase in x cause a proportional decrease in y and vice-a-versa. So, we can say y decreases by p%.

Correct option is (C) x/y = k

\(x\propto y\)

\(\Rightarrow\) x = ky

\(\Rightarrow\) \(\frac{x}{y}=k\)

Alternate :- If x is directly proportional to y then the ratio of x and y must be constant.

i.e., \(\frac{x}{y}=k\)

Correct option is  C) x/y = k

Correct option is (B) ₹ 3000

\(\because\) The cost of 65 packets of tea = Rs 2600

\(\therefore\) The cost of 75 packets of tea \(=Rs(\frac{2600}{65}\times75)\)

\(=Rs(\frac{200}{5}\times75)\)

\(=Rs(200\times15)\)

= Rs 3000

Correct option is  B) ₹ 3000

If more amount of sugar is bought the cost will also be more. Therefore it’s directly proportional.

Let the amount of sugar be x kg, 9/238.50 = x/371

238.50 × x = 9 × 371

238.50x = 3339

x = 3339/238.5.

x = 14

∴ 14kg of sugar can be bought for ₹371.

False

If one angle of a triangle is kept fixed, then the measure of the remaining two angles can’t vary inversely with each other.

e.g. In ΔABC, ∠A + ∠B + ∠C = 180° [sum of all angles of a Δ]

If ∠A = 50°, then ∠B + ∠C = 180° -50° = 130°

So, it is not depending on any proportion by applying angle sum properties of a triangle.