PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण

Punjab State Board PSEB 10th Class Hindi Book Solutions Hindi Grammar visheshan nirman विशेषण-निर्माण Exercise Questions and Answers, Notes.

PSEB 10th Class Hindi Grammar विशेषण-निर्माण

निम्नलिखित शब्दों के विशेषण बनाइए:
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 1
उत्तर:
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 2

प्रश्न 1.
इतिहास का विशेषण है
(क) इतिहासीय
(ख) ऐतिहासिक
(ग) इतिहासिक
(घ) ऐतिहासु।
उत्तर:
(ख) ऐतिहासिक

PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण

प्रश्न 2.
घर का विशेषण है
(क) घरेलू
(ख) घरीय
(ग) घरीला
(घ) घरेस्वी।
उत्तर:
(क) घरेलू

प्रश्न 3.
परिवार का विशेषण है
(क) परिवारी
(ख) परिवारू
(ग) पारिवारिक
(घ) परिवारीय।
उत्तर:
(ग) पारिवारिक

प्रश्न 4.
राष्ट्र का विशेषण है
(क) राष्ट्रीय
(ख) राष्ट्र
(ग) राष्ट्रिकता
(घ) राष्ट्रपन।
उत्तर:
(क) राष्ट्रीय

प्रश्न 5.
हृदय का विशेषण है,
(क) हृदयी
(ख) हार्दिय
(ग) हार्दिक
(घ) हार्दित।
उत्तर:
(ग) हार्दिक

प्रश्न 6.
नगर का विशेषण है
(क) नागर
(ख) नागरिक
(ग) नागर्कि
(घ) नागार्कयी।
उत्तर:
(ख) नागरिक

प्रश्न 7.
जिज्ञासा का विशेषण जिज्ञासु है (हाँ या नहीं में उत्तर लिखें)
उत्तर:
हाँ
प्रश्न 8.
भारत का विशेषण भारतीय है। (सही या गलत में उत्तर लिखें)
उत्तर:
सही

प्रश्न 9.
प्रकृति का विशेषण परकृतिय है (हाँ या नहीं में उत्तर लिखें)
उत्तर:
नहीं

प्रश्न 10.
साहित्य का विशेषण साहित्यिक है (सही या गलत में उत्तर लिखें)
उत्तर:
गलत।

PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण

वर्ष
निम्नलिखित में से किसी एक शब्द का विशेषण लिखिए
1. बाहर, पंजाब।
उत्तर:
बाहर = बाहरी
पंजाब = पंजाबी।

2. दिन, धन।
उत्तर:
दिन = दैनिक
धन = धनवान, धनवती।

3. बुद्धि, शहर।
उत्तर:
बुद्धि = बुद्धिमान
शहर = शहरी।

वर्ष
1. रंग, खामोश
उत्तर:
रंग = रंगीला
खामोश = खामोशी।

2. सप्ताह, ज्ञान
उत्तर:
सप्ताह = साप्ताहिक
ज्ञान = ज्ञानी/ज्ञानवान।

3. व्यापार, हित
उत्तर:
व्यापार = व्यापारिक
हित = हितवती/हितैषी।

प्रश्न 1.
विशेषण किसे कहते हैं?
उत्तर:
जिन शब्दों से संज्ञा या सर्वनाम की विशेषता व्यक्त होती है, उन्हें विशेषण कहते हैं; जैसेभोली राधा चतुर कृष्ण की बातों में आ गई। इस वाक्य में भोली राधा का तथा चतुर कृष्ण का विशेषण है।

प्रश्न 2.
मूल रूप से कौन-से शब्द विशेषण हैं?
उत्तर:
मूल रूप से विशेषण शब्द अच्छा, कोमल, बुरा, लाल, पीला, विद्वान, पुराना, कठोर, नया, निपुण, मज़बूत आदि हैं।

प्रश्न 3.
विशेषणों का निर्माण कैसे होता है?
उत्तर:
विशेषणों का निर्माण संज्ञा, सर्वनाम, क्रिया तथा अव्यय शब्दों में प्रत्यय लगाकर तथा कहीं-कहीं कुछ आवश्यक परिवर्तन करके किया जाता है; जैसे-

  1. संज्ञा शब्द ‘पालन’ में ‘अक’ प्रत्यय लगाकर तथा अंतिम वर्ण को हटाकर विशेषण शब्द ‘पालक’ बनता है।
  2. सर्वनाम शब्द ‘वह’ में ‘सा’ प्रत्यय लगाने से विशेषण शब्द ‘वैसा’ बनता है।
  3. क्रिया शब्द ‘पढ़ना’ में ‘आक् प्रत्यय लगाने तथा अंतिम ‘ना’ हटाने से विशेषण शब्द ‘पढ़ाकू’ बनता है।
  4. अव्यय शब्द ‘भीतर’ में ‘ई’ प्रत्यय लगाने से विशेषण शब्द ‘भीतरी’ बनता है।

PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण

प्रश्न 4.
संज्ञा शब्दों से विशेषण शब्दों का निर्माण कीजिए।
उत्तर:
संज्ञा शब्द से विशेषण शब्द का निर्माण
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 3
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 4
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 5

PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण

प्रश्न 5.
सर्वनाम शब्दों से विशेषण शब्दों का निर्माण कीजिए।
उत्तर:
सर्वनाम शब्द से विशेषण शब्द निर्माण
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 6

प्रश्न 6.
क्रिया शब्दों से विशेषण शब्दों का निर्माण कीजिए।
उत्तर:
क्रिया शब्द से विशेषण शब्द निर्माण
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 7

प्रश्न 7.
अव्यय शब्दों से विशेषण शब्दों का निर्माण कीजिए।
उत्तर:
अव्यय शब्द से विशेषण शब्द निर्माण
PSEB 10th Class Hindi Vyakaran विशेषण-निर्माण 8

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

1. Solve the following fractions :

Question (i).
4 + \(\frac {7}{8}\)
Answer:
4 + \(\frac {7}{8}\)
= \(\frac{4 \times 8+7}{8}\)
= \(\frac{32+7}{8}\)
= \(\frac {39}{8}\)
= 4\(\frac {7}{8}\)

Question (ii).
\(\frac{9}{11}-\frac{4}{15}\)
Answer:
\(\frac{9}{11}-\frac{4}{15}\)
= \(\frac{9 \times 15-4 \times 11}{11 \times 15}\)
= \(\frac{135-44}{165}\)
= \(\frac {91}{165}\)

Question (iii).
\(\frac{11}{16}-\frac{2}{5}+\frac{8}{10}\)
Answer:
\(\frac{11}{16}-\frac{2}{5}+\frac{8}{10}\)
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 1a
LCM of 16, 5 and 10
= 2 × 2 × 2 × 2 × 5
= 80
= \(\frac{11 \times 5-2 \times 16+8 \times 8}{80}\)
= \(\frac{55-32+64}{80}\)
= \(\frac {87}{80}\)
= 1\(\frac {7}{80}\)

Question (iv).
\(2 \frac{1}{5}+6 \frac{1}{2}\)
Answer:
\(2 \frac{1}{5}+6 \frac{1}{2}\)
= \(\frac{11}{5}+\frac{13}{2}\)
= \(\frac{11 \times 2+13 \times 5}{5 \times 2}\)
= \(\frac{22+65}{10}\)
= \(\frac {87}{10}\)
= 8\(\frac {7}{10}\)

Question (v).
\(8 \frac{1}{2}-3 \frac{5}{8}\)
Answer:
\(8 \frac{1}{2}-3 \frac{5}{8}\)
= \(\frac{17}{2}-\frac{29}{8}\)
= \(\frac{17 \times 4-29}{8}\)
= \(\frac{68-29}{8}\)
= \(\frac {39}{8}\)
= 4\(\frac {7}{8}\)

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

Question (vi).
\(\frac{9}{10}-\frac{9}{100}+\frac{9}{1000}\)
Answer:
\(\frac{9}{10}-\frac{9}{100}+\frac{9}{1000}\)
= \(\frac{9 \times 100-9 \times 10+9}{1000}\)
= \(\frac{900-90+9}{1000}\)
= \(\frac {810}{1000}\)

2. Arrange the following in ascending order :

Question (i).
\(\frac{2}{17}, \frac{10}{17}, \frac{3}{17}, \frac{16}{17}, \frac{5}{17}, \frac{8}{17}\)
Answer:
Ascending order of \(\frac{2}{17}, \frac{10}{17}, \frac{3}{17}, \frac{16}{17}, \frac{5}{17}, \frac{8}{17}\) is:
\(\frac{2}{17}, \frac{3}{17}, \frac{5}{17}, \frac{8}{17}, \frac{10}{17}, \frac{16}{17}\)

Question (ii).
\(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
Answer:
\(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 2a
L.C.M of 5, 7 , 10 = 2 × 5 × 7
= 70
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 3a

3. The three sides AB, BC and CA of a triangle ΔABC are \(\frac {5}{6}\)cm, \(\frac {2}{3}\)cm and \(\frac {7}{10}\) cm respectively. Find the perimeter of the triangle.
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 9
Answer:
Slides of ΔABC are
AB = \(\frac {5}{6}\) cm,
BC = \(\frac {2}{3}\)
CA = \(\frac {7}{10}\)
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 4a
L.C.M. (6, 3, 10) = 2 × 3 × 5 = 30
Perimeter of ΔABC = AB + BC + CA
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 5a

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

4. Ramesh studies for 5\(\frac {2}{3}\) hours daily. He devotes 2\(\frac {4}{5}\) hours of his time for science devotes for other subjects ?
Answer:
Total daily time for all subjects
= 5\(\frac {2}{3}\) hours = \(\frac {17}{3}\) hours
Time for science and mathematics
= 2\(\frac {4}{5}\) hours = \(\frac {14}{5}\) hours
Time for other subjects
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 6a

5. Sonia jogs once around the rectangular park of sides 10\(\frac {2}{3}\)m and 12\(\frac {1}{2}\)m. Find the total distance covered by the Sonia.
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 7a
Answer:
Length of rectangular park
= 12\(\frac {1}{2}\)m = \(\frac {25}{2}\)m
Breadth of rectangular park
= 10\(\frac {2}{3}\)m = \(\frac {32}{3}\)m
Total distance covered by Sonia = 2 [Length + Breadth]
= \(2\left(\frac{32}{3}+\frac{25}{3}\right) \mathrm{m}\)
= \(2\left(\frac{32 \times 2+25 \times 3}{3 \times 2}\right) \mathrm{m}\)
= \(2\left(\frac{65+75}{6}\right) \mathrm{m}\)
= \(\frac {278}{6}\) m
= \(\frac {139}{3}\) m
= 46\(\frac {1}{3}\) m

6. Ritu coloured a picture in \(\frac {7}{12}\) hours. Vaibhav coloured the same picture in \(\frac {3}{4}\) hours. Who worked for a longer time and by what fraction ?
Answer:
Time taken by Ritu to colour
= \(\frac {7}{12}\) hours
Time taken by Vaibhav = \(\frac {3}{4}\) hours
= \(\frac {3}{4}\) × \(\frac {3}{3}\)
= \(\frac {9}{12}\) hours
Since 9 > 7
∴ \(\frac {9}{12}\) > \(\frac {7}{12}\)
∴ Vaibhav worked for more time.
Difference between time taken by
Vaibhav and Ritu = \(\frac{3}{4}-\frac{7}{12}\)
= \(\frac{3 \times 3-7}{12}\)
= \(\frac{9-7}{12}=\frac{2}{12}\)
= \(\frac {1}{6}\) of an hour.

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

7. Multiple Choice Questions :

Question (i).
Fraction \(\frac {2}{5}\), \(\frac {7}{5}\) are :
(a) Like fractions
(b) Unlike fractions
(c) Equivalent fractions
(d) None of these
Answer:
(a) Like fractions

Question (ii).
What fraction do 8 hours of a day represents ?
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {8}{60}\)
(d) \(\frac {2}{3}\)
Answer:
(b) \(\frac {1}{3}\)

Question (iii).
Equivalent fraction of \(\frac {3}{5}\) is :
(a) \(\frac {13}{15}\)
(b) \(\frac {5}{3}\)
(c) \(\frac {9}{15}\)
(d) \(\frac {5}{13}\)
Answer:
(c) \(\frac {9}{15}\)

Question (iv).
Shaded area of given triangle represents the fractions:
(a) \(\frac {1}{3}\)
(b) \(\frac {3}{4}\)
(c) \(\frac {1}{4}\)
(d) \(\frac {2}{3}\)

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 8a
Answer:
(b) \(\frac {3}{4}\)

Question (v).
Sum of fractions \(\frac {2}{7}\), \(\frac {3}{4}\) is equal to :
(a) \(\frac {5}{28}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {5}{11}\)
(d) \(\frac {29}{28}\)
Answer:
(d) \(\frac {29}{28}\)

PSEB 7th Class Maths MCQ Chapter 1 Integers

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 1 Integers MCQ Questions

Multiple Choice Questions

Question 1.
The value of -| – 21| is :
(a) 21
(b) -21
(c) 1
(d) None of these.
Answer:
(b) -21

Question 2.
17 + (-18) =
(a) 35
(b) 1
(c) -1
(d) -35.
Answer:
(c) -1

Question 3.
(-15) × 0 is equal to :
(a) 0
(b) -15
(c) 15
(d) 1.
Answer:
(a) 0

Question 4.
The product of 3 × -1 is :
(a) 3
(b) – 3
(c) 1
(d) -1.
Answer:
(b) – 3

Question 5.
(-8) ÷ (-1) is equal to :
(a) 8
(b) 1
(c) -8
(d) -1.
Answer:
(a) 8

PSEB 7th Class Maths Solutions Chapter 1 Integers MCQ

Fill in the blanks

Question 1.
0 is greater than every …………….. integer
Answer:
Negative

Question 2.
25 – 10 = -10 + ……..
Answer:
25

Question 3.
15 × ……… = 0
Answer:
0

Question 4.
369 ÷ ……… = 369
Answer:
1

Question 5.
20 ÷ ……… = -2.
Answer:
-10

PSEB 7th Class Maths Solutions Chapter 1 Integers MCQ

Write True or False

Question 1.
Sum of two integeres is also integer.
Answer:
True

Question 2.
(-7) + 3 = 3 + (-7) (True/False)
Answer:
True

Question 3.
-2 + 2 = 0 (True/False)
Answer:
True

Question 4.
1 ÷ a = 1 (True/False)
Answer:
False

Question 5.
a ÷ 1 = 0. (True/False)
Answer:
True

PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4

1. Evaluate each of the following :
(i) 76 ÷ 19
(ii) (-156) ÷ (-12)
(iii) (-125) ÷ (-1)
(iv) (125) ÷ (-25)
(v) 0 ÷ (-5)
(vi) (-15) ÷ (15)
Answer:
PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4 1
PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4 2

2. Write all even integers between -18 and 0.
Answer:
All even integers between – 18 and 0 are :
-16, -14, -12, -10, -8, -6, -4, -2.

3. Write all odd integers between -9 and 9.
Answer:
All odd integers between -9 and 9 are :
-7, -5, -3, -1, 1, 3, 5, 7.

PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4

4. By what number should (-240) be divided to obtain 16.
Answer:
Let the required number be x
∴ -240 ÷ x = 16
PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4 3
Hence, the required number is -15

5. Find the value of :

Question (i).
125 ÷ [5 ÷ (-1)]
Answer:
125 ÷ [5 ÷ (-1)] = 125 ÷ (-5)
= -25 Ans.
PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4 4

Question (ii).
[169 ÷ 13] ÷ [26 ÷ 2]
Answer:
[169 ÷ 13] ÷ [26 ÷ 2]
= [13] ÷ [13] = 1 Ans.
PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4 5

Question (iii).
[(-105) ÷ 3] ÷ 7
Answer:
[(-105) ÷ 3] ÷ 7
= [-35] ÷ 7
= -5.
PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4 6

6. Simplify : 12 – [8 + 27 ÷ (2 × 8 – 7)]
Answer:
12 – [8 + 27 + (2 × 8 – 7)]
= 12 – [8 + 27 ÷ (16 – 7)]
= 12 – [8 + 27 ÷ (9)]
= 12 – [8 + 3] = 12- 11
= 1

PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4

7. Simplify : 10 – [8 – {11 + 30 ÷ (4 + 2)}]
Answer:
10 – [8 – {11 + 30 (4 + 2)}]
= 10 – [8 – {11 + 30 ÷ 6}]
= 10 – [8 – (11 + 5)]
= 10 – [8 – 16]
= 10 – [-8]
= 10 + 8 = 18

8. Multiple Choice Questions :

Question (i).
(-8) ÷ 2 =
(a) -16
(b) -4
(c) 4
(d) -8.
Answer:
(b) -4

Question (ii).
(-7) ÷ (-7) =
(a) -1
(b) 49
(c) -49
(d) None of these.
Answer:
(d) None of these.

Question (iii).
0 ÷ 2 =
(a) 1
(b) 2
(c) -2
(d) 0.
Answer:
(d) 0.

PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.4

9. The quotient of two integers is always an integer. (True/False)
Answer:
False.

10. If a and b are two unequal non-zero integers then a ÷ b = b ÷ a. (True/False)
Answer:
False.

PSEB 6th Class Computer Notes Chapter 8 Output Devices

This PSEB 6th Class Computer Notes Chapter 8 Output Devices will help you in revision during exams.

PSEB 6th Class Computer Notes Chapter 8 Output Devices

Introduction
Computer is an electronic device which gets input, processes it and give the output. Input is given to computer by input devices. The result which is provided to the user is called output. This output is given to the user using some devices. These devices are known as output devices.

Output Devices:
Output devices are those devices which are used to get output from the computer. These devices display the information into human readable form. These devices are connected to computer using some wires or wireless media. These devices can show the output in text, audio, video are graphic form. There are a lot of output devices available these days.

Uses of output devices: Output devices are used for following purposes:

  • These provide information from the computer.
  • These can be used to get hard copy as well as soft copy.
  • These give the information whether the computer has completed its work or not.

Types of Output Devices
There are many types of output devices available these days. Output devices can be categorized into following categories:

  • Monitor
  • Printer
  • Speaker
  • Headphone
  • Plotter
  • Projector

PSEB 6th Class Computer Notes Chapter 8 Output Devices 1

Monitor:
PSEB 6th Class Computer Notes Chapter 8 Output Devices 2
Monitor is also called visual display terminal. It is used to get soft copy of the output. It is most common output device. It looks like a television screen.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 3
There are different kinds of monitors available these days.

  • Cathode ray tube.
  • Flat panel display

1. Cathode ray tube monitor: These monitors were used in earlier days. These monitors use cathode ray technology to display the information. They look like bulky television sets. The size of these monitors is measured by the diagonal length on the screen. Monitors are available in 15, 17, 19 and 21 inches screens. Initially they were black and white but now colored monitors are also available.

2. Flat panel display monitors: These are the monitors which have a thin display portion. They are reduced in volume, weight and Pi requirement. These monitors can be hung up on the walls. We can see these monitors on calculators, video games, laptop computers etc. These monitors consume less power. Some examples of flat panel display monitors are LCD, LED and plasma.

PSEB 6th Class Computer Notes Chapter 8 Output Devices

Differences between CRT monitors and Flat panel monitors:

CRT Monitor Flat Panel Display Monitors
(i) CRT monitors are big in size. (i) Flat panel monitors are small in size.
(ii) CRT monitors are heavy. (ii) Flat panel monitors are light weight.
(iii) They produce large amount of heat. (iii) Flat panel produce very less heat.
(iv) They use more electricity. (iv) They use less electricity.
(v) CRT monitors are cheaper.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 4
(v) These monitors are costly.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 5

Speaker
Speaker is an output device. Speakers are used to get audio output from the computer. They are connected to the sound card of the computer. Speakers produce audio output in the form of sound waves. Any type of sound can be produced by computer using the speaker.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 6

  • These speakers are required if the user wants to listen songs or watch a movie on a computer. There are many types of speakers available in the market.
  • These ranges from small size to very big size. Normally a set of two speakers is attached to the computer. These speakers are connected to computers using two wires, one for data supply and second for power supply.

Headphone
Headphones are also an output device. These are nothing but smaller versions of speakers. Headphones are also known as earphones. These devices are portable devices. These devices can be plugged into the computer directly or to the speaker attached to the computer. These devices are used when the user alone wants to listen music.

Headphones are similar to the headphones which we use on our mobile phones. They are also available in a variety of models.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 7

Printer:
Printer is an output device which gives a hard copy of the output. The output given by the printer is permanent in nature.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 8
This output can be preserved in the form of paper. There are a lot of printers available in the market. Printer can be colored as well as black and white.

Printer can be divided into three main categories:

  1. Dot Matrix printer
  2. Inkjet printer
  3. Laser printer

1. Dot Matrix printer: As the name suggests, these printers print any character by combination of various dots. These printers are not very much costly. The speed of these printers is also slow. The printing quality of this printer is not very good. These produce a lot of sound while working. The resolution of printing is also very low. These printers fall in the category of impact printer. These printers are not used in large quantities these days.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 9

2. Inkjet printer: Inkjet printer can be called a non-impact version of Dot Matrix printer.
Similar to Dot Matrix printer inkjet printer also print the text or graphics in the form of small dots. These printers can be colored printers. The printing cost of these printers is very low. The main advantage of these printers is that they do not produce sound while working. These printers are faster than Dot Matrix printers. These printers have small dots of ink on the paper. That is why they are called inkjet printers.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 10
3. Laser printer: Laser printers are the most commonly used printers these days. They use laser technology to print on paper. This printer is black and white as well as colored. The speed of printing in this printer is very high. Their printing quality is also very high. These printers do not produce any sound while working. This printer is costly but the per paper cost is not very much.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 11

Plotter
Plotters also give hard copies of output. These are similar to the printer but plotters are used to print on big sized paper or canvas. Plotters are used in computer aided designs. These are used for some specific areas of application such as engineering design graphical design etc. Plotters use pens for drawing pictures in the media. The most common use of plotter is the big banners which we see in the market. These banners are printed with the help of plotters. Plotter can print black and white as well as color. Plotters are very costly devices. They cannot be awarded by single person.

These are of following types:

  • Drum Plotter
  • Flatbed Plotter
  • Inkjet Plotter.

1. Drum Plotter: In drum plotter, a drum is used to rotate in vertical motion. It contains one or more Horizontal pen holders. These pens are used to draw lines on the paper. Drum and pen draw the design by working together simultaneously. Each pen is program selectable. Pens use different colours to draw.

2. Flatbed Plotter: It draws on flat paper. This paper is spread on flat bet table. Paper is not rotated. The pen moves as per the drawing. It contains many pen holdings of different colours. The plot area is as per bed area. It can range from A4 to 50 feet or more.

3. Inkjet Plotter: These plotters uses inkjet technology instead of pens. They can draw in colour. These plotters are usually faster than other types of plotters.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 12

PSEB 6th Class Computer Notes Chapter 8 Output Devices

Projector
Projector is an output device which is used to get visual output. They can be connected to computers.
The projector gives the output big in size. The projectors are mainly used to display something to a large number of people. They are used in office meetings or in classroom teaching by the teachers.
PSEB 6th Class Computer Notes Chapter 8 Output Devices 13
Other Output Devices

Example:

  • Digital camera
  • Pen Drive
  • CD/DVD
  • Modem
  • Fax

Difference Between Input and Output Devices
Following are the differences between input and output devices:

Input devices Output devices
(i) Input devices are used to give input to computers. (i) These devices are used to get output from the computer.
(ii) Data and instructions are given as input using these devices. (ii) After processing information is taken using these devices.
(iii) Input devices are available in large categories. (iii) The categories of availability of in output devices is less.

PSEB 9th Class Science Solutions Chapter 8 Motion

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 8 Motion Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 8 Motion

PSEB 9th Class Science Guide Motion Textbook Questions and Answers

Question 1.
An athelete completes one round of circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 min 20 s?
Solution:
PSEB 9th Class Science Solutions Chapter 8 Motion 1
Diameter of circular track (d) = 200m
Radius of circular track (r) = \(\frac{d}{2}\) = \(\frac{200m}{2}\) = 100m
Length of circular track (circumference) = 2πr = 2 × \(\frac{22}{7}\) × 100
= \(\frac{4400}{7}\)m
Time taken to complete 1 round (t) = 40 s
Total time = 2 minutes 20 seconds
= (2 × 60 + 20) seconds
= (120 + 20) seconds
= 140 s.
Distance covered in 40 s = \(\frac{4400}{7}\) m = (Circumference of 1 complete circular track)
Distance covered in 1 s = \(\frac{4400}{7×40}\) m
Distance covered in 140 s = \(\frac{4400}{7×40}\) × 140 = 2200 m
An athelete starting from A and going in clockwise direction returns to point A in 3 rounds and reaches point B in 3.5 rounds.
∴ Displacement in 3.5 rounds = AB = shortest distance between initial and final position = 200 m from A to B.

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 2.
Joseph jogs from end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B (b) from A to C?
Solution:
(a) Length between end point A and end point B (AB) = 300 m
Time taken (t) = 2 min. 30 s
= (2 × 60 + 30) s
= (120 + 30) s
= 150 s.
PSEB 9th Class Science Solutions Chapter 8 Motion 2
Average speed = Average velocity
= \(\frac{Total distance between A and B(AB)}{Total time(t)}\)
= \(\frac{300m}{150s}\)
= 2ms-1

(b) Length from end A to end B + Length on return from B to point C.
= AB + BC
= 300 m + 100 m
= 400 m
Total Time = 2 min 30 s + 1 min
= 3 min 30 s
= (3 × 60 + 30) s
= (180 + 30) s
= 210 s
PSEB 9th Class Science Solutions Chapter 8 Motion 3

Question 3.
Abdul while driving to school, computes the average speed for his trip to be 20 km h-1. On his trip along the same route, there is less traffic and average speed is 40 km h-1. What is the average speed for Abdul’s trip?
Solution:
PSEB 9th Class Science Solutions Chapter 8 Motion 4
PSEB 9th Class Science Solutions Chapter 8 Motion 5

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time?
Solution:
Here, initial velocity of motorboat (u) = 0 [Starting from rest]
Acceleration (a) = 3.0 m s-2
Time (t) = 8.0 s
Distance covered by the motorboat (S) =?
We know, S = ut + \(\frac{1}{2}\)at2
= 0 × 8 + \(\frac{1}{2}\) × 3 × (8)2
= 0 + \(\frac{1}{2}\) × 3 × 8 × 8
∴ S = 96 m.
In other words, the motorboat covers a distance (S) = 96 m.

Question 5.
A driver of a car travelling at 52 kmh-1 applies the brakes and accelerates uniformly in opposite direction. The car stops in 5 s. Another driver going at 3 km h-1 applies his brakes slowly and stops in 10 s. On the same graph paper plot the speed versus time graph for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
In the figure AB and CD represent velocity-time graphs of two cars which have their speeds 52 kmh-1 and 30 kmh-1 respectively.
PSEB 9th Class Science Solutions Chapter 8 Motion 6
PSEB 9th Class Science Solutions Chapter 8 Motion 7
In this way, after applying brakes the second car would cover more distance than the first car.

Question 6.
Fig shows the distance-time graphs of three objects A, B and C. Study the graph and answer the following questions:
PSEB 9th Class Science Solutions Chapter 8 Motion 8
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Solution:
(a) Velocity of A = Slope of PN
\(\frac{10-6}{1.1-0}\)
\(\frac{40}{11}\) = 3.63 kmh-1
PSEB 9th Class Science Solutions Chapter 8 Motion 9
PSEB 9th Class Science Solutions Chapter 8 Motion 10
Because slope of object B is maximum of all therefore, it is moving fastest.
(b) Since all the three graphs do not intersect at any point therefore, all the three do not meet ever at the same point on the road.
(c) When the object B passes A at point E (at 1.4 hr) then at that time the object C will be at F i.e. 9.3 km away from the origin O.
(d) B passes C at G after covering 8 km.

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity it will strike the ground?After what time will it strike the ground?
Solution:
u = 0 ms-1
S = 20 m
a = 10 ms-2
υ = ?
t = ?

Using υ2 – u2 = 2as
υ2 – (0)2 = 2 × 10 × 20
υ2 = 4000
∴ υ = \(\sqrt{400}\)
= \(\sqrt{20 \times 20}\)
= 20 m s-1
Now υ = u + at
20 = 0 + 10 × t
or t = \(\frac{20}{10}\)
∴ t = 2 s

Question 8.
Speed-time graph for a car is shown in the fig.
PSEB 9th Class Science Solutions Chapter 8 Motion 11
(a) Find how far the car travelled in first 4 s. Shade the area on the graph that represents the distance travelled by car during this period.
(b) Which part of the graph represents uniform motion of the car?
Solution:
PSEB 9th Class Science Solutions Chapter 8 Motion 12
(a) 5 small squares of x axis = 2s
3 small squares of y axis = 2 ms-1
Area of 15 small squares = 2s × 2 ms-1 = 4m
∴ Area of 1 small square = \(\frac{4}{15}\)
Area of velocity-time graph under 0 to 5s = 57 complete small squares + \(\frac{1}{2}\) × 6 small squares.
= (57 + 3) small squares
= 60 small squares.
Distance covered by car in 4 s = 60 × \(\frac{4}{15}\) m
= 16 m

(b) After 6 s the car has uniform motion.

Question 9.
State which of the following situations are possible and give an example for each of these.
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
(a) Yes, this situation is possible.
Example: When an object is projected upwards, its velocity at the maximum height is zero although acceleration on it is 9.8 ms-2 i.e. equal to g.
PSEB 9th Class Science Solutions Chapter 8 Motion 13

(b) Yes, at the maximum height of projection the velocity is in the horizontal direction and its acceleration is perpendicular to the direction of motion as shown in figure.

Question 10.
An artificial satellite is moving in a circular path orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
Radius of circular path of artificial satellite (r) = 42,250 km
Angle formed (subtended) at the centre of earth (θ) = 2π radian
Time taken by the satellite to complete 1 revolution (t) = 24hrs
= 24 × 3600s
= 86400 s
PSEB 9th Class Science Solutions Chapter 8 Motion 14

Science Guide for Class 9 PSEB Motion InText Questions and Answers

Question 1.
An object has moved through a distance. Can it have zero displacements? If yes,support your answer with an example.
Answer:
Yes, a body can have zero displacement, if fhis body While moving occupies its final position coinciding with its initial position.
Example: Suppose a body starting its motion from initial position O covers some distance and reaches a position A. If this body while moving returns to its initial position O then in that situation its displacement will be zero.
PSEB 9th Class Science Solutions Chapter 8 Motion 15
But distance covered by the body = OA + AO
= 60 km + 60 km
= 120 km

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Solution:
Total distance round the boundary of field once (i.e. circumference)
= AB + BC + CD + DA
= 10 m + 10 m + 10 m + 10 m = 40 m
Time taken to go round the field once = 40 s
PSEB 9th Class Science Solutions Chapter 8 Motion 16
Total time taken = 2minutes 20 seconds
= (2 × 60 + 20) seconds
= (120 + 20) seconds
= 140 seconds.
Time taken by fanner to complete 3 rounds of field = 3 × 40 s = 120 s
Time left after completing 3 rounds of field = (140 – 120)s = 20 s
∴ Distance covered by farmer in 40 s = 40 m
∴ Distance covered in 1 s = 1 m
Distance that would be covered in 20 s = 20 m
In other words farmer starting from point A and while going along the boundary of the field and after completing 3 rounds in 2 min 20 s would reach the point C.
∴ Displacement = AC
(the shortest distance between initial and final position)
PSEB 9th Class Science Solutions Chapter 8 Motion 17

Question 3.
Which of the following is true for displacement?
(a) It cannot be zero
(b) Its magnitude is greater than the distance travelled by the object.
(c) Its magnitude is less than or equal to distance travelled by the object.
Answer:
(c) Its magnitude is less than or equal to distance travelled by the object.

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 4.
Distinguish between speed and velocity.
Answer:
Distinction between Speed and Velocity:

Speed Velocity
1. It is defined as the rate of a change of a position of a body i.e. the distance covered by a body per unit time. It is defined as the rate of change of displacement of a body. i.e. it is the speed in a particular direction.
2. It is a scalar quantity and can be completely represented by its magnitude only. It is a vector quantity. To represent it completely it requires both magnitude and direction.
3. Speed of an object is always positive. Velocity of an object can be both positive and negative.

Question 5.
Under what condition(s) is the magnitude of average velocity of an object is equal to its average speed?
Answer:
We know, Average speed = Total distance travelled / Total time taken
and Average velocity = Displacement /Total time
When a body travels in a straight line with variable motion in the same direction then total distance covered and displacement are equal in magnitude. In this case the average speed and average velocity are equal.

Question 6.
What does the odometer of an automobile measure?
Answer:
The odometer of an automobile measures the distance covered by it.

Question 7.
What does the path of an object look like when it is in uniform motion?
Answer:
When an object is in uniform motion, it moves along a straight line. But an object can also move with uniform motion along a circular path.

Question 8.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station?The signal travels at a speed of light that is 3 × 10s ms-1.
Solution:
Time taken by the signal to reach the ground station from spaceship (t) = 5 min = 5 × 60 s = 300 s
Speed of Signal (υ) = Speed of light = 3 × 108 ms-1
Distance of the spaceship from earth (s) = ?
Distance of spaceship from ground (s) = speed of signal (υ) × Time (t)
= 3 × 108 × 300
= 3 × 108 × 3 × 102
= 9 × 108 × 102
= 9 × 1010 m

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 9.
When will you say a body is in:
1. uniform acceleration?
2. non-uniform acceleration?
Answer:
1. Uniform Acceleration. When a body travels in a straight line and its velocity changes by equal amounts in equal intervals of time then it is said to travel with uniform acceleration.
2. Non-Uniform Acceleration. When the velocity of a body changes by unequal amounts in equal intervals of time then the body is said to travel with non-uniform acceleration.

Question 10.
A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s. Find the acceleration of the bus.
Solution:
PSEB 9th Class Science Solutions Chapter 8 Motion 18
PSEB 9th Class Science Solutions Chapter 8 Motion 19
Hence, the bus has negative acceleration (retardation).

Question 11.
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-1 in 10 minutes. Find its acceleration.
Solution:
PSEB 9th Class Science Solutions Chapter 8 Motion 20

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 12.
What is the nature of the distance-time graphs (x – t) for uniform and non-uniform motion of an object?
Answer:
When a body covers equal distances in equal intervals of time, then it is said to travel with uniform motion. In this situation, the distance covered by the body is directly proportional to the time taken. Therefore, distance-time (x – t) graph for uniform motion is a straight line.
PSEB 9th Class Science Solutions Chapter 8 Motion 21
Distance – time (x – t) graph for non-uniform motion may be a curved graph of any shape because a body travels unequal distances in equal intervals of time.
PSEB 9th Class Science Solutions Chapter 8 Motion 22

Question 13.
What can you say about the motion of object whose distance – time graph is a straight line parallel to time axis?
Answer:
PSEB 9th Class Science Solutions Chapter 8 Motion 23
The object whose distance-time (x – t) graph is a straight line parallel to the time axis will be at rest with respect to the surroundings.

Question 14.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to time axis?
Answer:
PSEB 9th Class Science Solutions Chapter 8 Motion 24
The object whose speed – time (u – t) graph is a straight line parallel to time axis shows that it is in motion with uniform speed.

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 15.
What is the quantity which is measured by the area occupied below velocity-time graph?
Answer:
PSEB 9th Class Science Solutions Chapter 8 Motion 25
The area occupied below velocity-time graph measures displacement of the body.

Question 16.
A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for two minutes. Find (a) the speed acquired (b) the distance travelled.
Solution:
(a) Initial speed of the bus (u) = 0 (Starting from Rest)
Acceleration of the bus (a) = 0.1 m s-2
Time taken (t) = 2 minutes
= 2 × 60 s
= 120 s
Final speed of the bus (υ) = ?
Distance travelled by the bus (S) =?
We know, υ = u + at
υ = 0 + 0.1 × 120
υ = 1 × 12
υ = 12 ms-1

(b) Again, using S = ut + \(\frac{1}{2}\) at2
S = 0 × 120 + \(\frac{1}{2}\) × 0.1 × (120)2
= 0 + \(\frac{1}{2}\) × 0.1 × 120 × 120
= \(\frac{1}{2}\) × 1 × 12 × 120
= 720 m/s

Question 17.
A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will move before it is brought to rest?
Solution:
Initial speed of train (υ) = 90km h-1
= 90 × \(\frac{5}{18}\) m s-1
= 5 × 5 ms-1
= 25 ms-1
Uniform acceleration (a) = – 0.5m s-2
Final speed of the train (υ) = 0
Distance moved by the train (S) =?
We know, υ2 – u2 = 2as
(0)2 – (25)2 = 2 × (-0.5) × S
– 25 × 25 = -1 × S
∴ S = 625 m

PSEB 9th Class Science Solutions Chapter 8 Motion

Question 18.
A trolley, while going down an inclined plane has an acceleration of 2 cm s~2. What will be its velocity 3 s after the start?
Solution:
Here initial velocity of trolley (u) = 0 [∵ starting from rest]
Acceleration (a) = 2cm s-2
Time (t) = 3 s
Final velocity of trolley (υ ) = ?
We know, υ = u + at
υ = 0 + 2 × 3
∴ Final velocity of trolley (υ) = 6 cm s-1 Ans.

Question 19.
A racing car has uniform acceleration of 4 ms-2. What distance will it cover in 10 s after start?
Solution:
Acceleration of racing car (a) = 4 ms-2
Initial velocity of racing car (u) = 0
Time (t) = 10 s
Distance covered by the car (S) = ?
We know, S = ut + \(\frac{1}{2}\)at2
S = 0 × 10 + \(\frac{1}{2}\) × 4 × (10)2
S = 0 + 2 × 10 × 10
∴ Distance covered by racing car (S) = 200 m

Question 20.
A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s-2 in the downw ard direction. What will be the height attained by the stone and how much time will it take to reach there?
Solution:
Here, initial velocity (u) = 5 m s-1
Acceleration (a) = – 10 ms-2
[∵ it moves upward against the gravity]
Final velocity of stone (υ) = 0 [At the highest point it is brought to rest]
Height attained (S = h) = ?
Time taken (t) =?
We know,
υ = u + at
0 = 5 + (-10) × t
0 = 5 – 10 × t
10 × t = 5
or t = \(\frac{5}{10}\)
∴ Time taken (t) = 0.5 s
Again, using υ2 – u2 = 2as
(0)2 – (5)2 = 2 × -10 × h
– 5 × 5 = – 20 × h
or h = \(\frac{-25}{-20}\) = \(\frac{5}{4}\)
∴ Height attained (h) – 1.25 m

PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint

This PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint will help you in revision during exams.

PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint

Introduction:
MS Paint or Microsoft paint is an application software. This software is used to draw objects and shapes. The user can work with colours in this software. The drawings can be saved and printed. The drawing can also be used in other application software such as Microsoft Word, Microsoft PowerPoint.

What is MS Paint?
It is an application software developed by Microsoft. This software is provided by a company with Microsoft Windows operating system. It is the default software which is used to develop non commercial paintings. There are many different tools available in Paint. This software is very helpful for new users and children. The user can draw paintings in colour or black and white. This painting can be saved as bitmap files or other format. These paintings can also be printed on paper using a colour printer. These paintings can be set as wallpaper on the computer. These paintings can also be pasted in other applications like MS Word and MS PowerPoint. Save its painting in various formats such as JPG, GIF, BMP etc.

How to Start MS Paint?
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 1
Or

  • Click on the start button on taskbar/ super bar. The Start menu will appear.
  • Click on All Programs, another menu will appear.
  • Click on the Accessories option in this menu. Another menu will appear. This menu has a Paint option.
  • Click on the Paint option.
    Or
  • Click on the start button and type “Paint” in the search bar. Click the Icon from the list and press enter key. Paint window will appear.

PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 2

Parts of a Paint Window
Paint window is shown in above figure. It has following main parts:
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 3
1. Title Bar: The title bar is present at the top of the paint window. At the left end of the title bar, the first item shown is a little paint palette. If we click this button, a standard window menu opens having options Restore, Move, Size, Minimize, Maximize and Close. Another thing we will see the title of our picture followed by the name of the program-Paint. If we haven’t saved our picture, the name will be shown as “Untitled”.

  • Quick Access Toolbar: The next four items make up the Quick Access Bar offering buttons for Save, Undo, Redo and Customize.
  • Minimize, Maximize/Restore, Close: Title bar has three buttons on its right corner. They are:
    (a) Minimize Button: Used for minimizing the paint window onto the taskbar.
    (b) Maximize/Restore button: Used for maximizing or restoring the paint window.
    (c) Close Button: Used for closing the paint window.

2. Quick Access Toolbar: It is a toolbar present in the title bar by default. This bar provides us with frequently used commands. Its position can be changed both to below or above the ribbon and icons can be added and removed as per the user’s requirement.

PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint

To Move Quick Access Toolbar below the Ribbon:
If we prefer to show Save, Undo and Redo buttons below the ribbon, Click on the “customize quick access bar” button and a menu will appear. Near the bottom of the menu that appears, we will see Show below the Ribbon. Click Show below the Ribbon.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 4
The Quick Access Toolbar will move below the Ribbon. We can add more options such as New, Open, and Print Preview etc. to the Quick Access Toolbar with the help of Customize icon.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 5
Move Quick Access Toolbar below the Ribbon

Here are commands and their functions discussed below:

Name of Command Function Shortcut Key
New Creates a new/blank image file. Ctrl + N
Open Opens a dialog box to open an existing image file. Ctrl + 0
Save Saves changes to the current file. Ctrl + S
Print Print the current picture. Ctrl + P
Print Preview Displays the image on screen as it will appear after printing on paper.
Send in e-mail Send a copy of the picture in an e-mail as an attachment.
Undo Repeat or Reverse the last action. Ctrl + Z
Redo Restores previous undo action. Ctrl + Y
Show below/ above the ribbon Shows Quick Access Toolbar below or above the ribbon
Minimize the ribbon Toggle the ribbon On/Off.

Adding Ribbon items to the Quick Access Toolbar: Many other items from the ribbon can also be added to the Quick Access Toolbar. On the Ribbon, right click on anything we like to add. A menu will appear which includes the option “Add to Quick Access Toolbar”. Click on this option.

Menu Bar
The Menu bar has three tabs named as Paint Button, Home tab ribbon and View tab ribbon. On the right side of the menu bar, the Help button appears as shown in the figure below.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 6

1. Paint Button: This Button appeared at the beginning of Menu bar. When we click on this button and the following Menu Appears.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 7
The various commands given in Paint button are explained below:

Name of Command Functions
Save As Save changes to the new file with a different file name. It asks for a new name every time. We can change format of the new file too. Such as PNG, JPEG, BMP, GIF etc.
From scanner and camera Import picture from scanner or camera.
Set as desktop background Set the current picture as our desktop background.
Properties Change the properties of the picture. The Properties dialog will give us information about the picture .
Exit To close the paint window.

2. Home Tab Ribbon: All tools, shapes, colour palette and most of the commands are grouped together in the ribbon except Save, Undo and Redo commands which are shown at title bar or in the Quick Access Toolbar. Drop-down arrows below each item in the ribbon will give us other options for the tool. Most of the tools used for drawing or other tasks are present in Home Tab Ribbon.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 8

There is also an option to minimize the ribbon. If we choose this, the ribbon d sappears entirely, but pops into view if you click on the Home tab.

3. View Tab Ribbon: We can use the View tab by clicking on it. The options such as zoom in, zoom out, show or hide and display are there in the View tab. Zoom in or out can be used alone or in conjunction with the Zoom Tool. We can also use the status bar for Zoom in or Zoom out purposes.

PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint

Scroll Bar
Scroll bars are used to move the screen. These are of two types:

  • Horizontal Scroll bar: It is present at the bottom of the Paint window. It moves the screen left and right.
  • Vertical Scroll bar: It is present at the right side of the Paint window. It moves the screen up and down.

Status Bar
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 9
1. Cursor Position: It gives the Cursor Position, which is helpful when we want to position any picture precisely.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 10

2. Selection Size: It shows the size of a selection we are making or size of an object we are drawing.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 11

3. Image Size: It shows the size of our entire picture, even if the picture is very large and is not visible completely in the window. If we have not changed the units in the Properties dialog box, the measurement will be displayed in pixels. We can change measurement to inches or centimetres.

4. Disk Size: Once we have saved our picture, this option will show the size or drawing on Disk. If the paint window is very small, this figure might not be shown.

5. Zoom Slider: The Zoom Slider is convenient if we are working in a zoomed in view and want to zoom out. However, we cannot zoom in on a particular spot, as we can do with the Magnifier.

Work Area
Free space of the Paint window is called the work area. It is used for making drawings. This area is usually between Ribbon and status bar.

Saving Our Drawing
It is good to save our picture as soon as we begin to work. We must click on the Save button on the Quick Access Toolbar every few minutes. This prevents loss of work if the program closes unexpectedly, as in a power failure.
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 12

When we click the Save for the first time, we will find a dialog box where we have to type a name for the picture. Type a desired name in the file name text box and click the Save button.

Save as:
PSEB 6th Class Computer Notes Chapter 4 Introduction to MS Paint 13
Click Save as in file menu.
With the help of Save as option we can save a Copy of a picture with another file name. Go to the Paint button and open the menu.
In the dialog box, just change the existing name then click the Save button.

PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers

This PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers will help you in revision during exams.

PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers

Introduction
A computer is a combination of hardware and software. We can say that a computer is a machine in itself. There are always some rules which should be followed to work with a machine. Similarly if the user has to work with a computer he must have basic knowledge of working, with computers.

Operating System
Operating system is a system software without which a computer cannot work. IT acts as an interface between the hardware and the user. Each computer has an operating system installed on it. Operating system is displayed in the form of screens, menus, dialogue box, icons and widgets etc.

Definition of Operating System

  • An operating system is an interface between a computer user and computer hardware. It makes computer hardware work by controlling all the internal processes of the computer.
  • There are many types of operating systems. Each operating system is developed for different types of machines. Each operating system has a different set of commands and it understands the machine architecture very well.

Windows
Windows is an operating system developed by Microsoft. It is a graphical user interface. It means it is easy to work with pointing devices such as mice in windows. Most of the computers in this world are using windows operating systems. Everything is displayed in a rectangular frame on the screen in this operating system. That is why operating system is named as windows. It is the most popular operating system in the world.

DOS
DOS stands for Disc Operating System. This operating system is also developed by Microsoft. It is a character user interface operating system. This operating system is not very powerful. The user types commands to instruct the computer. This operating system was used on computers with small memory and low speed hardware. The user has to remember a lot of commands to work in it. That is why it was difficult to work in this operating system. Also the user cannot do a variety of work simultaneously in this operating system.

Android
The Android Operating System is one of the latest operating systems in digital word. It is the operating system which is used in mobile phones. This operating system was developed by Google. This operating system is developed for devices which have touchscreen in them. It provides a very beautiful and convenient interface which can be used by finger touch. The user can do the tasks like pinching, swiping and typing. Now the Android operating system is also used in televisions, cars , wrist watches and many other digital devices. Each of these devices have different user interfaces.

PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers

Starting Computer System
The first task to work with an operating system is to make it work. This process is called starting the computer system. To start a computer system is not a difficult task. It is as simple as to start any other machine.

Powering on the Computer System
To start a computer it must be switched on first. Each computer system has a power button on its system case. The user has to press that power button to switch on the computer system. Similarly there is a power button on the monitor also. If the monitor is not powered on then the user has to switch on the monitor.
PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers 1
After switching on the computer, the computer starts to place the operating system in its memory. This process is known as booting. This process takes some time depending on the memory and the hardware attached to the computer system. After the booting process screen is displayed on the monitor.

Logging-in
After the booting process the initial windows screen is displayed. This screen provides the opportunity for the user to connect to the main operating system. This process is known as login.

On the screen all the users are displayed which are created on this computer. To login to the screen the user must know who is username and password.

  • To Login to the computer , click on the username.
  • Windows will ask to enter the password for you.
  • Type in the password of that username.
  • Windows will let you in.
  • It will display the next screen which is called desktop.

Desktop and its Components
When the user logged into the Windows operating system, he is taken to a screen. This screen is known as the desktop of the computer. It is basically a dashboard of a computer system. All the operations are performed from this screen on. It has various parts. The parts of a desktop are discussed below.
PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers 2
Desktop

Desktop has following three parts:

  • Icons
  • Taskbar
  • Shortcut

PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers 3

Icons
Icons are the small pictures on the desktop. These icons represent a file or a program or folder on the computer. The user has to click on this icon to open.dat file or program. The following types of icons are available on Windows desktop.
PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers 4

1. My Computer: This icon is used to assess the hard disc of the computer. This icon opens a window on the computer screen which displays all the hard disks attached to the computer and other removable media such as pen drives or CDs. The user can work in this window.

2. Network: This icon is used to connect to the network location attached to this computer system. This icon is used when the computer is connected to a local area. This icon displays icons of all the computer systems which are connected to users’ computers through Local area networks. The user can go to the other computer and work on files.

3. Recycle Bin: Recycle Bin is a folder which works like a dustbin. In this folder contains all the deleted files, folder, icons on the computer. When anything is not required on the computer system the user can delete that item. That item is shifted to the recycle bin folder. If the user deletes an item accidentally he can restore that item from the recycle bin.

4. Users File: This icon is used to assess the current user data. Thus it can hold all the default locations for the different types of files. It may include my documents, my pictures, my videos desktop etc. The name of this icon changes as per the user logged in.

PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers

Shortcuts
A shortcut is to reach the original file. A shortcut represents a link to the program file folder or any other item on the computer. It is an easy way to assess that item. The program can be opened using that shortcut. If the shortcut is deleted then the original program is not deleted or removed from the computer system. The user can create many shortcuts for a single program or file. The shortcut icon is identified by a small Arrow on a normal icon.
PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers 5
Shortcut

Wallpaper
Wallpaper is an image which is displayed behind the desktop screen. It is also called a desktop background. The user can place a screen as per his requirement. This screen can be a picture colour or pattern. There are some preloaded wallpapers available in the Windows operating system. The user can change the wallpaper as per his requirement.

The following steps are used to change a wallpaper.

  • Right-click on wallpaper displaying on Desktop.
  • Choose the “Personalize” option from the popup menu.
  • A new window will appear.
  • Select the required theme from the window. We can alternatively click on the “Desktop wallpaper” option from the bottom and choose the wallpaper.

Taskbar
Taskbar is located at the bottom of the screen. This bar is a part of the Operating System. It allows us to start a program using the Start menu. This bar always remains visible during working in any application. We can navigate through Active programs using the taskbar. The area on the right side of the taskbar is called “Notification Area”. This area allows us to check date and time, items running in the background etc. The taskbar was first introduced with Microsoft Windows 95 and is found in all subsequent versions of Windows. We can have a look of taskbar and it’s all parts as under:
PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers 6
Taskbar

All these parts of the taskbar are having their own functions. Let’s discuss the use of each one of them. It has a logo of windows on it. It is the first item on Taskbar.

Its icon is:

  • Start Button: We can start any application and program with the help of the Start button.
  • Quick Launch Bar: This section of the taskbar enables us to launch
    programs without locating them from the Start menu. It is located next to the Start button.
  • System Tray: It is located at the right side of Taskbar. It contains miniature icons for easy access to system functions such as fax, printer, modem, volume etc.
  • Notification Area: This area is a part of the taskbar that provides notifications and status of devices. It can also be used to display icons for system and program features that are not on the desktop.
  • Clock: At the end of taskbar, Clock is displayed where Current Time and Date can be seen. We can change Time and Date by clicking on it. It requires Administration access to change the Time or Date.
  • Active Programs: This area of taskbar is between quick access bar and System Tray area. In this area of taskbar, all the active programs appeared as an icon and we can easily navigate among them.

Working with Windows Applications
Microsoft Windows 7 provides some of the inbuilt applications which are used for different purposes. These applications are the most basic applications and each one is of a different type. Let’s discuss some of these applications.

Notepad
This is a basic text editor of Microsoft Windows. This is mainly used to type character data. It is the most simple application for creating files with text data only. This application is mainly used for writing codes (Source Files) of programming languages like HTML, Java etc.

Running Notepad

  • Start → All Programs → Accessories → Notepad
    Or
  • Click on the start button and type “Notepad ” in the search bar. Click the Icon from the list and press enter key.

Wordpad
This is a Rich Text Format word processor. We can create a document with formatting using this application. This application can be used to write letters or to create documents with texts data only. Graphics like charts, pictures, clip-art etc. cannot be inserted in it.

Running WordPad

  • Start → All Programs → Accessories → WordPad
    Or
  • Click on the start button and type “WordPad” in the search bar. Click the Icon from the list and press enter key.

Paint
This is inbuilt graphical applications of Microsoft Windows 7. We can create drawing or other basic graphical objects using this application. It is also known as MS Paint. MS Paint has several tools for drawing and colouring. We can edit an existing picture also using this application.

Running Paint

  • Start → All Programs → Accessories → Paint
    Or
  • Click on the start button and type “Paint” in the search bar. Click the Icon from the list and press enter key.

Calculator
This application of Microsoft Windows 7 is used to do calculation related work. We can use this application as we use a real calculator. Its layout and buttons are also similar to real calculators. We can perform calculations up to 16 digits using this application.

Running Calculator

  • Start → All Programs → Accessories → Calculator
    Or
  • Click on the start button and type “Calculator” in the search bar. Click the Icon from the list and press enter key.

Using Run Box
Run box allows the user to open a program by name. This option of windows is very useful for fast opening of programs. Instead of navigating through menus, we can directly type the name of the program to run. Each program exists in the computer’ as an executable file. For example, MS Paint is named as “pbrush”, Notepad as “notepad”, WordPad as “Wordpad” and Calculator as “Calc”. We must write the correct name of that program which is to start. We can also start any file by typing it’s full path. The picture below shows how the Run box looks like in Microsoft Windows 7.

To open the Run Box in microsoft windows 7, we can use the following steps:
Press Window
PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers 7
button + R together.
Or
Click on the start button and search for “Run”. Click the Icon from the list and press enter key.

PSEB 6th Class Computer Notes Chapter 3 Basics of Working with Computers

Search Box of Start Menu
We can use the Search Box of Microsoft Windows 7 to search files or programs of computers. It can find given file from hard drive. It is located at the bottom of Start Menu of Windows 7. If this bar is missing from the start menu then we can re-enable it from Control Panel. The Search Box of start menu looks as given in the picture bellow.
If this part of start menu is missing then we can use following steps to show Search Box.

Steps to show “Search Box”

  • Open the Start menu and click “Control Panel.”
  • Click “Uninstall A Program” under Programs.
  • Click “Turn Windows features on or off.”
  • Click on the checkbox named ‘Window Search”.
  • Click “OK” to save your changes.
  • Search Box will start showing in Start Menu after restarting your computer.

Shutting Down Computer System
When the user completes its work on the computer system it is necessary to shut it down. The process of shutting down a computer system is as simple as starting it. All the tasks are performed by the computer system itself to shut it down. The user has only to start the procedure. Shutting down a computer is necessary to avoid the data loss. When the computer is in workable condition many files have been opened by it. The process of shutting down closes all those files, saving their data and returning them back to their inactive state.

Windows have given commands to shutdown the computer system. These commands are selected from the start menu of the window.

Following steps are used to shutdown a window based computer.

  • Click on the start button.
  • Click on the arrow available with the shutdown button.
  • It will display many options such as sleep shutdown log of restart lock switch user.
  • Window will start the process as per the options selected.

The detail of these options are given below:

Sleep:
This option of Power Off menu can be used when we want to leave a computer for some time. During this mode, the power of Monitor/LCD gets OFF and all your data is kept safe. When we put a computer into sleep mode then its power remains ON and its power light starts blinking. This blinking Red-Colour LED on the CPU shows that the computer is in sleep Mode. We can press the power button to resume the computer from this mode.

Shut Down:
This option of Power menu can be used when we have finished all our work. When we shut-down a computer, all parts of computer systems are turned off and no power remains active in the Computer System. We can switch off the main power-supply of the computer when it is shut down. This process can take some time and we must wait till it is over. This process may take time according to the size of data being used and number of programs currently running.

Log Off:
Windows 7 allows us to access a computer differently among users. We can keep our data secure from other users with the help of user accounts in it. If we have finished our work and want to leave a computer but another user is there to access the same computer for own work in a different user account then we can use the Log-off option of Power Menu.

Restart:
This option of power menu can be used when any new program is installed or any updation in the system is done. Sometimes when a new device or hardware is attached to the computer, it is required to restart our computer. In such a case, we can use the restart option of the Power menu to shut down our computer and start it again. When the restart button is pressed then the computer automatically gets started after being shut-down.

PSEB 6th Class Computer Notes Chapter 2 Computer Components

This PSEB 6th Class Computer Notes Chapter 2 Computer Components will help you in revision during exams.

PSEB 6th Class Computer Notes Chapter 2 Computer Components

Introduction to Basic Components of Computer System:
Computer is a device which gets input, processes it and gives output to the user. Input processing and output are called basic tasks of a computer system. For doing these tasks computers need various components. For this computer needs Standard Input Output devices and a Processing Unit.

Standard Input devices include keyboard and mouse. Both of these devices are necessary for a computer. If the keyboard is not attached to the computer, the computer will show an error message at the time of booting. The keyboard is used to enter text messages and the mouse is a pointing device. The most common Output device used in computers is a monitor. It is required to see the result of our work.

The combination of these systems is called the Standard Input Output unit.
PSEB 6th Class Computer Notes Chapter 2 Computer Components 1

How Computer Works?
The process of processing the data is called the processing cycle of a computer. It is the way of processing within a computer system.

Three steps are involved in this cycle.
1. Input: This part is related with input of data and instruction. Computers cannot work without proper in. To enter the necessary data and instructions in the computer input unit is required. Man can enter the data and instruction in its own language and that is converted into computer languages by this input unit.

There are many types of input devices used for input. Keyboard is the most common input device, used to enter textual data. The second common device for input is the mouse. This device is used to point a special part on the screen or to select a command your mouse pointer. The other common devices which are used for input are microphone touch screen scanner etc.

2. Processing: Processing is the process of acting according to given instructions. This is the main task done by computer as it gets instructions from the computer. The main aim of Computer is to work as per the given instructions. This task is done by the processor of the computer. The processor is just like the brain of a human being. This processor is also known as the Central Processing Unit.

3. Output: After processing the data is converted into some information. The user of the computer needs that information from the computer as its result. This result is shown to the user using output devices. The most common output device is a monitor. The other output devices are speaker, printer, plotters.
PSEB 6th Class Computer Notes Chapter 2 Computer Components 2

PSEB 6th Class Computer Notes Chapter 2 Computer Components

Block Diagram of Computer:
A block diagram of a computer is a Graphical representation of various steps involved in working of a computer. The main parts shown in this block diagram are input unit, processing unit and output unit. The processing unit is divided into three parts namely memory unit, control unit and arithmetic logic unit.
PSEB 6th Class Computer Notes Chapter 2 Computer Components 3

Parts of Central Processing Unit
The Central Processing Unit of a computer system is divided into three parts.

  1. Memory Unit
  2. Control Unit
  3. Arithmetic and Logical Unit

PSEB 6th Class Computer Notes Chapter 2 Computer Components 4
1. Memory Unit: This unit is also known as storage unit. This unit holds the data and instruction in it. It can store data and instruction temporary and permanently. Many types of memories are used in this unit. Type of memory is used for special purposes. These memories vary in size, storage capacity nature of operation, speed and cost.

2. Control Unit: Control unit is meant for controlling all the activities of the computer. This unit works as the brain of a computer. It performs all the operations given in the form of instructions. It also controls the functioning of other components of the computer. It controls both input and output units also.

3. Arithmetic and Logical Unit: This is the third part of the Central Processing Unit. This unit is related to performing all the arithmetic and logical operations. A lot of calculations are done by computer in each type of work. All these calculations are done by this unit of CPU. Part of Central Processing Unit is integrated within the processor.

Types of Computer Memories
Computers can store data in many ways. This is possible due to the use of different types of memory in computers.

Computer memory can be classified into two types.

  1. Primary Memory
  2. Secondary Memory

PSEB 6th Class Computer Notes Chapter 2 Computer Components 5

Primary Memory:
Primary memory is also called main memory. It is directly accessible by the Central Processing Unit of the computer. All the instructions and data are placed in this memory before processing. This memory is also known as main memory.

Primary memory is of two types.

  1. Random Access Memory
  2. Read Only Memory

1. Random Access Memory: This memory is also known as RAM. This memory is very fast. All the data and instructions are stored in this memory when the computer is working. It is a volatile type of memory. It means that when the computer is switched off its contents gets destroyed. This memory cannot store the data permanently. But without this memory the computer cannot work.

2. Read Only Memory: It is also known as ROM. It is a permanent type of memory. This memory is placed on computers and other electronic devices in the form of microchips. The ordinary user cannot write in this memory. He can only read from this memory. That is why this memory is called Read Only Memory. Normally this memory stores the data and information which is required to start a computer. Only users cannot write in this memory. Only the manufacture of this memory can write in it.

PSEB 6th Class Computer Notes Chapter 2 Computer Components

Secondary Memory
Secondary memory is also known as auxiliary memory or external memory. This memory is used to store data and instructions for a long time. This memory is a non-volatile memory. The storage capacity of secondary memory is higher than primary memory. This memory is not directly accessible from the CPU. Different types of memory storage devices are used in this type such as hard disk, CD, DVD, pendrive, Magnetic tape etc.

Categories of Computer:
The computers can be categorised according to their speed, memory processing capability, storage capacities and generations.

There are four main types of computer.

  1. Micro computer or personal computer
  2. Mini computer
  3. Mainframe computer.
  4. Supercomputer

PSEB 6th Class Computer Notes Chapter 2 Computer Components 6
1. Micro Computer or Personal Computer: Micro or Personal Computers are the most common computers. These are the computers which are used in homes, schools and offices. These computers are small in size and the cost of this computer is very less. These computers can be afforded by small businesses, small schools and individuals also. The computer on which we work falls in this category. Examples of computers of this category are desktop computer, laptops, palmtops, tablets, mobile phones etc.

2. Mini Computer: This type of computer has more processing capabilities than a microcomputer.
These computers are also larger in size.
These computers are used by multiple users at a time. The main features of these computers is highest storage capacity sharing of resources and high processing speed. Micro computers are mainly used as small or mid range servers. These computers are used in large businesses, large team based companies and scientific applications.
PSEB 6th Class Computer Notes Chapter 2 Computer Components 7

3. Mainframe Computer: The processing speed of Mainframe computers is higher than micro and mini computers. These computers are capable of handling large amounts of data within less time. These computers are capable of performing operations on large amounts of data with high processing speed. Mainframe computers are used in patience such as government offices, banks etc.
PSEB 6th Class Computer Notes Chapter 2 Computer Components 8

4. Supercomputer: Supercomputer is the most powerful computer invented till now. The processing speed of supercomputers is very fast. It can store a large amount of data. These computers are specially used for complex application. Supercomputers are the most powerful and expensive computers till now. The application areas of these types of computers are weather forecasting climate research, oil and gas exploration and processing of astronomical data.
PSEB 6th Class Computer Notes Chapter 2 Computer Components 9

PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer

This PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer will help you in revision during exams.

PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer

Introduction to Computer:
Computers are used everywhere these days. All the offices, banks, school, colleges use computers for their daily work. Let us try to understand the meaning of computers.

Meaning of Computer:
A computer is an electronic device which gets the data and instructions from the user. It processes the data as per given instructions and gives the result in the desired form. Computer can be defined as:

A computer is an electronic machine that accepts data as input from the user and processes the data under the control of a set of instructions and gives the result as an output.
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 1
Literally, the word computer is taken from Latin word compute which means to calculate. So in simple words, we can say that a computer is an electronic device which was made to calculate. Earlier their definition was correct as the main work of the computer was to calculate. Now the computer is doing a lot of work to decide the calculation.

PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer

Uses of Computer:
Following are the uses of computer in Modern Times:

  • We can do mathematical calculations on the computer.
  • We can play games on the computer.
  • We can draw pictures on the computer.
  • We can listen to songs and watch films on the computer.
  • We can use computers to print books and newspapers.
  • We can use computers to book our tickets to travel in trains, buses and airplanes.
  • We can check the arrival and departure time of trains, buses and airplanes with the help of a computer.
  • We can check the weather conditions of any place before travelling.
  • We can print reports, results of our school or time table.
  • We can store our data into a computer for future use.

Applications of Computer
Computers are used in many areas. Some of the application areas of computers are given below:
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 2

1. Education: Students and teachers are using computers in their education. Teachers use computers to teach the students and students use computers to prepare their notes making, drawing and making projects etc. The internet is used to search, study material, results, time table and various reports are prepared using the computer. Computers are also used for record keeping, online data, processing etc. Punjab Education Department is using computer on their web portal like www.epunjabschool.gov.in www.pseb.ac.in

2. Entertainment: Computers are a good source of entertainment also. The user can watch movies, listen song and camp. Beautiful presentations can be made on the computer. All the films made in these days use computers for their special effects.

3. Sports: Computers are also used in the sports field. They are used to improve the performance of players to keep data and to evaluate the performance and to display information about various tournaments which will happen in future. Digital scoreboards are also a part of games today. The players can improve their by analysing their record.

4. Communication: Computers have given new ways of communication. The user can make video calls, audio calls, chat using computer data and send email to their friends and relatives. Any type of data can be shared using smartphone, tablet, laptop and computers.

5. Shops: Shopkeepers are also using computers for their various works. They are using computers to manage their stock to generate purchase orders to generate bills to advertise their shop. The use of computers in shops save a lot of time for the shopkeeper and the customer.

6. Hospitals: Hospitals are using computers to maintain patient records and to monitor the health of patients. The medical history of a patient can be stored and communicated by using a computer. Now the doctor is also using a computer to check their patient from a remote place. Computers are also used in Laboratories to do 2x and data analysis.

7. Banks: The hanking system is working by using computers. All types of records of customers are maintained by using computers. Then use a computer to maintain data of accounts withdrawal of money deposit of money, loans etc. Customers can now use e-banking services to do their work from their home. ATMs are totally computer based.

8. Government Offices: Government offices are using computers in governance, bill payment, Income Tax application processing. Most of the government’s work is online these days. The citizens can apply for any government service with the help of using computer and internet from their homes. They can pay to build Taxes and avail other governmental services using the internet.

9. Travel: Computers are also used for travel purposes. The user can book their tickets, get knowledge of some geographical areas, check availability of means of transport, cost, estimation and regional culture etc. The user can also book hotels from their home by using the internet.

Characteristics of Computer
The main characteristics of computer are as given below:
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 3
Characteristics of Computer
Speed Accuracy Reliability Diligence Automotion Storage
1. Speed: Computer is a very fast machine. It can do calculation in milliseconds. It can complete days work in seconds. That is why it has reduced human effort.

2. Accuracy: Computer is an accurate machine. It never makes any mistakes. If a mistake is performed in any calculation that is always due to human error. A computer can do any task repeatedly for thousands of times without any error.

3. Reliability: Reliability means the ability of being trusted. A computer is a reliable machine. We cannot doubt the results given by computers because
they are always correct. There is no need of rechecking the results. It gives the same result every time.

4. Diligence: Computer never gets tired. It can to do work for many days or even for years. During this time its efficiency does not decrease and neither it commits errors after long working. In fact, a computer is a hardworking machine.

5. Automation: Automation means to do the work automatically. Computers can be instructed to do the work without any interaction. He can be given the instructions and the computer can do work without asking anything. We just need to start the computer and ask him to do the particular work. After that it will do the whole work without asking anything.

6. Storage: The storage capacity of computers is very large. It can attach many storage devices such as hard disk, pen drive at 17 to it. It can store any type of data such as text, audio, video, animation in it. We can also store the storage capacity of computers. Computers store data for a very long time.

Portable Computing Devices
Portable devices are those devices which can be easily carried from one place to another. Technology has given us many portable devices. Some of them are explained below:
1. Mobile Phone or Smartphone: It is the most common portable computing device used these days. Each one is using their mobile phones these days. Smartphones can do a lot of calculation work and processing.
Their speed is also very fast. Modern mobile phones are equipped with camera large storage capacity and High processing speed.
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 4
They can be used in education, business trading, entertainment, gaming etc. Mobile phones are very small devices. It can be carried in our pockets. There are a lot of mobile phones available in the market with different features.

2. Tablet Computer: It is a small computer.
It is mainly known as a tablet. If the battery is used in this tablet. It has a touch screen which is used to give input and get the output. Tablets can be used for many purposes. Most of the students are using tablets in their education at home as well as at their schools.
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 5

3. Palmtop Computer: Palmtop computer is a small size computer which can be placed on our palm. This computer has a small screen and small keyboard. It is just like a mobile phone having a keyboard. These computers are not very popular in these days. These computers were used for limited purposes only
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 6

4. Laptop Computer: Laptop computer is the most famous computer these days. It is a computer with a big screen and keyboard attached to it. It can be easily taken from one place to another. If the battery is used to give power to the laptop computer. All it had also a touchpad and many other features built in it.
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 7

5. Note Book: It is a portable computer. It is smaller than laptop. It is also lighter in weight. It has a battery to work. The battery can be charged again and again. It can be easily taken from one place to another.
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 8

PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer

Limitations of Computer:
PSEB 6th Class Computer Notes Chapter 1 Introduction to Computer 9
Computers are capable of doing a lot of work. But, after all this is a machine and unable to do some tasks. These kinds of tasks are considered as limitations of computers.

Some of the main limitations of computer are as under:

  • A computer cannot take decisions by itself.
  • A computer cannot correct wrong instructions.
  • Computer cannot do any work without instruction from the user.
  • It does not have feelings or IQ (Intelligence Quotient)
  • It does not have knowledge and experience like a human being.
  • Computer can not start work until asked.