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Students can Download Maths Chapter 7 Tangents Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 7 Tangents Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 7 Solutions Malayalam Medium

Students can Download Physics Chapter 2 Magnetic Effect of Electric Current Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Physics Chapter 2 Magnetic Effect of Electric Current Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 2 Solutions Malayalam Medium

You can Download Second Degree Equations Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 4 Second Degree Equations Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 4 Second Degree Equations Notes

Textbook Page No. 81

Second Degree Equation Class 10 Kerala Syllabus Question 1.
When each side of a square was reduced by 2 metres, the area became 49 square metres. What was the length of a side of the original square?
Answer:
Let the length of each side of the original square be x, then

The length of each side of the original square = 9 cm

Second Degree Equation Class 10 Questions And Answers Kerala Syllabus Question 2.
A square ground has a 2 metre wide path all around it. The total area of the ground and the path is 1225 square metres. What is the area of the ground alone?
Answer:
’Each side of the ground = x
Each side of ground+path = x + 4
(x + 4)² = 1225 = 35²;
x + 4 = 35
x= 35 – 4 = 31

Area of the ground alone = 31² = 961 m²

Second Degree Equation Class 10 Extra Questions Kerala Syllabus Question 3.
The square of a term in the arithmetic sequence 2, 5, 8, ……., is 2500. What is its position?
Answer:
Let 2500 is the square of the n^th term of the arithmetic sequence 2, 5, 8, …………

It is the 17^th term.

To solve a system of linear equations with steps, use the system of linear equations calculator.

Sslc Maths Second Degree Equations Kerala Syllabus Question 4.
2000 rupees was deposited in a scheme in which interest is compounded annually. After two years the amount in the account was 2205 rupees. What is the rate of interest?
Answer:
Amount (P) = 2000
Compound interest (r %)
Year (n)

Textbook Page No. 86

Sslc Second Degree Equation Questions Kerala Syllabus Question 1.
1 added to the product of two consecutive even numbers gives 289. What are the numbers?
Answer:
Let the two consecutive even numbers be x, x+2
x, x+2
x(x+2) + 1 = 289
x² + 2x = 288
x²+ 2x – 288 = 0
x²+ 2x = 288
(x+1)² = 288 + 1
(x+1)² = 289
x+1 = ± 17
x = 16, – 18
The numbers are 16, 18

Second Degree Equation Questions And Answers Kerala Syllabus Question 2.
9 added to the product of two consecutive multiples of 6 gives 729. What are the numbers?
Answer:
Let the two consecutive multiples of 6 be x, x+6
x, x+6
x(x + 6)+ 9 = 729
x² + 6 x + 9 = 729
x² + 6x = 720
x²+ 6x – 720 = O
x² +6x = 720
(x+3)²= 720+9
(x+3)² = 729
x+3 = ± 27
x = 24, – 30
The numbers are 24, – 30

10th Maths Second Degree Equation Kerala Syllabus Question 3.
How many terms of the arithmetic sequence 5, 7, 9, …, must be added to get 140?
Answer:
First term f= 5,
Common difference = 2

10 terms should be added to get 140

Second Degree Equation Class 10 Model Question Paper Kerala Syllabus Question 4.
16 added to the sum of the first few terms of the arithmetic sequence 9, 11, 13, gave 256. How many terms were added?
Answer:
9, 11, 13,…………
First term f= 9,
Common difference d = 2
Sum of first n terms

Second Degree Equation Class 10 Notes Kerala Syllabus Question 5.
An isosceles triangle has to be made like this

The height should be 2 meters less than the base. The area of the triangle should be 12 square meters. What should be the length of its sides?
Answer:
Base = AB = x
Height= CD = x – 2

Sslc Maths Chapter 4 Questions And Answers Kerala Syllabus Question 6.
A 2.6-meter long rod leans against a wall, its foot 1 meter from the wall. When the foot is moved a little away from the wall, its upper end slides the same length down. How much farther is the foot moved?

Answer:

Textbook Page No. 91

Sslc Maths Chapter 4 Solutions Kerala Syllabus Question 1.
The product of a number and 2 more than that is 168, what are the numbers?
Answer:
Let the number be x
x(x+2) = 168
x² + 2x = I68
x² + 2x + 1 = I68 + 1
(x+1)² = 169
x + 1 = ± 13
x + 1 = 13
x + 1= – 13
x = 13 – 1 = 12
x = -13 – 1 = – 14
The number is 12 and 14 or – 12 and – 14

Maths Chapter 4 Class 10 Kerala Syllabus Question 2.
Find two numbers with sum 4 and product 2.
Answer:
Sum of the numbers = 4
If, First number = x
Then second number = 4 – x

Class 10 Maths Second Degree Equation Kerala Syllabus Question 3.
How many terms of the arithmetic sequence 99, 97, 95, … must be added to get 900?
Answer:

The spectrally extended signal produced by nonlinear function calculator 520 is likely to have a pronounced dropoff in amplitude as frequency increases.

Class 10 Maths Chapter 4 Kerala Syllabus Kerala Syllabus Question 4.
A rod 28 centimeters long is to be bent to make a rectangle.
i. Can a rectangle of diagonal 8 centimeters be made?
ii. Can a rectangle of diagonal 10 centimeters be made?
iii. How about a rectangle of diagonal 14 centimeters?
Calculate the lengths of the sides of the rectangles that can be made.
Answer:




∴ Cannot make a rectangle of diagonal 14 centimeters.

Textbook Page No. 97

Maths Second Degree Equation Kerala Syllabus Question 1.
The perimeter of a rectangle is 42 meters and its diagonal is 15 meters. What are the lengths of its sides?
Answer:

Sslc Second Degree Equation Questions And Answers Kerala Syllabus Question 2.
How many consecutive natural numbers starting from 1 should be added to get 300?
Answer:
Sum of natural numbers starting from 1:

Second Degree Equation Class 10 Kerala Syllabus Kerala Syllabus Question 3.
What number added to 1 gives its own square?
Answer:
Let the number be x

Second Degree Equation Kerala Syllabus Question 4.
In writing the equation to construct a rectangle of specified perimeter and area, the perimeter was wrongly written as 24 instead of 42. The length of a side was then computed as 10 meters. What is the area in the problem? What are the lengths of the sides of the rectangle in the correct problem?
Answer:
If length is 10 and perimter is 24
2 (l + b) = 24
2(10 + b) = 24
10 + 5 = 12
b = 2
Width= 2
Area= lb =10 x 2 = 20 sq.cm
Area in the correct problem =20
lb = 20
2(l + b) = 42
l + b = 21
l + \(\frac { 20 }{ l }\) = 21

Kerala Syllabus 10th Standard Maths Chapter 4 Kerala Syllabus Question 5.
In copying a second-degree equation to solve it, the term without x was written as 24 instead of – 24. The answers found were 4 and 6. What are the answers of the correct problem?
Answer:

Second Degree Equations Orukkam Questions and Answers

Worksheet 1

Second Degree Equations Exercises Kerala Syllabus Question 1.
Write two numbers whose square is 25?
Answer:
Let x be the number
x² = 25
x= √25 = ±5
numbers are –5, +5

Second Degree Equation Class 10 Formulas Kerala Syllabus Question 2.
When the square of a number is added to the number we get 30. What are the numbers?
Answer:
Let x be the number ,
x² + x = 30
x² + x – 30 = 0, (x + 6) (x – 5) = 0
x = – 6, 5
numbers are – 6, +5

Second Degree Equations Class 10 Kerala Syllabus Question 3.
Find the side of the square whose area and perimeter are numerically equal.
Answer:
Let x be the length of side.
Perimeter = Area
x² = 4x
x = 4
Length of side = 4

Question 4.
How many odd numbers from 1 makes the sum 961?
Answer:
Sum of continuous n odd numbers = n²
n² = 961
n = √961 = ± 31
Sum of continuous 31 odd numbers

Question 5.
A man’s age after 15 years will be the square of his age 15 years ago. What is his present age?
Answer:

Worksheet 2
Form the equation

Question 6.
The sum of a number and its square is ten times that number
Answer:
Let x be the number
x² + x = 10 x
x² – 9x = 0

Question 7.
The sum of a number and its square root is 6.
Answer:
Let x be the number
x + √x = 6
x – 6 = √x
(x – 6)² =(–√x)2
x² – 12x + 36 = x
x² – 13 x + 36 = 0

Question 8.
The sum of first n natural numbers is 210.
Answer:
\(\frac { n(n+1) }{ 2 }\) = 210
n² + n = 420
n² + n – 420 = 0

Question 9.
The area of a rectangle whose length is 5 more than its width
Answer:
Let x be the length of side.
other side = x + 5
x (x + 5) = 150
x² + 5x – 150 = 0

Question 10.
The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\)
Answer:
Let x be the number

Question 11.
The sum of even numbers from 2 in an order is 240
Answer:
2 + 4 + 6 + + 2n = 420
2(1 + 2 + 3 + ….+ n) = 420

Question 12.
A man’s age after 15 years will be the square of his age 15 years ago.
Answer:
Let x be the present age
(x – 15)² = x + 15
x² – 30 x + 225 = x +15
x² – 31x + 210 = 0

Worksheet 3

Question 13.
When 8 times a number is added to its square we get 8. Find the number by making the equation properly.
Answer:
x² + 2 x = 8
x² + 2x + 1 = 8 + 1
(x + 1)² =9
x + 1 =3
x = 2
Number = 2

Question 14.
Which term in the sequence 2, 5, 8 …….. gives its square 2500?
Answer:
nth term = 3n + (2 – 3) = 3n – 1
(3n – 1)² = 2500,
3n – 1 = 50,
n = 17
square of 17^th term is 2500

Question 15..
A man’s age after 15 years will be the square of his age 15 years ago. Find the age
Answer:
Let x be the age
x + 15 = (x – 15)²
x + 15 = x² – 30 x + 225
x² – 31 x = – 210

Question 16.
The length of a rectangle is 2 more than its width. The area of the rectangle is 80. Find length and breadth.
Answer:
Let x be the width, then length = x + 2
x (x + 2) = 80
x² + 2x = 80
x² + 2x + 1 = 80 + 1
(x + 1)² = 81
x + 1 =9
x = 8
length = 10
width = 8

Question 17.
The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\) Find the number.
Answer:
Let x be the number

Question 18.
The sum of some even numbers starting from 2 is 420. Find the number of even numbers added.
Answer:

Worksheet 4

Question 19.
Sum of the squares of three consecutive natural numbers is 110.
Answer:
If the numbers are x, x + 1, x + 2, then
x² + ( x + 1)² + (x + 2)² = 110
x² + x² + 2x + 1 + x² + 4x + 4 = 110
3x² + 6x – 105 = 0

Question 20.
The product of the digits of a two-digit number is 12.When 36 is added to the number we get a two-digit number in which the digits are reversed. Find the two-digit number .
Answer:
Let the two digit number be 10 x + y, then
x y = 12
y = \(\frac { 12 }{ x }\)
If we change the positions of digits 10 y + x
10 y + x = 36+ 10 x + y
Let’s take x be the digits in the position of 10.
Let’s take \(\frac { 12 }{ x }\) as the digits in the position of unit.
Number which digits are reversed \(\frac { 120 }{ x }\) +x

Question 21.
Serena and Johan had 45 diamond stones. They sold 5 stones. The product of the remaining stones is 124. Find the number of stones each had
Answer:
Let x be the stones Serena has and 45 – x be the stone Johan has. After selling 5
(x – 5)(45 – x – 5) = 124
(x – 5)(40 – x) = 124
x² – 45x + 200 = – 124
x² – 45x + 324 = 0


Number of diamond stones Serena has = 36
Number of diamond stones Johan has = 9

Question 22.
The sum of a number and its reciprocal is \(1 \frac{1}{2}\) . Find the number.
Answer:
Let x be the number

Question 23.
The sum of two numbers is 15. Sum of its reciprocals is \(\frac { 3 }{ 10 }\). Find the numbers.
Answer:
Let x, 15 – x be the numbers

Question 24.
A two-digit number is four times sum of its digits. The number is three times product of the digits. Find the number
Answer:
Let x, y be the numbers
10 x + y = 4 (x + y)
6 x = 3 y

Worksheet 5

Question 25.
A train travels a distance of 300km constant speed. If the speed of the train is increased by 5 km, the journey would have taken 2 hours less. Find the original speed of the train
Answer:
Let x km/hr be the speed.
Let the time required to travel in the same speed be \(\frac { 300 }{ x }\) hour. If the speed is increased by 5 km/ hr the time will decreased by 2 hours.

Question 26.
An express train takes 3 hours less than a passenger train for a journey of 600 km. If the speed of the passenger train is 10 less than the speed of the express train find the speeds of both trains (Use Pythagoras theorem in distance, not in speeds)
Answer:
Let x km/hr be the speed of the express.

Speed of the express 50 km/hr
Speed of the passenger 50 km/hr

Worksheet 6

Question 27.
One year ago a man’s age is eight times the age of his son. At present man’s age is the square of son’s age. Find the present age.
Answer:
Let x be the present age of son

Question 28.
A man’s age after 15 years will be the square of his age 15 years ago. Find the present age by forming a second degree equation
Answer:
Let x be the present age .
x + 15 = (x – 15)2
x + 15 = x² – 30x + 225
x² – 31x + 210 = 0

Present age = 21

Question 29.
The product of Layas’s age before 5 years and after 8 years is 30. Find the present age.
Answer:
Let x be the present age of Laya
(x – 5)(x + 8) = 30
x² – 3x – 40 = 30,
x² – 3x – 70 = 0

Present age of Laya = 7

Worksheet 7

Question 30.
Sravani teacher asked the students to construct a rectangle having area 5 square unit and perimeter 8. Jeevan, a wise student of the class, after making some calculations told that it is not possible to construct such a rectangle. Can you agree with him. Justify reasonably
Answer:
Let x be the length, then the area will be 5 cm², So the width will be \(\frac { 5 }{ x }\) cm. When the perimeter is 8 cm, 4 – x will be the width. The width which was obtained first will be \(\frac { 5 }{ x }\) and the width obtained now i.e., 4 – x will be equal.

– 4 < 0, x will not be obtained. Hence it is not possible to draw a rectangle with the given values.

Question 31.
The perimeter of a rectangle is 34 cm, area 60 square centimeter. Find the sides
Answer:
Let x be the length
Perimeter 34 cm, therefore width will be 17-x cm
Perimeter = 60 cm, therefore width be \(\frac { 60 }{ x }\) cm

x = 5 cm
length = 12 cm therefore width = 17 – 12 = 5 cm
length = 12 cm ,
width = 5 cm

Question 32.
The length of the rectangle is 4 more than its breadth. Area of the rectangle is 140 square centimeter. Calculate length and breadth
Answer:
Let x be the width of rectangle
length = x + 4 cm
x(x + 4) = 140
x² – 4 x – 140 = 0

width =10 cm
length = 14 cm

Question 33.
When the sides of a square are increased by 4, area become 256. Find the length of the first square.
Answer:
Let x be the side of square

x will not -20
The side of the first square = 12 cm

Question 34.
The area of a right-angled triangle is 60 square unit. The one of the perpendicular sides is 10 more than other. Find the sides of the triangle.
Answer:
Let x cm, x + 10 cm be the perpendicular sides

Question 35.
The area of an isosceles triangle is 60 square meters. One of the equal sides is is 13 cm. Find the third side. Take base x then h = \(\sqrt{13^{2}-x^{2}}\)
Answer:

Second Degree Equations SCERT Question Pool Questions and Answers

Question 36.
When the sides of a square are increased by 8 cm each, its area becomes 1225 sq. cm Frame an equation using the above data by taking the side of the smaller square as x cm. Find the sides of both the squares. [Score: 3, Time: 5 Minutes]
Answer:
One side of smaller square = x
One side of bigger square = x + 8
Area = (x + 8)² = 1225 (1)
x + 8 = 35
x = 35 – 8 = 27 (1)
Side of smaller square = 27 cm
Side of bigger square = 35 cm (1)

Question 37.
The difference of two positive numbers is 6. Their product is 216. Find the numbers. [Score: 3, Time: 4 Minutes]
Answer:
Let the numbers be x, x + 6
Products : x (x+6) = 216 (1)
x² + 6x = 216
x² + 6x + 9 = 216 + 9 = 225
(x + 3)² = 225 (1)
x + 3 = 15
x = 15 – 3 = 12
Numbers : 12, 18 (1)

Question 38.
In a right triangle one of the perpendicular sides is one less than 2 times the smaller side. Hypotenuse is one more than 2 times the same smaller side. By taking the smaller side as x cm, write the algebraic expression for the other two sides. Compute all the 3 sides of the right triangle. [Score: 4, Time: 5 Minutes]
Answer:
Smaller side = x
Perpendicular side = 2x – 1
Hypotenus = 2x + 1 (1)
x² + (2x – 1)² = (2x + 1)² (1)
x² + 4x² – 4x + 1 = 4x² + 4x + 1
x² – 8x = 0 (1)
x (x – 8) = 0
x = 0 or x = 8
Side=8 cm, 15 cm, 17 cm (1)

Question 39.
Find the sides of a rectangle whose perimeter is 100 metres and area 600 sq. metres. [Score : 4, Time : 5 Minutes]
Answer:
Perimeter = 100 m .
Length+Breadth = 50 m.
Length = 25 + x.
Breadth = 25 – x
Area = (25 + x)(25 – x) = 600 (1)
25² – x² = 600 (1)
x² = 625 – 600 = 25, x = 5 (1)
Sides, 25 + 5 = 30 m
25 – 5 = 20 m (1)

Question 40.
The one’s place of a two-digit number is 4. The product of the number and digit sum is 238.
a. If ten’s place digit is taken as x, Write the number.
b. Frame a second-degree equation and find the number. [Score : 4, Time : 6 Minutes]
Answer:
Digit in the ten’s place = x
Number = 10 x + 4 (1)
The product of the number and digit sum
= (x + 4) (10 x + 4) = 238 (1)

Question 41.
How many consecutive natural numbers from 1 should be added to get 465? [Score : 4, Time : 5 Minutes]
Answer:
Sum of first n natural numbers

Question 42.
The product of the digits of a two-digit number is 12. When 36 is added to this number, got a new number with digits reversed. Find the number. [Score : 5, Time : 7 Minutes]
Answer:

Question 43.
For a two-digit number, one’s place is 3 more than its ten’s place. The product of this number and its digit sum is square of double the digit sum. What is the number? [Score : 4, Time : 6 Minutes]
Answer:

Question 44.
A pavement of Width 2 metres is built around a square shaped garden. The area of the pavement alone is 116 square metres. Find one side of the garden.
[Score : 3, Time: 4 Minutes]
Answer:
Side of garden = x
Area of the pavement Area of the pavement Area of the pavement Side of garden including pavement = x + 4 (1)
Area of the pavement = (x + 4)² – x² = 116
x² + 8x + 16 – x² = 116 (1)
8x + 16 = 116,
\(x=\frac{116-16}{8}=12.5 \mathrm{m}\)
Side of garden = 12.5m (1)

Question 45.
A rectangle of Width 8 centimetres is cut off from a square sheet along its side. The remaining rectangular portion has an area 84 sq. metres. Calculate the side of the square. [Score : 4, Time : 6 Minutes]
Answer:
Let the side of square = x
Sides of rectangle = x, x – 8 (1)
Area = x(x – 8) = 84
x² – 8x = 84 (1)
x² – 8x + 16 = 84 + 16
(x – 4)²= 100, (1)
x – 4= 10, x= 14
Side of square = 14 centimetre

Question 46.
The lengh of a rectangle is 3 metre more than 3 times its breadth. Its diagonal is 1 metre more than the length. Find the lengh and breadth of the rectangle. [Score: 4, Time: 7 Minutes]
Answer:
Sidc of rectangle = x
Length =3x +3, Diagonal = 3x + 4 (1)
(3x + 4)² = x² + (3x + 3)² (1)
9x² + 24 x + 16 = x² + 9x² + 18x + 9
x² – 6x = 7, x² – 6x = 7 (1)
x² – 6x + 9 = 16, (x – 3)² = 16
x – 3 = 4, x = 4 + 3 = 7 (1)
length of rectangle = 3 × 7 + 3 = 24 metre breadth of rectangle = 7 metre

Question 47.
In the figure AB is the diameter of the circle. CD = 10 cm. BC is 15 cm less than Ac. Find AB?

Answer:

Question 48.
The chords AB and CD of a circle intersect at. If MA = 6 cm, MB = 8 cm and CD = 16 cm. Find MC and MD.

Answer:
CD = 16 centimetre MC = 8 – x, MD = 8 + x
then, MA x MB = MC x MD (1)
6 x 8 = (8 – x)(8 + x)
48 = 64 – x², x² = 16, x = 4 (1)
MC = 8 – 4 = 4 centimetre (1)
MD = 8 + 4 = 12 centimetre (1)

Question 49.
In the figure, AD is drawn perpendicular to the side opposite to the right angled vertex A. BC-13 cm and AD=6 cm.
a. Take BD = x and express DC in terms of x.
b. Frame a second-degree equation and find the lengths BD and DC. [Score : 4, Time: 5 Minutes]

Answer:

Question 50.
Can the sum of a number and it’s reciprocal be 2/3? [Score: 4, Time: 5 Minutes]
Answer:
Number = x , then its reciprocal = 1/x

Since the discriminant is negative. We do not get a solution.
∴ The sum of a numbrer and it’s reciprocal never gives 2/3. (1)

Question 51.
The sum of a number and its reciprocal is \(\frac { 13 }{ 6 }\) What is the number ? [Score: 4, Time: 5 Minutes]
Answer:
Let number = x, then its reciprocal = 1/x

Question 52.
The sum of a number is 12 and sum of its reciprocal is \(\frac { 3 }{ 8 }\) Find the numbers. [Score: 4, Time: 5 Minutes]
Answer:
If we take numbers as 6 + x, 6 – x (1)

Question 53.
Consider an arithmetic sequence with common difference 20. If the sum of reciprocals of two consecutive terms of this sequence is 1/24, find the first term of the arithmetic sequence. [Score: 4,Time: 6 Minutes]
Answer:
’Terms x – 10, x + 10 (1)

Question 54.
Prove that the difference of a number and its reciprocal will be always positive. [Score: 3, Time : 5 Minutes]
Answer:
Number = x, its reciprocal = 1/x (1)
then, \(x-\frac{1}{x}=k\) (1)

Since k being a positive number, k² + 4 also negative. (1)

Question 55.
In copying a second-degree equation, the number without x was written as -30 in¬stead of 30. The answers found were 15 and -2. What are the answers of the correct problem? [Score: 5, Time: 8 Minutes]
Answer:

Second Degree Equations Exam oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 56.
Find two numbers whose sum is 6 and product is 9.
Answer:
’First number = x
Second number = 6 – x
x ( 6 – x ) = 9
6x – x² = 9
0 = 9 – 6x + x²
x² – 6x + 9 = 0
(x – 3)² = 0
(x – 3 ) (x – 3) = 0
x = 3
Numbers are 3, 3.

Question 57.
Ammu is 7 years younger than Divya. If 2 is added to the product of their ages, we get 200. Find their ages.
Answer:
Age of Ammu = x
Age of Divya = x – 7
x(x – 7)+2 = 200
x² – 7x + 2 = 200
x² – 7x – 198 = 0 (x – 18) (x + 11) = 0
x= 18, x = – 11
(age will not be a negative number)
Age of Ammu = 18,
Age of Divya =18 – 7 = 11

Question 58.
In order to solve a quadratic equation, Arun did the following steps. Find the values of x by completing each step.

Answer:

Question 59.
The perimeter of a rectangle is 42 cm and its diagonal is 15 cm. Find the dimensions of the rectangle.
Answer:
Let breadth = x; perimeter = 2 (length + breadth) length + breadth = 42/2 = 21
∴ length = 21 – x
x² + (21 – x) = 152
x² + 441 – 42x + x² = 225
2x² + 216 – 42x =0
x² – 21x + 108 = 0
(x – 12) (x – 9) = 0
x = 12 or x = 9
breadth = 9 unit
length = 12 unit

Question 60.
If the equation 4x² – 5 x + k = 0 has two equal solutions. Find the value of k.
Answer:
If there is one solution, the discriminant = 0

Short Answer Type Questions (Score 3)

Question 61.
The product of a number and the number 8 more than it is 105.
a. What is the least number to be added to make the product a perfect square?
b. What are the numbers in the problem?
Answer:
a. Let the number be x, then second number is x+ 8
x(x + 8) = 105
x² + 8x = 105
The number to be added to get a perfect square is 16

Question 62.
On Arts day, 180 sweets were distributed equally among the students. Tasting one sweet Deepa said, “It is a pity that 9 of our friends are absent today.”
Hearing this Deepu said, “Because of that we got one sweet more.”
a. Find out the total number of students,
b. How many students were present on that day?
Answer:
a. Let the number of students = n
Number of sweets given to one student = \(\frac { 180 }{ n }\)
Since 9 students were absent = \(\frac { 180 }{ n-9 }\)

b. Total present = 45 – 9 = 36

Question 63.
One of the perpendicular side of a right angled triangle has length 4 cm more than twice of the other side. It has surface area 80 cm². Find out the lengths of the perpendicular sides.
Answer:
Let one of the perpendicular side = x
Second side = 2x + 4, Area = 80
1/2 × (2x + 4) = 80
2x² + 4x = 160
x² + 2x = 80
x² + 2x + 12 = 80 +12
(x + 1)² = 81
x +1 = ± 9
x + 1 = 9
x = 8
x = – 10 cannot be accepted as – ve sign cannot exist for the length of the side
∴ One side = 8cm
Second side = 2 × 8 + 4 = 20 cm

Long Answer Type Questions (Score 4)

Question 64.
Reji’s father bought several note books of the same price. Total price Rs. 360. If the price of each book were less by 2 rupees, he would get 2 books more.
a. Find the number of books if the price of one book is x.
b. Find the number of books if the price of one books is less by 2 rupees.
c. Form a quadratic equation.
d. Find the number of books bought.
Answer:
(a) Price of each book be x rupees,
No. of Notebooks = \(\frac { 360 }{ x }\)

(b) If the cost of each book is Rs (x – 2)
No. of books brought for Rs. 360 = \(\frac { 360 }{ x – 2 }\)

Question 65.
1235 Sacks of rice is to be brought from Trivandrum to Kottayam. For this a minitruck takes 6 trips more than that of an ordinary truck. An ordinary truck can carry 30 sacks more than that of a mini truck. Find the capacity of each truck.
Answer:
Let the capacity of mini truck be x sacks and ordinary truck be x + 30 sacks

Long Answer Type Questions (Score 5)

Question 66.
a. The sum of a number and it’s reciprocal is \(\frac { 25 }{ 12 }\) What is the number?
b. Prove that the sum of a positive number and it’s reciprocal is always greater than or equal to 2.
Answer:

Free Limit using Substitution Calculator – Find limits using the substitution method step-by-step.

Question 67.
Sum of the area of two squares is 500 m². If the difference of their perimeters is 40m, find the sides of the two squares.
Answer:
Let the side of the squares be x and y meters.
According to the condition,
x² + y² = 500 (1)
4x – 4y = 40
(x – y) = 10
y = x – 10
Substituting the value of y in (1), we get
x² + (x – 10)² – 500
2x² – 20x – 400 = 0
x² – 10x – 200 = 0
x = 20 or x = – 10
As the side cannot be negative, x = 20
Hence, side of the first square, x = 20 m
Side of the second square, y = 20 – 10 = 10 m

Question 68.
a By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450km. Find the original speed of the bus.
b. 250 Rupees is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Answer:
Let speed of the bus be x km/hr

Second Degree Equations Memory Map

The value of x in the second-degree equations can be found out by mainly 3 ways.
1. Factorization
2. Completing the square
3. By using equation

You can Download Gas Laws Mole Concept Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 2 Gas Laws Mole Concept Textbook Questions and Answers

SCERT Class 10th Standard Chemistry Chapter 2 Gas Laws Mole Concept Solutions

Gas Laws And Mole Concept Questions And Answers Kerala Syllabus 10th Text Book Page No: 33

→ Complete the table 2.1

Answer:

→ If a gas which is kept in a cylinder having a volume of 1 liter, is completely transferred to another 5-liter cylinder then what will be the volume of the gas?
Answer:
5 liter

→ Press the piston after closing the nozzle of the syringe. What will happen to the volume of air inside the syringe?
Answer:
Volume decreases

→ Explain it on the basis of the distance between the molecules of gas and their freedom of movement?
Answer:
Gases molecules are separated from each other by a large distance. As a result, there will be a lot of vacant spaces. So as the piston is pressed, the molecUles come closer and this volume decreases

Use the Molarity Calculator Chemistry to calculate the mass, volume or concentration required to prepare a solution of compound of known molecular weight.

Sslc Chemistry Chapter 2 Kerala Syllabus  Text Book Page No: 34

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

→ What assumption can be made regarding the possibility of collision between gas molecules?
Answer:
The molecules collide each other.

→ Which energy gained due to the movement of molecules? Potential energy/Kinetic, energy.
Answer:
Kinetic energy

→ When a gas is heated, temperature is increased. What happens to the movement of molecules if the temperature of the gas is increased?
Answer:
Speed of motion increased

→ Asa result, what happens to the energy of the molecules?
Answer:
Energy of molecules increase

→ Volume
Answer:
The space needed for a substance to occupy is its volume. The volume of solids and liquids are definite. But the volume of a gas is the volume of its container in which it is present.

→ Pressure
Answer:
The force exerted at unit area is pressure, Therefore, force at unit area/ pressure
\(=\frac{\text {Force exerted at the surface}}{\text {Area of the surface}}\)

→ Temperature
Answer:
The average Kinetic energy of all the molecules in a substance is its temperature

Sslc Chemistry Chapter 2 Questions And Answers Text Book Page No: 35

→ Is there any change in the number of molecules?
Answer:
No

→ What happens to the pressure when the volume is decreased?
Answer:
Pressure increases

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

Mass to Moles Calculator — The quantity of substance n in moles is equal to the mass m in grams divided by the molar mass M in g/mol.

→ What changes can you observe in the volume of the gas inside the syringe?
Answer:
Volume is formed to be decreasing.

→ What about decreasing the pressure?
Answer:
Volume increasing

→ What relation do you arrive at between pressure and volume of the gas?
Answer:
As pressure increases volume decreases when pressure is reduced volume increased.

Gas Laws And Mole Concept Kerala Syllabus 10th Text Book Page No: 36

→ The size of the air bubbles rising from the bottom of an aquarium increases. Can you explain the reason?
Answer:
As the bubbles move upward, the pressure on them decreases. This causes increase in volume. So as the bubbles move upward, their size increases.

→ What do you observe?
Answer:
Ink rises through the tube

→ What is the reason for the rising of the ink upwards?
Answer:
When the bottle is placed in hot water, air inside the bottle becomes hot. This causes expansion of air. This pushes ink in the tube. So ink rises through the tube.

→ What did you observe on cooling the bottle after taking it out? Why?
Answer:
Ink comes down. Because as air becomes cool, Its volume decreases.

→ What can you infer about the relation between the volume and temperature of a gas?
Answer:
When temperature is increased volume of gas increases. Similarly, when temperature is decreased, The volume decreases.

→ Complete the table 2.2

Answer:

Sslc Chemistry Chapter 2 Notes Kerala Syllabus Text Book Page No: 37

→ In which unit is the temperature stated?
Answer:
Kelvin (K)

→ What happens to the volume when the temperature is increased?
Answer:
Volume increases.

→ If an inflated ballon is kept in sunlight, it will burst. What may be the reason for this?
Answer:
When the ballon placed in sun light, temperature increases. so volume of air inside the ballon increases. Thus ballon expands and finally bursts.

→ What happens to the volume of the gas when its pressure is decreased or temperature is increased. volume increased/decreased.
Answer:
volume increased

→ If the temperature and pressure are kept constant how can we increase the volume?
Answer:
Fill some more gas.

→ Fill the cylinder with a little more gas. Does the number of molecules increase or decrease now?
Answer:
Then number of molecules increases

→ What is the relation between the volume and number of molecules?
Answer:
When the number of molecules increases, volume increases.

→ According to Avagadro’s law when the temperature and pressure remain constant on which factor does the volume of gas depend?
Answer:
Depends on the number of molecules.

Kerala Syllabus 10th Standard Chemistry Chapter 2 Text Book Page No: 38

→ If the mass of a coin 5g, then what will be the mass of thousand coins?
Answer:
5 x 1000 = 5000g

→ If the mass of coins in a bag is 50,000 g, then how many coins will be there?
Answer:
\(\frac { 50000 }{ 5 }\) = 10000

→ Like this we can calculate the number of coins on the basis of mass. Can‘t we?
Answer:
Yes, it becomes easy.

→ Is their any relation between the mass and the number, if the particles are of the same mass.
Answer:
Yes

→ What may be the method of stating the mass of atoms?
Answer:
The mass of an atom is compared to the mass, of another atom and expressed as a number which shows how many times it is heavier than the other atom. The atomic mass of elements are expressed by considering 1/12 mass of an atom of carbon-12 as one unit.

→ What do you understand from the statement that the atomic mass of Helium is 4?
Answer:
Atomic mass of Helium is 4. That is mass of one atom of Helium is 4 times of 1/12^th mass of carbon atom.

Enter the formula and press “calculate” to work out the molar mass calculator with steps, the number of moles in 1 g and the percentage by mass of each element.

Chemistry Class 10 Chapter 2 Kerala Syllabus Text Book Page No: 39

→ How many oxygen atoms combine with one carbon atom?
Answer:
2 Oxygen atoms

→ How many oxygen atoms combine with 1000. carbon atoms?
Answer:
2000 oxygen atoms.

→ How many atoms are present in 12g carbon?
Answer:
6.022 x 10B carb Answer:
2 × 6.022 × 10²³ oxygen atoms

→ What will be the mass of these tabs ?
Answer:
2 × 16 = 32g

Gas Laws And Mole Concept Extra Questions 10th Text Book Page No: 40

→ Complete the table 2.5

Answer:

→ 1GAM sodium means 23g sodium. This contains 6.022 × 1023 atoms. If so, how many GAM is present in 69 g sodium? How many atoms are present in it?
Answer:
\(\frac { 69 }{ 23 }\) = 3 GAM,
3 × 6.22 × 10²³ sodium atoms.

Gas Laws And Mole Concept Notes Pdf Kerala Syllabus 10th Text Book Page No: 41

→ How many GAMs are present in each the samples given below? Calculate the, number of atoms present in each of die sample ? (Atomic mass N= 14, O= 16)
1. 42g Nitrogen,
2. 80g Oxygen
Answer:
1. 42g Nitrogen:
No.of GAM = \(\frac { 42 }{ 14 }\) = 3GAM
No.of atoms = 3 × 6.022 × 10²³

2. 80g Oxygen:
No.of GAM = \(\frac { 80 }{ 16 }\) = 5GAM
No.of atoms = 5 × 6.022 × 10²³

→ Complete the table 2.6

Answer:

→ Calculate the molecular mass of glucose (C₆ H₁₂ O₆) and sulphuric acid (H₂ SO₄)
Answer:
Molecular mass of glucose
= 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g
Molecular mass of sulphuric acid
2 × 1+ 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 g

Gas Laws And Mole Concept Questions And Answers Pdf 10th Text Book Page No: 42

→ Complete the table 2.7

Answer:

→ What is the molecular mass of oxygen?
Answer:
32 g

→ How many GMM is present in 32g oxygen?
Answer:
1 GMM

→ How many molecules are present in it?
Answer:
6.022 × 10²³ oxygen molecules

→ How many GMM is present in 28 gm nitrogen?
Answer:
1 GMM

→ How many molecules are present in N₂?
Answer:
6.022 × 10²³ N₂, molecules

→ How many GMM is present in 18 gm water?
Answer:
1 GMM

→ How many H₂O molecules arf present in it?
Answer:
6.022 × 10²³ H₂O Molecules

→ Calculate the number of GMM present in 96g oxygen?
Answer:
\(\frac { 96 }{ 32 }\) = 3GMM

Sslc Chemistry Chapter 2 Notes Pdf Kerala Syllabus Text Book Page No: 43

How many GMM are present in each of the given samples? Calculate the number of molecules present in each sample ?

→ 360 g glucose (Molecular mass = 180)
Answer:
No.of GMM = \(\frac { 360 }{ 180 }\) = 2 GMM
No. of molecules = 2 × 6.022 × 10²³

→ 90g water (Molecular mass = 18)
Answer:
No.of GMM = \(\frac { 90}{ 18 }\) = 5 GMM
No. of molecules = 5 × 6.022 × 10²³

→ How many molecules of water are present . in one mole of water ?
Answer:
6.022 × 10²³ water molecules

→ What is its mass?
Answer:
18 g

→How many GMM is present in it?
Answer:
1 GMM

Sslc Chemistry 2nd Chapter Notes Kerala Syllabus Text Book Page No: 44

22.4 L of a gas at STP = 1 mole
44.8 Lofa gas at STP = = 2 mole
224 L of a gas at STP = \(\frac { 224 }{ 22.4 }\) = 10 mole

Gas Law And Mole Concept Kerala Syllabus 10th Text Book Page No: 45

Complete the flow chart given below, related to one mole of substance.

Answer:

Gas Laws Mole Concept Let Us Assess

Gas Laws And Mole Concept Notes Kerala Syllabus 10th Question 1.
Examine the date given in the table (Temperature and number of molecules of the gas are kept constant).
a. Calculate P × V
b. Which is the gas law related to this?
Answer:
a. 8L atm
b. Boyle’s law

Gas Laws And Mole Concept Pdf Kerala Syllabus 10th Question 2.
Analyse the situations given below and explain the gas law associated with it.
a. When an inflated balloon is immersed in water, its size decreases.
b. A balloon is being inflated
Answer:
a. Avogadro’s law
b. Boyle’s law

Gas Laws And Mole Concept Class 10 Kerala Syllabus Question 3.
Certain data regarding various gases kept under the same conditions of temperature and pressure are given below.

a. Complete the table?
b. Which gas law is applicable here?
Answer:
a.

b. Avogadro’s law

Hss Live Guru 10th Chemistry Kerala Syllabus Question 4.
a. Calculate the mass of 112 L CO₂ gas kept at STP (molecular mass = 44)
b. How many molecules of CO₂ are present in it?
Answer:

Sslc Chemistry Chapter 2 Gas Laws And Mole Concept Question 5.
Calculate the volume of 170g of ammonia at STP ? (Molecular mass 17)
Answer:
Number of moles = \(\frac { Given mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10 moles
Volume at STP = mole × 22.4 L = 10 × 22.4 L = 224 L

Hsslive Chemistry 10th Kerala Syllabus Question 6.
Find out the number of moles of molecules present in the samples given below (GMM-N₂=28g, H₂O= 18g)
a. 56g N₂
b. 90g H₂O
Answer:
a. Number of mol molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 56 }{ 28 }\) = 2

b. Number of mol molecules = \(\frac { 90 }{ 18 }\) = 5

10th Chemistry 2nd Chapter Kerala Syllabus Question 7.
The molecular mass of ammonia is 17.
a. How much is the GMM of ammonia?
b. Find out the number of moles of molecules present in 170g of ammonia.
c. Calculate the number of ammonia molecules present in the above sample of ammonia?
Answer:
a. GMM of ammonia (NH3) = 14 + 3 × 1 = 17g = 1GMM

b. Number of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10

c. Number Of molecules = Mole × 6.022 × 10²³
= 10 × 6.022 × 10²³

Class 10 Chemistry Chapter 2 Kerala Syllabus Question 8.
The molecule’s mass of oxygen is 32.
a. What is the GMM of O₂
b. How many moles of molecules are there in 64g of oxygen? How many molecules are there in it?
c. Calculate the number of oxygen atoms present in 64g of oxygen?
Answer:
a. GMM of O₂ =2 × 16 = 32 g

b. No. of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 64 }{ 32 }\) = 2
Number of molecules = mole × 6.022 × 10²³
= 2 × 6.022 × 10²³
c. c. Number of atoms = Number of molecules × number of atoms in one molecules Number of atoms in one molecules of oxygen (O₂) =2
∴ total number of atoms = 2 × 6.022 × 10²³ × 2
=4 × 6.022 × 10²³

Gas Laws Mole Concept Extended activities

Hss Live Guru Chemistry 10 Kerala Syllabus Question 1.
How many grams of carbon and oxygen are required to get the same number of atoms as in one gram of Helium?
Answer:
GAM of Helium = 4 g
Number of mole atoms in 4 g of Helium = 6.022 × 10²³
Number of atoms in 1 gofHelium = \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
GAM ofCarbon = 12 g
∴ Number of atoms in 12gofCarbon=6.022 x 10²³
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
Carbon atoms = \(3 \mathrm{g}\left(12 \times \frac{1}{4}\right)\)
GAM of Oxygen = 16 g.
∴ Number of atoms in 16 g of Oxygen = 6.022 × 10²³
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
Oxygen atoms = \(4 \mathrm{g}\left(16 \times \frac{1}{4}\right)\)

Kerala Syllabus 10th Standard Chemistry Guide Question 2.
Examine the samples given:
a. 20 g of He
b. 44.8 L of NH₃ at STP
c. 67.2 L of N₂ at STP
d. 1 mol of H₂SO₄
e. 180 g of water.
i. Arrange the samples in increasing order of the number of molecules in each.
ii. What will be the ascending order of the total number of atoms?
iii. What will be the masses of samples b, c, and d?
Answer:

Mole Concept Class 11 Question 3.
In 90 grams of water.
a. How many molecules are present?
b.What will be the total number of atoms?
c. What will be the total number of electrons in this sample?
Answer:
a. GMM= 18 g
∴ Number of molecules
\(\frac { 90 }{ 18 }\) × 6.022 × 10²³ = 5 × 6.022 × 10²³ 18

b. ∴ Number of atoms = 3 × 5 × 6.022 × 10²³ = 15 × 6.022 × 10²³

c. Electrons in 1 H atom = 1
Electrons in 1 O atom = 8
Total electrons in H₂O molecule? 10
Total electrons in 90 g H₂O = 10 × 5 × 6.022 × 10²³ = 50 × 6.022 × 10²³

Gas Laws Mole Concept Orukkam Questions and Answers

Question 1.
a. Complete the table based on the data given in the box.

b. Express atomic weight and molecular weight in grams. How many moles is this? Find out the number of Atoms or molecules in it?

Answer:
a.

b.

Question 2.
Complete the table based on the molecules given in the first column and then answer the question given below.

a. 10 Mole of water = …………. g …………… Molecules
5 mole of CaO = ……….. g ………… Molecules
2 Mole of H₂SO₄ = …………. g ………… Molecules
1/2 Mole of Al₂O₃= …………. g ………….. Molecules
Answer:

a. 10 Mole of water = 10 × 18 = 180g, 10 × 6.022 × 10²³ Molecules
5 mole of CaO = 280g, 5 × 6.022 × 10²³ Mol-ecules
2 Mole of H₂SO₄ = 196g, 2 × 6.022 × 10²³ Mol-ecules
2- Mole of AlO = 100g, 1/2 × 6.022 × 10²³

Question 3.
Complete the table.

Answer:

Question 4.
Complete the data.

Answer:

Question 5.
Based on the reaction given below, write the answers for the questions.
N₂ + 3H₂ → 2NH₃ ;
a. Write the ratio of reactant molecules and product molecules.
b. How many moles of Ammonia forms when we take 2 moles of Nitrogen and six moles of Hydrogen?
c. Two moles of Nitrogen and three moles of hydrogen are taken in jar? Will they react together?
d. How many moles of Nitrogen and Hydrogen is needed for rearing 20 moles of Ammonia?
Answer:
a. 1:3:2
b. 2 Mole
c. No
d. 10 Mole Nitrogen 30 Mole Hydrogen

Question 6.
Balance the given equation and then write down the answers for the questions given below.
CH₄+O₂ → CO₂ + H₂O
a. How many moles of C02 formed when 20 moles of Methane burns in air?
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
b. Based on the equation above, How many moles of CO₂ is formed when 10 moles of Ethane is burned in air ?
Answer:
a. When 1 mole of methane bum in air 1 mole of CO₂ gas is formed.
When 20 moles methane burn 20
moles of CO₂ are formed.
The molecular weight of 20 moles = 20 × 44 = 880 g

b. When 2 moles of ethane is burned 4 moles of CO2 is formed.
The number of moles of CO₂ when 1 mole of ethane is burned = 4/2 = 2 mole.
The number of moles when 10 moles of ethane bums = 2 x 10 = 20 mole.
Weight of 20 moles = 20 × 44 = 880 g

Question 7.
Based on the given equation write down the answers.
2H₂+ O₂ → H₂O
a. How much Oxygen and Hydrogen is needed for making 1800g of water vapor?
b. How many moles of Oxygen is needed for the reaction with one mole of Hydrogen?
Answer:
a. 36 g of water vapour can be made using 4 g hydrogen.
The amount of hydrogen required to make 1 g water vapour = 4/36
The amount of hydrogen required to produce 1800 g watervapour= 4/26 × 1800 = 200 g
Mass of oxygen = 1800 – 200 = 1600 g

b. 0.5 Mole

Gas Laws Mole Concept Evaluation Questions

Question 1.
Find out the number of moles of hydrogen and Oxygen atoms present in 10 moles of HCI.
Answer:
One mole HC₁ contains 1 mole of hydrogen and 1 mole of chlorine.
Hydrogen contained in 10 moles of HC₁ = 1 × 10 =10 mol,
Clatom = 10 × 1 = 10 mol

Question 2.
Find out the mass of Hydrogen atom and chlorine atom in 10 moles of HCI.
Answer:
Mass of 10 mole hydrogen atom = 10 × 1 = 10g
Mass of 10 mole chlorine atom =10 × 35.5 = 355 g.

Question 3.
a Find out the mass of one mole of CaCO3. How many moles of calcium present in 1000g CaCO₃?
b. How many moles of Oxygen present in 1000gms of CaCO₃?
Answer:
a. Mass of 1 mole of CaCO₃ = 40 + 12 + 48 = 100 gram.
No of moles ofCa in 1000 g CaCO₃ = \(\frac { 1000 }{ 100 }\) = 10 mol

b. No of moles of Ca in 10 moles CaCO₃ =10 × 1 = 10 mol
No of moles of oxygen in 10 moles of CaCO₃ = 10 × 3 = 30mol.

This Combined Gas Law Calculator can help you estimate either the pressure, temperature or the volume of gas.

Question 4.
Find out number of moles of water formed when 4gms of Hydrogen and 32 gms of Oxygen combined together. What is the result when 5 gms of Hydrogen and 32 gms of Oxygen combined together?
Answer:
Ans. When 4 g of hydrogen and 32g of oxygen are combined 36 g of water \(\frac { 37 }{ 18 }\) = 2 mol
5g H + 32 gO → 37g H₂O
No of moles in 3 7 g of water = \(\frac { 37 }{ 18 }\) = 2.055 mol

Question 5.
a How much grams of NaCl is needed for making 2 molar solution (NaCl – 58.5). Wh-at is the amount of water needed for this?
b. How will you change a two molar solution of Sodium Chloride into 5 major?
Answer:
a. Mass of 2 moles of NaCl = 2 × 58.5 = 117 g
1 liter water is required for this.

b. When 2 moles of NaCl is dissolved in 4 liter of water 5 molar solution is obtained.

Finally, you encounter how to find molar concentration step-by-step manually, and if your preference indulges with instant calculations.

Question 6.
How many moles of Cl₂ present in 11.2 L of same in STP? Find out the mass of this?
Answer:
No.of moles present in 11.2 litre of chlorine = \(\frac { 11.2 }{ 22.4 }\) = 0.5 mol
Mass of 0.5 moles of chlorine = 0.5 × 35.5 = 17.759 g.

Question 7.
Find out the mass of Oxygen atom in 44.8L of CO₂ in STP.
Answer:
No of moles in 44.8 litre of CO₂ \(\frac { 44.8 }{ 22.4 }\) 2 mol,
1 Mole of CO₂ contains 1 mole of C and 1 mole of O₂
∴ 2 Mole of CO₂ contains 2 mole of O₂ or 4 mole of oxygen atom.
Man of oxygen atom 4 × 16 = 64 g

Question 8.
Find out the amount of CO₂ formed when the burning of one mole of Ethane.
Answer:
2 moles of CO₂ is formed when 1 mole of ethane burns.
Mass of 2 moles of CO₂ = 2 × 44 = 88g

Question 9.
Why are atomic mass of some elements are infractions ?
Answer:
The atomic masses of some elements are infractions because they exist as a mixture of isotopes of different masses. The fractional atomic masses arise because of this mixture.
Average mass = \(\frac { Total mass of all atoms }{ numbers of atoms }\)

Gas Laws Mole Concept SCERT Questions and Answers

Question 1.
One GAM substance contains Avogadro number of particles in it.
a. How many particles are there in Avogadro number ?
b. Write the number of atoms present in each of the following.
i. 32g Sulphur
ii. 32g Oxygen
iii. 32g Carbon
(Atomic mass S = 32, O = 16, C = 12)
Answer:
a. 6.022 × 10²³
b. i.6.022 × 10²³
ii. 2 x 6.022 × 10²³
iii \(\frac { 32 }{ 12 }\) × 6.022 × 10²³

Question 2.
a. Group the following into pairs having same number of atoms.
A. 2g Hydrogen
B. 16g Oxygen
C. 14g Nitrogen
D. 8g Helium (Atomic mass H=1, O= 16, N =14, He=4)
b. How many atoms are present in each pair?
Answer:
a. A, D2g Hydrogen, 8g Helium
B, C 16g Oxygen, 14g Nitrogen

b. A, D – 2 × 6.022 × 10²³
B,C – 6.022 × 10²³

Question 3.
N₂ + 3H₂ → 2NH₃
a. What is the ratio between the reactant molecules in the above reaction?
b. Complete the following table.

Answer:
a. 1:3
b. a – 2 NH₃,
b – l H₂,
c – 12H₂,
d – 4NH₃

Question 4.
2H₂ + O₂ → 2H₂O
a. What is the ratio between the reactant molecules in the above reaction?
b How many O₂ molecules are required to react 100 H₂ molecules completely?
c. How many water molecules are formed when 1000 H₂ molecules are reacted completely
Answer:
a. 2:1
b. 50 O₂ molecules
c. 1000 H₂O molecules

Question 5.
Complete the following table. (All the elements given are diatomic. Atomic mass O=16, N=14, CI=35.5)

Answer:
a. 6.022 × 10²³
b. 6.022 × 10²³
c. 71 g
d. 14 g

Question 6.
A sample of substances are given.

Hint: Molecular mass NH3 = 17, N2 = 28, H2SO4 = 98, O2=32
a. Which of these samples have same number of molecules?
b. Which of these samples has least number of molecules?
Answer:
a. 68g NH₃, 128gO₃
b. 49 g H₂SO₄

Question7.
Pick out the correct statements from the following. Also, correct the incorrect statements.
a The number of molecules present in 1 mol hydrogen and 1 mol oxygen are same,
b. 2 mol chlorine contains 4 x 6.022 x 10²³ chlorine molecules.
c. The mass of 1/2 mol nitrogen gas is 14 g.
d. 0.5 mol water has the mass 9g. There are 6.022 × 10²³ H20 molecules in it. (Atomic mass H = 1, O = 16, CI= 35.5, N= 14)
Answer:
Correct statements – a, c
No. of molecules in 2 mol chlorine is 2 × 6.022 × 10²³
Mass of 0.5 mol water is 9g. So it contains 0.5 × 6.022 × 10²³ H₂O molecules.

Question 8.
Complete the following.

Answer:
a. 2 × 6.022 × 10²³
b. 1GMM
c. 6.022 × 10²³

Question 9.
67.2 L of Carbon dioxide gas is filled in a cylinder at STP.
a. Calculate the mass of CO₂ present in it. (Atomic mass- C = 12, O = 16)
b. Calculate the number of molecules present in the cylinder.
Answer:
a. Molecular mass of CO₂=12 × 1+16 × 2 = 12 + 32 = 44
No. of moles in 67.2L CO₂ at STP = \(\frac { 67.2L }{ 22.4L }\) = 3
b. 3 × 6.022 × 10²³

Question 10.

Answer:
a. 2
b. 2 × 6.022 × 10²³
c. 17 g
d. 51 g
e. 3
f. 3 × 22.4 L

Question 11.
CH₄ + 2O₂ → CO₂ + 2H₂O
The equation describes the combustion of methane in air.
a. How many moles of oxygen is required for the complete combustion of 16g CH₄?
b. Calculate the amount of CO₂ formed when 100g of CH₄ is completely burnt?
Answer:
a. 2 mol
b. Amount of CO₂ produced by the combustion of 16g CH4 = 44g
Amount of CO₂ produced by the combustion of 1 gm CH4 = \(\frac { 44 }{ 16 }\) g
Amount of CO₂ produced by the combustion of 100g CH4 = \(\frac { 44 }{ 16 }\) × 100g

Question 12.
45 g glucose is taken in a beaker and made into 1 L (MM = 180).
a. Calculate the molarity of the solution,
b. Above solution is made up to 2 L by adding more water. What will be the molarity of the resultant solution?
c. How will you prepare IM solution of glucose with the same quantity (45 g) of glucose?
Answer:
a. 0.25
b. M = \(\frac { n }{ v }\) = \(\frac { 0.25 }{ 2 }\)
c. Add 250 ml water in 45 g glucose

Question 13.
Two gases occupy equal volume at STP are shown below.
(Atomic mass S = 32, O =16, N = 14

a. Find the mass of the gas in B.
b. Calculate the number of molecules present in B.
Answer:
a. No. of moles in 320g SO₂ = \(\frac { 320 }{ 64 }\) = 5
Mass of 5 mol NO₂ = 5 × 46 = 230 g
b. No.of molecules present in B = 5 × 6.022 × 10²³

Question 14.
The balanced chemical equation of a reaction (at STP) is given below.
2H₂(g) + O₂(g) → 2H₂O(g)
a. Calculate the volume of oxygen required to combine completely with 224 L of the hydrogen at STP.
b. Calculate the mass of water formed as a result of the reaction (a).
Answer:
a. 112 L
b. Volume of water obtained when 224 L hydrogen completely reacts with oxygen = 224 L
No. of moles in 224 L water = \(\frac { 224 }{ 22.4 }\) = 10
Massof 10 mol water = 10 × 18 = 180 g

Question 15.
Complete the table.

Hint: (MM – CO₂ = 44, CH₄ = 16, SO₂ = 64)
Answer:
a. 67.2 l
b. 132 g
c. 1/4.
d. 4g
e. 11.2 L
f. 1/2.

Question 16.
Analyse the following equation
2NO(g) + O₂(g) → 2NO₂(g)
a. Calculate the number of the moles of NO required to combine completely with 112 L of Oxygen at STP.
b. Calculate the mass of NO₂ formed when 112L of oxygen reacts completely?
Answer:
a. 10 mol
b. 2NO(g) + O₂(g) → 2NO₂(g) (2 : 1: 2)
No. of moles in 112L O₂ = 5 mol
According to equation no. of moles of NO₂ obtained by reacting oxygen completely with nitric oxide = 2
No. of moles of NO₂ obtained by reacting 5 mol oxygen completely = 10
Massof 10 mol NO₂ = 10 × 46 = 460g

Question 17.
The chemical equation of the decomposition of calcium carbonate is given below.
CaCO₃→ CaO + CO₂
(HintMM: CaCO₃ – 100, CaO – 56, CO₂ – 44)
a. Calculate the mass of CaCO3 required to get 224 g of CaO?
b. Calculate the number of CO2 molecules fronted when 224g of CaO is obtained?
Answer:
a. CaCO₃ → CaO + CO₂
100g 56g 44g
I I : I
Amount of CaCO₃ required to get 56g of CaO = 100g
Amount of CaCO₃ required to get 1 g of CaO = \(\frac { 100 }{ 56 }\)
Amount of CaCO₃ required to get 224 g of CaO = \(\frac { 100 }{ 56 }\) × 224 = 400g
b. 4 × 6.022 × 10²³

Question 18.
You are requested to make 20 moles of NaCl into packets of 100g each. (Hint^Molecular mass of NaCl is 58.5)
a. How many packets of NaCl can be prepared?
b. Is there any NaCl remaining? If so, how much?
Answer:
a. Mass of 20 mol NaCl = 20 × 58.5 = 1170g
1170 g NaCl can be made into 11 packets with 100g each.
b. Remaining NaCl = 1170 – 1100 = 70g

Gas Laws Mole Concept Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
GAM of Hydrogen is 1 g.
a. How many number of atoms are there in 1 g of Hydrogen?
b. Find the mass of 1 atom of hydrogen.
Answer:
a. 1g hydrogen = 6.022 × 10²³ atoms
b. Mass of 1 atom of hydrogen
= \(\frac{1 g}{6.022 \times 10^{23}}\) = 1.66 × 10-24 g

Question 2.
Number of molecules of substance is 3.011 × 10²⁴.
a. What is the number of molecules of 1 mole of any substance?
b. Find the number of moles of 3.011 × 10²⁴ molecules.
Answer:
a. 6.022 ×10²³
b. Number of moles of molecules = \(\frac { Number of molecules }{ N_A }\)
= \(\frac{3.011 \times 10^{24}}{6.022 \times 10^{23}}=5\)

Short Answer Type Questions (Score 2)

Question 3.
Fill the patterns.

Answer:
a. GMM
b. 22.4
c. Number of molecules
d. Number of molecules
e. Volume in litres
f. Mass

Question 4.
Identify the incorrect statements from those given with respect to the arrangements of molecules in gases.
a. The minute molecules are present without any freedom of movement
b. Collision take place between the molecules,
c. Increasing the number of molecules at constant volume causes the decrease in number of collisions.
d. The energy of molecules are comparatively high.
Answer:
a. The statements (a) and (c) are not correct

Question 5.
The pressure of 20L of a gas kept at 300 K is found to be 2 atoms. If the pressure is increased to 3 atom at the same temperature, what will be the new volume?
Answer:
According to Boyles law, P_V = a constant
Therefore, P₁ V₁ = P₂ V₂
Here, P₁ = 2atm V₁ = 20L P₂ = 3atm V₂=?
∴ 2 × 20 = 3 × V₂
Thus, V₂ = \(\frac { 2 × 20 }{ 3 }\) = 13.3 L

Charles Law Calculator is a free online tool that displays the volume of gas that tends to expand when heated.

Question 6.
If the temperature of 5L of a gas at atmospheric pressure is changed from 200K to 50 K, what will be the volume?
Answer:
According to Charles law, \(\frac { V }{ T }\) = a constant

Question 7.
What will be mass of 89.6 L of ammonia (NH3) gas at STP?
Answer:
89.6 L of NH3 gas at STP = \(\frac { 89.6 }{ 22.4 }\) = 4 mol
GMM of NH3 = 14 + 3 = 17g
Mass = mole × GMM = 4 × 17 = 68g

Question 8.
8. a. What is molar volume?
b. What is the molar volueofa gas at STP?
Answer:
a. The volume of one mole of a gas is called molar volume,
b. 22.4 L

Question 9.
Look at the balanced equation given.
2NaOH + CO₂ → Na₂ CO₃ + H₂0
a. Find out the mass of NaOH needed for 264 g CO₂ to react completely.
b. Find out the total number of moles of water molecules when CO₂ reacts.
Answer:
a. GMM of CO₂ = 44 g
∴Number of moles in the molecule of 264 g CO₂ = \(\frac { 264 }{ 44 }\) = 6
According to the equation NaOH needed for the reaction of 1 mole CO₂= 2 moles
∴ NaOH needed for the reaction of 6 moles CO₂ = 2 × 6 = 12 moles
GMM of NaOH = 23 + 16 + 1 = 40 g
Total mass of NaOH = 12 × 40 = 480 g

b. H₂O formed when lmole CO₂ reacts = 1 mole
∴ Total number of moles of water molecules when 6 mole CO₂ reacts = 6 moles

Question 10.
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
This is the equation of ignition of cooking gas butane.
Calculate the volume of CO₂ in STP during the complete ignition of 14 kg of cooking gas.
Answer:
Mass of Butane (C₄H₁₀) = 14 kg = 1400g
GMM of C₄H₁₀ = 4 × 12 + 10 × 1 = 58 g
∴ Number of moles in molecules = \(\frac { 1400 }{ 58 }\) = 241.38
Amount of CO₂when 2 moles of C₄ H₁₀ ignites = 8 moles of C₄H₁₀ ignites = \(\frac { 8 }{ 2 }\) × 965.52 moles
∴ Volume of CO₂ formed in STP
= 965.52 × 22.4 L = 21627.65 L

Question 11.
Write down the preparation of 100 ml NaOH solution of 0.1 M.
Answer:
GMM of NaOH = 40 g
Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}=\frac{\mathbf{n}}{\mathbf{v}}\)
M = 0.1 V=100ml = 0.1 L
0.1 = n/0.1
∴ n = 0.1 × 0.1 = 0.01
Mass needed to prepare 100 ml NaOH in 0.1M = 0.01 × 40 = 0.4g
Take 0.4 g NaOH in a beaker. Dissolve it Hilly by adding a little amount of water. Then, again add water to make it 100 ml.

Question 12.
The molarity of250 ml of Na₂CO₃ solution is 0.5 M. Find the mass of Na₂CO₃.
Answer:

Question 13.
63 g HNO₃ is in the dilute solution of 200 ml HNO₃ (Nitric acid). Find the molarity.
Answer:
a.

Short Answer Type Questions (Score 3)

Question 14.
Some equations related to gas laws are given below.
i. V α P
ii. \(\frac { V }{ T }\) = a constant
iii. V α n
iv. P_v = a constant
a. Which of these are correct?
b. Write the gas law to which it is related for the correct equations.
Answer:
a. Equations (ii), (iii) and (iv) are correct

b. (ii) Charles law
(iii) Avogadro’s law
(iv) Boyles law

Question 15.
In 100 g of CaCO₃
a. Find out the number of moles of each element and atom.
b. Find out the total number of atoms of each element.
Answer:

Question 16.
Fill the blanks in the given table.

Answer:

Long Answer Type Questions (Spore 4)

Question 17.
See CO₂ gas is taken in a cylinder provided with a piston. The cylinder is dipped in hot water.
a. What happens to the movement of CO₂ molecules?
b. What change do you expect in the position of the piston?
c. What is the relation between temperature and the volume of a gas?
d State this gas law.
Answer:
a. As the temperature increase, the energy of molecules increases. This increases the speed of the motion of molecules.
b. Piston is pulled in the upward direction. So piston moves upward.
c. As temperature increases, volume increases.
d. At constant pressure, the volume of a definite mass of gas is directly proportional to the temperature in Kelvin scale. (Charles law)

Question 18.
The molecular formula of ammonium sulfate is (NH₄)₂SO₄.
a. Find the gram molecular mass (GMM) of ammonium sulfate.
b. Calculate the number of molecules and atoms in 1.32g of ammonium sulfate.
Answer:
a. GMM of (NH₄)₂SO₄
= (14+4) × 2 + 32 + 4 × 16 = 36 + 32 + 64
= 132 g

Question 19.
Fill in the blanks of the table given below.

Answer:

Question 20.
See the diagram given below.

Answer:
a. 196g
b. 2 × 6.022 × 10²³
c. 2 GMM
d. 2 × 6.022 × 10²³

Question 21.
Write in pairs, equal number of atoms from those given below,
a. 2g Hydrogen
b. 16 g Oxygen
c. 14 g Nitrogen
d. 8 g Helium
Answer:

Question 22.
Certain compounds and its masses are given,
i) 68 g NH₃
ii) 28 g N₂
iii) 9 g H₂O
iv) 128 g O₂
a. Which of these compounds have equal number of molecules?
b. How many molecules are there?
c. How many atoms are there in 9 g of water?
Answer:

Question 23.
368 g NO ₂gas is given. Find the answers of each one given below,
a. GMM of NO₂
b. Number of moles of molecules of 368 g NO₂
c. Number of molecules
d. Number of atoms
e. Volume in STP
Answer:
a. GMM of NO₂
= 14 + 2 × 16 = 46 g

b. Number of mole = \(\frac { 368 }{ 46 }\) = 8

c. Number of molecules = Number of moles × NA
= 8 × 6.022 × 10²³

d. Number of atoms = Number of atoms in one molecule × Number of molecules
= 3 × 8 × 6.022 × 10²³ = 24 × 6.022 × 10²³

e. Volume in STP = Number of moles × 22.4
L = 8 × 22.4 L = 179.2 L

Question 24.
Find out the GMMofthe following. Also find out the total number of atoms.
a. 20 g Nitrogen (H₂)
b. 88.75 g Chlorine (Cl₂)
c. 4 g Calcium (Ca₂) ,
d. 7.75 g phosphorus (p₄)
(H = 1, Cl =35.5, Ca =40, P=31)
Answer:
a. 20 g of hydrogen (H2): GMM = \(\frac { 20 }{ 2 }\) = 10
Number of molecules = 10 × 6.022 × 10²³
Total number of atoms = 2 × 10 × 6.022 × 10²³
= 20 × 6.022 × 10²³

b. 88.75g of chlorine
Number of GMM = \(\frac { 88.75 }{ 71 }\) = 1.25
Number of molecules = 1.25 × 6.022 × 10²³
Total number of atoms = 2 × 1.25 × 6.022 × 10²³
= 2.5 × 6.022 × 10²³

c. 4 g Calcium (Ca)
Number of GMM = \(\frac { 4 }{ 40 }\) = \(\frac { 1 }{ 10 }\) = 0.1
Number of molecules = 0.1 × 6.022 × 10²³
Total number of atoms = 1 × 0.1 × 6.022 × 10²³
= 0.1 × 6.022 × 10²³

d. 7.75 g Phosphorus Number of GMM = \(\frac { 7.75 }{ 31 }\) = \(\frac { 1 }{ 4 }\) = 0.25
Number of molecules = 0.25 × 6.022 × 10²³

You can Download Solids Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 8 Solids Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 8 Solids Notes

Textbook Page No. 191

Solids Class 10 Kerala Syllabus Kerala Syllabus Chapter 8 Question 1.
A square of side 5 centimetres, and four isosceles triangles!of base 5 centimetres and height 8 centimetres, are to be put together to make a square pyramid. How many square centimetres of paper is needed?
Answer:
Area of base = 5 × 5 = 25 cm² Area of one triangle 1/2 × 5 × 8 = 20 cm² Curved surface area = 4 × 20 5cm = 80 cm²
Paper is needed to make a square pyramid = 25 + 80 = 105 cm²

Solids in Maths SSLC Question 2.
A toy is in the shape of a square pyramid of base edge 16 centimetres and slant height 10 centimetres. What is the total cost of painting 500 such toys, at 80 rupees per square metre?
Answer:
Surface Area of the toy
= 16 × 16 + 4 × 1/2 × 16 × 10
=256 + 320 = 576 cm²
Surface Area of 500 toys = 500 × 576 = 288000 cm²

Kerala Syllabus 10th Standard Maths Question 3.
The lateral faces of a square pyramid are equilateral triangles and the length of a base edge is 30 centimetres. What is its surface area?
Answer:
Lateral faces are equilateral surfaces Surface area
= Base Area + Curved surface area

= 900 + 900 √3
=900 + 1558.8 = 2458.8 cm² = 2459 cm²

The Chebyshev’s theorem calculator counts the probability of an event being far from its expected value.

Sslc Maths Chapter 8 Kerala Syllabus Question 4.
The perimeter of the base of square^»yra- mid is 40 centimetres and the total length of all its edges is 92 centimetres. Calculate its surface area.
Answer:
Base perimeter = 4a = 40
a=10cm
Total length of edges = 92 cm
Length of total laterals edge = 92 – 40 = 52
Length of one laterals edge = \(\frac { 52 }{ 4 }\) = 13 cm
Surface area of pyramid = a 2 + 2 al
= 102 + 2 × 10 × 13
= 100 + 260 = 360 cm2

KBPS full form, KBPS stands for, meaning, what is KBPS, description, example, explanation, acronym for, abbreviation, definitions, full name.

Surface Area of Kerala Question 5.
Can we make a square pyramid with the lateral surface area equal to the base area?
Answer:
Curved surface area = 2al
Area of base = a2
a2=2al
a = 21 ⇒ 1 = a/2
For making a square pyramid first we must determine its base, one side of the lateral will be the base. Other two sides make half of base by reducing the angle. That is angle at apex will be less than 90°.

Textbook Page No. 193

Sslc Maths Solids Kerala Syllabus Chapter 8 Question 1.
Using a square and four triangles with dimensions as specified in the picture, a pyramid is made.

What is the height of this pyramid?
What if the square and triangles are like this?

Answer:

Sslc Maths Solids Equations Kerala Syllabus Chapter 8 Question 2.
A square pyramid of base edge 10 centimetres and height 12 centimetres is to be made of paper. What should be the dimensions of the triangles?
Answer:
Base edge of a square pyramid = 10 cm
Let h be the height
Base edge a = 10 cm
Height h= 12 cm
Slant height =

Sslc Maths Chapter 8 Solids Kerala Syllabus Question 3.
Prove that in any square pyramid, the squares of the height, slant height and lateral edge are in arithmetic sequence.
Answer:
Height = h,
Slant height = l,
Lateral edge = e

Solids in Maths Question 4. A square pyramid is to be made with the triangles shown here as a lateral face. What I would be its height? What if the base edge is 40 centimetres instead of 30 centimetres?

Answer:

It is impossible to make a square pyramid of base edge 40cm.

Textbook Page No. 195

Solids Chapter Class 10 Kerala Syllabus Question 1.
What is the volume of a square pyramid of base edge 10 centimetres and slant height 15 centimetres?
Answer:
a = 10, l = 15
Volume of pyramid =

Sslc Maths Chapter Solids Kerala Syllabus Chapter 8 Question 2.
Two square pyramids have the same volume. The base edge of one is half that of the other. How many times the height of the second pyramid is the height of the first?
Answer:

The height of the second pyramid is 4 times the height of the first pyramid.

Solids Class 10 Kerala Syllabus Chapter 8 Question 3.
The base edges of two square pyramids are in the ratio 1:2 and their heights in the ratio 1:3. The volume of the first is 180 cubic centimetres. What is the volume of the second?
Answer:

Solids In Maths Sslc Kerala Syllabus Chapter 8 Question 4.
All edges of a square pyramid are 18 centimetres. What is its volume?
Answer:
Length of base edge a = 18 cm

Class 10 Maths Solids Kerala Syllabus Chapter 8 Question 5.
The slant height of a square pyramid is 25 centimetres and its surface area is 896 square centimetres. What is its volume?
Answer:
l = 25 cm
Surface area = 896 cm
a2 + 2al = 896
a2 + 2a × 25 = 896
a2 + 50a – 896 = 0

Sslc Maths Chapter 8 Solutions Kerala Syllabus Question 6.
All edges of a square pyramid are of the same length and its height is 12 centimetres. What is its volume?
Answer:

Sslc Maths Solids Questions Kerala Syllabus Chapter 8 Question 7.
What is the surface area of a square pyramid of base perimeter 64 centimetres and volume 1280 cubic centimetres?
Answer:
Base perimeter 4a = 64
a= 16 cm

Surface area = a2 + 2al
= 162 + 2 × 16 × 17 = 256 + 544 = 800 cm²

Textbook Page No. 198

Maths Solids Class 10 Kerala Syllabus Chapter 8 Question 1.
What are the radius of the base and slant height of a cone made by rolling up a sector of central angle 60° cut out from a circle of radius 10 centimetres?
Answer:
R = Radius of the circle

Radius of cone = 1.66 cm
Radius of circular part = slant height of cone = 10 cm.

Sslc Solids Solutions Kerala Syllabus Chapter 8 Question 2.
What is the central angle of the sector to be used to make a cone of base radius 10 centimetres and slant height 25 centimetres?
Answer:
Central anglejof the sector
(x) = \(\frac { r }{ l }\) × 360,
r = 10 cm, l = 25 cm
\(=\frac{10}{25} \times 360\)
= 144°

Solids Maths Questions Kerala Syllabus Chapter 8 Question 3.
What is the ratio of the base-radius and slant height of a cone made by rolling up a semicircle?
Answer:
Radius of bigger circle = R
Radius of smaller circle = r
Radius of circular base of the pyramid = r = \(\frac { R }{ 2 }\)
Ratio between radius and slant height

Textbook Page No. 199

Sslc Maths Solutions Kerala Syllabus Chapter 8 Question 1.
What is the area of the curved surface of a cone of base radius 12 centimetres and slant height 25 centimetres?
Answer:

r = 12cm l = 25cm
Curved surface area = πrl
π × 12 × 25 = 300π = π × 12 × 25
= 300 × 3.14 = 314 × 3
= 942 cm2

Kerala Syllabus 10 Maths Solutions Kerala Syllabus Chapter 8 Question 2.
What is the surface area of a cone of base diameter 30 centimetres and height 40 centimetres?
Answer:
Radius = \(\frac { 30 }{ 2 }\)= 15 cm =r
Height = h = 40 cm
Total surface area = Base area + Curved surface
area = πr2 + πrl

= 865 πr = 2718.6 m²

Std 10 Kerala Syllabus Maths Solutions Chapter 8 Question 3.
A Conical firework is of. base diameter 10 centimetres and height 12 centimetres, 10000 such fireworks are to be wrapped in colour paper. The price of the colour paper is 2 rupees per square metre. What is the total cost?
Answer:
r = 5 cm
h =12 cm
t = 13 cm
Total surface area of one firework = Base area + Curved surface area

Question 4.
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area.
Answer:
Perimeter of base of a cone is equal to half of perimeter of large cone.
Radius of pyramid = R/2
Perimeter of base

= 2 × Base area , that is twice

Textbook Page No. 200

Question 1.
The base radius and height of a cylindrical block of wood are 15 centimetres and 40 centimetres. What is the volume of the largest cone that can be carved out of this?
Answer:
r = 15cm, h = 40cm
Volume of cone = \(\frac { 1 }{ 3 }\) πr² h

= \(\frac { 1 }{ 3 }\) π × 15 × 15 × 40
= 3000π = 3000 × 3.14
= 314 × 30
= 9420 cm³

Question 2.
The base radius and height of a solid metal cylinder are 12 centimetres and 20 centimetres. By melting it and recasting, how many cones of base radius 4 centimetres and height 5 centimetres can be made?
Answer:

No. of cones = Volume of cylinder/Volume of cone Volume of cylinder
= π × 12 × 12 × 20
Volume of cone = \(\frac { 1 }{ 3 }\) π × 4 × 4 × 5

Question 3.
A sector of central angle 216° is cut out from a circle of radius 25 centimetres and is rolled up into a cone. What are the base radius and height of the cone? What is its volume?
Answer:

Question 4.
The base radii of two cones are in the ratio 3:5 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?
Answer:

Question 5.
Two cones have the same volume and their base radii are in the ratio 4:5. What is the ratio of their heights?
Answer:

Textbook Page No. 203

Question 1.
The surface area of a solid sphere is 120 square centimetres. If it is cut into two halves, what would be the surface area 0f each hemisphere?
Answer:
Surface area of the solid sphere

Surface area of one hemisphere = 3 πr²
3 πr² = 3 π × \(\frac { 30 }{ π }\) = 90 cm²

Question 2.
The volumes of two spheres are in the ratio 27 : 64. What is the ratio of their radii? And the ratio of their surface areas?
Answer:
Ratio of volumes

Question 3.
base radius and length of a metalder are 4 centimetres and 10 centimetres, If it is melted and recast into spheres of radius 2 centimetres each, how many spheres can be made?
Answer:

Question 4.
A metal sphere of radius 12 centimetres is melted and recast into 27 small spheres. What is the radius of each sphere?
Answer:
Radius = 12cm
Volume of bigger sphere
\(=\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \times 12^{3}\)
If the radius of smaller sphere is ‘r’
Volume of 27 smaller spheres = Volume of the bigger sphere

Question 5.
From a solid sphere of radius 10 centimetres, a cone of height 16 centimetres is carved out What fraction of the volume of the sphere is the volume of the cone?
Answer:
Radius of sphere = 10 cm
Radius of cone

Question 6.
The picture shows the dimensions of a petrol tank.

How many litres of petrol can it hold?
Answer:
Length of circular cylinder = 4 m
Height = 4 m
Radius = 1m
Volume of circular cylinder = π × 12 × 4 = 4π
= 4 × 3.14 = 12.56 cm³
Volume of two hemisphere = Volume of a sphere

Litres of petrol the tank can hold = 12.56 + 4.19 = 16.73 m³ = 16750 litre

Question 7.
A solid sphere is cut into two hemispheres. From one, a square pyramid and from the other a cone, each of maximum possible size are carved out. What is the ratio of their volumes?
Answer:

Solids Orukkam Questions & Answers

Worksheet 1

Question 1.
The base edge of a square pyramid is Stem, height 3cm. Calculate slant height and lateral edge
Answer:
Base edge = 8 cm, height = 3 cm

Question 2.
Slant height of a square pyramid is 10cm, height 6cm .Calculate total length of the edges.
Answer:
Slant height =10 cm, height = 6 cm

Question 3.
The slant height of a square pyramid is 12 cm, lateral edge 13 cm. Calculate height
Answer:
Slant height = 12 cm
lateral edge = 13 cm

Question 4.
The length of base edge is 24 cm, slant height 13 cm. Find height and lateral edge
Answer:
Base edge = 24 cm
Slant height = 13 cm
Height \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{25}=5 \mathrm{cm}\)
Length of lateral edge = \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{313} \mathrm{cm}\)

Worksheet 2

Question 5.
A sector is folded in such a way as to get a cone. Radius of the sector is 12 cm, central angle 120°.Calculate radius and slant height
Answer:
Slant height of cone = radius of sector = 12 cm
Radius of cone = \(\frac { 120 }{ 360 }\) of radius of sector = 12 × \(\frac { 120 }{ 360 }\) = 4 cm

Question 6.
The central angle of a sector is 90°, radius 16cm, calculate slant height and radius
Answer:
slant height of cone = 16 cm

Question 7.
Slant height of a cone is 20 cm, radius 10 cm. What should be the radius and central angle of the sector?
Answer:
Radius of sector = slant height of cone = 20 cm
Central angle of the sector = radius of cone \(\times \frac{360}{R}=10 \times \frac{360}{20}=18^{\circ}\)

Question 8.
Radius of a cone is 4cm, slant height is 5/2 times radius. Calculate the radius and central angle of the sector.
Answer:
Radius of sector = slant height of cone =

Worksheet 3

Question 9.
The base edge of a square pyramid is 6cm, height 4cm, calculate slant height and total surface area.
Answer:
Length of base edge = 6 cm, height = 4 cm,
slant height = \(\sqrt{(3)^{2}+(4)^{2}}=\) \(\sqrt{9+16}=\sqrt{25}=5 \mathrm{cm}\)
Total surface area = base area + curved surface area=(6)2 + 2 × 6 × 5 = 36 + 60 = 96 cm².

Question 10.
The height of a square pyramid is 12cm, slant height 15cm , calculate total surface area and volume
Answer:
height =12 cm. slant height = 15 cm
\(a=2 \sqrt{15^{2}-12^{2}}\) = 2 × 9 = 18 cm²
Total surface area = base- area + curved surface area
= (18)2 + 2 × 18 × 15 = 324 + 540 = 864 cm²
Volume = \(\frac{1}{3}(18)^{2} \times 12=1296 \mathrm{cm}^{3}\)

Question 11.
The base perimeter of a square pyramid is 48cm. Slant height is 10cm. Calculate lateral surface area and volume.
Answer:
Base perimeter = 48cm
baseedge = \(\frac { 48 }{ 4 }\) = 12 cm, slant height = 10cm
height = \(\sqrt{(10)^{2}-(6)^{2}}=\sqrt{64}=8 \mathrm{cm}\)
Curved siuface area= 2 × 12 × 10 = 240 cm²
Volume = \(\frac { 1 }{ 3 }\) (12)2 × 8 = 384 cm³

Question 12.
The height of a square pyramid is 15cm, volume 1620cm. Calculate the total surface area.
Answer:
Volume of square pyramid =

Worksheet 4

Question 13.
The base area of a cone is 25 π cm, curved surface area 165 π. Calculate total surface area.
Answer:
Total surface area of cone = base area + curved surface area
= 25 π +165 π = 190 cm²

Question 14.
Base area of a cone is 81 π, height 12 cm. Calculate volume
Answer:
Volume of cone = \(\frac { 1 }{ 3 }\) × base perimeter ×
height = \(\frac { 1 }{ 3 }\) × 81 π × 12 = 324 π cm³

Question 15.
The height of a cone is 4cm, slant height 5cm. Calculate total surface area
Answer:
Height of cone = 4 cm
Slant height = 5 cm

Surface area = π (3)² + π x 3 x 5 = 9 π + 15 π =24 π cm²

Question 16.
Radius of a cone is 10cm, volume 3140 cubic centimeter. Calculate total surface area
Answer:
Radius of cone = 10 cm

Question 17.
Calculate the surface area and volume of a sphere of radius 3 cm.
Answer:
Surface area of sphere = 4 π (3)² = 36 π cm²
Volume = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \pi \times(3)^{3}=36 \pi \mathrm{cm}^{3}\)

Workshee 5

Question 18.
Calculate the volume of a sphere of surface area 144 π square centimetre.
Answer:

Solids SCERT Questions & Answers

Question 19.
The measurements of the lateral surface of a square pyramid are shown in the figure. Calculate die base edge and slant height of die pyramid. [Score: 2, Time: 3 minute]

Answer:
Base edge = 10cm (1)
The given figure can be divided into n two right-angled triangles their angles are 30°, 60°, 90°, so ratio of their side will be 1: √3: 2.
2x = 10, x = 5
Slant height = 5 √3 cm (1)³

Question 20.
Is It possible to construct a pyramid of base edge 24 cm and lateral edge 13 cm? Justify [Score: 2, Time: 3 minute]
Answer:
Since slant height is 5 cm, such a pyramid can’t be constructed. (1)
Slant height should be greater than half of the base edge. \(\sqrt{13^{2}-12^{2}}=5 \mathrm{cm}\) (1)

Question 21.
Lateral surface of a square pyramid is shown in the Figure. All angles are equal

Find the total length of all edges of the square: pyramid. Find the slant height What is the |atio between height and slant height [Score: 4, Time: 5 minutes]
Answer:
Sum of edges = 8 x 8 = 64 cm (1)
Slant height = 4√3 cm (1)

Question 22.
Devikamade a square pyramid having base edge 40cii and height 15cm. Unfortunately, one lateral face got separated from the pyramid. Check which figure given below shows the isosceles triangle that got separated. [Score: 3, Time: 5 minute]
Answer:

Square pyramid can’t be constructed since slant height 1 should be greater than half the base edge. (1)

If base edge = 40 and slant leight = 35, then height can’t be 15 (1)

Here height of pyramid is 15, so this is the isosceles triangle that got separated (1)

Question 23.
A tent constructed in the form of a square pyramid of base perimeter 80 metres and lateral edge 26 metres
a. Calculate the slant height of the tent
b. Calculate the area of tarpaulin sheet. required to cover the lateral faces of the tent. [Score: 3, Time: 5 minute]
Answer:
Base perimeter = 80 m, Base edge= 20 m Lateral edge = 26 m

= 960 sq.metre (2)

Question 24.
The triangle given in the figure is one lateral face of a square pyramid.
a. Calculate the slant height.
b. Find the lateral surface area of the pyramid, [Score: 3, Time: 4 minutes]

Answer:
Slant height \(=\sqrt{17^{2}-8^{2}}\)
\(=\sqrt{275}=15 \mathrm{cm}\) (2)

Question 25.
A square pyramid is made from a solid cube having edge 30cm. Calculate the surface area. [Score: 3, Time: 5 minute]
Answer:
Base edge of the square pyramid = 30 cm
Height = 30cm
Slant height = \(\sqrt{30^{2}+15^{2}}\)
\(=\sqrt{900+225}=\sqrt{1125}=15 \sqrt{5}\) (1)
Lateral surface area = \(=4 \times \frac{1}{2} \times 15 \sqrt{2} \times 30\)
= 60 x 15√5 = 900√5 sq.cm (1)
Total surface area = 900 + 900√5
= 900(1 + √5) sq.cm (1)

Question 26.
The lateral faces of a square pyramid are equilateral triangles Lateral, edge = 20 cm
a. Calculate the slant height
b. Find its surface area.
C. Find its volume. [Score: 5, Time: 6 minute]
Answer:
a. Slant height = 10 √3 cm (1)
b. Lateral surface area

Question 27.
Prove that the ratio between the base edge, slant height and height of a square pyrarmid having equal edges is 2: √3 : √2. [Score: 4, Time: 5 minute]
Answer:
Slant height = √3 a
Height = \(\begin{array}{l}{=\sqrt{(\sqrt{3} a)^{2}-a^{2}}} \\ {=\sqrt{2} a}\end{array}\)
Base edge: Slant height : Height
= 2a: √3a : √2 a = 2: √3 : √2 (1)

Question 28.
The ratio between the base edges of two square pyramids is 1: 2. The heights are also in the same ratio. If the volume of the first pyramid islO cubic centimeters, calculate the volume of the sec ond one. [Score: 3, Time: 4 minute]
Answer:
v₁ : v₂ = 1 : 8
v₁ = \(\frac { 1 }{ 3 }\) a₂ h
v₂ = \(\frac{1}{3}(2 a)^{2} \times 2 h, v_{1}: v_{2}=1: 8\) (1)
Volume of the second pyramid = 800 cm² (1)

Question 29.
Meera constructed a square pyramid of base edge 10cm and height 6cm Manu made a square pyramid having base edge 5cm and height 4cm. Find the volume of the pyramids and compare the measurements. [Score: 3, Time: 4 minutes]
Answer:
Volume of Meera’s pyramid = \(\frac { 1 }{ 3 }\) x Baste edge x height 3
= \(\frac { 1 }{ 3 }\) x 102 x 6 =200 cm3 (1)
Volume of Manu’s pyramid = \(\frac { 1 }{ 3 }\) x 52 x 24 = 200cm³
Volumes are equal (1)

Question 30.
The central angle of a sector is 288° If this sector is rolled up to make a cone, find the ratio between the radius and slant height of the cone. [Score :: 4, Time: 5 minutes]
Answer:
360 x \(\frac { 4 }{ 5 }\) = 288
∴ Radius of the cone 4 = \(\frac { 4 }{ 5 }\) x radius of the big circle (1)
∴ If r is the radius of the circle Radius of the
Radius of the cone = \(\frac { 4 }{ 5 }\) r (1)
But radius of the circle = slant height of cone
i.e., 1 = r (1)
∴Ratio between the radius of the cone and slant height
\(=\frac{4}{5} r: r=\frac{4}{5}: 1=4: 5\) (1)

Question 31.
The ratio between the radius and slant height of a cone is 2 : 3. Find the central angle of the sector to make the cone. [Score: 3, Time: 4 minutes]
Answer:
Ratio between the radius and slant height 2 : 3 (1)
Area length of the sector is equal to \(\frac { 2 }{ 3 }\) part of the circle perimeter. (1)
Central angle of the sector = 360 x \(\frac { 2 }{ 3 }\) = 240° (1)

Question 32.
The central angle of a circle is divided in the ratio 2 : 3 to form two sectors. Two cones are made by rolling up the two Rectors.
a. Find out the ratio between the base perimeters of the cones.
b. What is the ratio between the curved surface areas. [Score: 3, Time 6 minutes]
Answer:
a. Central angles are in the ratio 2 : 3 , so, let the perimeter of the two clones which made by rolling up this two sectors be \(\frac { 2 }{ 5 }\)
part and \(\frac { 3 }{ 5 }\) part of the perimeter of the circle (1)
That is perimeter of each sector be \(2 \pi r \times \frac{2}{5} \text { and } 2 \pi r \times \frac{3}{5}\)
Ratio between base perimeters of cone = \(2 \pi r \times \frac{2}{5}: 2 \pi r \times \frac{3}{5}=2: 3\)
b. Base perimeter of the cones will be \(\frac { 2 }{ 5 }\) and \(\frac { 3 }{ 5 }\) parts of the circumference of the 5 circle. (1)
Ratio between the perimeters of die cones = \(\pi r^{2} \times \frac{2}{5}: \pi r^{2} \times \frac{3}{5}=2 ; 3\) (1)

Question 33.
Find the ratio between the radius and slant height of a cone by roiling up a sector with central angle 120°. If the curved surface area is 108 π, find the radius and slant height of the cone. [Score:5,Time:7minutesv]
Answer:
Area of the sector with central angle 120° is one-third of the area of the circle. (1)
Curved surface area = 108 π
Ares of die; sector = 108 π,
which is one-third of the area of the circle Area of the circle = 108 π × 3
πr² = 324 π (1)
Radius of the circle r = 18 cm
Slant height = 18 cm (1)
Radius of the cone = 6 cm (1)

Question 34.
A wooden cone is has radius 30crn and height 40crn. Find its slant height. Calculate the cost to paint the face of 10 such cones at the rate of Rs.50/- per square metre. [Score: 5, Time: 7 minutes]
Answer:
Base Radius = 30 cm
Height =40 cm
Slant height = \(\sqrt{40^{2}+30^{2}}=50\) (1)
Surface area of the cone = πr² + πrl = π × 302 + π × 30 × 50 (1)
= 900π + 1500π = 2400π (2)
Total cost to paint 10 cones \(=\frac{2400 \pi \times 10 \times 50}{10000}\)
= \(=\frac{2400 \times 3.14 \times 10 \times 50}{10000}=377 \mathrm{Rs}\) (1)

Question 35.
Two cones are made using two sectors of central angles 60° and 120° of a circle. If the radius of the smaller cone is 5cm
a. Calculate the radius and base area of the smaller cone.
b. Find the surface area of the bigger cone. [Score: 5, Time: 8 minutes]
Answer:
a. Central angle of the small sector = 60°
\(\frac { 1 }{ 6 }\) part of the area of the circle Base radius of cone formed from above sector = 5 cm (1)
Radius of the circle = 5 × 6 = 30 (1)
Similarly, area of the sector of central angle 120° = \(\frac { 1 }{ 3 }\) of the area of the circle Base radius ofthe bigger cone = 30 × \(\frac { 1 }{ 3 }\) = 10 (1)
Base area of the bigger cone = π × 102 = 100× (1)
b. Curved surface area ofthe bigger cone = π × 10 × 30 = 300π (1)
Surface area = 100π + 300π = 400π cm² (1)

Question 36.
Three solids a square pyramid, a cone and a sphere have been carved out from three solid cubes of the same size. Find the volume of each solid. [Score: 5, Time: 7 minutes]
Answer:
Volume of square pyramid = \(\frac { 1 }{ 3 }\) a³ (1)
Volume of cone

Question 37.
A metal sphere is melted and recasted into a cone. Both have same radii
a. Find the relationship between the height of the cone and the radius of the sphere.
b. Which solid has greater surface area? Justify. [Score: 5, Time: 7 minutes]
Answer:
a. If r is the radius of the sphere, then its volume = \(\frac { 4 }{ 3 }\) πr³ (1)
If h is the height of the cone, then its volume = \(\frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi r^{3}\)
h = 4r (1)
Height of pyramid is four times the radius of sphere.
Surface area of the sphere = 4πr²
Slant height of the cone = \(\sqrt{(4 r)^{2}+r^{2}}=\sqrt{17 r^{2}}=\sqrt{17} r\) (1)
Surface area of the cone = \(\pi r^{2}+\pi r \sqrt{17} r=\pi r^{2}(1+\sqrt{17})\)
b. .Surface area of the cone is greater (1)

Question 38.
A hemisphere and a cone with same radii are attached to get a solid as given in the figure. Radius of the hemisphere is 9 cm. The height of the two solids together is 21 cm.
a. Find the height of the cone.
b. Find the volume of the cone
c. Find the volume of the solids [Score: 4, Time: 6 minutes]
Answer:
Height of the cone 21 – 9 = 12 cm (1)

Solids Exam Oriented Questi0ns& Answers

Short Answer Type Questions (Score 2)

Questi0n 39.
Total surface area of a solid hemisphere is 675 π sqcm. Find the curved surface area of the solid hemisphere.
Answer:
3 πr² = 675π cm²
r2 = 225
The CSA of the solid hemisphere,
CSA = 2πr² = 2π × 225 = 450π cm²

Questi0n 40.
The volume of a solid right circular cone is 4928 cm3. If its height is 24 cm, them find the radius of the cone.
Answer:
V = 4928 cm³ and h = 24 cm

Questi0n 41.
The figure given below has the total length 20cm height and common diameter 6cm. Find the volume of the figure.

Answer:
volume of solid

Questi0n 42.
The circular plate of radius 12cm is cut out into six sectors having same size. Calculate the slant height and radius of circular cone used to make one sector.
Answer:
slant height = 12cm
centre angle = \(\frac { 360 }{ 60 }\) = 60
radius of square pyramid = 12 × \(\frac { 60 }{ 360 }\) = 2 cm

Questi0n 43.
A circus tent is in the shape of a square pyramid. The area of the base is 1600m² and its height is 375m then, how much canvas would be needed for this and also find its perimeter?
Answer:

Questi0n 44.
Height of a cone is 40cm. Slant height is 41cm.
a. Find diameter of its base,
b. Find volume
Answer:

Questi0n 45.
Radius and slant height of a solid right circular cone are 35cm and 37 cm respectively. Find the curved surface area and total surface area of the cone.
Answer:
r= 35 cm
l = 37 cm
CSA = πrI = π(35 x 37)=4070 cm²
TSA = πr (1 + r )
\(=\frac{22}{7} \times 35 \times(37+35)=7920 \mathrm{cm}^{2}\)

Questi0n 46.
Surface area of a wooden sphere is 40 cm2. It is cut into two identical hemispheres. Find
a. The area of the plane surface of one of the hemispheres,
b. Its surface area.
Answer:
a. Surface area of the sphere = 4 πr²
Here, 4 πr² = 40cm²
πr² = 10cm²
∴ Area of plane surface = πr² = 10cm²
b. Surface area of one piece (hemisphere)
= 3 πr² = 3 x 10 = 30cm²

Short Answer Type Questions (Score 3)

Questi0n 47.
A toy in the shape of a square pyramid has base edge 16 cm and slant height 10 cm. 500 of these are to be painted and the cost is 80 rupees per square meter. What would be the total cost?
Answer:
Surface area = Curved surface area + base area
4 × 1/2 × 16 × 10+ 162 = 320 + 256 = 576cm
Total surface area of 500 square prism is = 500 × 576 = 288000 cm²

Questi0n 48.
The base radius of a circular cone is 9cm and its height is 12cm. What is the radius and central angle of die sector used to make it?
Answer:
base radius r =9 cm
height h =12cm
slant height l = \(\begin{array}{l}{=\sqrt{12^{2}+9^{2}}=\sqrt{144+81}} \\ {=\sqrt{225}=15 \mathrm{cm}}\end{array}\)
radius of the sector = 15cm
centre angle of sector = 360 × \(\frac { 9 }{ 15 }\) = 216°

Questi0n 49.
The central angle of a sector is 120°. What is the ratio of radius and slant height of a circular cone made by it? What is the radius and slant height of a cone if its curved surface area is 108 π cm².
Answer:
The ratio of radius and slant height = \(\frac { 120 }{ 360 }\) = \(\frac { 1 }{ 3 }\) = 1 : 3
curved surface area (πrl) =108π
rl=108; r × 3 r = 108
3 × r² = 108; r² = 36
radius (r) = 6 cm; slant height (l) = 3r = 3 × 6 = 18cm

Questi0n 50.
For constructing a square pyramid, Rabiya cut of four triangles and a square. Figure given below shows the measures of these triangles and square. Can you make a square pyramid by using these measures? Explain the reason.

Answer:
Base edge = 42 cm
Slant edge = 29 cm
Slant height = \(\sqrt{29^{2}-\left(\frac{42}{2}\right)^{2}}=\sqrt{29^{2}-21^{2}}\)
\(\sqrt{841-441}=\sqrt{400}=20 \mathrm{cm}\)
Slant height is less than the half of the base edge 21cm so not possible for making a square pyramid

Long Answer Type Questions (Score 4)

Questi0n 51.
A sector shown in the figure is rolled up and made a cone. Find its
a. Slant height
b. Base radius
c. Volume

Answer:
a. Slant height = radius of the
sector = 30cm


\(\frac{2000 \sqrt{2 x}}{3} \mathrm{cm}^{3}\)

Questi0n 52.
Paddy is filled in a cylindrical shaped vessel. Then it has the following shape. How many litres of paddy does it contain?

Answer:

Questi0n 53.
A petrol tank is in the shape of a cylinder with hemisphere of the same radius as the base of the cylinder attached to both ends. If the total length of the tank is 5 meters and the base radius of the cylinder is 1 metre, how many litres of petrol can it hold?
Answer:
Volume of cylinder = πr²h = 3πcm³
r=1, h=3
Volume of hemisphere = 2/3 πr³

Long Answer Type Questions (Score 5)

Questi0n 54.
A toy is in the shape of a hemisphere attached to one end of the cone Total height of the toy is 14.5cm, and common diameter is 7cm.
a. Draw a rough figure based on this fact,
b. Find the volume of the toy.
Answer:

Questi0n 55.
Prove that:
i. ∠BAT = ∠BPA
ii. ∠BAS = ∠AQB

Answer:

Solids Memory Map

Students can Download Chemistry Chapter 1 Periodic Table and Electronic Configuration Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

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SCERT 10th Standard Chemistry Textbook Chapter 1 Solutions Malayalam Medium

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SSLC Maths Chapter 5 Trigonometry Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 5 Solutions Malayalam Medium

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SSLC Chemistry Chapter 6 Nomenclature of Organic Compounds and Isomerism Questions and Answers Malayalam Medium

SCERT 10th Standard Chemistry Textbook Chapter 6 Solutions Malayalam Medium

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SSLC Physics Chapter 4 Reflection of Light Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 4 Reflection of Light Solutions

Text Book Page No. 80

Sslc Physics Chapter 4 Kerala Syllabus Question 1.

→ Which is the incident ray?
Answer:
AO

→ Which is the reflected ray?
Answers:
QB

→ Is there any relation between the angle of incidence and the angle of reflection?
Answer:
The angle of incidence and angle of reflection are equal.

→ Are the incident ray, reflected ray and normal to the mirror at the point of incidence in the different planes?
Answer:
In the same plane.

Reflection Of Light Class 10 Kerala Syllabus Question 2.

→ What difference is seen between the surfaces of the two objects?
Answer:
In the first figure the surface is smooth, in the second figure surface is rough.

→ Fig. 4.2(b) are the rays of light travelling parallel after reflection?
Answer:
No. When light falls on a rough surface, it undergoes an irregular reflection. This is scattered reflection.

Reflection of Light Class 10 Question 3.
In Fig. 4.2(a) regular reflection is depicted. Can you give a definition for such reflections by observing the figure?
Answer:
When light falls on a smooth surface, it undergoes an regular reflection, the rays of light travelling parallel after reflection This is regular reflection.

Text Book Page No. 81

Sslc Physics Chapter 4 Notes Kerala Syllabus Question 4.
Record in your science diary the following features about the images formed here.

→ The distance from the mirror to the object and the image from the mirror.
Answer: Equal

→ Is the image real or virtual?
Answer: Virtual

→ The size of the image.
Answer:
Same size as object.

Sslc Physics Reflection Of Light Kerala Syllabus Question 5.
How many images can we see at a time busying two mirrors?
Answer:
The number of images changes accordance with the relation between the angle between the mirrors.

Text Book Page No. 82

Sslc Physics Chapter 4 Reflection Of Light Kerala Syllabus Question 6.
Table 4.1.

Answer:

→ How many images can be seen when viewed from A and B?
Answer:
3

→ What if viewed from other positions in between the mirrors?
Answer:
3

→ How much is the angle between the mirrors?
Answer:
90°

→ What is the relation between the angle between the mirrors and the number of images?
Answer:
Number of images = \(n=\frac{360}{\theta}-1\)

Class 10 Physics Chapter 4 Reflection Of Light Kerala Syllabus Question 7.
Table 4.2.

Answer:

Text Book Page No. 83

Class 10 Physics Chapter 4 Kerala Syllabus Question 8.

Answer:

Text Book Page No. 84

Sslc Physics Chapter 4 Solutions Kerala Syllabus Question 9.
Draw a straight line on a table as shown in the figure. At one end of the line, place a concave mirror of focal length 20 cm. Mark principal focus (F) and center of curvature (C) on the line.

Fix a burning candle on the principal focus in such a way that it is at a slight distance from the center of curvature. Arrange a screen in such a way that a clear image is obtained on the screen.

→ What is the position and features of the image?
Answer:
Between F and C, small, real, inverted.

→ Observe the change in position of image and features on changing the position of the candle
Answer:

Physics Chapter 4 Class 10 Kerala Syllabus Question 10.
Table 4.4.

Answer:

Physics Class 10 Chapter 4 Kerala Syllabus Question 11.
Aren’t the focal length of \(\frac{u v}{u+v}\) the mirror and the average value of obtained from the table the same?
Answer:
Yes

Text Book Page No. 85

Sslc Physics Chapter 4 Reflection Of Light Notes Kerala Syllabus Question 12.
Record the measurements shown in the figure (4.6) using the New Cartesian Sign Convention.

→ Distance to the object from the mirror (u) =
Answer:
Negative.

→ Distance to the image from the mirror (v) =
Answer:
Negative.

→ Height of object (OB) =
Answer:
Positive.

→ Height of image (IM) =
Answer:
Negative.

Text Book Page No. 86

Kerala Syllabus 10th Standard Physics Chapter 4 Question 13.
The given figure shows the image formation by a concave mirror. Analyse the figure and write down different measures using New Cartesian Sign Convention.

Answer:

Sslc Physics Chapter 4 Notes Pdf Kerala Syllabus Question 14.
An object is placed in front of a concave mirror 20 cm away from it. If its focal length is 40 cm, locate the position of image and its nature.
Answer:
f = – 40 cm, u = – 20 cm
f = \(\frac { uv }{ u+v }\)
– 40 = \(\frac { – 20 }{ – 20 + v }\)
– 40 ( – 20 + v ) = – 20 v
– 20 + v = \(\frac { – 20 }{ – 40 }\) = \(\frac { 2 }{ v }\)
– 20 = [/latex] = \(\frac { v }{ 2 }\) – v = – \(\frac { v }{ 2 }\)
∴ – \(\frac { v }{ 2 }\) = – 20
v = 40 cm
Features of image:
At behind the mirror, very large, virtual, direct.

10th Physics Chapter 4 Kerala Syllabus Question 15.
Find out the height of object (ho), height of image (hi), position of object(u) and position of image (v) using New Cartesian Sign Convention and tabulate them. (The height of image hi can be directly measured by fixing a graph paper on the screen).
Answer:

Text Book Page No. 87

Physics Class 10 Chapter 4 Kerala Syllabus Question 16.
\(\frac{h_{i}}{h_{o}}\) is magnification. Does it have any relational with the value of \(\frac{v}{u}\) ?
Answer:
\(m=\frac{-h_{i}}{h_{o}}=\frac{-v}{u}\)

Text Book Page No. 88

Question 17.
An object is placed 8 cm away in front of a concave mirror of focal length 5 cm. Find out the position of image and magnification. Find out whether the image is inverted or erect by drawing the ray diagram on a graph paper.
Answer:
f = – 5 cm, u = – 8 cm

Features of images : Beyond C, big, real, inverted

Question 18.
What are the features of an image that is obtained from magnification?
Answer:

• When magnification is 1, the size of the image and the size of the object are equal.
• When magnification is more than 1, the size of the image is greater than the size of the object.
• When magnification is less than 1, the size of the image is smaller than the size of the object.
• When the magnification is positive, image is real and inverted.
• When the magnification is negative, image is virtual and erect.

Question 19.
Observe the given figures and complete the table using New Cartesian Sign Convention.

Answer:

Text Book Rage No. 89

Question 20.

From the above table, find out which mirror always gives an erect and diminished image and write it down.
Answer:
The image formed by a convex mirror is always erect and diminished.

Question 21.
Why it is written on rearview mirrors that “Objects in the mirror are closer than they appear”.
Answer:
The image formed by a convex mirror is always erect and diminished. Hence the driver who sees the image of vehicles on the mirror develops a feeling that the vehicles coming from behind are at a greater distance. This may turn out to be dangerous.

Reflection of Light Let Us Assess

Question 1.
A dental doctor uses a mirror of focal length 8 cm. To see the teeth clearly what should be the maximum distance between the teeth and the mirror? Justify your answer. Which type of mirror has been used by the doctor?
Answer:
The dental doctor uses a concave mirror. Effect and enlarged image can be obtained using the mirror. Such images are formed when the object is kept in between the main focus and pole of a concave mirror. So the minimum distance between the mirror and the teeth must be in 8 cm to view the teeth clearly.

Question 2.
Imagine that a spherical mirror gives an image magnified 5 times at a distance 5 m. If so determine whether the mirror is concave or convex. How much will be the focal length of the mirror?
Answer:

Image formed is larger than the object, so the used mirror is concave.

Question 3.
A motorcyclist observes a car coining from behind with a magnification 1/6. If the actual distance between the car and the bike is 30 m calculate the radius of curvature of the mirror.
Answer:

Question 4.
A shaving mirror of focal length 72 cm is kept in a beauty clinic. A man uses it standing 18 cm away from the mirror. At what distance will the image be formed? Is the image real or virtual? What is the magnification of the image?
Answer:
f =-72 cm, u = -18cm

Image formed on the 24 cm away from the mirror. Virtual image.
Magnification.
m = \(\frac{-v}{u}=-\frac{24}{18}\) = – 1.33
This is a concave mirror.

Question 5.
Wrap a rubber ball of diameter 12 cm completely with an aluminum foil and make the surfaces smooth. Where will be the image of an object kept 12 cm away from the center of the ball? Is the image real or virtual?
Answer:
Surface of a rubber ball is similar to a convex mirror.
Diameter = 12 cm
⇒ Radius = 6 cm
2f = R
f = \(\frac { R }{ 2 }\) = \(\frac { 6 }{ 2 }\) = 3 cm
u = \(\frac { 12 }{ 2 }\) = 6 cm

Mirror is a convex, so image is virtual.

Question 6.
We are able to read a book since light falling on a surface gets reflected from the book and reaches the eye. But on such occasions, we cannot see our images like that from a mirror. Explain why?
Answer:
Irregular reflection takes place in the book. So images not formed.

Question 7.
Is the image formed by a plane mirror real or virtual? Write an instance when such a mirror gives an inverted image.
Answer:
Image formed by a plane mirror is always direct and virtual. The image will be at the same distance at the object.

If the objects placed erect as shown in the figure then the image will be inverted.

Reflection of Light Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
1. Classify the following statements as to those related to concave mirrors and convex mirrors and tabulate them accordingly.
a. to view the face.
b. as makeup mirror.
c. as reviewer mirrors in vehicles.
d. in solar concentration.
e. in periscopes.
f. as shaving mirror.
Answer:
Concave mirror:

• In solar concentration.
• Makeup mirror.
• Shaving mirror.

Convex mirror:

• In rearview mirrors of vehicles.
• In Searchlights.

Plane mirror:

• In periscopes.
• To see face.

Question 2.
Find out the relation and All in the blanks.
Small, virtual and direct images:
Convex mirror.
Size same as object, virtual and direct image :………………
Answer:
Plane mirror.

Question 3.
Which mirror gives rectifiable whatever the distance between the object and mirror is
i. Plane mirror.
ii. Convex mirror.
iii. Concave mirror.
iv. Plane or convex mirror.
Answer:
Plane or convex mirror.

Question 4.
Relation between angle of incidence (i) and angle of reflection (r) of light is………..
i. i > r
ii. i < r
iii. i = r
iv. None of these
Answer:
iii. i = r

Question 5.
Find the odd one in the group.
More scattered reflection, virtual image, direct image, large image
Answer:
Large image.

Very Short Answer Type Questions (Score 2)

Question 6.
Radius of curvature of a concave mirror is 24 cm. Then find the focal length of the mirror?
Answer:
R = 24 cm
f = \(\frac{R}{2}=\frac{24}{2}\) = 12 cm

Question 7.
Find the radius of curvature of a convex mirror whose focal length is 0.6 m.
Answer:
f = \(\frac{\mathrm{R}}{2}\)
∴ R=2f
0.6 = \(\frac{R}{2}\)
R = 2 × 0.6 = 1.2 m

Question 8.
Write the characteristics of the image formed by an object placed at the center of curvature of a concave mirror?
Answer:
Position: At the center of curvature at the same side
Size: Same as the size of the object
Nature: Real, Inverted

Question 9.
i. Which are the different types of mirror?
ii. What are the peculiarities are images obtained in the face mirror?
Answer:
i. Convex, concave and plane mirror.
ii. Image will be formed behind the mirror and will be equal distance from the mirror to the object. It will be direct and virtual.

Question 10.
State whether the below-given statement are true or false. If false correct them.
i. Real images are always produced by concave mirror.
ii. The image formed from a convex mirror is direct and enlarged.
Answer:
i. False. If the position of object is between focus and pole, virtual image is produced by concave mirror.
ii.False. The image formed by convex mirror will be smaller than the object and erect.

Very Short Answer Type Questions (Score 3)

Question 11.
Write three differences of real image and virtual image which is made by spherical mirrors.
Answer:
Real images:

• Inverted.
• Can be shown on a screen.
• Can measure the length of the image and distance to the image.

Virtual images:

• Cannot be taken on a screen.
• Not able to be measured direct.

Question 12.
Write the uses of concave mirror?
Answer:

• Used as a shaving mirror.
• Used as a makeup mirror.
• Doctors used as head mirrors.
• On film projectors.

Question 13.

a. Examine the figure and find the magnification.
b. What is the height of the object if height of the image is 4 cm when the object is placed on the same position in front of the mirror.
Answer:
a. Magnification = \(=\frac{\mathrm{hi}}{\mathrm{ho}}=\frac{– 10}{5}=– 2\)

Question 14.
Write the uses of convex mirror?
Answer:

• As reflectors in street lights.
• As rear view mirrors in vehicles.
• In searchlights.

Question 15.
Examine the position of the object given in the figure and table the following peculiarities.

a. Position of image.
b. Size of image.
c. Features of image.
Answer:
a.Behind the mirror.
b.Larger than obj etc.
c. Direct, virtual.

Question 16.
When an object of height 4 cm is placed in front of a concave mirror an image of height 8 cm is formed. Find the magnification.
Answer:
Magnification
hi = 4 cm, ho =  – 8 cm
Magnification m = \(\frac { hi }{ h0 }\) = \(\frac { –8 }{ 4 }\) = – 2

Very Short Answer Type Questions (Score 4)

Question 17.
Complete the table.

Answer:
a. Passes through principal focus.
b.Seem to come from the principal focus.
c, d. Returns parallel to the principal axis.
e, f. Returns through the same path.
g, h. Reflects in the same angle of incident ray.

Question 18.
Complete the ray diagram.


Answer:

Question 19.
Complete the table.

Answer:
a. 30°
b. 40°
c. 50°
d. 60°

Question 20.
An object of 5 cm is kept in front of a convex mirror having radius of curvature 30 cm at a distance of 10 cm. Calculate the position, size, and feature of object.
Answer:
u = – 10 cm
v = ?
Focal length = \(\frac{r}{2}=\frac{30}{2}\) = + 15 cm

The image is formed 60cm behind the concave mirror. The’ image will be virtual and direct.

You can Download Geometry and Algebra Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 9 Geometry and Algebra Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 9 Geometry and Algebra Notes

Textbook Page No. 213

Geometry and Algebra Class 10 Solutions Question 1.
What are the coordinates of the fourth vertex of the parallelogram shown on the right?

Answer:
Coordinate of the fourth vertex is = (6, 6)

Geometry and Algebra Class 10 Question 2.
The Figure shows a parallelogram with the coordinates of its vertices:
Prove that x₁ + x₃ = x₂ + x₄ and y₁ + y₃ = y₂ + y₄.

Answer:
Let P be the; point of intersection of diagonals of a parallelogram, which is the midpoint of AB and OC.

From eqn (1)
Comparing y coordinates

Geometry And Algebra Class 10 Chapter 9 Kerala Syllabus Question 3.
A parallelogram is drawn with the lines joining (x₁, y₁) aid (x_2, y₂) to the origin as adjacent sides. What are the coordinates of the fourth vertex?

Answer:
Coordinates of the fourth vertex = (x₁ + x₂, y₁ + y₂)
Let P be the point of intersection of diago¬nals of a parallelogram, which is the midpoint of AB and OC.

Sslc Maths Chapter 9 Kerala Syllabus Question 4.
Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals.
Answer:

OA² + OB² = x₁² + y₁²  + x₂²  + y₂²
OA = BC and OB = AC
OA² + OB² + OC² + AC² = 2(x₁² + y₁² + x₂² + y₂² )
AB² = (x₂ – x₁ )² + (y₂ – y₁ )²
OC² = (x₁ – x₂)² + (y₂ – y₃)²
OB² + OC² = 2(x₁² + x₂² + y₁² + y₂²)
OA² + OB² + AC² + BC² = AB² + OC²

Sslc Maths Geometry And Algebra Chapter 9 Kerala Syllabus Question 5.
In this picture, the midpoints of the sides of the large triangle are joined to make a small triangle inside. Calculate the coordinates of die vertices of the large triangle.
Answer:

textbook page no: 217

Sslc Geometry And Algebra Chapter 9 Kerala Syllabus Question 1.
A circle is drawn with the line joining (2, 3) and (6, 5) as diameter. What are the coordinates of the centre of the circle?
Answer:
Midpoint of the line joining (2, 3) and (6, 5) is

Sslc Maths Chapter 9 Solutions Kerala Syllabus Question 2.
The coordinates of two opposite vertices of a parallelogram are (4, 5) and (1, 3). What are the coordinates of the point of intersection of its diagonals?
Answer:
Midpoint of line joining the points (4, 5) and (1, 3) is

The coordinates of the point of intersec¬tion of its diagonal is (3, 4).

Geometry And Algebra Class 10 Solutions Chapter 9 Kerala Syllabus Question 3.
The coordinates of the vertices of a quadrilateral, taken in order, are (2, 1), (5, 3), (8, 7), (4, 9).
i. Find the coordinates of the midpoints of all four sides.
ii. Prove that the quadrilateral got by joining these midpoints is a parallelogram.
Answer:


Since the opposite sides of the quadrilateral PQRS are equal it is a parallelogram.

Geometry And Algebra Sslc Chapter 9 Kerala Syllabus Question 4.
In the figure, the midpoints of the large quadrilateral are joined to form the small quadrilateral within:
i. Find the coordinates of the fourth vertex of the smaller quadrilateral.
ii. Find the coordinates of the other three vertices of the larger quadrilateral.

Answer:

Midpoint of AB is P

Coordinate of the fourth vertex of smaller quadrilateral -(6, 2)
ii. Coordinates of other three vertices of the bigger quadrilateral (10, 3),(8, 7),(4, 5)

Geometry And Algebra Class 10 Kerala Syllabus Chapter 9 Kerala Syllabus Question 5.
The coordinates of the vertices of a triangle are (3, 5), (9, 13), (10, 6). Prove that this triangle is isosceles. Calculate its area.
Answer:
IfA (3, 5), B (9, 13), C(10, 6)
are the vertices of the triangle ABC

As BC = AC the triangle is an isosceles triangle

Geometry And Algebra Class 10 Pdf Chapter 9 Kerala Syllabus Question 6.
The centre of a circle is (1, 2) and a point on it is (3, 2). Find the coordinates of the other end of the diameter through this point
Answer:
Midpoint of the diameter

textbook Page no: 220

Geometry And Algebra Class 10 Questions And Answers Chapter 9 Kerala Syllabus  Question 1.
The coordinates of the points A and Bare (3, 2) and (8, 7). Find the coordinates of
i. the point P on AB with AP: PB = 2 : 3
ii. the point Q on AB with AQ: QB = 3: 2
Answer:

Geometry And Algebra Class 10 Scert Chapter 9 Kerala Syllabus Question 2.
Find the coordinates of the points which divide the line joining (1, 6) and (5, 2) into three equal parts.
Answer:

Coordinate of Q is \(\left(\frac{11}{3}, \frac{10}{3}\right)\)

Algebra Solutions Question 3. The coordinates of the vertices of a triangle are (-1, 5), (3, 7), (1, 1). Find the coordinates of its centroid.
Answer:
Centroid of the triangle

Question 4.
Calculate the coordinates of the point Pin the picture:

Answer:
Bisector of an angle divides the opposite side in the ratio of the sides containing the angle.

So, the coordinates of the point P which divides the line joining (4, 0) and (0, 3) in the ratio 4 : 3.
BP : PC = 4 : 3, so coordinate of P is Coordinate of x \(=4+\frac{4}{7}(0-4)=4-\frac{16}{7}=\frac{12}{7}\)
Coordinate of y \(=0+\frac{4}{7}(3-0)=0+\frac{12}{7}=\frac{12}{7}\)
\(\text { Coordinate of } P \text { is }\left(\frac{12}{7}, \frac{12}{7}\right)\)

Textbook Page No. 227

Question 1.
Prove that the points (1, 3), (2, 5), (3, 7) are on the same line.
Answer:
Coordinates of slope of a line, which joins the points (1, 3) and (2, 5) = \(\frac{5-3}{2-1}=2\)
Slope of a line, which joins the points (2, 5) and (3, 7) \(=\frac{7-5}{3-2}=2\)
both have same slope, so they make equal angles with the x-axis.
∴ They all lie on the same line.

Algebra 10th Standard Question 2. Find the coordinates of two more points on the line joining (-1, 4) and (1, 2).
Answer:
Change in x is 2 and y is -2
When change in x be 1 then change in y = \(\frac{-2}{2}=-1\)
Let change in x be 1 then change in y will be 3.
Coordinate of y = 4 + 3 x – 1 = 1
Point (2, 1)
Let change in x be 3 then change in ywillbe 4.
∴ Coordinate of y = 4 + 4 x – 1 = 0
∴ Point (3, 0)
Two points are (2, 1), (3, 0)

Question 3.
x₁, x₂, x₃, ……… and y₁, y₂, y₃, …….. are arithmetic sequences. Prove that all the points with coordinates in the sequence (x₁, y₁), (x₂, y₂), (x₃, y₃) ……… of number pairs are on the same line.
Answer:
If the line which contains the points (x₁, y₁), (x₃, y₃) also contains (x, y), then

Therefore, (x₁, y₁), (x₂, y₂) …………. will be in the same line

Question 4.
Prove that If the points (x₁, y₁), (x₂, y₂), (x₃, y₃) are on a single line, then the points (3x₁ + 2y₁, 3x₁ – 1y₁), (3x₂ + 2y₂, 3x₂ – 2y₂), (3x₃ + 2y₃, 3x₃ – 2y₃) are also on a single line. Would this be true if we take some other numbers instead of 3 and 2?
Answer:


Yes it is possible.

Textbook Page no: 231

Question 1.
Find the equation of the line joining (1, 2) and (2, 4). For points on this line with consecutive natural numbers 3, 4, 5, ……… as the x coordinates, what is the sequence of y coordinates?
Answer:
Equation of line joining the points (1, 2), (2, 4)

Question 2.
Find the equation of the line joining (–1, 3) and (2, 5). Prove that if the point (x, y) is on this line, so is the point (x+3, y+2).
Answer:

Question 3.
Prove that whatever number we take as, the point (x, 2x + 3) is a point on the line joining (–1, 1) and (2, 7).
Answer:

= \(\frac{2 x+3-7}{x-2}=\frac{2 x-4}{x-2}=2\)
The slopes are same hence die point (x, 2x+3) is passing through the points (–1, 1), (2, 7)

Question 4.
The x coordinate of a point on the slanted (blue) line is 3:
i. What it its y coordinates?
ii. What is the slope of the line?
iii. Write the equation of the line.

Answer:
As we get a right triangle on joining die three points’, with angles 30 : 60: 90 and ratio of their sides 1 : √3: 2

Question 5.
In the picture here, ABCD is a square. Prove that for any point on the diagonal BD, the sum of the x and coordinates is zero.

Answer:
B(2, -2), D(-2, 2) are the vertices of the diagonal BD: Sum of x,y coordinates of B = 2 + –2 = 0
Sum of x,y coordinates of D = –2 + 2 = 0
If (x, y)is a point on BD, then

Question 6.
Prove that for any point on the line intersecting the axes in the picture, the sum of the x and y coordinates is 3.

Answer:

∴ the sum of the and y coordinates of the point is 3.

Question 7.
Find the equation of the circle with center at the origin and radius 5. Write the coordinates of eight points on this circle.
Answer:
Equation of the circle
= (x – 0)² + (y – 0)² = 5²
= x² + y² = 25
x² = 25 – y²
x = \(=\sqrt{25-y^{2}}\)
y = 0 x = 5 (5, 0), (–5, 0)
y = 5 x = 0 (0, 5), (0, –5)
y = 3 x = 4 (4, 3), (4, 3)
y = 4 x = 3 (3, 4), (3, –4)
Coordinates of points on the circle
(5, 0), (–5, 0), (0, 5), (0, -5), (4, 3), (4, 3), (3, 4), (3, 4).

Question 8.
Prove that if (x, y) be a point on the circle with the line joining (0, 1) and (2, 3) as diameter, then x² + y² – 2x – 4y + 3 = 0. Find the coordinates of the points where this circle cuts they axis.
Answer:
As (0, 1), (2, 3) are the points joining the diameter of the circle, the center will be

Let (x, 0) be the coordinate which intersect
the x axis then, x² – 2x +3=0
\(x=\frac{2 \pm \sqrt{4-12}}{2}=\frac{2 \pm \sqrt{-8}}{2}\)
This circle can not be intersect the x axis. Let (0, y) be the coordinate which intersect the y axis then, y² – 4y + 3 = 0,
(y – 3)(y – 1) = 0 Coordinates are (0, 1) and (0, 3)

Question 9.
What is the equation of the circle shown here?

Answer:

Geometry and Algebra Orukkam Questions & Answers

Worksheet 1

Question 1.
If A(2, –1), B(3, 4), C(–2, 3) are the vertices of a triangle find the fourth vertex.
Answer:
Difference between x coordinates of A and B = 3 – 2 = 1, Coordinate of y = 4 + 1 = 5
x coordinates of D = –2 – 1 = -3
y coordinates of D = 3 – 5 = –2
Coordinates of D = (–3, –2)

Question 2.
In triangle ABC (4, 2) is the midpoint of AB. The midpoint of BC is (5, 4), the midpoint of AC is (3, 3). Find the vertices of the triangle.
Answer:

2x² = 12
∴ x² = 6
∴ y² = 3
From(1) y₁ = 4 – 3 = 1
x = 8 – 6 = 2, y₃ = 6 – 1 = 5
From(3) x₃ = 6 – 2 = 4 ,
Coordinates of A, B and C A(2, 1), B(6, 3), C(4, 5)

Question 3.
If A(2, –2), B(14, 10), C(11, 13) are the three vertices of a parallelogram ABC D write the coordinates of the fourth vertex
Answer:
A(2, –2), B(14, 10), C(11, 13).
Difference of x coordinates of A and B = 14 – 2 = 12.
x coordinate of D = 11 – 12 = 1.
Difference of y coordinates of A and B= 10 – (–2) = 12
y coordinates of D =13 – 12 = 1
Coordinates of D = (–1, 1)

Worksheet 2

Question 4.
Find the coordinates of the point which divides the line joining(4, –3), (9, 7) in the
Answer:

Question 5.
Find the coordinates of the midpoint of the line joining the points(1, –2), (–3, 4)
Answer:
Coordinates of midpoint =

Question 6.
The points A(6, 1), B(8, 2), C(9, 4), D(p, 3) are the vertices of a parallelogram. Find the value of p using the concept that the diagonals of a parallelogram bisect each other.
Answer:

The x coordinate of O using the concept that the diagonals of a parallelogram bisect

Question 7.
One end of the diameter of a circle is (1, 4). The center of the circle is (3, –4). Find the coordinates of other end
Answer:
One end of the diameter = (x₁, y₁) = (1, 4)
Center of the circle = (x, y) =(3, -4)
If other end of the diameter = (x₂, y₂)

Other end of the diameter (5 – 12)

Question 8.
Find the coordinates of the points P and Q which trisect the line joining (2, –3), and (4, –1) P,Q
Answer:

P can divide the line which joins the points (2, –3),(4, –1)in the ratio of 1 : 2

Question 9.
Prove thatA(6, 4), B(5, –2), C(7, –2) are the vertices of an isosceles triangle. If D is the midpoint of the side BC ,find the coordinates of D. Calculate the length of this median. Also, find the coordinates of centroid
Answer:

Worksheet 3

Question 10.
A-line makes an angle 45° with x-axis.Find the slope of the line
Answer:
Slope of the line = tan 45° = 1

Question 11.
The points on a line are (–1, 1), (3, 1), (5, 1)What is the angle made by this line with x-axis? What is the slope of this line
Answer:
This line is parallel to the x-axis so the slope is 0.

Question 12.
Find the slope of the line passing through (1, –3) (3, –5).
Answer:
Slope = \(\frac{-5+3}{3-1}=\frac{-2}{2}=1\)

Question 13.
A line passing through a point at a distance 4 from the right of origin on x-axis. If (3, –5) is a point on this line, find the equation of the line.
Answer:
Points are (4, 0) (3, –5)
Slope = \(\frac { -5 }{ -1 }\) = 5

Question 14.
If a line cut x-axis at (5, 0) and y-axis at (0, –3) Find the slope of the line given below.
Answer:
Slope of the line = \(\frac{-3-0}{0-5}=\frac{-3}{-5}=\frac{3}{5}\)

Question 15.
Find the slope of the line given below.

Answer:
Slope of the line = tan 60° = √3

Worksheet 4

Question 16.
Prove that the points (1, 3) (2, 5) (3, 7) are on a line
Answer:

Three points are lies on the same line.

Question 17.
The numbers in the sequence 2, 5, 8,11 and the numbers in the sequence 7, 11, 15, 19 …. are joined pairwise as given below(2, 7), (5, 11), (8, 15) Prove that these are on a line.
Answer:
Distance between (2, 7) and (5, 11) = \(=\sqrt{9+16}=5\)
Distance between (5, 11) and (8, 15) = \(=\sqrt{9+16}=5\)
Hence the distances are same, so the points are lie on the same line.

Question 18.
Find the slope of the line passing through (–2, 3) (5, 7) Write the slope of the fine parallel to it.
Answer:
Slope of the line which passing through the points (–2, 3) and (5, 7) = \(\frac{7-3}{5+2}=\frac{4}{7}\)
The line which joins the points (–2, 2) and (5, 6) is parallel to the first line.
Slope of the line which passes through the points (–2, 2) and (5, 6) = \(=\frac{6-2}{5+2}=\frac{4}{7}\)
Slope of the line which passes through the points (–7, 4) and (0, 8) = \(=\frac{8-4}{0+7}=\frac{4}{7}\)
The lines are parallel because their slopes are same.

Worksheet 6

Question 19.
Find the equation of the line passing through (5, 1), (1, –1), (11, 4)
Answer:
Equation of the line which passes through the points (5, 1), and (1, –1)

Equation is x – 2y – 3 = 0
Let find (11, 4) be on the equation 11 – 2 x 4 – 3 = 11 – 8 – 3 = 0
Equation is x – 2y – 3 = 0

Question 20.
(3, 0) is a point on the line joining the points (3x², 6x), (3y², 6y) then prove that xy = –1
Answer:
Slope of the line which passing through the points (3x², 6x) and (3y², 6y)

That is equal to the slope of the line which joins the points (3, 0) and (3x², 6x).

Worksheet 7

Question 21.
If a line is passing through (1, 1). This point divides the segment between axes in the ratio 3 : 4. Find the equation of this line.
Let consider the line 4x + 3y = 7, which cuts x axis at (a, 0) and y axis at (0, b)

Geometry and Algebra SCERT Questions & Answers

Question 22.
In the parallelogram ABCD A(2, 3); B(7, 3); and D(4, 7). Find the coordinates of C. [Score: 4, Time: 6 minutes]

Answer:
In the figure ΔAED, ΔBFC
AD = BC. DE = CF (1)
AE = BF = 2 unit (1)
∴ x coordinate of C
= 7 + BF = 7 + AE = 7 + 2 = 9 (1)
y coordinate of C = 3 + FC = 3 + ED = 3 + 4 = 7 (1)
C has the coordinates (9, 7)

Question 23.
In parallelogram ABCD A(2, 3); B(6, 5); and D(4, 7). Find the coordinates of C [Score: 4, Time: 6 minutes]

Answer:
Draw AE, DF parallel to x-axis. (1)
Draw BE, CF parallel to y-axis.
In right triangles Δ ABE, Δ DCF (1)
AB=DC, ∠BAE = ∠CDF
∠ABE = ∠DCF
∴ AE = DF= 6 – 2 = 4
BE = CF =5 – 3 = 2
x coordinate of C = 4 + DF = 4 + 4 = 8
y coordinate of C = 7 + CF = 7 + 2 = 9
Coordinates ofC (8, 9) (1)

Question 24.
In parallelogram ABCD A(x₁, y₁),B(x₃, y₂), D(x₂, y₃) Find the coordinates of C. [Score: 4, Time: 6 minutes]

Answer:
Draw AE, DF parallel to x-axis
Draw BE, CF parallel to y-axis
In right triangles ΔAEB, ΔDFE
AB=DC (1)
∠BAE = ∠CDF ∠ABE = ∠DCF
∴ AE = DF = x₂ – x₁ (1)
BE = CF = y₂ – y₁ (1)
x coordinate of C =
x₃ + DF = x₃ + x₂ -x,
y coordinate of C = y₃ + CF = y₃ + y₂ – y₁
∴ C has the coordinates =(x₂ + x₃ – x₁, y₂ + y₃ – y₃) (1)

Question 25.
Find the coordinates of the fourth vertex of the parallelogram shown here. [Score: 2, Time: 6 minutes]

Answer:
If A(x₁, y₁); B(x₃, y₂); D(x₂, y₃) are vertices of a parallelogram ABCD, then C has the coordinates (x₃ + x₂ – x₁, y₃ + y₂ – y₁) (1)
x coordinate of fourth vertex = 8 + 5 – 3 = 10
y coordinate of fourth vertex = 7 + 4 – 2 = 9
Coordinate of fourth vertex (10, 9) (1)

Question 26.
In the parallelogram PQRS, P(-3, 2) Q(2, 7), S(1, 9) are three vertices. Find the length of the diagonal PR. [Score: 4, Time: 6 minutes]

Answer:
x coordinate of R = 2 + 1 – (–3) = 6 (1)
y coordinate of R = 9 + 7 – 2 = 14 (1)
(6, 14) is the coordinate of R

Question 27.
A ABC A(6, 8); B(3, 4); C(–2, 2) are its vertices. The bisector of ∠A cuts BC at D.
a Find BD: CD.
b. Find the coordinates of D. [Score: 4, Time: 8 minutes]
Answer:

Question 28.
A circle is drawn with AB as diameter whose endpoints are A(3, 1) and 5(9, 10)
a. Find the coordinates of the center of the circle.
b. If another circle with diameter one-third of the above circle is drawn with the same center, what are the points that the circle cuts AB? [Score: 5, Time: 8 minute]
Answer:
a. Coordinates of the center OA

Geometry and Algebra Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 29.
Find the midpoint of the line segment joining the points(3, 0), (–1, 4).
Answer:
Mid point of the line segment joining the line segment joining the points (3, 0), (–1, 4).
\(=\left(\frac{3-1}{2}, \frac{0+4}{2}\right)=(1,2)\)

Question 30.
Find the equation of the line joining (1, 3), (2, 7).
Answer:
If (x, y) be any point on this line, then
\(\frac{(y-4)}{(x-1)}=2\)
y – 4 = 2(x – 1)
y – 4 = 2x – 2
2x + y + 2 = 0

Question 31.
What is the distance between the origin and the point (–2, -3)? What is the relation between this distance and the distance between the origin and the point (2, 3)?
Answer:
The distance to the points (–2,–3) is \(\sqrt{(-2)^{2}+(-3)^{2}}=\sqrt{13}\)
The distance to the points (2, 3) is \(\sqrt{2^{2}+3^{2}}=\sqrt{13}\)
Both the distances are equal.

Question 32.
Mark the points A(2, 3), 6(6, 6) by drawing the axes. Draw a rectangle with AB as di-agonal and sides are parallel to the axes. Find the coordinates of the other vertices of the rectangle. Find the length and breadth of the rectangle. Find the length of the diagonal AB.
Answer:
P(6, 3) Q(2, 6)
length = 4, breadth = 3
AB = \(\sqrt{4^{2}+3^{2}}-\sqrt{25} \div 5\)

Question 33.
What is the equation of the circle? Centre (2, 1), radius √5.
Answer:
Equation of the circle
(x – 2)² + (x – 1)² = 5
x₂ – 4x + 4 + y₂ – 2y + 1 = 5
x₂ + y₂ – 4x – 2y = 0

Question 34.
If (7, 3), (6, 1), (8, 2) and (p, 4) are the vertices of a parallelogram taken in order, then find the value of p.

Answer:
The midpoint of the diagonal AC and the diagonal BD coincide.

Question 35.
Find the equation of the line joining (–1, 3), (2, 5).
Answer:

Short Answer Type Questions (Score 3)

Question 36.
A is a point on the Y axis, equidistant from (3, 5) and (2, 6). Draw a rough sketch. If the Y co-ordinate of Ais P. What are the coordinates of A, in terms of P? Calculate the Value of P and hence the coordinates of A.
Answer:
The coordinate of A is (0, P). Since A is a point on the y-axis.

P2 – 12P + 40 = P2 – 10P +34
P2 – P2 – 12P + 10P = 34 – 40 – 2P = -6
P=3
∴ Coordinates of A are (0, 3)

Question 37.
The points A(0, 0), B(10, 0), C(5, 5√3 ) are the vertices of the triangle ABC. Find AB, BC, AC and prove that the triangle is equilateral.
Answer:
A(0, 0), B(10, 0), C (5, 5√3)
AB = 10 unit


AB = BC = AC
∴ ΔABC is equilateral

Question 38.
Prove that the points (2, 3), (7, 5), (9, 8), (4, 6) are the corners of a parallelogram.
Answer:

Since opposite sides are parallel, ABCD is a parallelogram.

Question 39.
a. What is the point at which the line 2x + 4y – 1 = 0 cuts the x-axis?
b. What about the y-axis?
Answer:
a. If the line 2x + 4y – 1 = o cut the x -axis at (x, 0), then 2x – 1 = 0
x = 1/2
The line intersect x axis is ( 1/2, 0 )
If the line 2x + 4y – 1 = 0 cut the y-axis at (0, y), then
4y – 1 = 0
y = 1/4
The line intersect y axis is (0, 1/4).

Long Answer Type Questions (Score 4)

Question 40.
Find the slope of the line joining (2, 3) and (3, –1). Check whether this line passes through the point (5, 6). What about the point (5, -9)?
Answer:

∴ Slope of the line joining (2, 3) and (3, –1)
\(=\frac{3-1}{2-3}=\frac{3+1}{-1}=-4\)
Slope of the line is –4
For point (5, –6)
Slope = \(\frac{6-3}{5-2}=\frac{3}{3}=1\)
Since slope = 4,
this line does not pass through(5,6)
For point (5, –9) Slope = \(\frac{-9-3}{5-2}=\frac{-12}{3}=-4\)
The line (5, –9) pass through the point.

Question 41.
Write the coordinates of each points on the lines 2x – y + 1 = 0 and 3x – 2y + 3 = 0. Find the co-ordinates of the point of intersection of the lines.
Answer:

Question 42.
a. Find the slope of the line joining the points (5, 3) and (4, 1).
b. Which are the points where this line meets x-axis and y-axis?
c. Write the coordinates of two other points on the line.
Answer:
a. Let (5, 3) = (x₁, y₁) and (4, 1) = (x₂, y₂),
= slop = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-3}{4-5}=\frac{-2}{-1}=\frac{2}{1}=2\)
b. If the line meets x axis at the point (x, 0), then slope = \(\frac{a-3}{x-5}=2 \Rightarrow 2(x-5)=-3\)
2x – 10 = –3,
2x = –3 + 10 = 7,
x = 7/2 = 3.5,
point is = (3.5, 0)
y axis \(\frac{y-3}{0-5}=2 \text { point }(0,-7)\)

c. slope is (2/1) means when x coordinate increase/decreases by 1, y co-ordinate increases decreases by 2. Another point on the line is (5 + 1), (3 + 2) = (6, 5)
A third point = (6 + 1) (5 + 2) = (7, 7)

Long Answer Type Questions (Score 5)

Question 43.
In the figure, the center of the circle is the origin and its radius is 1 unit. The points A,B are on the circle with ∠AOP = 30°, ∠AOB = 15°.
a. Find the coordinates of A and B.
b. Find the relation between the slope of the lines OA, OB and the tan measures of angles they make with the x-axis.
Answer:

Slopes of the lines and die tangent values-of the angles made by the lines with the axis are equal.

Question 44.
A circle is drawn with origin as the center and radius 10 units. Show that the point (–8, –6) is a point on the circle. Find out whether the point (9, –1) lies within the circle or outside the circle. Why?
Answer:
The distance between the centre and the point (-8,-6) is = \(=\sqrt{(-8)^{2}+(-6)^{2}}=\sqrt{100}=10\)
It is equal to the radius. So the point (-8, -6) is a point on the circle.
The distance between the centre, and the point (9,-1) is = \(\sqrt{9^{2}+(-1)^{2}}=\sqrt{81+1}=\sqrt{82}\)
√82 is smaller than 10.
So the point (9, –1) is a point inside the circle.

Geometry and Algebra Memory Map

Distance between (x_1,y₁) and (x₃, y₂) is \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
Distance of (x, y) from (0, 0) is \(\sqrt{x^{2}+y^{2}}\)

You can Download Electro Magnetic Induction Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Physics Chapter 3 Electro Magnetic Induction Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 3 Electro Magnetic Induction Solutions

Text Book Page No. 45

Sslc Physics Chapter 3 Kerala Syllabus Question 1.
You know that electrical energy can be converted into many other forms. Write down some examples.
Answer:

Text Book Page No. 46

Electromagnetic Induction Class 10 Kerala Syllabus Question 2.
Record your observations in the table given below.

Answer:

Text Book Page No. 47

Sslc Physics Chapter 3 Notes Kerala Syllabus Question 3.
Using magnet of high strength, and increasing the number of turns in the solenoid, the experiment is repeated. On the basis of the experiments complete the table (3.2).

Answer:

Physics Chapter 3 Class 10 Kerala Syllabus Question 4.
→ Why did the galvanometer needle deflect in the experiment?
Answer:
Whenever there is a change in the magnetic flux linked with a coil, an emf is induced in the coil.

→ Which were the instances in which there was a flow of current through the solenoid?
Answer:
Whenever there is a relative motion between the magnet and the solenoid, there is flow of electricity.

→ Which were the instances in which the current increased?
Answer:
Number of turns increased.Strong Magnet used Magnet / Solenoid moves with greater speed.

Text Book Page No. 48

Sslc Physics Chapter 3 Solutions Kerala Syllabus Question 5.
Observe the figure given below.
(The figure indicates the two stages of doing the experiment)

→ On which instance will the magnetic flux linked with the solenoid be less?
Answer:
Magnet is moved away from it solenoid.

→ On which instance will the magnetic flux linked with the solenoid be greater?
Answer:
When magnet is moved into the solenoid.

→ On which instance does a change in magnetic flux linked with the solenoid occur? (while it is moving/ while it is stationary)
Answer:
while it is moving.

Electromagnetic Induction Class 10 Questions And Answers Kerala Syllabus Question 6.
What may be the factors affecting the induced emf?
Answer:

• Number of turns of the coiled conductor.
• Magnetism.
• Movement of magnet and solenoid.

Class 10 Physics Chapter 3 Solutions Kerala Syllabus Question 7.
Which are the factors on which the direction of induced current in electromagnetic induction depend?
Answer:

• Direction of magnetic field
• Direction of conductor
• Direction of induced current

Text Book Page No. 49

Physics Chapter 3 Class 10 Kerala Syllabus Question 8.
Table 3.3

Answer:

Text Book Page No. 50

10th Physics Chapter 3 Kerala Syllabus Question 9.
The current obtained from the cell is unidirectional and is of the same magnitude. But what are the peculiarities of the current obtained by electromagnetic induction?
Answer:

• Direction changes.
• Magnitude changes.

Class 10 Physics Chapter 3 Kerala Syllabus Question 10.
We use mechanical energy in generators to move the magnet or coiled conductor continuously. In that case what shall be the energy change in a generator?
Answer:
Mechanical energy → Electrical energy

Class 10 Physics Chapter 3 Notes Pdf Kerala Syllabus Question 11.
Observe Fig 3.5(a) and write down the parts given in Fig. 3.5(b)

Answer:
ABCD: Armature.
B1, B2: Brush.
R1, R2: Slip rings.

Hsslive Guru 10th Physics Chapter 3 Kerala Syllabus Question 12.
ABCD indicates one turn of the armature coil. When the coil rotates about the axis in the clockwise direction, the portion AB moves upward and the portion CD moves downward.
Then according to the Fleming’s Right-Hand Rule.

→ What is the’direction of induced current in the portion AB? ( from A to B/ from B to A).
Answer:
from A to B.

→ What is the direction of induced current in the portion CD? (from C to D/from D to C).
Answer:
from C to D.

→ What is the direction of induced current in the coil ABCD? ( from A to D/from D to A).
Answer:
from A to D.

→ What is the direction of induced current in the external circuit? (through the galvanometer) (from B2 to B1/ from B1 to B2.
Answer:
from B2 to B1.

Class 10 Physics Chapter 3 Notes Kerala Syllabus Question 13.
What will be the positions of AB and CD when the armature completes 180° or one-half rotation?
Answer:
AB will be near the south pole of the magnet and CD will be near the north pole.

Physics 3rd Chapter Class 10 Kerala Syllabus Question 14.
Depict this stage of rotation of the armature in the science diary. At this instance,

→ What is the direction of movement of AB?
Answer:
Direction of AB is downward.

→ What is the direction of movement of CD?
Answer:
Direction of CD is upward.

→ What is the direction of current in the armature?
Answer:
from D to A.

→ What is the direction of current through the external circuit (through the galvanometer)?
Answer:
from B1 to B2.

Text Book Page No. 52

Kerala Syllabus 10th Standard Physics Chapter 3 Question 15.

Answer:

Text Book Page No. 52

Hss Live Physics 10th Chapter 3 Kerala Syllabus Question 16.
When 50 Hz AC is used, how many times will the direction of current change in the circuit?
Answer:
100 times.

Electromagnetic Induction Hsslive Kerala Syllabus Question 17.
Current is induced when the armature of a generator rotates. Slip rings and brushes are the ways and means by which this current is brought to the outer circuit. Is this arrangement necessary if the magnet in generator is made to rotate?
Answer: No

Hss Live Guru 10th Physics Kerala Syllabus Question 18.
Record in your science diary, the other ways by which mechanical energy can be made available for the working of a generator.
Answer:
From coal, naphtha, water from dam, exciters and big batteries. Field magnets are used as electromagnets. Mechanical energy also obtained from heat released by nuclear fusion.

Sslc Physics Chapter 3 Notes Pdf Kerala Syllabus Question 19.
Now write down in your science diary how the power for operating the generator which Babu saw near the stage was generated.
Answer:
The device which converts mechanical energy into electrical energy. When the armature rotates on its axis in the magnetic field, the magnetic flux linked within the coil changes. As a result, current is induced in the coil.

Sslc Physics 3rd Chapter Notes Kerala Syllabus Question 20.
Is it possible to produce DC (Direct Current) using a generator?
Answer:
Possible. If split-ring commutator is used in a generator instead of slip rings, we will get DC.

Text Book Page No. 54

Hsslive Guru 10th Physics Kerala Syllabus Question 21.
What are the similarities between the DC motor that we saw in the previous chapter and a DC generator?
Answer:

• Permanent magnet.
• Armature.
• Brushes.

Physics 10th Class Chapter 3 Kerala Syllabus Question 22.
Connect the output of a small DC generator to a galvanometer and rotate the armature continuously.

→ How is the needle deflected?
Answer:
Same direction.

→ Is the direction of current changing?
Answer:
No.

→ Is the magnitude of current the same?
Answer:
No. emf increases and decreases.

Physics Class 10 Chapter 3 Kerala Syllabus Question 23.

Answer:


Text Book Page No. 55

Physics Class 10 Chapter 3 Notes Kerala Syllabus Question 24.
Turn on & turn off the switch continuously. What do you observe?
Answer:
Bulb glows and then goes off.

Physics Class 10 Chapter 3 Kerala Syllabus Kerala Syllabus Question 25.
If the switch is kept in the on position what do you observe?
Answer:
Bulb does not glow continuously.

Question 26.
On what occasions do the flux change?
Answer:
When the switch is made to go on or off continuously.

Question 27.
What are the occasions when current flows through the second coil?
Answer:
When the switch in the first coil is kept on or off.

Question 28.
Can you suggest a method by which change can be brought in magnetic flux without switching on and off continuously?
Answer:
If AC is given to the primary coil instead of j DC, emf will be continuously induced in the secondary coil.

Text Book Page No. 56

Question 29.
Examine the diagrams of step up and stepdown transformers and list out the differences in their designs.
Answer:

Text Book Page No. 57

Question 30.

Answer:

Question 31.
A transformer working on a 240 V AC supplies a voltage of 8 V to an electric bell in the circuit. The number of turns in the primary coil is 4800. Calculate the number of turns in the secondary coil.
Answer:
V_p = 240V
V_s = 8V
N_p = 4800
\(\begin{array}{l}{\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}} \\ {N_{s}=\frac{V_{s} \times N_{p}}{V_{p}}=\frac{8 \times 4800}{240}=160}\end{array}\)

Question 32.
The input voltage of a transformer is 240V AC. There are 80 turns in the secondary coil and 800 turns in the primary. What is the output voltage of the transformer?
Answer:
V_p = 240V
N_s = 80
N_p = 800
\(\begin{array}{l}{\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}} \\ {V_{s}=\frac{N_{s} \times V_{P}}{N_{p}}=\frac{80 \times 240}{800}=24 \mathrm{V}}\end{array}\)

Question 33.
If voltage and current are known, what is the formula for finding power?
Answer:
Power = Voltage x Current

Text Book Page No. 58

Question 34.
In a transformer, if the voltage of the primary is V_p and the current in the primary is I_p, voltage in the secondary is V_s and the current I_s, write down the formulae connecting them.
Answer:
Power in the primary = V_p x I_p
Power in the secondary = V_s x I_s

Question 35.
In a transformer without any loss in power, there are 5000 turns in the primary and 250 turns in the secondary. The primary voltage is 120 V and the primary current is 0.1 A. Find the voltage and current in the secondary.
Answer:
N_p = 5000
N_s = 250
V_p = 120
I_p = 0.1A

Question 36.
Categorize the following relations appropriately as step up or step down transformers.

Answer:

Question 37.
As a result of the flow of current through a solenoid, is there any chance of inducing an electric current in the same solenoid?
Answer:
Yes. By self-induction.

Text Book Page No. 59

Question 38.

→ In which circuit does the bulb give a light with low intensity?
Answer:
Second circuit

→ Why does the intensity of light de-crease in that circuit?
Answer:
Back emf more

→ In which circuit is a magnetic field developed around the solenoid?
Answer:
On both circuits

→ If so in which circuit is a continuous emf induced?
Answer:
On second circuit

→ Have you understood the reason behind the decrease in the intensity of light in the second circuit? Write it down in the science diary.
Answer:
When AC passes through a solenoid, a changing magnetic field is generated around it. Due to this, an induced emf is generated inside the solenoid. This induced emf is in a direction opposite to that applied on the coil. Hence this is a back emf. This back emf reduces the effective voltage in the circuit.

Text Book Page No. 60

Question 39.
Inductors are widely used in AC circuits. Why?
Answer:
Inductors are used in the electronic circuits, to control and decrease current without power loss.

Question 40.
If resistors are used instead of inductors, what will be the disadvantage?
Answer:
Electric energy is lost in the form of heat.

Question 41.
Inductors are not used in DC circuits. Find out the reason and write it down in the science diary.
Answer:
Back emf is not produced as the flux formed by the current has no variation. So current control by inductor in DC is not possible.

Question 42.
Which are the main parts of a moving coil microphone?
Answer:
Diaphragm, Permanent magnet and Voice coil.

Question 43.
Which is the moving part in it?
Answer:
Diaphragm and Voice coil.

Question 44.
If a sound is produced in front of a movable diaphragm, what will happen to the diaphragm?
Answer:
Vibrates corresponding to the sound signals.

Question 45.
What happens to the voice coil then?
Answer:
Vibrates.

Question 46.
What will be the result?
Answer:
Creates electric signal corresponding to the sound.

Text Book Page No. 61

Question 47.
Find out the similarities and differences between a moving coil microphone and a moving coil loudspeaker and write them down in the science diary.
Answer:
Similarities:

• Both have voice coil.
• Have permanent magnet.
• Have diaphram.

Differences:

• Loudspeaker:
  Electrical energy → sound energy.
• Microphone:
  Sound energy → Electrical energy.

Question 48.
What is the energy transformation that takes place in a moving coil microphone?
Answer:
Sound energy → Electrical energy.

Text Book Page No. 62

Question 49.
How do we get mechanical energy for such generators?
Answer:

• Water from dam.
• Nuclear energy.
• Heat produced during the combustion of Naphta, Coal, Lignite ….etc

Question 50.
Write down the name of some power stations in Kerala.
Answer:

• Idukki – Moolamattom.
• Idukki – Pallivasal.
• Alappuzha – Kayamkulam.

Question 51.
Heat is generated in accordance with the equation H = I2Rt. In that case

→ What are the methods to reduce the heat generated?
Answer:

• Reduce current.
• Reduce resistance.
• Reduce the time taken.

→ If current (I) is reduced to half, how much will be the reduction in heat? (half, one fourth)
Answer:
one fourth.

→ If current (I) is reduced to 1/10 time show much will be the reduction in heat?
Answer:
1/100

→ How can we reduce the current without change in power? Find out on the basis of the equation P = V x I.
Answer:
by increasing voltage.

Text Book Page No. 63

Question 52.
→ What is the method to reduce the transmission loss?
Answer:
Power can be increased using a step-up transformer so that power loss can be minimized during power transmission.

→ Which type of transformer is there in a power station?
Answer:
Step-up transformer.

→ Which type of transformer is there in a substation?
Answer:
Step down transformer.

→ Which type of transformer is a distribution transformer?
Answer:
Step down transformer.

→ If a person standing on the earth touches a phase line, will she get an electric shock? Why?
Answer:
The person will get an electric shock because there is a potential difference (400V) between phase and earth.

→ Which are the lines essential for household electrification?
Answer:
Neutral, Phase and Earth lines.

Text Book Page No. 64

Question 53.
Analyze Fig.3.14 and find the answers to the questions given below.

→ To which device is the electric line reaching our home connected first?
Answer:
Watt-hour meter.

→ From where does the earth line start?
Answer:
From the main switch.

→ What is the use of a watt-hour meter?
Answer:
The electricity used is measured using a Watthour meter.

→ In which line are the fuses connected?
Answer:
Phase line.

→ What is the function of the main switch? Where is its position in the circuit?
Answer:
The electricity reaches the electrical appliances through the main switch. Therefore instead of turning each switch OFF, the main switch can be switched OFF. The main switch is placed just after the Watthour meter.

→ In the household electrical circuit, which is the third line, other than the phase and the neutral?
Answer:
Earth line.

→ What are the colors used for wires in phase, neutral and earth lines?
Answer:

• Phase – Red.
• Neutral – Black.
• Earth – Green.

→ Where is the earth wire connected in a three-pin socket?
Answer:
Pin E.

→ How are the household devices connected? Series/parallel.
Answer:
Parallel.

Question 54.
The advantages of connecting devices in parallel. Write them down.
Answer:

• Devices work according to the marked power.
• Devices can be controlled using switches as per need.
• The bulbs will get the required voltage.

Text Book Page No. 65

Question 55.
In a house, 5 CF lamps each of 20 W, works for 4hours, 4 fans each of 60 W work for 5 hours and a TV of 100 W works for 4 hours in a day. What will be the daily consumption shown by the watt-hour meter?
Answer:
4hour’s consumption of 5 CFL having 20 W
\(\frac { 5 × 20 × 4 }{ 1000 }\) = 0.4 kWh
5 hour’s consumption of 4 Fan having 60 W
\(\frac { 4 × 60 × 5 }{ 1000 }\) = 1.2 kWh
4 hour’s consumption of one T.V having 100 W
\(\frac { 1 × 100 × 4 }{ 1000 }\) = 0.4 kWh
Total unit = 2 kWh

Text Book Page No. 66

Question 56.
→ Which are the circumstances that lead to the flow of excess current in a house hold circuit?
Answer:
Over loading or Short circuit.

→ What happens to the circuit when there is an excess current in a circuit?
Answer:
More heat is generated. So appliances and circuit will get damaged.

→ How does a safety fuse protect a circuit?
Answer:
The current in the circuit may increase due to reasons such as short circuit, overload, excess flow of current or any problems in the insulation. As the higher temperature produced in the circuit due to these reasons makes the fuse wire to melt and the flow of current stops. Thus the circuit and the appliances are protected.

Text Book Page No. 67

Question 57.
What are the differences between ordinary fuse and MCB?
Answer:
Ordinary fuse works making use of heating effect of electricity while MCB works mak-ing use of heating and magnetic effects of electricity. The current in the circuit may increase due to reasons such as short circuit, overload, excess flow of current or any problems in the insulation.

As the higher temperature produced in the circuit due to these reasons makes the fuse wire to melt and the flow of current stops. Thus the circuit and the appliances are protected. MCB automatically breaks the circuit whenever there is an excess flow of current due to short circuit or overloading. After rectifying the circuit we can switch on the MCB and make the circuit as it was.

Question 58.
What is the advantage of MCB over a safety fuse?
Answer:
MCB is more sensitive to current than fuse, In case of MCB, the faulty zone of electrical circuit can be easily identified, With MCB it is very simple to resume to the supply, MCB is reusable and hence has less maintenance and replacement cost.

Question 59.
What is the function of ELCB/RCCB in the circuit?
Answer:
ELCB helps to break the circuit automatically whenever there is a current leak due to insulation failure or any other reason. Hence a person touching the electric circuit or a device does not get an electric shock. Nowadays RCCB, which ensures more safety than ELCB is made use of. In an ELCB one end of the relay coil is connected to the outer metallic cover and the other end is earthed.

If current leaks into the earth due to insulation failure or any other reason, a potential difference is developed between the ends of the relay coil. As a result, ELCB gets tripped off if the current exceeds a particular limit. This is with the help of a relay. In RCCB there is a provision to distinguish between the phase current and neutral current and to break the circuit if needed.

Question 60.
In the figure, which are the lines that are connected to the coil of the electric iron?

Answer:
Earth, Phase and Neutral lines.

Question 61.
If the phase line comes into contact with the body of the appliance due to defects in the insulation, what happens to the person who touches the body of the appliance?
Answer:
The person will get shock.

Question 62.
How can safety be ensured using a three-pin plug?
Answer:
The pin E of a three-pin plug comes into contact with the earth line. This pin is now connected to the body of the appliance. If the body comes into contact with an electrie connection, electricity flows to the earth through the earth wire. The flow of current to the earth through a circuit of low resistance increases the current. Hence heat generated in the fuse wire increases, the fuse wire melts and the circuit is broken. This will protect the instrument and the person handling it.

Question 63.
Which line comes into contact with the pin E.
Answer:
Earthline.

Question 64.
Howdoestheearth pin differs from the other pins? Why is it made different in this way?
Answer:
The earth pin is thick and long. As it is long it first comes in contact with the earth. Similarly, it loses the contact with the earth only at the end. As it is thick it offers less resistance. It carries more current than the amount of current coming from the phase.

Question 65.
Which part of the instrument is connected to the earth line?
Answer:
The earth line is connected to the metal casing of the electrical appliances.

Question 66.
Is TV working on AC or DC?
Answer:
on AC

Question 67.
We get DC from a mobile phone battery. But we use AC for charging it. What may be the reason?
Answer:
Most of the devices that work only on DC work by converting AC into DC. Mobile charger is a device that converts AC into DC.

Question 68.
Classify the devices known to you as those working in AC and DC.

Answer:

Text Book Page No. 71

One way switch:
Use to start/stop current flow and to change the direction of current flow.

Two-way switch:
Having two or more switches in different locations to control one electrical device.

Three-pin socket:
Electricity can be supplied to electrical appliances safely.

Ceiling rose:
Ceiling rose helps in that one live wire that is needed for ceiling light or fan to go in continuation.

ELCB:
Helps to break the circuit automatically whenever there is a current leak due to insulation failure or any other reason.

Regulator:
Automatically maintain a constant voltage level.

Indicator:
A device which provides visual or remote indication of a fault on the electric power system.

RCCB:
Distinguish between the phase current and neutral current and to break the circuit if needed.

MCB:
Automatically breaks the circuit whenever there is an excess flow of current due to short circuits or overloading.

Kit Kat Fuse:
Protecting a circuit against excess current.

Switchboard:
To control the flow of power.

Meter:
Instrument used to measure the magnitude of a quantity.

Main switch:
The main power supply of the RC models.

Bulb holder:
Holding a light bulb or lamp.

Clamp Ammeter:
The vector sum of the currents flowing in all the conductors passing through the probe.

Multimeter:
Can measure voltage, current, and resistance.

AC Voltmeter:
To measure the AC voltage across any two points of electric circuit.

Wire stripper:
Used to strip the electrical insulation from electric wires.

Screwdriver (*):
Tool, usually hand-operated, for turning screws with shaped slotted heads.

Screwdriver (-):
Tool, usually hand-operated, for turning screws with shaped slotted heads.

Tester:
Touched to the conductor being tested.

Plier:
For gripping something round like a pipe or rod, some are used for twisting wires.

Gloves:
Leather protector gloves – Worn over rubber insulating gloves to help provide the mechanical protection needed against cuts, abrasions and punctures.

Insulation tape:
To insulate electrical wires and other materials that conduct electricity.

Wire ( cable):
To carry electrical currents.

PVC Pipe fittings:
Used for plumbing and drainage.

PVC channel:
In Control Panels and Electrical Cabinets.

PVC Pipe:
Used as the insulation on electric wires.

Text Book Page No. 72

Question 69.
Construct such a circuit in aplywood as shown in Fig 3.19. The list of materials required and the number of items are given in the table.

Answer:

• Main switch.
• Fuse.
• MCB.
• Main fuse.
• Switchboard.
• ELCB.
• Fan.
• Bulb.

Electro-Magnetic Induction Let Us Assess

Question 1.
Write down the names of some devices that work based on the principle of electromagnetic induction.
Answer:

• Electric generator,
• Transformer,
• Moving coil galvanometer,
• Induction cooker etc.

Question 2.
What are the components essential for proving electromagnetic induction experimentally?
Answer:
Magnet, soft iron core connecting wire and Galvanometer.

Question 3.
Which are the factors that affect the induced emf in electromagnetic induction?
Answer:

• Number of turns in the armature.
• Strength of magnetic field.
• Speed of armature or field magnet.

Question 4.
Take a used cell from a calculator or remote control and connect it to a galvanometer as shown in the figure. What do you observe?

Answer:
The galvanometer needle deflects to one direction because there is a slight current flow.

Question 5.
Write down the names of DC sources.
Answer:
Cell battery, Solar cell, Generator, Thermocouple, etc.

Question 6.

a. Write down the names of parts numbered.
b. State the working principle of this device.
Answer:
a. 1. Field magnet,
2.Armature,
3. Slip rings.

b.Generators convert mechanical energy into electrical energy. Flux is changed and current is produced in the coil when the armature rotates from an axis. So current is induced due to flux change.

Question 7.
Write down the special features of AC and DC.
Answer:
Direct current (DC): Direction constant, emf increases and decreases.
Alternating current (AC): Direction changes continuously, emf increases and decreases.

Question 8.
Analyze the given graph and find out the instances at which the emf is maximum and minimum.

Answer:
Maximum: T/4, 3T/4
Minimum : 0, T/2, T

Question 9.
There is only one type of generator AC generator. Write down your responses about this statement.
Answer:
According to the nature of output, generators can be classified into 2 types AC generator and DC generator. Though ac current is produced in a DC generator with the help of split-ring commutator ac is converted into dc current.

Question 10.
Line diagrams of a generator are given.

a. What is the specialty of the electricity reaching the galvanometer if the armatures of both the generators are made to rotate?
b. What is the speciality of the electricity reaching the galvanometer if the field magnets of both the generators are made to rotate?
c. Draw the graphical representation of electricity obtained in both.
Answer:
a. DC in first and AC in second.
b. AC in both.

Question 11.
Electromagnetic induction is
a. charging a substance.
b. process of developing a magnetic field around a coil by passing electricity through a coil.
c. process of rotating the armature of a generator.
d. process of making electricity by the relative motion of a magnet or a coiled conductor.
Answer:
d. process of making electricity by the relative motion of a magnet or a coiled conductor.

Question 12.
Which is the device used to generate electricity?
a. generator.
b. galvanometer.
c. motor.
d. ammeter.
Answer:
a. generator.

Question 13.
Write down the similarities and differences in the structure of an AC generator and a DC generator.
Answer:
Similarities:

• Field magnet and Armature are present.
• Works on the principle of electromagnetic induction.
• AC produced on armature.

Differences:

Question 14.
A conductor hung horizontally in the north-south direction is connected to a galvanometer. The conductor is situated in a magnetic field acting in the East-west direction, In which direction should you move the conductor if maximum current is to be induced in the conductor in the north-south direction? Justify your answer.
a. towards east.
b. downwards.
c. upwards.
d. towards north.
Answer:
b. downwards. According to Fleming’s Right-hand rule.

Question 15.
Copper wires of the same length and thickness are connected to points A and B in all the three circuits. In-circuit (a) copper wire is not coiled. In circuits (b) and (c), the copper wire is coiled. Observe the circuits and answer the following questions.

a. When circuit (a) is switched on, what do you observe?
b.When circuit (b) is switched on, what change do you observe in the intensity of light? Justify your answer.
c. When circuit (c) is switched on, what change do you observe in the intensity of light? Justify your answer.
Answer:
a. Bulb glows.
b. No change in the intensity of light. There is no back emf produced as DC current is used in the circuit.
c. Intensity of light decreases. Back emf is produced in the circuit as A.C current flows through it and current is decreased.

Question 16.
The current in the secondary coil of a transformer is 1A and that in the primary is 0.5 A.
a. What type of transformer is this?
b.If 200 V is available in the secondary coil of this transformer, what is the voltage in the primary?
c. Explain the working principle of a transformer.
Answer:
a. Step down Transformer.

c.Mutual Induction:
Keep two sets of coils side by side. When the direction of current flow or its intensity is changed in one, the magnetic flux around it also changes. An emf and current is induced in the secondary coil. This phenomenon is called mutual induction.

Question 17.
In connection with the working of a microphone, a few statements are given inboxes. Arrange them in the proper sequence.

Answer:
Sound is produced → diaphragm vibrates → Voice coil vibrates → electrical signals are produced in the voice coil.

Question 18.
Thick insulated copper wires are used in the primary coil of a step-up transformer and in the secondary of a step-down transformer. What is the necessity of this?
Answer:
The power in the primary and secondary.of a transformer is equal. Current is more where voltage is less because P = VI. Thick wires are needed to flow more quantity of current. In the primary of step up and in the secondary of step down, quantity of current is more.

Question 19.
Which situation causes short circuit?
Answer:
When a phase and neutral comes into direct contact in a household circuit or when two-phase come in contact or when a phase and neutral comes into direct contact in a distribution line.

Question 20.
What is the role of earthing wire in a household circuit?
Answer:
The pin E of a three-pin plug comes into contact with the earth line. This pin is now connected to the body of the appliance. If at all the body comes into contact with an electric connection, electricity flows to the; earth through the earth wire.

The flow of current to the earth through a circuit of low resistance increases the current. As a result heat generated in the fuse wire increases and the circuit gets broken. This ensures the safety of instrument and the person handling it.

Question 21.
Why do we say that metallic devices should be earthed?
Answer:
Chances of electric shock are” more when metallic devices are used. When metallic devices are earthed, chances of electric shock can be prevented.

Question 22.
An electric heater calibrated 1.5 kW, 230 V is connected to a household branch circuit having 5 Afuse wire and is, made to work. What will happen?
Answer:
Current produced the circuit when the device works is
I = \(\frac { P }{ V }\) = \(\frac { 1500 }{ 230 }\) = 6.52 A
So 5 Amperage fuse bums.

Question 23.
Which are the devices connected in series in a household circuit?
Answer:
Fan, Switch, Regulator, etc.

Question 24.
What can be done to save electrical energy in schools and houses?
Answer:

• Use devices of good efficiency.
• Use electricity only when required.
• Reduce cooking using electricity.
• Reduce the use of air conditioners and refrigerators.
• Use LED bulbs instead of other bulbs.

Question 25.
Why do some mobile phones use three-pin plugs?
Answer:
In mobile phones, instead of earth pin, a plastic pin is used. For safety purpose, the phone neutral socket will have caps. Only if the pin enters the earth socket the other two sockets will open. It will be difficult to connect two-pin in such sockets. To solve this problem in mobile charges three-pin plug can be used.

Electro Magnetic Induction Extended Activities

Question 1.
Michael Faraday, the Father of electricity, did not even get elementary education. Are j you not inspired by the achievements of Faraday in the field of science? Conduct a seminar on “Contributions df Faraday and the hard work behind it.”
Answer:
Michael Faraday, an eminent scientist in the fields of physics and chemistry. He made his first invention in 1821. He proved that when a wire is kept in the magnetic field and current is passed through it, the wire moves.

By the series of experiments conducted in 1831 by using magnetic power Faraday found the important theory of production of current. He was called the father of electric current. He made valuable contributions in the field of chemistry also. It is wonderful to know that he didn’t get any formal education.

Question 2.
Energy is precious, especially electrical energy. Society must be convinced of the necessity of reducing the consumption of electrical energy. Prepare and propagate posters for this purpose.
Answer:

• Control the consumption of current, or else we will be in darkness.
• Next-generation also deserves current.
• Electricity is only for essential purpose.
• Use only efficient machinery to use minimum current.

Question 3.
Compare the induced current obtained when the armature coil rotates once in between the poles of a magnet, and the induced current obtained when the experiment using a magnet and coil was performed.
Answer:

Question 4.
Exhibit a model of electrical distribution network.
Answer:

Question 5.
Draw an electrical circuit containing the electrical appliances required for your classroom.
Answer:

Question 6.
How can the earthing be done in order to ensure safety in electrical circuits? Discuss and prepare a note.
Answer:
Overloading or short-circuiting may occur during the working of electrical devices due to some problems in the mains or within the circuit. The excess current produced is discharged to the ground through the earth wire. This is known as earthing.

Thus earthing provides safety in the electrical circuits. In household electrical circuits for the effective functioning of the earth wire, the earthing system should be proper.

• The pit in which the earth wire is placed should be filled with salt and coal.
• During summers it is required to wet the places where it is earthed.

Question 7.
Observe and record the meter reading in your house for 10 consecutive days. Based on this, find out the average consumption per day. Find out methods to reduce consumption and record them. Present your findings in the Energy Club.
Answer:
Average usage = \(\frac { Total usage }{ no.of days }\)
Methods to reduce the usage of electricity.

• Use devices of good efficiency.
• Use electricity only when required.
• Reduce cooking using electricity.
• Reduce the use of air conditioners and refrigerators.

Electro Magnetic Induction Orukkam Questions and Answers

Question 1.
Observe the figure.

a. Identify the device in figure 1?
b.What is Rj and Bj indicate in the figure?
c. What is the working principle of that device?
d. Identify the graph which indicate the emf produced by that device?
e. Suggest a way of method to increase the value of emf produced in armature coil of the device?
f. Which law is related with the direction of the induced emf in the armature? Explain?
g. Which part is used as rotor?
h.Which part is used as stator?
i. What are the advantages of armature used as stator?
j. Field magnets in power generator are electromagnet. Why?
k. Identify the device in figure 2?
l. Identify the graph given above of emf which is produced by that device?
m.What is the angle between the armature coils of a three-phase generator?
n. Some statements are given below which related to single-phase generator and three-phase generator. Classify them.
a. For each field magnet, there is only one armature.
b. There are three sets of armature coils for each field magnet.
c. AC generated in all the armature coils will be of the same phase at the same time.
d. AC generated in all the three armature coils will be of different phase at the same time.
Answer:
a. AC generator.
b. R1 – Slip Rings,
B1 – Brushes.
c. Electromagnetic Induction.
d. Graph 1.
e. Increase the number of turns.
f. Refer page no: 39
g. Field magnet.
h. Armature.
i. We can avoid rings and graphite brushes when the armature is used as stator. So there is no possibility of producing sparks.
j. Refer page no: 72 Question No: 1
k. Three-phase generator.
l. Graph 2.
m. 120°.
n. Single-phase generator – a, c.
Three phase-generator – b,d.

Question 2.
Observe the graph and answer the following.

a. In which angles are the rate of changes of flux is maximum?
b. In which angles are the induced emf maximum?
c. What is meant by period of AC?
d. What is meant by frequency of AC?
e. What is the aim of increasing number of magnetic pole and armature coil in a power generator?
Answer:
a. 90° and 270°
b. 90° and 270°
c. The time taken by the armature coil for a full rotation is called Period.
d. The number of cycles per second is the frequency of AC.
e. To increase the amount of electricity produced.

Question 3.
Observe the figures.

a. What kind of transformer is in figure 1?
b. What kind of transformer is in figure 2?
c. What is the working principle of a transformer?
d. Some statements are given below related to step-up transformer and step down transformer. Classify them. Number of turns in primary is less than secondary.

• Thick wires are used in secondary.
• Thick wires are used in primary.
• Voltage in secondary is high.
• Voltage in secondary is low.

e. Some relations are given about step-up transformer and stepdown transformer. Classify them.
Answer:
a. Step-up transformer.
b. Step down transformer.
C. Mutual induction.

Question 4.
Construct the circuit as in the figure. Observe the intensity of light when switch in the circuit is turn on.

a. What change is observed when 6V battery is connected instead of 6V AC in the circuit?
b. What change will occur in magnetic flux when 6V battery is connected to the circuit and 6V AC connected to the circuit?
c. What is the peculiarity of the induced emf when AC source is connected in the circuit instead of DC.
d. What is the name of this induced emf?
e. Name the phenomenon caused to reduce the intensity of light in the circuit.
f. What is the change in intensity of light will be occurred when soft iron core is inserted into the inductor?
g. What is the reason for the change in the intensity of light?
h. What is the phenomenon which caused to glow the LED?
i. Explain the phenomenon.
Answer:
a. Intensity of light decreases
b. When 6V AC is given, there is a change in the magnetic flux linked with the solenoid. Due to the change in flux, an induced emf is developed.
c. Emf will be in the opposite direction.
d. Back emf.
e. Self Induction.
f. Intensity of light decreases.
g. Change in flux increased.
h. Mutual Induction.
i. When the strength or direction of the current in one coil changes, the magnetic flux around it changes. As a result, an emf is induced in the secondary coil. This phenomenon is called Mutual induction.

Question 5.
Match the following related to ideas of different kind of power stations are given below.
Pallivasal, coal, Nuclear power station, Flowing water, Thermal power station. Kalpakam, Hydroelectric power station, Kayamkulam, Nuclear energy.
Answer:
Pallivasal :
Flowing water – Hydroelectric power station.

Kalpakam:
Nuclear energy – Nuclear power station

Kayamkulam:
Coal – Thermal power station.

Question 6.
There are some terms in connection with various power stations given. Fill in the blanks suitably.

Answer:
a. Nuclear Power station.
b. Kinetic energy.
c. Electrical energy.
d. Chemical energy.
e. Mechanical energy.
f. Heat energy.
g. Electrical energy.

Question 7.
Power stations are centers that generate and distribute large quantities of electricity. Power transmission is the process of sending electricity to distant places through wires from the power stations.
a. What is the voltage at which electricity is generated at the power station?
b. What is the voltage of electricity supplied for domestic consumption?
c. In which stage of power transmission is the step-up transformer used?
d. Which are the stages in which step down transformers are used in power transmission?
e. What are the problems arises during distant energy transmission?
f. What are the factors which depend when electricity passes through the conductor?
g. What are the methods to reduce the resistance?
h. What change will occur in power when intensity of current is reduced in the circuit?
i. How is it possible to reduce the intensity of current without power loss?
j. What will be the change in secondary current when secondary voltage of a step-up transformer is increased 10 times?
Answer:
a. 11 kV or 11000 V.
b. 230 V.
c. Phase I
d. Phase ii, iii, iv.
e. Voltage drop, Power loss.
f. Current, Resistance, Time.
g. Increase the thickness of the conductor.
h. Power decreases
i. Increase the voltage.
j. Become less by 1/10.

Question 8.
Step-up or Step down transformer in the power distribution system erected for the purpose of household distribution.
a. How many lines reach the input of the distribution transformer?
b.What is the potential of each phase line?
c. How many lines go out of the distribution transformer? Which are they?
d.Which method is adopted for connecting secondary coil in the distribution transformer?
e. Three-phase lines from output of distribution transformer are connected into a common point. What is the name of the common point?
f. What is the potential difference between a phase line and neutral line?
g. What is the potential difference between any two-phase line?
h.What is the potential difference between the earth and the neutral line?
i. What is the potential difference between a phase line and the earth line?
Answer:
a. 3 lines.
b. 400 V.
c. 4 lines come out, 3 phase and one neutral.
d.Star connection.
e. Neutral.
f. 230 V.
g. 400 V.
h.0 V.
i. 0 V.

Question 9.

a. How are the bulbs arranged in figure 1?
b. How are the bulbs arranged in figure 2?
c. Which circuits has more resistance?
d. All the switches in both the circuits may be turned on, in which circuit do the bulb give brighter light?
e. In which circuit current flowing through each bulb is maximum?
f. Which mode of connecting devices in your household electrical circuit is advisable?
Answer:
a. Series.
b. Parallel.
c. Series Connection.
d. Parallel Connection.
e. Parallel Connection.
f. Parallel Connection.

Electro Magnetic Induction Some more activities

Question 1.
Three-pin plug is used in the circuit of power devices like electric iron.
a. In three pin plug, what each pin indicate?
b. Which part of the device is connected to the earth pin?
Answer:
a. E-Earth line, N-Neutral line, P-Phase line.
b. To the body of the device.

Question 2.
What is the reason for massaging the body of a person who got electric shock?
Answer:
To increase the body temperature.

Electro Magnetic Induction SCERT Questions and Answers

Question 1.
Four identical solenoids and magnets are, depicted in the diagram. Magnets are shown as moving in the figures 1 and 2 and solenoids are shown as moving in figures 3 and 4.

a. Which are the figures showing inducing of emf? How?
b. Pair the pictures in which the galvanometer needle deflects in the same direction. What does the deflection indicate?
c. Draw the picture of a device that can produce continuous electricity using the same principle as shown above for producing electricity.
Explain its working.
Answer:
a. Emf will be induced in all the 4 coils in the diagrams; flux change happens in all the four activities.
b. Pairs (1) and (4), (2) and (3). deflection of galvanometer needle indicates the direction of the flow of current.

Question 2.
Certain factors that influence the induced emf in electromagnetic induction are listed below:
i. Increasing the speed of magnet.
ii. Decreasing the strength of magnet.
iii. Increasing the number of turns.
iv. Reducing the speed of rotation of the magnet.
v. Increasing the strength of magnetic field.
vi. Using solenoid with less number of turns.
Which of the above ideas can be used to get maximum emf by electromagnetic induction?
Answer:
i, iii, v.

Question 3.
The diagram shown below is an arrangement for producing 10V AC using electromagnetic induction. Observe the diagram carefully and answer the questions.

a. Which law is used to find out the direction of induced emf when the armature coil ABCD starts rotating?
b. In which direction the side CD should move (up/down) to produce a flow of current from x to y? What should be the direction of movement of the side AB (up/down) to produce a flow of current in the same direction?
C. Find out the frequency by drawing a time emf graph if the armature completes 10 cycles in 5 seconds.
Answer:
a. Fleming s Right Hand Rule.
b.Upward, side AB moves down.

Question 4.
Parts of an AC generator are given below. Field magnet, Armature, Slip rings, Brush.
a. Explain the position of these parts in an AC generator.
b. Write down the functions of any two.
Answer:
a. Armature rotates in the magnetic field of the field magnet. Springs are soldered to the ends of armature coil, Graphite Brush is connected in such a way as to have constant contact with springs.
b. Armature – Generate emf.
Field magnet – creates magnetic flux.

Question 5.
The various stages of rotation of an armatures coil while completing one rotation in a magnetic field and the graph of the emf produced by the coil are shown below:

a. Select the appropriate points from the pictures m,n,o,p corresponding to the positions 1,2,3,4 and 5 in the graph.
b. Explain the scientific basis for the pairing.
c. Which two pictures show the maximum flux changes? Point out one difference between the two.
Answer:
a. 1-m, 2-0, 3-n, 4-p, 5m OR 1-n, 2-p, 3-m,4-0, 5-n
b. The change in direction and intensity of magnetic flux in armature file.
c. n and o. Magnetic flux change takes place in the opposite directions in these two.

Question 6.
Maximum voltage produced in an AC generator completing 60 cycles in 30 seconds is 250 V.
a. What is the period of the armature?
b. How many cycles are completed in T/2 seconds?
c. What is the maximum emf produced when the armature completes 1800 rotation?
Answer:
a. Period = Time taken to complete one full rotation.
Period = \(\frac { Time taken for rotation }{ Number of rotations }\) = \(\frac { 30 }{ 60 }\) = \(\frac { 1 }{ 2 }\) = 0.5s
When the Armature ABCD rotates, change in the magnetic flux depends on its direction and intensity.
b. only \(\frac { 1 }{ 4 }\) of a rotation. T \(\frac { 1 }{ 2 }\),\(\frac { T }{ 2 }\),\(\frac { 1 }{ 4 }\)
c. 250V

Question 7.
In single-phase generators armature functions as rotor whereas in power generators armature functions as stator.
a. List out the other differences between single-phase and three-phase generators?
b. What are the advantages of using armature as stator in power generators?
Answer:
a.

b.

• Can increase the no.of turns of the armature coil.
• Can avoid slip rings and also the subsequent sparks.

Question 8.
If a rotor in a power generator completes 3000 rotations in a minute.
a. Which type of AC is produced here, single-phase or three-phase?
b. What will be the frequency of the AC produced if a pair of magnetic poles are ) used in the generator?
c. What would be the time interval to produce the same emf in the two phases if j the rotor completes 20 rotations in a minute?
Answer:
a. Three phase AC
b. Frequency = \(\frac { 3000 }{ 1 × 60 }\) = 50Hz
c. 20 rotations in a minute -3 seconds for completing one rotation. So the time interval to produce same emf is 1 second.

Question 9.

a. Observe the picture and identify the parts labeled as a, b, and c in the picture shown.
b. What is the energy change taking place in a moving coil microphone? What are the roles of the parts a,b, and c?
c. Why is the output signal of a moving coil microphone sent to the loudspeaker only after feeding it directly into an amplifier?
Answer:
a.1. Diaphragm.
2. Permanent magnet.
3. Voice coil.
b.
Mechanical energy to electrical energy Diaphragm – Vibrates in accordance with the sound falling on it.
Permanant magnet – Creates a magnetic field.
Voice coil – Creates electric signals corresponding to the sound.
c. The weak signals obtained from the microphone are strengthened by an amplifier.

Question 10.
Coils wound around a soft iron core connects two bulbs, B1 and B2 of 6 V. Analyse the figure and answer the questions.

a. If 6 V, DC is given in the coil A and the switch is on, which of the bulbs, B, & B2 will glow? Why?
b. If AC is given to the coil A instead of DC, which of the bulbs will glow B1 or B2?
c. Was there any variation in the brightness of Bulb B1 when AC and DC were supplied? Why?
d. Write down the names of two instruments that use the principle of mutual induction.
Answer:
a. Only B1 because electricity flows only through B1
b. B1 and B2 because of mutual induction.
c. Light of B1 will be reduced when AC is passed through it. This is due to the back emf produced due to self-induction.
d. Transformer, Relay switch.

Question 11.

Two solenoids A and B made of insulated copper wire are shown in the diagram. Observe the diagram and answer the questions below:
a. Write down different methods to light up the bulb connected in the coil B using the principle of electromagnetic induction.
b. What is the scientific principle applied to light up the bulb B?
Answer:
a.i. Switching S1 on and off by giving a DC supply.
ii. Keeping switched on after giving AC supply to the coil A.
iii.When coil A is moved relatively after supplying DC.
b. Mutual induction.

Question 12.
A transformer without power loss carries 4000 turns in the primary and 2000 turns in the secondary. If the primary voltage is 120 V and the current is 0.2 A.
a. What is the output voltage of transformer?
b. What is the current in the secondary?
c. What is the maximum output power?
d. Where are the thick wires to be used in this transformer in the primary or secondary? Why?
Answer:

d. In the secondary. Since P = V1 current is more in the secondary.

Question 13.
In a step up transformer, primary power is 50 W and output voltage is 100 V. If the current in the primary is 1 A.
a. What is the primary voltage?
b. What will be the current in the second ary?
c. \(\frac { Secondary power }{ Primary power }\) =
Answer:
a. P = V_p × I_p
500 W = V_p × 1A, V_p = 500 V
b. I_s = \(\frac { I_p × V_p }{ V_s }\) = \(\frac { 1 × 500 }{ 100 }\) = 5A
c. Primary and secondary powers are equal
\(\frac { Secondary power }{ Primary power }\) = 1

Question 14.
Power Station A :
Potential energy → Kinetic energy → Mechanical energy → Electrical energy.
Power Station B :
Chemical energy → Heat energy → Mechanical energy → Electrical energy.
Power Station C :
Nuclear energy → Heat energy → Mechanical energy → Electrical energy.
a. Identify the above power stations A, B, and C based on the energy change taking place in them.
b. Write down the merits and demerits of the power stations A, B, and C.
c. Name other three energy sources that can be used instead of these.
Answer:
a. A- Hydroelectric power station.
B- Thermal power station.
C- Nuclear power station.

b.1. Merits: No pollution, Energy of waterfall is used.
Demerits: Deforestation, Extinction of rare species of flora and fauna.

2. Merits: less area needed, danger possibility is less.
Demerits: Pollution, High consumption of fossil fuels.

3. Merits: Energy production on large scale, Less area needed.
Demerits: Pollution, Problems of radiation.

c. (1) Solar energy.
(2) Tidal energy.
(3) Windmills.

Question 15.
Names of power stations and statements related to them are given below in two groups. Match them suitably.
A. (1) Uses the energy of water stored at a height.
(2) Uses the heat generated by the burning of fuels.
(3) Uses nuclear energy.
B. (a) Pallivasal Power Station.
(b) Ramagundam Power Station.
(c) Kalpaklcam Power Station.
Answer:
a – 1,
b – 2,
c – 3.

Question 16.
Different stages of power production and transmission are given. Arrange them in the proper sequence.
a. Distribution transformer reduces the voltage from 11 kV to 230 V.
b. Voltage is reduced from 220 kV to 66 \ kV to supply electricity to large scale industries.
c. Turbine is rotated to operate the generator.
d. Transmission of electricity from the Power Station starts at 220 kV.
e. Electricity is produced at 11 kV.
f. Domestic consumers get electricity.
Answer:
c,
e,
d,
b,
a,
f.

Question 17.
Different stages of power transmission and distribution are given in a flow chart.

i. Choose the suitable terms from the bracket given below and arrange them in the given boxes in the correct order. (Power generator, substation, power transformer, star connection, distribution transformer, household connection, power grid)
ii. What do you mean by power grid? Give two advantages of power grid.
Answer:
i.
a. Power generator.
b. Power transformer.
c. Substation.
d. distribution transformer.
e. Star connection.
f. household connection.
ii. Power grid is the network that connects different power generating center and distribution systems to one another. Due to this arrangement, if any defect occurs either at the generator or transmission lines, electricity can be taken from any other power generating center through another set of transmission lines.

Question 18.
We use 230 V for our household purpose.
But the power produced at 11 kV is transmitted to distant places after increasing the voltage.
a. Why is electricity transmitted to distant places at high voltage?
b. Write down one disadvantage of transmitting electricity at high voltage.
c. What are the use of substations? Which type of transformer is used here?
Answer:
a. To reduce energy loss in the form of heat high voltage is used for long-distance transmission.
b. Transmission through populated areas is difficult. Transmission lines must be at a great height to ensure safety.
c. To reduce high transmission line voltage as and when required. Step down transformers are used for this purpose.

Question 19.
Power stations are centers where electricity is produced and distributed on a large scale.
a. What is the voltage of power production in our country?
b. Calculate the current if the power produced here in 11 MW.
Answer:
a. 11kV
b. p = VI
I = \(\frac { P }{ V }\) = \(\frac { 11 MV }{ 11 kV }\)

Question 20.
Identify the relation and fill up the blank.
Step up transformer: power transformer
Step down transformer:
Answer:
Distribution transformer.

Question 21.
Given diagram shows the star connection drawn by a child in his science diary.

a Can you draw a neutral line from the diagram? Why?
b.Draw the correct diagram of star connection.
Answer:
a. No: Three-phase lines do not meet at a single point.

Question 22.
Identical five bulbs of 230V and 40W are connected using star connection in the diagram given below.

a. What do you call the point N which is earthed in the circuit?
b. State the voltage differences between two phases and between a phase and neutral line in a star connection.
c. Which of the connected bulbs will glow? Why?
d. Two houses have electric connection from the same distribution transformer, but unfortunately one of the houses suffers power failure frequently; Explain it on the basis of household distribution.
Answer:
a. Neutral point.
b. 400 V in between two phases.
230 V in between a phase and neutral.
c. B4 glows because it is connected between the lines phase and neutral B, and B2 glows with low brightness. Each bulb gets 200 V as it is connected in series with 400 V.
d. Three phases in a transformer have separate fuses. So power supply problem in a line need not affect the other phases.

Question 23.
Examine the given circuits and answer the questions.

a. Identify series and parallel connection from the given circuits,
b. Parallel mode of connected devices is advisable for household electrification? Why?
Answer:
a. Figure A in series connection.
Figure B in parallel connection.
b. If the appliances are connected in parallel it will help to control each appliance with separate switches. It gives the needed voltage and current to appliances.

Question 24.
Four 60W, 230V bulbs are connected between a phase and a neutral as shown in the diagram. (Line voltage is 230 V)

a. How are the bulbs B, and B, are connected? What about B1 and B2?
b. Which bulb will glow at 60W power?
c. Which bulbs will glow with low power? Why?
d. Which bulbs in the circuit can be controlled using separate switches.
Answer:
a. B, and B, in series B₁ and B₂ in parallel.
b. B, and B₂.
c. B₃ and B₄ because each bulb gets less voltage as they are in series. So the bulb’s intensity will be less.
d. B₁ and B₂.

Question 25.
A flowchart of art household wiring is given in the diagram below. Fix the suitable devices in the space provided by picking from the given list.

(ELCB, MCB, Main switch, Watt-hour meter, Three-pin plug)
Answer:
(A) Main switch.
(B) ELCB.
(C) MCB.
(D) Three-pin plug.

Question 26.
We know that earth is an electron bank and that it has zero potential. But quite often we get electric shock from computer despite proper earthing.
a. Can electric shock from electric devices be avoided by earthing alone?
b. Proper earthing will give us safety from electric shock. What is meant by proper earthing? How can the earthing arrangement by done properly to ensure safety?
Answer:
a. No
b. Earth wire should carry all charge from the metal body of an electrical device to the earth effectively.