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Kerala State Syllabus 8th Standard Maths Solutions Chapter 5 Money Maths

Money Maths Text Book Questions and Answers

Textbook Page No. 90

Money Maths Class 8 Kerala Syllabus Chapter 5 Question 1.
Sandeep deposited Rs 25000 in a bank which pays 8% interest compounded annually. How much would he get back after two years?
Solution:
Interest in the first year
= 25000 × \(\frac{8}{100}\)
=Rs 2000
Principal in the second year
= 25000 + 2000
= 27000
Interest in the second year
= 27000 × \(\frac{8}{100}\)
= 2160
Total amount gets back at the end of 2 year
= 27000 + 2160
= 29160

Money Math Class 8 Kerala Syllabus Chapter 5 Question 2.
Thomas took out loan of 15000 rupees from a bank which charges 12% interest, compounded annually. After 2 years, he paid back 10000 rupees. To settle the loan, howmuch should he pay at end of three years?
Solution:
Amount borrowed = 15000
Interest for the first year
= 15000 × \(\frac{12}{100}\) = 1800
Loan for the second year
= 15000 + 1800 = Rs.16800
Interest for the second year
= 16800 × \(\frac{12}{100}\) = 2016
Loan for the end of second year
= 16800 + 2016 = 18816
Amount payback at the end of second year = 10000
Loan for the third year
= 18816 – 10000 = 8816
Interest for the 3rd year
= 8816 × \(\frac{12}{100}\) = 1057.92
Amount to be repair at the end of third year
= 8816 + 1057.92 = Rs.9873.92

Money Maths Class 8 Solutions Kerala Syllabus Chapter 5 Question 3.
Rs 200 was got as interest from a bank for 2 years at the rate 5%. Compute the compound interest for the same principal for 2 years at the same rate of interest.
Solution:
Interest for 2 years = Rs 200
Interest for one year = Rs 100
Excess amount in the 2nd year as the compound interest
= 100 × \(\frac{5}{100}\) = 5
Compound interest for two years = Rs. 205

Textbook Page No. 92

Class 8 Money Maths Kerala Syllabus Chapter 5 Question 1.
Anas deposited Rs 20000 in a bank where compound interest in computed 6% annually. How much amount he will get at the end of 3rd year?
Solution:
Amount Anas gets at the end of third year

Money Maths Class 8 Questions And Answers Kerala Syllabus Question 2.
Diya Deposited Rs. 8000 in a bank where compound interest is computed annually at 10%. Rs. 5000 was withdrawn by her at the end of 2 years. How much amount will he there in the account of Diya after 1 more year?
Solution:
Amount deposited = 8000
Rate of interest = 10%
Amount in the account of Diya at the end of 2 years
= 8000 × (1 + \(\frac{10}{100}\))²
= 8000 × \(\frac{110}{100}\) × \(\frac{110}{100}\) = Rs.9680
Amount withdrawn =Rs. 5000
Principal for the 3rd year
= 9680 – 5000 = Rs.4680
Amount she get at the end of 3 rd year
= 4680 (1 + \(\frac{10}{100}\))
= 4680 × \(\frac{110}{100}\)
= Rs. 5148

Money Maths Class 8 Pdf Kerala Syllabus Chapter 5 Question 3.
Varun borrowed rupees 25000 from a bank where compound interest in computed at 11% annually. He repaid Rs 10000 at the end of 2 years. How much amount he has to repay after one more year?
Solution:
Amount borrowed = 25000
Rate of interest = 11%
Total liability after 2 year

Amount he repaired = Rs.10000
Principal for the third year
= 30802.50 – 10000
= Rs.20802.50
Amount he has to repay at the end of third year
= 20802.50 × (1 + \(\frac{11}{100}\))
= 20802.50 × \(\frac{111}{100}\)
= Rs. 23090.78

Textbook Page No. 93

Hss Live Guru 8 Maths Kerala Syllabus Chapter 5 Question 1.
Arun deposited Rs 5000 in a bank where compound of interest is computed half yearly. Mohan deposited RS 5000 in a bank where compound interest is computed quarterly she rate of interest at both the banks is 6%. Money was withdrawn by both of them after an year. How much amount Mohan got more than Arun.
Solution:
Amount deposited by Arun = 5000.
Rate of interest = 6%
Amount Arun gets after an year
= 5000 × (1 + \(\frac{8}{100}\))²
= 5000 × \(\frac{103}{100}\) × \(\frac{103}{100}\)
= 5304.50
Amount deposited by Mohan
= Rs 5000
Rate of interest = 6%
Amount Mohan gets after 1 year

= Rs 5306.82
Amount Mohan gets more than
Arun = 5306.82 – 5304.50
= Rs 2.32

Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 5 Question 2.
Rs. 16000 was borrowed by a man from a bank where compound interest is computed quarterly. Annual interest rate is 10%. How much he has to pay after 9 months to clear his liabilities?
Solution:
Amount borrowed = Rs 16000
Interest rate = 10 %
9 Months = 3 quarterly years
Amount to be repaired after 9 months
= 16000 × (1 + \(\frac{2.5}{100}\))³
= 16000 × (\(\frac{102.5}{100}\))³
= 16000 × \(\frac{102.5}{100}\) × \(\frac{102.5}{100}\) × \(\frac{102.5}{100}\)
= 17230.25

Hss Live Guru 8th Maths Kerala Syllabus Chapter 5 Question 3.
Manu deposited Rs 15000 in a financial institution. Interest is calculated in every 3 months and added to the amount. Rate of interest in 8 %. How much he gets after 1 year.
Solution:
Amount deposited = Rs 15000
Rate of interest = 8%
Amount he gets after 1 year

Kerala Syllabus 8th Standard Notes Maths Chapter 5 Question 4.
John deposited Rs 2500 on 1st January in a co-operative bank. Bank computes compound interest half yearly. Annual interest rate is 6%. Again he deposited Rs 2500 on 1st July? How much amount he will have in his account at the end of the year?
Solution:
Amount John deposits on 1st January = Rs.2500
Principal after the half year up to July
1 st = 2500 (1 + \(\frac{3}{100}\))
= 2500 × \(\frac{103}{100}\)
= Rs.2575
Amount deposited in July 1st
= Rs 2500
Principal after July 1st= 2575 + 2500 = 5075
Amount he gets at the end of the year 3
= 5075 × (1 + \(\frac{3}{100}\))
= 5075 × \(\frac{103}{100}\)
= Rs 5227.25

Hss Live Guru Class 8 Maths Kerala Syllabus Chapter 5 Question 5.
Ramlath deposited Rs 30,000 in a financial institution where compound interest is computed in every four months. The ann¬ual interest rate in 9%. How much amount Ramlath gets after 1 year?
Solution:
Amount deposited =Rs 30,000
Rate of interest = 9 %
Amount gets after 1 year
= 30000 (1 × \(\frac{3}{100}\))³
= 30000 × \(\frac{103}{100}\) × \(\frac{103}{100}\) × \(\frac{103}{100}\)
= Rs.32781.81

Textbook Page No. 95

Hss Live Guru 8 Kerala Syllabus Chapter 5 Question 1.
The e-waste increases by 15 % every year, according to the study report. There was around 9 crore tonnes of e-waste in 2014. Then how many tonnes of e-waste will be there in 2020.
Solution:
E-waste in 2014 – 9 crore tonnes
Rate of increase = 15 %
The E-waste in 2020 = 9 × (1 + \(\frac{15}{100}\))⁶
= 9 × (\(\frac{115}{100}\))⁶
= 20. 82 crore tonnes

Hss Live 8 Maths Kerala Syllabus Chapter 5 Question 2.
A T.V manufacturer reduces the price of a particular model by 5% every year. The current price of this model is Rs. 8000. What would be the price after 3 years?
Solution:
The present price of the T.V = Rs. 8000
rate of reduction = 5% ,
Price after 2 years = 8000 (1 – \(\frac{5}{100}\))²
= 8000 × (\(\frac{95}{100}\))²
= 8000 × \(\frac{95}{100}\) × \(\frac{95}{100}\)
= Rs.7220

Class 8 Maths Hsslive Kerala Syllabus Chapter 5 Question 3.
Tiger is our national animal. Ac-cording to a statistics the number of tigers is reduced annually by 3%. There are 1700 tigers in 2011 according to the senses or tiger protection authority. Then how many tigers will be there in 2016?
Solution:
The number of tigers in 2011 = 1700
Reduction rate = 3%
No. of tigers in 2016 after 5 years of the senses

= 1459.84
= 1460

Money Maths Additional Questions and Answers

Kerala Syllabus 8th Standard Maths Notes Chapter 5 Question 1.
Nirmala deposited Rs. 20,000 in a bank where compound interest of rate 7% is computed annually Rs. 5000 was with drawn after one year. How much amount she will get after 2 years.
Solution:
Amount deposited in the bank = Rs. 2000
Rate of interest = 7%
principal after one year
= 20000 + [20000 × \(\frac{7}{100}\)] = 20000 + 1400 = 21400
Amount withdrawn after one year = Rs. 5000 .
Principal for the 2-nd year
= 21400 – 5000 = Rs. 16400
Amount she gets after 2 years
= 16400 + [16400 × \(\frac{7}{100}\)] = 16400 + 1148
= Rs. 17548

Question 2.
Aswathy deposited Rs. 10,000 in a bank where compound interest is calculated at the rate of 10%. Rs.5000 was deposited in the beginning of the 2-nd year and Rs.5000 in the beginning of the 3-rd year. How much amount she gets after 3 years?
Solution:
Amount deposited in the first year = Rs. 10,000
rate of interest = 10%
Amount at the end of first year
= 10000 + [10000 × \(\frac{10}{100}\)] = 10000 + 1000 = Rs. 11000
Principal’for the 2-nd year
= 11000 + 5000 = Rs. 16000
Amount at the end of the 2-nd year
= 16000 + [16000 × \(\frac{10}{100}\)] = 16000 + 1600 = Rs. 17600
Principal for the 3-rd year
= 17600 + 5000 = Rs.22600
Amount at the end of 3-rd year
= 22600 + [22600 × \(\frac{10}{100}\)] = 22600 + 2260 = Rs. 24860
Amount Aswathy will get at the end of 3-rd year = Rs. 24860

Question 3.
Raj an borrowed Rs. 15000 from a co-operative bank for business purpose. The bank computes 9% interest. How much money she has to repay after 5 months.
Solution:
Amount borrowed = Rs. 15000
Rate of interest = 9%
Time duration = 5 month months
Interest .
= 15000 × \(\frac{9}{100}\) × \(\frac{5}{12}\) = 562.50
Amount he has to repay
= 15000 + 562.50
= Rs 15562.50

Question 4.
Rasiya deposited Rs. 20,000 in a bank where compound interested is computed half yearly. If the rate of interest is 11%, How much amount she will get after an year?
Solution:
Capital for first year = Rs. 20,000
Rate of interest = 11 %
Interest of first half year
= 20000 × \(\frac{11}{100}\) × \(\frac{1}{2}\) = Rs. 1100
Principal of 2-nd half year
= 20000 + 1100
= 21100
Interest of the 2-nd half year
= 21100 × \(\frac{11}{100}\) × \(\frac{1}{2}\) = Rs. 1160.5
Amount she gets after one year
= 21000 + 1160.5
= Rs. 22160.5

Question 5.
The price of a car is rupees 5 lakh and it depreciates by 6% every year. What would be the price after 2 year ?
Solution:
Here the price every year is 6% less than the previous years price.
First year’s price = Rs 500000
First year’s depreciation
= 500000 × \(\frac{6}{100}\) = Rs. 300000
Second year’s price = Rs 4,70000
Second year’s depreciation
= 470000 × \(\frac{6}{100}\) = 28200
The price of the car after 2 years
= Rs 4,70,000 – Rs 28200
= Rs 441800

Question 6.
The simple interest of an amount is Rs 50 for two years and the compound interest is Rs 55. The rate of interest is same in both the cases. Find the rate? Find the amount?
Solution
The simple interest of the amount for two years = Rs 50
The simple interest of the amount for 1 year = Rs 25
The compound interest of the amount for 2 years = Rs 55
The interest in the 2 nd year
= 55 – 25 = Rs 30
Interest for Rs 25 = Rs 5

Question 7.
A company which manufactures computers increases its production by 10% every year. In 2009 the company produced 80,000 computers. How many computers would it produced in 2011?
Solution:
Here the number of computers produced every year is 10% more than the number produced the year before. So starting from 80,000. We have to find the number of computers produced every year after that for two years.
Number of computers produced in 2009 = 80,000
Number of computers produced in 2010
= 80000 + 80000 × \(\frac{10}{100}\)
= 80000 + 8000 = 88000
Number of computers produced in 2011
= 88000 + 88000 × \(\frac{10}{100}\)
= 88000 + 8800 = 96800

Question 8.
Ramu borrowed Rs 50,000 from a bank at the interest rate of 10% for agricultural purpose. The interest rate will be reduced by 5% if he repays the amount properly within 2 years. If he fails to repay in time there will be fine as 1%. Ramu could not repay the amount in time. How much amount Ramu repair?
Solution:
Amount borrowed =Rs 50,000
Rate of interest = 10%
Interest rate including fine = 10 + 1=11%
Amount to be repaid
= 50000 + [50000 × \(\frac{11}{100}\) × 2]
= 50000 + 11000 = 61000
= Rs 61000

Question 9.
Raju has Rs 800 with him. He spent 25 % of it. How much amount left with him?
Solution:
Total amount = Rs 800
Amount spent = 800 × \(\frac{25}{100}\) = 200
Amount left with Raju = 800 – 200
= Rs 600

Question 10.
Balu and Ramu decided to borrow Rs 15000 each for a joint business. Balu borrowed Rs 15000 from a financier Who imputes Rs 5 per month for Rs 100 as interest. Ramu borrowed Rs 15000 from a bank where compound interest of 12 % is computed. How much money both of them have to repay after 2 years?
Solution:
Interest Balu has to pay after 2 years
= 15000 × \(\frac{60}{100}\) × 2 = Rs. 18000
(Interest for Rs 100 per month in Rs 5. Interest for Rs 100 in an year = 5 × 12
= Rs. 60.
The rate of interest = 60%)
Amount Balu has to repay after 2 years
= 15000 + 18000
= Rs 33000
In the case of Ramu interest is computed as compound interest.
Principal for first year = Rs 15000
Interest for 1 st year
= 15000 × \(\frac{12}{100}\) × 1 = Rs 1800
Principal for 2 nd year
= 15000 + 1800
= Rs 16800
Interest for the second year = 16800 × \(\frac{12}{100}\) × 1 = Rs. 2018
Amount Ramu has to repay
= 16800 + 2018
= Rs 18816

Question 11.
The population of Kerala increases by 3% every year. The current population is 5 crore. What would he the population after 2 years.
Solution:
Current population = 50000000
Percentages increase in every year = 3%
The population in Kerala after 3 year
= 50000000 [1 + \(\frac{3}{10}\)]²
= 50000000 × \(\frac{(103)^{2}}{10000}\)
= 530,45000

Question 12.
A financial company claims that it charges only 20% interest on loans. But if a person takes out a loan of Rs 100 he would get only Rs 80, after subtracting the annual interest of rupees 20 at the outset. And he has to pay back Its 100 after 1 year. How much is their real interest?
Solution:
A borrower gets only Rs 80 when he borrows Rs 100
He has to repay Rs 100.
ie. Rs.20 as additional.
Real interest
\(\frac{20}{80}\) × 100 = 25%

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

Kerala State Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures

Circle Measures Textual Questions and Answers

Textbook Page No. 131

Circle Measures Class 9 Kerala Syllabus Question 1.
Prove that the circumcentre of an equilateral triangle is the same as its centroid.
i. Calculate the length of a side of an equilateral triangle with vertices on a circle of diameter 1 centimetre.
ii. Calculate the perimeter of such a triangle.
Answer:
Perpendicular bisector of sides of a triangle that can meet at a point is its circumcenter

Since the triangle is equilateral the perpendicular bisectors of the sides are also median. Since the triangle is equilateral the perpendicular bisectors of the sides are also centroid. That is circumcentre of an equilateral triangle is the same as its centroid.
i. Length of one side of an equilateral triangle is.

Circle Measures Class 9 Scert Chapter 9 Kerala Syllabus Question 2.
Calculate the perimeter of a square with vertices on a circle of diameter 1 centimetre.
Answer:
In the square ABCD
AO = 1/2 cm
In A AEO, ∠EAO = 45°
∠AEO = 90°
Since AO = 1/2 cm
AE = \(\frac{1}{2 \sqrt{2}} \mathrm{cm}\)
(Since Δ AEO is a isosceles right triangle, hypotenuse is √2 times of a perpendicular side.)

Perimeter = \(=4 \times \frac{\sqrt{2}}{2}=2 \sqrt{2} \mathrm{cm}\)

Circle Measures Class 9 Chapter 9 Kerala Syllabus Question 3.
Calculate the perimeter of a regu¬lar hexagon with vertices on a circle of diameter 1 centimetre.
Answer:
If we draw diagonals through centre of circle inside the regular hexagon it divides the regular hexagon into 6 equilateral triangles.
Let diameter be 1 cm
OA = 1/2 cm
OA = OB = AB.
therefore
One side = 1/2 cm
Perimeter of regular hexagon = 6 × 1/2 = 3 cm

Textbook Page No. 134

Hss Live Guru 9th Maths Chapter 9 Kerala Syllabus Question 1.
The perimeter of a regular hexagon with vertices on a circle is 24 centimetres.
i. What is the perimeter of a square with vertices on this circle?
ii What is the perimeter of a square with vertices on a circle of double the diameter?
iii. What is the perimeter of an equilateral triangle with vertices on a circle of diameter half that of the first circle?
Answer:
Perimeter of a regular hexagon = 24 cm
Length of one side of a regular hexagon = 24/6 = 4 cm
Length of one side of a regular hexagon is equal to the radius of the circle.
i. Diagonal of a square = 8 cm Diagonal of a square is equal to the diameter of the circle.
Let a be the side of the square
a² + a² = 8²
2a² = 64
a² = 64/2 = 32
a= 4√2
∴ One side of a square = 4√2 cm
Perimeter of a square = 4 × 4√2 cm = 16√2 cm

ii. The perimeter of a square with verti¬ces on a circle of double the diameter 2 × 16√2 = 32 √2 cm
(The perimeters of circles are scaled by the same factor as their diameters.)

iii. One side of an equilateral triangle with vertices on a circle of half the diameter of the first circle
\(=2 \sqrt{2^{2}-1^{2}}=2 \sqrt{3} \mathrm{cm}\)
Perimeter = 3 × 2√3 = 6√3 cm

Circles Class 9 State Syllabus Chapter 9 Kerala Syllabus Question 2.
A wire was bent into a circle of diameter 4 centimetres. What would be the diameter of a circle made by bending a wire of half the length?
Answer:
The ratio between the perimeters are equal to the ratio between their diameters. The perimeter of the first circle is twice the perimeter of the second circle.
Therefore diameter of the second circle is half of the diameter of the first circle. Diameter of the second circle.
= 4/2 = 2 cm

Kerala Syllabus 9th Standard Maths Solutions  Question 3.
The perimeter of a circle of diameter 2 metres was measured and found to be about 6.28 metres. How do we compute the perimeter of a circle of diameter 3 metres, without measuring?
Answer:
If diameter is 2 meters, perimeter is 6.28 meter.
If the diameter is 1 metre, perimeter is 6.28/2 meter.
If the diameter is 3 metres, perimeter = \(\frac{6.28}{2} \times 3=9.42 m\)

Textbook Page No. 137

Hss Live Guru Maths 9 Chapter 9 Kerala Syllabus Question 1.
In the pictures below, a regular hexagon, square and a rectangle are drawn with their vertices on a circle. Calculate the perimeter of each circle.

Answer:
a. AB = 2 cm
In the figure triangle are
equilateral triangles,
therefore
radius OA = 2 cm
Perimeter of circle = 2 πr
= 2 × π × 2 = 4π cm

b. ABCD is a square
AB = BC = 2 cm, ∠5 = 90°
AC = \(\sqrt{2^{2}+2^{2}}=\sqrt{8}=2 \sqrt{2}\)

Radius of circle = 1/2 × 2√2
= √2 cm
Perimeter of circle = 2π × √2 cm
= 2√2 π cm

c. PR = \(\sqrt{2^{2}+(1.5)^{2}}\)
= \(\sqrt{6.25}=2.5 \mathrm{cm}\)
Radius of circle = 1/2 × 2.5 = 1.25 cm
Perimeter of circle = 2 × π × 1.25
= 2.5 π cm

Kerala Syllabus 9th Standard Maths Notes Question 2.
An isosceles triangle with its vertices on a circle is shown in this picture.

What is the perimeter of the circle
Answer:
Consider the centre of circle as O and triangle as ABC
OC = Radius of circle = r
OD = 4 – r
AD = 2 cm
AO = r

Therefore, in triangle AOD
(AO)² = (AD)² + (OD)²
r²=2² + (4 – r)²
r²= 4 +16 – 8r + r²
8r=20;
r = 20/8 = 5/2 = 2.5 cm
∴ Perimeter = 2π × r = 2π × 2.5 cm
= 5π cm

Circles Class 9 Kerala Syllabus Chapter 9 Question 3.
In all the pictures below, the centres of the circles are on the same line. In the first two pictures, the small circles are of the same diameter.

Prove that in all pictures, the perimeters of the large circle is the sum of the perimeters of the small circles.
Answer:
a. Smaller circles have same diameters. Consider the diameter as d, perimeter of smaller circle
= π × diameter = π d
Perimeter of two small circles = 2πd

Diameter of the large circle = d + d = 2d
Perimeter of the large circle
= π × 2d = 2πd
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

b. Let d be the diameter of one small circle, then perimeter = π d
Sum of perimeter of three small circles =3 π d
Diameter of the large circle = d + d + d = 3d
Perimeter of the large circle = π × 3d = 3 π d
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

c. In figure diameter of three circless are different, let consider the diameters of small circles are p, q and r.

Perimeter of first small circle = πp
Perimeter of second small circle = πq
Perimeter of third small circle = πr
Sum of perimeters of three small circles
= πp + πq + πr = π (p + q + r)
Diameter of large circle = p + q + r
Perimeter of large circle = π (p + q + r)
Therefore also here the perimeters of the large circle is the sum of the perimeters of the small circles.

Hss Live Class 9 Maths Chapter 9 Kerala Syllabus Question 4.
In this picture, the circles have the same centre and the line drawn is a diameter of the large circle. How much more is the perimeter of the large circle than the perimeter of the small circle

Answer:
If ‘r’ is the radius of the small circle,
radius of the large circle = r + 1
Perimeter of the small circle = 2 π r
Perimeter of the large circle = 2 π (r + 1) = 2 π r + 2 π
i.e., perimeter of the large circle is 2 π units more than the perimeter of the small circle.

Textbook Page No. 141

Hss Live Guru 9 Maths Chapter 9 Kerala Syllabus Question 1.
In the pictures below, find the difference between the areas of the circle and the polygon, up to two decimal places.

Answer:
i. Radius of the small circle = 2cm
Area = π × 2² = 3.14 × 4
= 12.56 cm²
Daigonal of the square = 4 cm
One side of the square = 4/√2 cm
Area of the square = \(\frac{4}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=\frac{16}{2}=8 \mathrm{cm}\)
Differences between the areas = 12.56 – 8 = 4.56 cm

ii. Radius of the circle = 2 cm
Area= π × 2² = 3.14 × 4 = 12.56 cm²
One side of the regular hexagon=2 cm
Area of the regular hexagon = \(6 \times \frac{\sqrt{3} \times 2^{2}}{4}=6 \sqrt{3}\)
= 6 × 1.73 = 10.38 cm²
Differences between the areas = 12.56 – 10.38 = 2.18 cm²

9th Std Kerala Syllabus Maths Solutions Question 2.
The pictures below show circles through the vertices of a square and a rectangle

Calculate the areas of the circles
Answer:
i. One side of a square is 3 cm, therefore its diagonal is 3√2 cm
Diameter of the circle = 3√2 cm

ii. Rectangle inside the circle having length 4 cm and breadth 2 cm.
Diagonal = \(\sqrt{4^{2}+2^{2}}=\sqrt{16+4}\)

Diameter of the circle = √20 cm
Radius = \(\frac{\sqrt{20}}{2} \mathrm{cm}\)
Area = \(\pi \times\left(\frac{\sqrt{20}}{2}\right)^{2}=\pi \times \frac{20}{4}\)
= 5 π cm²

Hsslive Guru 9th Maths Chapter 9 Kerala Syllabus Question 3.
Draw a square and draw circles cen¬tered on the corners, of radius half the side of the square. Draw another square formed by four of the first square and a circle just fitting into it.

Prove that the area of the large circle is equal to the sum of the areas of the four small circles
Answer:
In the figure length of one side of the square is 2r
Radius of one small circle = r
Perimeter of one small circle = π r²

Radius of four small circles = 4 π r²
One side of a square in the second figure = 2r + 2r = 4r
Radius of the circle in the figure
= 4r/2 = 2r
perimeter of the circles in the figure = π × (2r)²
= π × 2r × 2r = 4πr²
The area of the large circle is equal to the sum of the areas of the four small circles

Hsslive 9th Maths Chapter 9 Kerala Syllabus Question 4.
In the two pictures below, the squares are of the same size.

Prove that the green regions are of the same area.
Answer:
Let ‘a’ be the side of the square in the picture.
Area of the square = a²
If the four sectors in the vertices are joined together, a circle is formed because the radius of each sector is a/2.
Area of the shaded part = Area of the square – area of the circle.

ii. In the second picture diameter of the
circle = a Radius = a/2
Area of the shaded part = Area of the square – area of the circle.

i.e., Areas of the shaded portions are equal.

Hsslive Guru 9 Maths Chapter 9 Kerala Syllabus Question 5.
Parts of circles are drawn inside a square as shown in the picture below. Prove that the area of the blue region is half the area of the square.

Answer:
Let ‘a’ be the side of the square
Area of the blue part in the half portion of the square is equal to half of the area of the circle having diameter ‘a’.
Area of the blue part in the half portion of the square \(=\frac{1}{2} \times \pi\left(\frac{a}{2}\right)^{2}=\frac{\pi a^{2}}{8}\)
We must subtract area of two circles
having diameter a/2 from the half the
area of the square to get the area of remaining blue shaded part.
Area of remaining blue shaded part.
= \(\frac{a^{2}}{2}-2 \times \frac{1}{4} \times \pi\left(\frac{a}{2}\right)^{2}\)
Area of blue shaded part.
= \(\frac{\pi a^{2}}{8}+\frac{a^{2}}{2}-\frac{\pi a^{2}}{8}=\frac{a^{2}}{2}\)
i.e., area of the blue part is equal to half the area of the square.

Hss Guru 9 Maths Chapter 9 Kerala Syllabus Question 6.
In the figure, semicircles are drawn with the sides of a right triangle as diameter.

Prove that the area of the largest semicircle is the sum of the areas of the smaller ones.
Answer:

In ΔABC, ∠5 = 90°
According to Pythagoras principle,
AB² + BC² = AC² ………. (1)
Radius of the semicircle with diameter
AB = AB/2
Area of the semicircle with diameter AB

= \(\frac{1}{2} \times \pi \times \frac{A C^{2}}{4}=\frac{\pi}{8} A C^{2}\)
Sum of the areas of the smaller semicircles = \(\frac{\pi}{8} A B^{2}+\frac{\pi}{8} B C^{2}=\frac{\pi}{8}\left(A B^{2}+B C^{2}\right)\)
= π/8 AC² = Area of the largest semicircle

Textbook Page No. 148

Hsslive Maths Class 9 Chapter 9 Kerala Syllabus Question 1.
In a circle, the length of an arc of central angle 40° is 3 π centimetres. What is the perimeter of the circle? What is its radius?
Answer:

Question 2.
In a circle, the length of an arc of central angle 25° is 4 centimetres.
i. In the same circle, what is the length of an arc of central angle 75°?
ii. In a circle of radius one and a half times the radius of this circle, what is the length of an arc of central angle 75°?
Answer:
i. Length of the arc having central angle 25° = 4 cm
Three times of 25 is 75.
Length of the arc having central angle 75° = 4 x 3 = 12 cm

ii. Length of the arc having central angle 75° and radius r = 12 cm
Length of the arc having central angle 75° and radius 1 1/2 r
= \(12 \times 1 \frac{1}{2}=18 \mathrm{cm}\)

Question 3.
From a bangle of radius 3 centimetres, a piece is to be cut out to make a ring of radius ^ centimetres.
i. What should be the central angle of the piece to be cut out?
ii. The remaining part of the bangle was bent to make a smaller bangle. What is its radius?
Answer:
Perimeter of the bangle having radius 3 cm = 6 π cm.
Perimeter of the bangle having radius 1/2 cm = π cm.
π is the 1/6 part of 6 π.
Therefore the central angle of the piece to be cut out = 360 × 1/6 = 60°
ii. Length of the remaining part of the bangle = 6π – π = 5π cm
Radius of the other small bangle = 5 π / 2π = 2.5 cm

Question 4.
The picture shows the parts of a circle centred at each vertex of an equilateral triangle and passing through the other two vertices.

What is the perimeter of this figure?
Answer:
Since the triangle is equilateral, each angle is 60°. There are in each side is in a circle of radius 4 cm and the central angle is 60°.

Question 5.
Parts of circles are drawn, centred at each vertex of a regular octagon and a figure is cut out as show below:

Calculate the perimeter of the fig¬ure.
Answer:
Sum of the angles in an octagon
= (n-2) × 18o° = 6x 180°= 1080° One angle of the regular octagon
= 1080 – 8 = 135°
One side of the regular octagon = 2 cm. Radius if the sectors having centre is each vertices of the circle= 1 cm The second picture shows the cut-down form of 8 sectors having centre angle 135° and radius 1 cm.
The perimeter is found by calculating the length of 8 arc having radius 1 cm and central angle 135°.

Textbook Page No. 151

Question 1.
What is the area of a sector of central angle 120° in a circle of radius 3 centimetres? What is the area of a sector of the same central angle in a circle of radius 6 centimetres?
Answer:
Area of the sector in the circle having radius 3 cm and angle of center 120°
π × 3² × \(\frac { 120 }{ 360 }\) = 3π cm²
Area of the sector in the circle having radius 6 cm and angle of center 120°
= π × 6² × \(\frac { 120 }{ 360 }\) = 12π cm²

Question 2.
Calculate the area of the green coloured part of this picture.

Answer:
In the picture area of the shaded part is the difference between the area of the two sectors.
Area of the large sector

Area of the shaded part = 9.42 – 4.19 = 5.23 cm²

Question 3.
Centred at each corner of a regular hexagon, a part of a circle is drawn and a figure is cut out as shown below:

What is the area of this figure?
Answer:
The area of the cut-down portion = Area of the regular hexagon – Area of 6 sectors Area of the regular hexagon

Area of the sector = 120°/360° part of area of the circle
(One angle of a regular hexagon is 120°)

Area of the 6 sectors = \(\frac{6 \times \pi}{3}=2 \pi \mathrm{cm}^{2}\)
Area of cut down portion = 6 √3 – 2 π cm²

Question 4.
The picture below shows two circles, each passing through the centre of the other:

Calculate the area of the region common to both.
Answer:
Consider the picture given below, we can divide the circle into two part by using the line AB, the area of the two portions are same.
That is we get the total area by multi¬plying area of the sector by 2.

We can find out the area of part above the line AB .
Given AB = 2cm
Circles having equal
radius. So,
AC = BC = 2 cm.
AABC is an equilateral triangle so angles are 60° each.
Add the area of sectors having centre A and B .
The area of ΔABC include twice, so we will subtract it once.
Area of sectors having centre A

Area of sectors having centre B

Area of ΔABC

Area of the part above the line AB = 2.09 + 2.09 – 1.7 = 2.45 cm²
The area of the region common to both = 2 × 2.45 = 4.90 cm²

Question 5.
The figure shows three circles drawn with their centres on each vertex of an equilateral triangle and passing through the other two ver¬tices;

Find the area of the region common to all three.
Answer:
The area of the common part = Area of three sectors – 2 × area of an equilateral triangle having side 2 cm

= 2 × 3.14 – 2 × 1.73
= 6.28 – 3.46
= 2.82 cm²
The area of the region common to all three = 2.82 cm²

Circle Measures Exam Oriented Questions And Answers

Question 1.
In the picture PQRS is a square of side 10 cm. A, B, C and D are midpoints of the sides of the square

Semi circles are drawn inside the square.
a. Compute the area of the square.
b. Compute the area of the semi-circle.
c. Compute the area of the shaded part.
Answer:
a. Since one side of the square is 10 cm,
Area = side x side = 10 × 10= 100 cm²

b. The diameter of one semicircle = half of the side of the square
Diameter = 5 cm
∴ Radius = 5/2 cm
Since the four semicircles are equal. Area of the 4 semicircles
= Area of 2 circles = 2 × πr²

c. Area of the shaded part = Area of the square – Area of four semicircles.

Question 2.
If a circular shaped dining table has an area of 31400 cm², find its radius. What will be its perimeter ?
Answer:
Area of the table = 31400 cm²

Question 3.
In the given picture shown two semicircular shaped iron bar can be cut down from a rectangular shaped iron bar. Calculate the area of the remaining shaded portion.

Answer:
Area of the rectangle = 24 × 14 = 336 cm²
Area of two semicircle = Area of a complete circle
Diameter of the circle = 14 cm
Radius = 7 cm
Area of the circle = π r²
= π × 72 =49
π = 153.86 cm2 Area of the remaining portion
= 336 – 153.86= 182.14 cm²

Question 4.
In the figure A, B, C and Dare the points on the square which touches the circle. If the radius of the circle is 6.
a. What is the length of one side of the square?
b. Find the area of the circle.
c. What will be the area of the shaded portion?

Answer:
a.Diameter of the circle = 6 × 2 = 12 cm
Side of the square = 12 cm
b. Area of the circle = n r²
= n × 6 × 6 = 36n =36 × 3.14 =113.04 cm²
c. Area of the square = 12 × 12 = 144 cm²
Area of the shaded portion = 144 – 113.04 = 30.96 cm²

Question 5.
In the ACB is the arc drawn by taking O as the centre and OA as the radius. Then find the area of the shaded region.

Answer:
The area of sector which has 7 cm radius and 90° central angle =

Area of the right angled triangle AOB
\(=\frac{7 \times 7}{2}=24.5\)
Area of the shaded part = 38.455 – 24.5 = 13.965 cm²

Question 6.
In the pictures given below, find the area of the shaded part?

Answer:
’The radius of the sector in the picture (1) is 5cm and its central angle is 40°.
Area of the shaded part \(=\pi \times 5^{2} \times \frac{40}{360}=\frac{25}{9} \pi \mathrm{cm}^{2}\)
The radius of the sector in the picture (2) is 6cm and its central angle is 300° (360 – 60).
Area = \(\pi \times 6^{2} \times \frac{300}{360}=36 \pi \times \frac{5}{6}\)
= 30 π cm²

Question 7.
What amount of reeper is needed to enclose a circular shaped dining table of area 6.28 cm²?
Answer:

Question 8.
A wheel which has 20 cm radius is rotating forward. After 10 rotations what distance will the wheel travel forward?
Answer:
Radius of the wheel = 20 cm
When the wheel rotates once it will travel the distance same as its area
Perimeter of the wheel = 2 × π × radius = 2 × 3.14 × 20 = 125.64cm
The distance travelled forward when the wheel rotates once = 125.64 cm
The distance travelled forward when the wheel rotates 10 times
= 125.64 × 10 = 1256.4 cm

Question 9.
In the picture the central angles of both the sectors are equal. Sum of the radii of the sectors is 18 cm.
Area of the shaded part is 18 π cm². Find the radii of the sector.

Answer:
‘Let ‘r’ be the radius of the small sector and ‘R’ be the radius of the large sector.
R + r = 18 …………. (1)
Area of the shaded portion = Area of the large sector – area of the small sector

Question 10.
In the figure O is the radius of the circle and OABC is a rectangle. OA = 8 cm, OC = 15 cm. Hence find the Area of the shaded portion

Answer:

Area of the shaded portion

Question 11.
The wheel of a vehicle has a diameter 60 cm. For this vehicle travels a distance of 200 m, how many times must this wheel rotates.
Answer:
‘The distance travelled when the wheel is rotated once =2 π r = 2 × π × 30
= 188.4 cm = 1.884 m.
The time required for the wheel to travel a distance of 200m = 200/1.884 = 106.16 = 106

Question 12.
In the below-given figures, there are two circles with the same centre. Then find the area of the second region.

Answer:
a. Area of the shaded portion =Area of the outer circle – Area of the inner circle
= π R² – π F= π × 10² – π × 8²
= 100π – 64π = 36π
=36 × 3.14 = 113.04 cm²

b. Outer radius = 7 + 2 = 9 cm Inner radius = 7 cm
Area of the shaded portion = π × 92- π × 72 = 81π – 49π = 32π = 32 × 3.14 =100.48 cm²

c. Outerradius = 10.5
Inner radius = 10.5 – 1 = 9.5
Area of the shaded portion = π × (10.5)² – π × (9.5)²
= π × (10.5² – 9.5)²
= π × (10.5² – 9.5²)
= π × (10.5 + 9.5) (10.5 – 9.5) = π × 20 × 1 = 62.8 cm²

d. Outer radius = \(\frac { 16π }{ 2π }\) = 8
Inner radius = \(\frac { 14π }{ 2π }\) = 7
Area of the shaded portion

Question 13.
Two semicircular pieces are cut out from a rectangular sheet. Find the area of the remaining portion.

Answer:
Area, of the rectangle = 50 × 20 = 1000 cm2 If the two semicircles cut out are joined it becomes a circle
Its radius = 20/2 = 10 cm
Area of the portion cut out
= πr² = π × 10 × 10 = 3.14 × 10 × 10 = 314 cm²
Area of the remaining portion
= 1000 – 314 = 686 cm²

Question 14.
If a square, equilateral triangle, regular hexagon and circle have the same perimeter. Which of these has the largest area ?
Answer:
If a square, equilateral triangle, regular hexagon and circle has a perimeter of 12 cm.
Equilateral triangle
One side = 12/3 = 4


Circle has the largest area.

Question 15.
In the figure, if ACB is the arc of circle having O as the centre and OA as the O, radius. Then find the area of the shaded portion

Answer:
Radius of the sector = 8cm Central angle = 90°

OBA is a right angled triangle
∴ Area of ΔOBA = 1/2 × 8 × 8 = 32 cm²
Area of shaded portion = 50.24 – 32 = 18.32 cm²

Question 16.
The area of an equilateral triangle is 17300 cm². Draw circles with radius half the length of one side of the triangle and vertices as the centre of the circle. Calculate the area of the shaded portion.
Answer:
Area of the equilateral triangle

One side = 200 cm,
Radius of the circle = 100 cm
Central angle of each sector = 60°
Area of one sector = \(\frac{\pi r^{2} \times 60^{\circ}}{360}\)
Area of the three sectors

Question 17.
In the figure, a side of the regular hexagon has length 20 cm. Find the area of the shaded portion.

Answer:
We have learnt that length of a side of the regular hexagon drawn inside a circle will be equal to the radius of the circle.
So the radius of the circle will be 20 cm.
Area of the circle = πr² = π × 20 × 20 = 400π = 1256 cm²
Area of the regular hexagon \(=\frac{6 \times \sqrt{3} \times a^{2}}{4}\)

Area of shaded portion = 1256 – 1039.2 = 216.8 cm²

Question 18.
If radius of a sector is 7 cm and its perimeter is 2.5 cm. Then find its area.
Answer:
Area of sector
= 2 × radius + length of the arc = 25
Length of the arc = 25 – 2 × radius = 25 – 14 = 11 cm
Let us check the ratio between the length of the arc and its area
The length of the arc: Area of the sector

11 : area of the sector = 2:7
Area of two sectors =7 × 11
Area of the. sector = \(\frac{7 \times 11}{2}\) = 38.5 cm²

Question 19.
In the figure if the length of one side of a square is 12 cm, then find the area of the shaded portion.

Answer:
Area of the square = 12² = 144
Diameter of the circle = 12; r = 6 Area of the circle
= πr² = π × 6 × 6 = 3.14 × 6 × 6 = 113.04 cm²
Area of shaded portion = 144 – 113.04 = 30.96 cm²

Kerala State Syllabus 10th Standard Social Science Notes Chapter 2 World in the Twentieth Century

Twentieth century is the period that influenced world history greatly. Imperialism was the developed form of capitalism which emerged in Europe after the Industrial Revolution. When capitalism developed into imperialism, it faced many crises. When the imperialist powers entered into mutual competitions in order to conquer the world, conflict became widespread. The international problems which surfaced during this period caused mutual mistrust and enmity. The conflicts among the imperialist powers ultimately led the entire world to a war. The growth of Fascism and Nazism, Second World War, international efforts for peace, cold war, non-alignment and globalisation are the other topics discussed in this unit.

→  Industrial Revolution: The changes which took place in England from the 18th century onwards came to be known as the Industrial Revolution. The basic feature of this was that human labour was substituted by machines.

→  Capitalism : The economic system in which production and distribution are controlled by capitalists with the aim to increase profit.

→  Imperialism : The practice of extending a nation’s political, economic and cultural dominance on another nation is imperialism.

→  Triple Alliance and Triple Entente : Military alliances that fought in the First World War.

→  Nationalism : A nation is defined as a people settling in a definite territory, speaking a common language, having a common culture and historical tradition. The ideology and programme of action based on this concept is called nationalism.

→  Aggressive nationalism : The policy of invading neighbouring countries, considering one’s nation as supreme and justifying whatever be the actions of the nation.

→  Pan-Slav Movement : The movement started under the leadership of Russia to unite the Slavic People of Serbia, Bulgaria, Greece, etc. in Eastern Europe.

→  Pan-German Movement : The movement started under the leadership of Germany to establish her dominance in Central Europe and Balkan provinces and to unite the Teutonic people.

→  Revenge Movement : The movement started under the leadership of France to regain her territories of Alsace-Lorraine which were captured by Germany in the Franco-Prussian war of 1871.

→  Treaty of Versailles : The treaty signed with Germany by the victorious powers after the First World War.

→  World Economic Depression : The economic crisis that started in 1929 and affected the whole world.

→  League of Nations : The international organisation formed after the first world war to maintain peace in the world.

→  Fascism, Nazism : The political ideology that supported dictatorship, racial superiority, aggressive nationalism and single party rule.

→  Munich Pact : The agreement that approved the claim of Germany over Sudetanland, a part of Czechoslovakia.

→  Policy of appeasement: Capitalist countries like Britain and France considered Soviet Union, being a socialist country, as their chief enemy and did not prevent fascist attacks. This policy which encouraged fascist attacks is known as policy of appeasement.

→  Non-Aggression Pact : The agreement signed between Germany and Soviet Union in 1939, by which they agreed not to attack each other and to partition Poland.

→  Teutonic People : The Germanic people are also called Teutonic peoples. Originally they belonged to Northern Europe. They spoke languages of the Germanic branch of the lndo-European language family.

→  Pearl Harbour Attack : The attack of Japan in 1941 on Pearl Harbour, the American naval base in the islands of Hawaii.

→  Hiroshima, Nagasaki : Japanese cities where atom bombs were dropped in 1945 by USA.

→  Hibakusha: The surviving victims of the atomic bombings of Hiroshima and Nagasaki.

→  United Nations Organisation : The international organisation formed after the Second World War to prevent war and maintain peace in the world.

→  Decolonization : The process of the colonies of Asia and Africa securing freedom from imperialist control.

→  Cold War : The enmity based on ideological conflict and diplomatic confrontations between US bloc and Soviet bloc.

→  Bipolar politics : USA led the Capitalist bloc and Soviet Union led the Socialist bloc after the Second World War. This ideological division between the power blocs is called bipolar politics.

→  Military Pacts : Military agreements formed among capitalist bloc and socialist bloc after the Second World War.

→  Non-alignment : The policy adopted by the newly independent countries of Asia and Africa not to join the power blocs and to follow an independent foreign policy.

→  Zionism: The movement that started to establ ish a homeland for the Jews.

→  PLO : Palestinian Liberation Organisation was a movement with the objective of establishing a nation for the Palestinians.

→  Oslo Pact: The pact signed between Israel and Palestine under the mediation of USA. By this, Israel agreed to recognise Palestine as a free nation.

→  Glasnost : The administrative reform started under Mikhail Gorbachev in Soviet Union to implement openness in political processes.

World In The Twentieth Century Notes

→  Perestroika: The administrative reform started under Mikhail Gorbachev to restructure the economic system of Soviet Union.

→  Unipolar world : USA emerged as a global power and centre of world politics following the disintegration of Soviet Union. The world order dominated by the USA is called unipolar world.

→  Neo imperialism: The multinational companies began to interfere in the economic, social and cultural sectors of the newly independent countries of Asia and Africa and Latin America for serving the interests of capitalist countries. This is known as neo imperialism.

→  Globalisation: The policy of transfer of products, ( services, raw materials, capital, latest technology and human resources across the borders of countries without any restriction.

→  Liberalisation: The policy of adoption of liberal regulations and taxation systems to facilitate the import of multinational products to domestic markets.

→  Privatisation : The policy of privatisation of public sector undertakings to promote private sector. (The process of reducing the role of public sector in the economy and increasing the role of private sector is known as privatisation).

World in the Twentieth Century – Famous Persons

→  Francis Ferdinand : The heir to the throne of Austria who was assassinated in June 1914 at Sarajevo. This was the immediate cause for the First World War.
→  Benito Mussolini : Leader of fascist reign in Italy.
→  Adolf Hitler: Leader ofNazi reign in Germany.
→  Matteotti: Eminent socialist thinker of Italy who opposed fascism. ‘
→  Woodrow Wilson : The US President who gave leadership to the formation of League of Nations.
→  Nelson Mandela ; Leader of anti – imperialist struggle in South Africa.
→  Quami Nkrumah : Leader of anti – imperialist struggle in Ghana.
→  Jomo Kenyatta : Leader of the freedom movement in Kenya.
→  Bernard Baruch : The American economist who first used the word ‘cold war’.
→  Architects of Non : Aligned movement:

• Jawaharlal Nehru – India
• Gamal Abdul Nasser – Egypt
• Marshal Tito-Yugoslavia .
• Ahmed Sukarno – Indonesia

→ Yasser Arafat: Founder President of Palestinian Liberation Organisation.

→  Mikhail Gorbachev : The last President of Soviet Union.

World in the Twentieth Century – Important Years and Events

• 1871 – Franco Prussian War
• 1904 – Moroccan Crisis
• 1912 – Balkan Crisis
• 1914 – Assassination of Francis Ferdinand
• 1914-18 – First World War
• 1919 – Paris Peace Treaty: Treaty of Versailles
• 1924 – Mussolini in power
• 1929 – World Economic Depression
• 1933 – Hitlar as Chancellor of Germany
• 1938 – Munich Pact
• 1939-45 – Second World War
• 1941 – Pearl Harbour Attack
• 1945 – Atom bombs dropped at Hiroshima and Nagasaki in Japan by USA
• 1945 – Formation ofUNO (October24,1945)
• 1948 – Formation of Israel
• 1955 – Emergence of Non-Aligned Movement
• 1991 – Disintegration of Soviet Union
• 1993 – Oslo Pact

Kerala Syllabus 10th Standard Social Science Notes

You can Download Soldiers of Defense Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard BiologySolutions Chapter 5 Soldiers of Defense

Soldiers of Defense Text Book Questions and Answers

Sslc Biology Chapter 5 Kerala Syllabus Question 1.
Our surrounding are full of microorganisms. Most of them are pathogens too. Though we live in the midst of germs are we susceptible to diseases? What may be the reason?
Answer:
Numerous germs are present in our surroundings, that have the capacity to cause diseases. We are often in contact with them. There are several mechanisms in the human body which prevent the entry of germs. So we don’t get infected always.

Biology Chapter 5 Class 10 Kerala Syllabus Question 2.
What are the mechanisms in the body which prevent the entry of pathogens?
Answer:

• A protein called keratin in skin, sebum, and acids.
• Mucus in the trachea
• Cilia in the bronchus
• Hydrochloric acid in the stomach.
• Cough and sneezing.
• The wax in the ear.
• The enzyme lysozyme in tears and saliva.
• Blood, Lymph.

Sslc Biology Chapter 5 Notes Kerala Syllabus Body Coverings And Secretions

Biology Class 10 Chapter 5 Kerala Syllabus Question 3.
Skin is referred to as a “fort of resistance”, why?
Answer:
Keratin makes the skin a thick fort which prevents germs from entering it. So the skin is referred to as a fort of resistance.

10th Class Biology 5th Chapter Kerala Syllabus Question 4.
What is the function of cilia and mucus in the respiratory tract?
Answer:
Mucus in the trachea prevents the entry of germs into the lungs. The cilia in the bronchus wipe out dust that enters it.

Kerala Syllabus 10th Standard Biology Chapter 5 Question 5.
What are the methods in ears, eyes, and saliva to prevent germs?
Answer:
The enzyme lysozyme present in the tears and saliva are fight against germs. The wax in the ear prevents pathogens.

Class 10th Biology Chapter 5 Notes Kerala Syllabus Question 6.
What is the role of hydrochloric acid in the stomach to prevent germs that enter the body through food?
Answer:
Since hydrochloric acid is present stomach, the germs that enter through food and water are destroyed.

Class 10 Biology Chapter 5 Notes Kerala Syllabus Question 7.
Which are the secretions that help to defend pathogens? Analyze illustration and complete the table.


Answer:

Body Fluids And Defense

• Body fluids like blood and lymph play an important role in defense mechanisms.
• Controlling the entry of germs into the body.
• Neutralizing germs and the toxic substances they produce, preventing their multiplication.

White blood cells and Defense actions

Inflammatory Response

Lysozyme Antibodies Question 8.
Based on the indicators, analyze the following illustration. Write your inference in the science diary.

Answer:
The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessel thereby increasing the blood flow. Blood plasma and white blood cells reach the wound site. This is the reason for the swelling of the wound site. This defense mechanism is known as inflammatory response.

Question 9.
What is the advantage of the dilation of blood vessels at the wound site?
Answer:
The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessels thereby increasing the blood flow.

Label the Specializations of the Plasma Membrane Question 10.
Is inflammatory response a defense activity? Why?
Answer:
Inflammatory response is a defense activity. Inflammation formed in the body due to the changes in the wall of blood capillaries in a part of the body that affected a wound. When germs enter through – the wound, changes occur in the capillary wall of that part. It leads to inflammation. Flow of blood through these capillaries increases and as a result more leucocytes come out from the capillaries and destroy the germs by engulfing them. The affected parts swell and become red-colored due to the arrival of more blood at the affected part of the capillaries.

Question 11.
Prepare the flowchart which showing the stages of inflammatory response.
Answer:
Germs enter through wound → Produces chemical messages → Blood vessels dilate → White blood cells from the blood vessel reach the wound site → White blood cells destroy the germs.

Phagocytosis

Phagocytosis is the process of engulfing and destroying germs. The cells engaged in this process are called phagocytes. (Phago – to engulf, cyte – cell) Monocytes and neutrophils are phagocytes.

Question 12.
Stages of phagocytosis
Answer:

Question 13.
Complete the flow chart by analyzing illustration showing the stages of phagocytosis.

Answer:

Blood Clotting

Blood clotting is a defense mechanism to prevent the loss of blood through wounds. In this process fibrin, the plasma protein forms a fibrous network. Blood cells get entangled in the network to form a blood clot.

Question 14.
Analyze the following illustration that details the stages of blood clotting

Answer:
When a person gets a cut or wound the blood that flows out from the wound changes from a liquid to a gel, forming a clot which plays the wound and prevents further bleeding.

When the platelets come into contact with the atmospheric air, they burst and liberate thromboplastin. It converts prothrombin in the plasma to thrombin. This thrombin converts soluble fibrinogen molecules into insoluble fibrin. This fibrin filament form-fine network over the wound and trap blood corpuscles and platelets to form a clot. The clot seals the wounds and stops bleeding.

Healing Of Wounds

Healing of the wound is a stage after inflammatory response and blood clotting. When wound occurs new tissues are formed in place of the tissues damaged by the wound. In such situations, the wound scar does not remain. In cases when new tissues cannot be formed, the connective tissue heals the wound. In such situations, the wound scar remains.

Question 15.
In some situations, the wound scar remains. Why?
Answer:
When wound occurs, new tissues cannot be formed, then connective tissues heal the wound. In such situations, the wound scar remains.

Fever, A Defense Mechanism

Question 16.
The normal body temperature is 37°c (98.6°F). Body temperature rises during fever. Is it a disease or a symptom. Analyze the flowchart given and write your inferences in the science diary.

Answer:
Fever is not a disease. But it is a type of resistance activity. Though the body can control the multiplication of germs through mechanism like raising body temperature. The chemical substances produced by the white blood cells raises the body temperature. If the rise in body temperature persists for a long time, it may badly affect the internal organs including the brain. Hence it is necessary to seek medical assistance immediately.

Question 17.
Fever is the rise in the body temperature. Is it beneficial to the body?
Answer:
Our normal body temperature is 36.9°c. This temperature is suitable for the multiplication of germs. When infection occurs body rises the temperature through fever to reduce the capacity of multiplication of them.

Lymphocytes – The Warrior

Specific defense is the system which identifies and destroys pathogens. White blood cells known as lymphocytes are capable of destroying the pathogens in this way. Lymphocytes are of two types namely B lymphocytes and T lymphocytes. B lymphocytes mature in them bone marrow. T lymphocytes mature in the thymus gland.

B – Lymphocytes

B lymphocytes produce certain chemical substances to act against antigens. The chemical substances which act against antigens are called antibodies.

Antibodies destroy the pathogens in three different ways

1. Destroy the bacteria by disintegrating their cell membrane.
2. Neutralize the toxin of the antigens.
3. Destroy the pathogens by stimulating other white blood cells.

T – Lymphocytes

T lymphocytes stimulate other defense cells of the body. Moreover, these cells are capable of destroying cancer cells and cells affected by virus.

Question 18.
Complete illustration showing the defense mechanisms of blood.

Answer:

• Inflammatory response
• Phagocytosis
• Blood clotting

Lymph And Defense

The lymph formed from the blood and reabsorbed into blood has a prominent role in defense mechanisms, lymph contains plenty of lymphocytes. They destroy the disease-causing bacteria in lymph nodes and spleen.

Immunization

Defense mechanisms become slow when germs enter the body. This causes the spread and multiplication of germs. Immunization is the artificial method to make the defense cells alert against the attack of pathogens.

Question 19.
What are vaccines?
The substances used for synthesizing antibodies are called vaccines.

Question 20.
Which components of vaccine act as antigens?
Answer:
The components from alive or dead or neutralized germs neutralized toxins or cellular parts of the pathogens will be the component of each vaccine.

Question 21.
How do vaccines induce immunity?

Answer:
In induced immunity antibodies which can act against pathogens or toxins produced by them are synthesized in the body itself. The body prepares antibodies to act against these foreign bodies.

Treatment – Final Defense

Question 22.
Which are the different methods of treatment that we depend on?
Answer:

• Ayurveda
• Sidda
• Unani
• Naturopathy
• Homeopathy
• Allopathy (Modern medicine)

Question 23.
Observe the pictures given below and write the name and use of this equipment for diagnosis.

Answer:
A. Stethoscope – to measure heartbeat
B. Thermometer – To measure body temperature
C. Sphygmomanometer – To record blood pressure.

Question 24.
Given below is the table including a few other modern equipments and their uses.
Answer:

Laboratory Tests

The report of a test showing the quantity of different factors in blood

Question 25.
Identify the specializations in medicine and the related areas and complete the table

Answer:

Antibodies

The scientist Alexander Fleming, who first synthesized antibiotics in 1928. Antibiotics are used to resist bacterial diseases.

Question 26.
Antibiotics are very helpful but use of it should be with great care. Why?
Answer:

• Regular use develops immunity in pathogens against antibiotics
• Destroys useful bacteria in the body.
• Reduces the quantity of some vitamins in the body.
• Some of them cause allergy, problems to stomach, bones, and kidneys.

Question 27.
Is it proper to use antibiotics without recommendation by a doctor? Why? Discuss. Write your inferences in the science diary.
Answer:
No. Though antibiotics are effective medicines, their regular use brings many side effects. Therefore use medicines only by the instruction of the doctor. Doctors prescribe medicine by considering the dose, method of use, period of use, age of the patient, etc. indiscriminate use of them causes health problems. Regular use develops immunity in pathogens against antibiotics, destroys useful bacteria in the body and reduces the quantity of some vitamins. So self-treatment without the instruction of the doctor is not good.

First Aid

Question 28.
Observe figures A, B and C and identify the instance in which the following type of first aid is given.

Answer:

• A – Breathing has stopped but heartbeat has not as in drowning electric shock, choking gas, suffocation, etc.
• B – Bone and ligament injuries; and fractures
• C – Choking occurs when an object swallowing

Blood Transfusion

The transfer of blood from one person to another is called blood transfusion. Certain instances such as blood is lost excessively in accidents, affected with diseases like blood cancer and surgical operations require blood transfusion.

Different Types of Blood Group

A, B, AB, O are the main blood groups. Carl Landsteiner proposed blood grouping on the basis of the presence or absence of A, B antigens seen on the surface of the red blood cells. The blood group in which Rh factor is present are positive blood groups and those without Rh factor are negative blood groups.

Question 29.
Can a patient receive blood from any person?
Answer:
No. Blood of certain persons cannot be received by others. The antigen present in the received blood and antibody in the recipient’s blood will react each other to form blood clot.

Question 30.
Observe the table and identify the various types of blood group, antigens, and antibodies present in them
Answer:

Question 31.
Prepare posters on the greatness of donating blood
Answer:

Defense Mechanisms In Plants

Question 32.
Complete the illustration by including different defense mechanisms in plants.

Answer:

Let Us Assess

Question 1.
Which among the following is not included in non-specific body defense?
a) production of sebum
b) action of hydrochloric acid in the stomach
c) action of B lymphocytes
d) action of lysozyme in saliva
Answer:
c) action of B lymphocytes

Question 2.
Write the functions of the two types of lymphocytes in the defense mechanism of the body.
Answer:
B lymphocytes produce antibodies and it destroys the antigens, T- lymphocytes stimulate white blood cells and also destroys cancer cells.

Question 3.
What is the basis of grouping blood Into different types? Everybody cannot receive blood of all groups. Why?
Answer:
The basis of blood grouping is the presence of antigen seen on the surface of the red blood cells. When an antigen reaches one s blood, it stimulates defense activity to produce antibody. The antigen and antibody react each other and form a blood clot. Hence everyone cannot receive blood from aH blood groups.

Soldiers of Defense More Questions And Answers

Question 1.
Identify the diagram and mention the defense process taking place in A

Answer:
Skin.
A is the outermost Keratin layer. Keratin is a protein, it blocks the entry of germs.

Question 2.
Respiratory track is always free from germs. Why?
Answer:
Mucus in the trachea prevents the entry of germs into the lungs. The cilia in the bronchus wipe out dust that enters it. Cough and sneezing help to expel foreign bodies from the respiratory tract. So respiratory tract is always free from germs.

Question 3.
Complete the table suitably.

Answer:
a) Ear
b) Hydrochloric acid
c) eye / mouth
d) The outermost layer blocks the entry of germs/ Keratin / Sebum / Acids.

Question 4.
Is swelling of the wound site helpful or not? why?
Answer:
Yes, It is helpful. The wounds and cuts occur in the skin, that area swells and blood vessels dilate It increases the blood flow and more white blood cells can come out through the enlarged pores and destroy the germs.

Question 5.
Generally, bacteria are useful but some of them are pathogenic. How?
Answer:
After entering the body they multiply by binary fission and produce certain toxic substances, which either disrupt the cellular activities or destroy the cell itself.

Question 6.
Following are certain steps of a defense process identify the process.
1. Phagocytes reach near the pathogens.
2. Engulf pathogens in the membrane sac
3. Membrane sacs combine with lysosome.
4. The enzyme in the Iysosome destroys the pathogens.
5. Expels the remnants from phagocyte.
Answer:
Phagocytosis

Question 7.
Observe the illustration and identify the process.

Answer:
Phagocytosis

Question 8.
What are the factors needed for blood clotting?
Answer:
Prothrombin, Fibrinogen in plasma, Calcium ions, Vitamin K, Red blood cells, Platelets.

Question 9.
Blood clotting is a defense mechanism to prevent the loss of blood through wounds. Mention the different stages of this process.
Answer:

• Tissues of the wounded part degenerate to form the enzyme thromboplastin.
• Thromboplastin converts prothrombin in the plasma to thrombin.
• Thrombin converts the fibrinogen in the plasma to fibrin.
• Blood clot is formed by the entangling of platelets and red blood cells in the fibrin network.

Question 10.
Complete the flowchart showing the blood clotting.

Answer:
A) Thrombin
B) Fibrin

Question 11.
“Fever is not a disease, it is a defense mechanism.” Analyze the statement.
Answer:
Yes. The presence of toxin produced by the pathogens stimulate the white blood cells and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body temperature reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 12.
How is antibody destroy germs?
Answer:
Antibody destroys the bacteria by disintegrating their cell membrane and neutralize the toxin of the antigens by stimulating other white blood cells.

Question 13.
Define the following
1. Antigen
2. Antibody
3. Antibiotic
Answer:
Any foreign body that stimulates the defense mechanism is called an antigen. The chemical substance produced by the lymphocytes act against antigen is an antibody. Antibiotics are medicines used to resist bacterial diseases.

Question 14.
The graph shown below represents the difference in the number of two bacteria when taken a particular antibiotic by a patient.

Answer:
a) This antibiotic is effective and reduces the number of pathogens. But its antibiotics. It also destroys useful bacteria in the body.
b) Further use of these antibiotics is not effective because the harmful bacteria got resistance against it.

Question 15.
Complete the boxes according to the given hint.

Answer:
A – Regular use develops immunity in pathogens against antibiotics
B – Destroys useful bacteria in the body.
C – Reduces the quantity of some vitamins in the body.

Question 16.
Can a person with ‘A’ group blood receive blood from ‘B’ group person? Or it take place vice versa? Give reason for this.
Answer:
A person with ‘A’ group blood cannot receive or donate blood with ‘B’ group person. Because the antigen present in the received blood and antibody in the recipient’s blood will react each other and form a blood clot (Coagulation).

Question 17.
Give more examples of vaccine
Answer:

Question 18.
“Germs, both alive and dead are used to get immunity”. Substantiate the statement with vaccines used for rabies and tuberculosis. (March 2015)
Answer:
Germs, both alive and dead are used as vaccines. ‘ Dead germs are utilized in rabies vaccine which acts against rabies. Live, but inactivated vaccines are used in BCG vaccine against tuberculosis.

Question 19.
Observe the following figure and answer the given questions.

a) Label A and B.
b) How did they protect our body? (March 2015)
Answer:
a) A – small hair, B – Sebaceous gland
b) Since sebum is oily, water does not stick on to the skin. Covering of hair protects the body from cold and heat and also prevents the entry of foreign bodies.

Question 20.
Whichever be the type of germs infected, the initial symptom appear in human body will be the fever. Give reason. (March 2014)
Answer:
Bacterial infection produces toxic substances in body, body temperature is suitable for bacterial growth, in order to control the growth of bacteria body rises the temperature, fever is not a disease.

Question 21.
Constant use of antibiotics is not good for health. This is the opinion of Rahim.
a) Do you agree with his opinion? Why?
b) Give two specific examples for justifying your answer. (Model 2013)
Answer:
a) I agree with this. Constant use of antibiotics results a few side effects.
b) Constant use of antibiotics may destroy useful bacteria in the body, develop resistance in bacteria against antibiotics or reduces the level of certain vitamins in the body.

Question 22.
How does the influence of the following action blocking germs.
a) Rise in body temperature.
b) Low oil content on skin.
c) Swelling occurs near wound.
d) Lymphocytes produce Antibodies. (March 2013)
Answer:
a) To resist the strengthening or increasing of causative organisms.
b) Waterproof and oily, germs cannot grow.
c) Flow of blood through the capillaries increases and more leucocytes comes out from the capillaries and destroy the germs by engulfing them.
d) Lymphocytes produce antibodies to destroy germs.

Question 23.
Whichever be the type of germs infected, the initial symptom appear in human body will be the fever Give reason. (March 2013)
Answer:
Bacterial infection produces toxic substances in body body temperature is suitable for bacterial growth, in order to control the growth of bacteria body rises the temperature, fever is not a disease.

Question 24.
The graph representing the difference in the number of two bacteria, while applying a particular antibiotic on a patient is shown below

A. Analyze the graph and record the findings.
B. Is this antibiotic effective against the bacteria? Why? (Model 2012)
Answer:
A. The harmful bacteria decrease in number in the first few weeks. Later they increase in number. Number of useful bacteria are decreasing gradually.
B. The antibiotic is not effective because the harmful bacteria got resistance against it. Moreover, number of useful bacteria decreases.

Soldiers of Defense Questions and Answers

Question 1.
Which among the following is the odd one? Why? Lymphocyte, Monocyte, Neutrophil, Basophil, Eosinophil (Question Pool-2017)
Answer:
Lymphocyte – Involved in specific defense

Question 2.
Skin is the largest sense organ of the body. It helps! us to sense heat, cold, touch, pressure, etc and it j acts as a soldier of defense of the body,
a) Does the skin have significance in defense as mentioned above? Justify. (Question Pool – 2017)
Answer:
Yes. The outermost Keratin, the protein layer blocks the entry of germs; sebum and some acids in the: skin-are disinfectants,

Question 3.
A table indicating primary level defense is given below. Arrange column B based on column A. (Question Pool – 2017)

Answer:
i) – c
ii) – d
iii) – a
iv) -b

Question 4.
Which among the following is the odd one and why? (Question Pool – 2017)
a) The Mucus of trachea destroys the pathogens.
b) The wax in the ear destroys pathogens.
c) Neutrophil destroys pathogens by engulfing them.
d) Lysozyme present in Saliva destroys pathogens
Answer:
C, Secondary defense

Question 5.
Nimisha’s hand got injured in an accident. After some time the wound area got swollen.
a) What is this type of activity known for?
b) Is it a defense mechanism? Why?
Answer:
a) Inflammatory response
b) Yes
Secondary level defense
Process to destroy pathogens in the body

Question 6.
Using the following statements,-prepare a flow chart of inflammatory response. (Question Pool-2017)
a) Production of chemical messages.
b) White blood cells destroy pathogens.
c) Blood vessels dilate.
d) Pathogens enter into the wound.
e) White blood cells come out from blood vessels.
f) Blood flow increases
Answer:
d
a
c
f
e
b

Question 7.
The given illustration includes white blood cells which act as a part of nonspecific defense. Fill up the blanks and complete the word web. (Question Pool -2017)

Answer:
A – Neutrophil /Monocyte
B – Stimulates other white blood cells / dilates blood vessels
C – Eosinophil
D – Engulfs and destroys germs

Question 8.
When there is an injury or wound, the blood vessel of that part dilates. (Question Pool – 2017)
a) What is its benefit?
b) Which white blood cell dilates the blood vessel?
Answer:
a) The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessels thereby increasing the blood flow. Blood plasma and white blood cells reach the wound site and it destroys the germs,
b) Basophil

Question 9.
Observe the given illustration and answer the following questions. (Question Pool-2017)

Answer:
a) Which is the process indicated in the illustration?
b) Which are the.white blood cells involved in the process?
c) Is it a specific defense mechanism? Justify
Answer:
a) Phagocytosis
b) Neutrophil, Monocyte
c) No
does not identify and destroy pathogens that enter to the body.

Question 10.
The flow chart given below indicates a type of defense mechanism occurring in the body. (Question Pool – 2017)

a) Complete the flow chart
b) Which process is it related to?
Answer:
a) i) Engulfs pathogen in the membrane sac
ii) The enzyme in the lysosome destroys the pathogens
iii) Expels the remnants
b) Phagocytosis

Question 11.
Blood clotting is a defense mechanism. Analyze the statement. (Question Pool – 2017)
Answer:

• Prevents the entry of germs through wound
• Prevents bleeding through wounds

Question 12.
Prepare the flow chart of the clotting of blood using the following statements. (Question Pool-2017)
a) Thromboplastin converts prothrombin to thrombin.
b) Blood flows from the wound.
c) Blood clot is formed.
d) Thrombin converts fibrinogen to fibrin.
e) Tissues degenerate to form the enzyme called thromboplastin.
f) The red blood cells and platelets entangle in the fibrin network.
Answer:
d
a
c
f
e
b

Question 13.

a) Identify A
b) B is a vitamin and C is an enzyme. Name them.
c) How does the lack of B or C affect the consequent chemical process? (Question Pool – 2017)
Answer:
A – Prothrombin
B – Vitamin K, C – Thromboplastin
C – Thrombin not formed
fibrinogen not converted to fibrin

Question 14.
Blood clot is formed by the entangling of red blood cells and platelets in the fibrin network.
White blood cells are not involved in this process. What explanation will you give for this? (Question Pool-2017)
Answer:

• White blood cells do not have a definite shape
• They come out through the fibrin network

Question 15.
One of the scars of the wound obtained by Binu while playing football remained even after 10 years. What explanation will you give for the scar remaining as such? (Question Pool-2017)
Answer:
When wound occurs, new tissues cannot be formed, then connective tissues heal the wound. In such situations, the wound scar remains

Question 16.
Fever is a defense mechanism. Is the statement correct? Justify your answer. (Question Pool-2017)
Answer:
Yes. The presence of toxin produced by the pathogens stimulate the white blood cell and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body temperature reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 17.
Complete the illustration (Question Pool-2017)

Answer:
a) B – lymphocyte
b) T – Lymphocyte
i) Stimulates white blood cell and destroys pathogens
ii) Destroys the bacteria by disintegrating their cell membrane
iii) Stimulates defense cells
iv) Destroys the cell which is affected by virus.

Question 18.
After attending a class on immunity, Arun raised a question to his teacher. (Question Pool-2017)
“In spite of so many defense mechanisms in the body, why are we still affected by diseases?
a) What explanation will you give for Arun’s doubt?
Answer:

• Bad habits
• Unhealthy food habits
• Unhygienic
• Excess pathogens

Question 19.
The use of some modern equipment are given below. Identify the equipment. (Question Pool-2017)
a) To record electric waves in the brain.
b) To record electric waves in the heart muscle
c) To understand the structure of internal organs using ultrasonic sound waves.
Answer:
a) EEG
b) ECG
c) Ultrasound scanner

Question 20.
The doctor prescribed antibiotics to Sunil who is affected with cholera, but not to Anil who is affected with chickenpox. What is the reason? (Question Poo1 -2017)
Answer:
Antibiotics are used to prevent bacterial diseases. Chickenpox is a viral disease. Cholera is a bacterial disease. So the doctor prescribed antibiotics to Sunni

Question 21.
Enlist the demerits of antibiotics for Jose who is preparing for a seminar on the topic “The merits and demerits of Antibiotics. (Question Pool -2017)
Answer:

• The frequent use of antibiotics produces disease defense in pathogens.
• Destroys useful bacteria in the body.
• Reduces the level of some vitamins in the body.

Question 22.
Ashiq who met with an accident was in need of blood. Antigen A and D and Antibody b was identified in his blood. (Question Pool -2017)
a) Name his blood group?
b) Whose blood, among the following can be accepted by ashiq?
(i) Venu = A⁺
(ii) Amal- AB⁺
(iii) Suhara – AB^–
(iv) Anoop – A^–
Answer:
a ) A⁺
b) (i)VenuA⁺
(ii) Anoop A^–

Question 23.
The table given below indicates blood groups.

Answer:
i) A
ii) a
iii) AB
iv) No
v) O
vi) No

Question 24.
Box A includes the major components of vaccines and box B includes the diseases against which they are used. Match them appropriately. (Question Pool -2017)

Answer:
i) -d
ii) -c
iii) -b
iv) -a

Question 25.
Ravi prepared an illustration showing defense mechanisms in plants. Complete it. (Question Pool – 2017)

Answer:
a) Prevents the entry of germs which have crossed the cell wall, through cell membrane.
b) Bark
c) Cuticle in leaves
d) Cell wall

Question 26.
“This mode of treatment is a lifestyle in tune with nature rather than a mere method of treatment” This is a statement regarding a well-known mode of treatment.
a) Name the treatment.
b) Apart from this, name any two well-known modes of treatment. (Question Pool – 2017)
Answer:
a) Ayurveda
b) 1. Allopathy
2. Homeopathy/ etc.

Question 27.
Prepare two suitable placards to conduct an awareness rally in association with World Blood Donation day. (Question Pool-2017)
Answer:
2 placards contain appropriate concepts
Example: Donate blood Donate Life
Blood donation – Nothing to loose profits – Life

Question 28.
Match the following pairs (Question Pool – 2017)
a) T-lymphocyte: Thymus gland
B – lymphocyte:………………
b) EEG: to record electric waves in brain
………… to record electric waves in heart muscles
c) First Antibiotic: Alexander Flemming
First vaccine:……………..
d) Heartbeat: Stethoscope
Blood pressure:……………..
e) Antigen: Red blood cells
Antibody:…………………
Answer:
a) Bone marrow
b) ECG
c) Edward Jenner
d) Sphygmomanometer
e) Plasma

Question 29.
Given below is an equipment used for disease diagnosis. (Question Pool-2017)

a) Identify the equipment
b) What is its use?
c) Name another equipment that works on the same principle.
Answer:
a) ECG
b) To record electric waves in heart muscles
c) EEG

Question 30.
“it is possible to build up a healthy society with hospitals, doctors, and medicines” This is Bashir’s opinion. Evaluate it (Question Pool -2017)
Answer:

• The opinion of Bashir is wrong
• Nutritious food
• Healthy lifestyle
• Hygiene, These are the factors which build up a healthy society.

Question 31.
“It is not necessary to detect blood groups if we can accept blood from anyone”. This was an argument put forward by Sivaprasad in a discussion on blood transfusion.
a) What is the base of blood group determination?
b) Can a person receive any blood from anyone? Why? (Question Pool – 2017)
Answer:
a) The presence of antigens A and B on the surface of Red blood cells,
b) 1. Not possible
2. When a foreign antigen reaches one’s blood, it stimulates the defense activity
3. The antigen present in the received blood and the antibody in the recipients, blood will react each other to forms a blood clot (coagulation)

Question 32.
Given below is the picture of white blood cells which are parts of specific defense? 32 (Question Pool – 2017)

a) Identify A and B
b) What is the role of A in specific defense?
c) Give anyone difference between A and B
Answer:
a) A – T- lymphocyte
B – B lymphocyte
b) 1. Stimulates other defense cells
2. Destroys cancer cells and virus affected cells.
c) B lymphocytes matured at bone marrow T lymphocytes matured at thymus

Question 33.
Analyze the following statement and answer the following questions. 33 (Question Pool-2017)
When there is a wound, the body temperature rises.
a) What is the significance of white blood cells in this activity?
b) How does immunity become possible through a rise in temperature?
Answer:
a) The chemical substances produced by white blood cells rise body temperature
b) The presence of toxin produced by the pathogens stimulate the white blood cell and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body tempera¬ture reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 34.
Statements related to nonspecific defense and specific defense are given below. Identify the type of the defense and mark them using the letters N and S respectively.
(Orukkam – 2017)
a) The cilia in bronchus wipe out dust that enters it.
b) Destroy the bacteria by disintegrating their cell membrane.
c) The blood vessels near the wound diabetes.
d) The rise in body temperature reduces the rate of multiplication of pathogens.
e) B lymphocytes produce certain chemical substances against antigens.
f) Eosinophil produces chemical substances needed for inflammatory responses.
g) T lymphocytes destroys cancer cells.
h) The enzyme lysozyme present in tears destroys germs.
i) T lymphocytes destroy cancer cells.
j) Phagocytes engulf and destroy germs.
Answer:
a) N
b) S
c) N
d) N
e) S
f) N
g) S
h) N
i) S
j) N

Question 35.
Our body has the capacity to destroy germs those enter the body by breaking the first level defense Write your comment on this statement? (Orukkam – 2017)
( Hints: inflammation, different types WBCs and their functions, phagocytosis)
Answer:
The statement is true. When germs enter to the body the blood vessels diates (inflammation). This helps the white blood cells to act at the site A few white blood cells destroy germs by phagocytosis.

Question 36.
The basis of blood grouping is the presence of antigens in red blood cells. Complete the table given below based on this statement. (Orukkam-2017)
Blood groups

Answer:

a) B lymphocytes produce antibodies against antigens.
b) We can use antigens as vaccines for the formation of antibodies in advance.
c) Neutralized toxins – Diphtheria
Alive but neutralized germs – Measles
Cellular parts of pathogens – Hepatitis B Killed germs – Cholera

Question 37.
Complete the illustration suitability related to antibiotics (Orukkam – 2017)

Answer:
A) Prevent bacterial diseases
B) Prolonged use may develop immunity in germs
C) Destruction of useful bacteria
D) Deficiency of certain vitamins in the body

Question 38.
The wound scar does not remain always. Write reason? (Orukkam – 2017)
Answer:
If the wound is filled with same tissue, no wound scar occurs there.

Question 39.
Fill in the blanks by observing the relationship in the first pair. (Orukkam – 2017)
a) EEG: To record electric waves in the brain
ECG:……………………………..
b) Rabies: Killed germs
Typhoid:………………………..
Answer:
a – Records electric waves in the heart muscles
b – Alive but neutralized germs.

Question 40.
Name the first vaccine? Who developed this? Write the situation which leads to the development of vaccine? (Orukkam – 2017)
Answer:
Smallpox vaccine, developed by Edward Jenner. He observed no smallpox disease in people who had affected cowpox earlier.

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Kerala State Syllabus 10th Standard Social Science Notes Chapter 6 Eyes in the Sky and Data Analysis

Our ancestors made maps after collecting information about the surface of the earth and survey of earth’s surface which lasted for a long time. Aerial photography which developed later made map making quite easy. Satellite remote sensing and geographical information system which developed in the 1960’s as a result of the progress in science and technology paved the way for map making efficient and fast. The chapter helps you to understand how the launching of satellites and modern computer softwares help in the analysis of geophysical data.

→ Remote sensing : The method of collecting information about an object or phenomenon without actual physical contact.

→ Scanners : Sophisticated equipments that can detect electromagnetic radiation.

→ Sensors : The instruments used for data collection through remote sensing.

→ Active Remote Sensing : Remote sensing made with the help of artificial sources of energy is known as active remote sensing.

→ Passive Remote Sensing : Remote sensing carried out with the help of solar energy is known as passive remote sensing.
→
Platform : The carrier on which sensors are fixed is platform.

→ Terrestrial Photography : The method of obtaining earth’s photographs using cameras from the ground.

→ Aerial Remote Sensing : The method of obtaining photographs of the earth’s surface continuously from the sky using cameras mounted on aircrafts.

Social Science Short Notes For Class 10 Kerala Syllabus

→ Satellite Remote Sensing : The process of collecting information using sensors fitted on artificial satellites is called satellite remote sensing.

→ Stereoscope: The instrument used for obtaining three dimensional view from the stereo pairs is called stereoscope.

→ Stereo pair: Two aerial photographs of adjoining areas or two adjacent aerial photographs with overlap.

→ Overlap: Each aerial photograph contains about 60% of the area shown in the previous photograph is termed as overlap.

→ Stereoscopic vision : The three dimensional
view obtained while viewing a stereo pair through a stereoscope. ‘

→ Geostationary satellites : Artificial satellites that orbit the earth at a height of about 36000 km in accordance with the earth’s rotation.

→ Sun Synchronous satellites: Artificial satellites that revolve around the earth along poles at a height of 900 km from the earth’s surface.

→ Spectral Signature : The amount of reflected energy by each object.

→ Spatial Resolution : The size of the smallest object on the earth’s surface that a satellite sensor can distinguish.

→ Geographic Information System : Geographic Information System is a computer based information management system by which data collected from the sources of information like maps, aerial photographs, satellite imageries, tables and surveys are incorporated into the cqmputer using softwares, which are retrieved, analyzed and displayed in the form of maps, tables and graphs.

→ Spatial Data : Each feature on the earth’s surface has its own latitudinal and longitudinal location. Such information is known as spatial data.

→ Attributes: The additional information about the characteristics of each spatial data on the earth’s surface are called attributes.

→ Layers : The thematic maps prepared and stored in Geographic Information System for analytical purpose are called layers.

10th Geography Notes

→ Network Analysis : GIS analysis that deals with linear features on a map such as roads, railway and rivers.

→ Buffer Analysis : GIS analysis used for analyzing activities around a point feature or at a definite distance along a linear feature.

→ Overlay Analysis : GIS analysis used to identify the interrelationship of various surface features on earth and the changes they have undergone over a period of time.

→ Global Positioning System : A system using signals received from the artificial satellites to display the latitude, longitude, height and time of a place.

→ IRNSS : Indian Regional Navigation Satellite System is the satellite based navigation system developed by India as an alternative to GPS. It g is designed to provide accurate position information service.

Kerala Syllabus 10th Standard Social Science Notes

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The Biology of Movement Textual Questions and Answers in Malayalam

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टूटा पहिया Text Book Activities & Answers

टूटा पहिया अभ्यास के प्रश्न

Sslc Hindi Chapter 4 Notes Kerala Syllabus प्रश्ना 1.
“वहाँ उसके और ममी के बीच में बहुत सारी चीजें आ जाती हैं। वह बंटी की अपनी दुनिया थी।…..” इस हालत में बंटी की चिंताएँ क्याक्या हो सकती हैं? लिखें।
उत्तर:
बंटी की चिंताएँ
माँ मुझे यहाँ क्यों ले आती हैं? यह जगह मुझे बिलकुल अपरिचित है। यहाँ मेरा दम घुटता है। माँ तो यहाँ आकर माँ नहीं रह जाती हैं। उसका प्रिंसिपलवाला यह चेहरा मुझे बिलकुल पसंद नहीं आता। यहाँ माँ व्यस्त ही व्यस्त हैं। अपने लिए कुछ समय निकाल नहीं पातीं। मैं अपनी माँ को चाहता हूँ। किसी प्रिंसिपल को नहीं।

टूटा पहिया Summary in Malayalam and Translation

टूटा पहिया शब्दार्थ