Study Point

2023-06-19T22:07:00.001-07:00

Youtube earning methods

Question: YouTube pe product placement ka istemal karke paise kaise kamaye ja sakte hain?

Answer: YouTube par product placement se paise kamane ke liye, aap neeche diye gaye steps ko follow kar sakte hain:

1. Product selection: Sabse pehle, aapko apne YouTube channel ke niche aur audience ke interests ke hisab se ek product choose karna hoga. Dhyan rakhe ki product aapke viewers ko pasand aaye aur unke liye relevant ho.

2. Partner with brands: Ek baar jab aap product decide kar lete hain, toh aapko uss product se sambandhit brands ke saath collaboration karna hoga. Brand se contact karke unhe apne channel ke viewership, engagement aur demographics ke baare mein bataye aur unhe convince kare ki aap unke product ko promote karne ke liye ek sahi platform hain.

3. Integrate product in your content: Product placement mein aapko apne videos mein product ko natural tarike se integrate karna hoga. Ismein aap uss product ka use karte hue, uski features aur benefits ke baare mein baat kar sakte hain. Dhyan rakhe ki product ko jyada jhatka na lage, balki content ko enhance kare.

4. Disclose the partnership: YouTube ke guidelines ke mutabik, aapko hamesha apne viewers ko batana chahiye ki aapne product placement ke liye ek partnership ki hai. Ise aap video ke description ya video mein verbally bata sakte hain. Transparency viewers ki trust build karti hai.

5. Monitor performance and engagement: Apne product placement campaigns ke performance ko regularly monitor kare. Views, likes, comments, aur audience engagement ki analysis kare. Isse aap samajh payenge ki product placement kitna successful raha hai aur future mein improvements kaise kar sakte hain.

Yaad rahe, product placement se paise kamana ek lamba process hai. Aapko apne channel ko grow karna hoga, viewership aur engagement badhana hoga, aur brands ke saath consistent collaborations banana hoga. Dhire-dhire aapko opportunities aur earnings badhne lagenge.

write a question and answer for "What are the best platforms for offering public speaking services on YouTube?" in easy hinglish

Question: YouTube par public speaking services offer karne ke liye sabse acche platforms kaun se hain?

Answer: YouTube par public speaking services offer karne ke liye kuch acche platforms hain, jo neeche bataye gaye hain:

1. YouTube Live: YouTube Live ek powerful platform hai, jahan aap live public speaking sessions kar sakte hain. Aap apne subscribers ke saath real-time interaction kar sakte hain aur unke sawaalon ka jawab de sakte hain. Iske through aap apne audience ke liye valuable content create kar sakte hain.

2. YouTube Shorts: YouTube Shorts ek short video format hai, jisme aap apne public speaking skills ko demonstrate kar sakte hain. Aap chhoti motivational clips, powerful quotes ya impactful speeches create kar sakte hain, jisse logon ko inspire mile.

3. YouTube Premieres: YouTube Premieres aapko apne audience ke saath video release experience share karne ka mauka deta hai. Aap apne recorded public speaking sessions ko scheduled time par release kar sakte hain aur live chat ke dwara audience se interact kar sakte hain. Isse aap engagement aur anticipation create kar sakte hain.

4. Collaborations with other YouTubers: Dusre popular YouTubers ke saath collaboration karke bhi aap apne public speaking services ko promote kar sakte hain. Aap dusre channels par guest appearances kar sakte hain, interviews de sakte hain ya unke saath panel discussions kar sakte hain. Isse aap apne audience reach ko expand kar sakte hain.

5. Social Media Promotion: YouTube ke saath-saath aap apne public speaking services ko dusre social media platforms jaise Facebook, Instagram, aur Twitter par promote kar sakte hain. Yahaan par aap short snippets, quotes aur highlights share karke apni presence badha sakte hain.

In platforms ka sahi istemal karke aap apne public speaking skills ko showcase kar sakte hain aur apne audience tak pahunch sakte hain. Dhyan rahe ki aap consistent aur valuable content provide kare, apne audience ke feedback ko sunen aur interaction maintain kare

2023-04-30T23:57:00.003-07:00

Real Numbers (Part - 1)

Introduction Of Real Numbers

Real numbers are a fundamental concept in mathematics that forms the basis of much of our understanding of the world around us. In simple terms, real numbers are all the numbers that can be represented on the number line, including both rational and irrational numbers.

Rational numbers are numbers that can be expressed as the ratio of two integers, such as 1/2, 3/4, or -7/5. Irrational numbers, on the other hand, are numbers that cannot be expressed as the ratio of two integers, such as pi (3.14159265359...) or the square root of 2 (1.41421356...).

Real numbers are used extensively in many branches of mathematics, including algebra, calculus, geometry, and number theory. They are also used in many real-world applications, such as physics, engineering, and finance.

Real numbers are typically represented using decimal notation, which uses a decimal point to separate the integer and fractional parts of the number. For example, the real number 3.14159 represents pi to five decimal places. Real numbers can also be represented using scientific notation, which expresses the number as a power of 10 multiplied by a decimal number.

Euclid division algorithm

The Euclidean division algorithm is a method of dividing one integer (the dividend) by another integer (the divisor), and obtaining both the quotient and the remainder. The algorithm is based on the division algorithm, which states that given two integers a and b, with b not equal to zero, there exist unique integers q and r such that:

a = bq + r

where q is the quotient and r is the remainder.

The Euclidean division algorithm starts with the dividend and divisor, and repeatedly subtracts the divisor from the dividend until the remainder is less than the divisor. The last nonzero remainder is the greatest common divisor of the original dividend and divisor.

Here are the steps of the Euclidean division algorithm:

Let a be the dividend and b be the divisor.
Divide a by b to get the quotient q and remainder r.
If r is zero, then the algorithm stops, and the result is q.
If r is not zero, then set a = b and b = r, and go back to step 2.

The algorithm will terminate after a finite number of steps, because the remainder r decreases with each step, and is always positive. The greatest common divisor of the original dividend and divisor is the last nonzero remainder obtained in the algorithm.

Euclid's Division Lemma

Euclid's Division Lemma is a fundamental concept in number theory that states that given two positive integers "a" and "b", there exist unique integers "q" and "r" such that:

a = bq + r

where "q" is the quotient and "r" is the remainder obtained when "a" is divided by "b" and 0 <= r < b. In other words, "q" gives the number of times "b" divides "a" completely, and "r" gives the remaining part that is not divisible by "b".

For example, let's say a = 20 and b = 3. We want to find q and r such that 20 = 3q + r.

We start by dividing 20 by 3 to get:

20 = 3 × 6 + 2

Here, 6 is the quotient, and 2 is the remainder, which is less than 3. Therefore, we can write:

20 = 3 × 6 + 2

This means that 3 divides 20 six times completely, leaving a remainder of 2.

Euclid's Division Lemma is used in various mathematical proofs and applications, including the greatest common divisor (GCD), least common multiple (LCM), and modular arithmetic.

Surface area and Volume ( Part – 7 )

2019-11-22T21:53:00.001-08:00

Surface area and Volume ( Part – 7 )

1. A metal pipe is 63 cm long . The inner diameter of a cross section is 4 cm , the outer diagram being 4.4 cm . Finds its

a). Inner curved surface area ,

b). Outer curved surface area ,

c). total surface area

Sol. a). radius of metal pipe r = 4 / 2 cm = 2 cm and height h = 63 cm

Inner curved surface area of pipe = 2 ∏ r h

Inner curved surface area of pipe = 2 * 22 / 7 * 2 * 63 cm^2

Inner curved surface area of pipe = 44 * 18 cm^2

Inner curved surface area of pipe = 792 cm^2

b). Outer radius of pipe R = 4.4 / 2 = 2.2 cm and height h = 63 cm

Outer curved surface area of pipe = 2 ∏ R h

Outer curved surface area of pipe = 2 * 22 / 7 * 2.2 * 63 cm^2

Outer curved surface area of pipe = 44 * 19.8 cm^2

Outer curved surface area of pipe = 871.2 cm^2

c). Inner radius of pipe r = 2 cm , outer radius R = 2.2 cm and height h = 63 cm

Area of upper ring of pipe = ∏ ( R^2 – r^2 )

Area of upper ring of pipe = ∏ ( 2.2^2 – 2^2 )

Area of upper ring of pipe = ( 22 / 7 ) ( 4.84 – 4 )

Area of upper ring of pipe = ( 22 * 0.12 )

Area of upper ring of pipe = 2.64 cm^2

Area of lower ring of pipe = 2.64 cm^2

Total surface area of pipe = Inner curved surface area + Outer surface area + Area of upper ring + Area of lower ring

Total surface area of pipe = 792 + 871.2 + 2.64 + 2.64 cm^2 = 1668.48 cm^2

Hence , the total surface area of pipe is 1668.48 cm^2 .

maths ncert solutions

2. A metal pipe is 70 cm long . The inner diameter of a cross section is 4 cm , the outer diagram being 4.4 cm . Finds its

a). Inner curved surface area ,

b). Outer curved surface area ,

c). total surface area

Sol. a). radius of metal pipe r = 4 / 2 cm = 2 cm and height h = 70 cm

Inner curved surface area of pipe = 2 ∏ r h

Inner curved surface area of pipe = 2 * 22 / 7 * 2 * 70 cm^2

Inner curved surface area of pipe = 44 * 20 cm^2

Inner curved surface area of pipe = 880 cm^2

b). Outer radius of pipe R = 4.4 / 2 = 2.2 cm and height h = 70 cm

Outer curved surface area of pipe = 2 ∏ R h

Outer curved surface area of pipe = 2 * 22 / 7 * 2.2 * 70 cm^2

Outer curved surface area of pipe = 44 * 22 cm^2

Outer curved surface area of pipe = 968 cm^2

c). Inner radius of pipe r = 2 cm , outer radius R = 2.2 cm and height h = 70 cm

Area of upper ring of pipe = ∏ ( R^2 – r^2 )

Area of upper ring of pipe = ∏ ( 2.2^2 – 2^2 )

Area of upper ring of pipe = ( 22 / 7 ) ( 4.84 – 4 )

Area of upper ring of pipe = ( 22 * 0.12 )

Area of upper ring of pipe = 2.64 cm^2

Area of lower ring of pipe = 2.64 cm^2

Total surface area of pipe = Inner curved surface area + Outer surface area + Area of upper ring + Area of lower ring

Total surface area of pipe = 880 + 968 + 2.64 + 2.64 cm^2 = 1853.28 cm^2

Hence , the total surface area of pipe is 1853.28 cm^2 .

Assignment – 2 (Motion)

2019-11-21T04:28:00.000-08:00

Assignment – 2 (Motion)

Q.1. If a car is moving with a speed of 120 km/h for one hour and for next one hour car is moving with a speed of 80 Km/h. Find the average speed of car in km/h and meter/sec ?

Q.2. If a car is moving with a speed of 56 km/h for one hour and for next one hour car is moving with a speed of 42 Km/h. Find the average speed of car in km/h and meter/sec ?

Q.3. If a car is starting moving with a speed of 26 m/sec and if the acceleration is 4 m/sec. Find the speed after 10 sec and also find the distance after 10 sec?

Q.4. If a car is start moving from rest and get a speed of 26 m/sec. Find the distance after 10 sec?

science ncert solutions

Q.5. If a car is moving with a speed of 42 km/h and after applying break the speed decreases 18 km/h in 5 sec. Find the acceleration.

Q.6. If a ship cross a river in 20 sec with a average speed of 42 m/sec and return in 40 sec with same speed. Find the distance covered by ship and also find the Average Velocity of the ship.

7. A bullet of mass 25 gram hits a target with 50 m/sec and stop in 0.02 sec.

a). The distance covered by bullet before stopping.

b). Force exerted by target on bullet.

8. A bullet of mass 25 gram hits a target with 150 m/sec and stop in 0.04 sec.

a). The distance covered by bullet before stopping.

b). Force exerted by target on bullet.

9. If a car is moving with initial velocity of 84 km/h and sudden applying break and velocity decreases to 54 km/h in 10 sec. Find the acceleration of car.

10. If a car is moving with initial velocity of 112 km/h and sudden applying break and velocity decreases to 72 km/h in 20 sec. Find the acceleration of car.

11. A body of mass 72 Kg start moving from rest and covered a distance of 1600 mm. with a acceleration of 20 m/ sec^2. Find final velocity of body and also find the momentum of body.

12. A ball of mass 225 g moving with velocity of 20 m/sec and hits another ball of mass 4kg (rest). Find the combined velocity of balls after collision.

13. A car of mass 56 kg moving with velocity of 20 m/sec for 20 sec and hits tree wall. Find the distance covered by car and also find the impact force.

14. A ball of mass 280 g moving with velocity of 20 m/sec and hits another ball of mass 4kg (rest). Find the combined velocity of balls after collision.

15. A car of mass 75 kg moving with velocity of 20 m/sec for 20 sec and hits tree wall. Find the distance covered by car and also find the impact force.

16. If a car is moving with initial velocity of 140 km/h and sudden applying break and velocity decreases to 90 km/h in 10 sec. Find the acceleration of car.

Surface area and Volume ( Part – 6 )

2019-11-20T02:08:00.000-08:00

Surface area and Volume ( Part – 6 )

1. The curved surface area of a right circular cylinder of height 7 cm is 44 cm^2 . Find the diameter of the base of the cylinder ?

Sol. Curved surface area of cylinder 44 cm^2 and height h = 7 cm

Let , the radius of base of cylinder = r cm

Curved surface area of cylinder = 2 ∏ r h

44 = ( 2 * 22 * r * 7 ) / 7

44 = 44 r

r = 1 cm

Hence , the diameter of base of cylinder = 2 r = 2 * 1 = 2 cm

2. The curved surface area of a right circular cylinder of height 21 cm is 132 cm^2 . Find the diameter of the base of the cylinder ?

Sol. Curved surface area of cylinder 132 cm^2 and height h = 21 cm

Let , the radius of base of cylinder = r cm

Curved surface area of cylinder = 2 ∏ r h

132 = ( 2 * 22 * r * 21 ) / 7

132 = 132 r

r = 1 cm

Hence , the diameter of base of cylinder = 2 r = 2 * 1 = 2 cm

3. The curved surface area of a right circular cylinder of height 28 cm is 176 cm^2 . Find the diameter of the base of the cylinder ?

Sol. Curved surface area of cylinder 176 cm^2 and height h = 28 cm

Let , the radius of base of cylinder = r cm

Curved surface area of cylinder = 2 ∏ r h

176 = ( 2 * 22 * r * 28 ) / 7

176 = 176 r

r = 1 cm

Hence , the diameter of base of cylinder = 2 r = 2 * 1 = 2 cm

maths ncert solutions

4. It is required to make a closed cylindrical tank of height 1 m and base diameter 70 cm from a metal sheet . How many square meters of the sheet are required for the same ?

Sol. Radius of base of cylinder r = 70 / 2 = 35 cm = 0.35 m and height h = 1 m

Total surface area of cylinder = 2 ∏ r ( r + h )

Total surface area of cylinder = ( 2 * 22 * 0.35 ) / 7 * ( 0.35 + 1 )

Total surface area of cylinder = 2.2 * 1.7

Total surface area of cylinder = 3.74 m^2

Hence , 3.74 square meter metal sheet is required to make the closed cylindrical tank .

5. It is required to make a closed cylindrical tank of height 1 m and base diameter 210 cm from a metal sheet . How many square meters of the sheet are required for the same ?

Sol. Radius of base of cylinder r = 210 / 2 = 105 cm = 1.05 m and height h = 1 m

Total surface area of cylinder = 2 ∏ r ( r + h )

Total surface area of cylinder = ( 2 * 22 * 1.05 ) / 7 * ( 1.05 + 1 )

Total surface area of cylinder = 6.6 * 2.05

Total surface area of cylinder = 13.53 m^2

Hence , 13.53 square meter metal sheet is required to make the closed cylindrical tank .

6. It is required to make a closed cylindrical tank of height 1 m and base diameter 280 cm from a metal sheet . How many square meters of the sheet are required for the same ?

Sol. Radius of base of cylinder r = 280 / 2 = 140 cm = 1.40 m and height h = 1 m

Total surface area of cylinder = 2 ∏ r ( r + h )

Total surface area of cylinder = ( 2 * 22 * 1.40 ) / 7 * ( 1.40 + 1 )

Total surface area of cylinder = 61.6 * 2.40

Total surface area of cylinder = 147.84 m^2

Hence , 147.84 square meter metal sheet is required to make the closed cylindrical tank .

Why do we fall ill ( Part – 4 )

2019-11-19T02:37:00.001-08:00

Why do we fall ill ( Part – 4 )

Infectious and Non – Infectious Causes

It is very important to keep public health and community health factor in mind , if we are thinking about that we have to think about the cause of disease . One group of cause is the infectious agents mostly microbes or micro – organism . In us the disease where the microbes is the immediate cause , are called infectious diseases . This is because is the microbes can be spread in the community and the disease they cause will spread with them . On the other hand the disease that does not cause microbes , it can not be spread in the communities from external . Instead , these are mostly internal , non – infectious causes .

For example :- Some cancers are caused due to the genetic abnormalities . High blood pressure is caused by high weight and lack of exercise .

The spread of the disease and the treatment of the disease is different for different disease . Whether is also a factor of Infectious and Non – Infectious Causes .

science ncert solutions

Infectious Diseases

We can say that the entire diversity of living things is classified into different groups . It can be classified on the basis of common characteristics in the organisms . These organisms are found in wide range . These organisms may be virus , bacteria , fungi or single – celled animals or protozoan . Some disease is also called by multi cellular organisms , such as worms of different kinds .

Common example of disease caused by the viruses are the common cold , dengue fever . Disease like typhoid , cholera is caused by bacteria . Many types of skin diseases is caused by fungi . Malaria and Kalaazar disease is caused by protozoan microbes .

We have to know about these is very essential because by categories these disease , we are able to get the treatment for that . Members of each one of these groups virus , bacteria , and so on have many biological characteristics in common .

Virus , bacteria and fungi multiple them very fastly . In other hand worms multiply themselves very slowly These are closely related to each other . This means many important life process is similar in the bacteria groups but not shared with the virus group . As a result , drugs that block one of these life process in the members of the group is likely to be affected many other members of the group .

For example :- Let us take antibiotics . They commonly block the biochemical path ways important for bacteria . Growing bacteria is unable to make cells and die easily . Human cells do not make a cell walls anyway . So penicillin can not have such an effect on any bacteria that use such processes for making cell walls . Similarly , many antibiotics work against many species of bacteria rather than simply working against them .

Surface area and Volume (Part - 5)

2019-11-18T02:03:00.000-08:00

Surface area and Volume (Part - 5)

1. Ramesh wanted to make a temporary shelter for her car , by making a box – like structure with tarpaulin that covers all the four sides and the top of the car ( with the front face as a flap which can be rolled up ) . Assuming that the stitching margins are very small , and therefore negligible , how much tarpaulin would be required to make the shelter of height 2.5 m , with base dimensions 4 m * 4 m ?

Sol. Length of shelter l = 4 m , breadth b = 4 m

and height h = 2.5 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 4 * 4 + 4 * 2.5 + 2.5 * 4 ) – 4 * 4 m^2

Area of four walls and top of shelter = 2 ( 16 + 10 + 10 ) – 16 m^2

Area of four walls and top of shelter = 2 ( 36 ) – 16 m^2

Area of four walls and top of shelter = 72 - 16 m^2

Area of four walls and top of shelter = 56 m^2

Hence , 56 m^2 tarpaulin is required to make this shelter .

2. Ramesh wanted to make a temporary shelter for her car , by making a box – like structure with tarpaulin that covers all the four sides and the top of the car ( with the front face as a flap which can be rolled up ) . Assuming that the stitching margins are very small , and therefore negligible , how much tarpaulin would be required to make the shelter of height 2.5 m , with base dimensions 3 m * 3 m ?

Sol. Length of shelter l = 3 m , breadth b = 3 m

and height h = 2.5 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 3 * 3 + 3 * 2.5 + 2.5 * 3 ) – 3 * 3 m^2

Area of four walls and top of shelter = 2 ( 9 + 7.5 + 7.5 ) – 9 m^2

Area of four walls and top of shelter = 2 ( 24 ) – 9 m^2

Area of four walls and top of shelter = 48 - 9 m^2

Area of four walls and top of shelter = 39 m^2

Hence , 39 m^2 tarpaulin is required to make this shelter .

maths ncert solutions

3. Ramesh wanted to make a temporary shelter for her car , by making a box – like structure with tarpaulin that covers all the four sides and the top of the car ( with the front face as a flap which can be rolled up ) . Assuming that the stitching margins are very small , and therefore negligible , how much tarpaulin would be required to make the shelter of height 2 m , with base dimensions 4 m * 3 m ?

Sol. Length of shelter l = 4 m , breadth b = 3 m

and height h = 2 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 4 * 3 + 3 * 2 + 2 * 4 ) – 4 * 3 m^2

Area of four walls and top of shelter = 2 ( 12 + 6 + 8 ) – 12 m^2

Area of four walls and top of shelter = 2 ( 26 ) – 12 m^2

Area of four walls and top of shelter = 52 - 12 m^2

Area of four walls and top of shelter = 40 m^2

Hence , 40 m^2 tarpaulin is required to make this shelter .

4. Ramesh wanted to make a temporary shelter for her car , by making a box – like structure with tarpaulin that covers all the four sides and the top of the car ( with the front face as a flap which can be rolled up ) . Assuming that the stitching margins are very small , and therefore negligible , how much tarpaulin would be required to make the shelter of height 1.5 m , with base dimensions 4 m * 3 m ?

Sol. Length of shelter l = 4 m , breadth b = 3 m

and height h = 1.5 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 4 * 3 + 3 * 1.5 + 1.5 * 4 ) – 4 * 3 m^2

Area of four walls and top of shelter = 2 ( 12 + 4.5 + 6 ) – 12 m^2

Area of four walls and top of shelter = 2 ( 22.5 ) – 12 m^2

Area of four walls and top of shelter = 45 - 12 m^2

Area of four walls and top of shelter = 33 m^2

Hence , 33 m^2 tarpaulin is required to make this shelter .

Assignment (Motion)

2019-11-16T02:13:00.000-08:00

Assignment (Motion)

Q.1. If a car is moving with a speed of 100 km/h for one hour and for next one hour car is moving with a speed of 80 Km/h. Find the average speed of car in km/h and meter/sec ?

Q.2. If a car is moving with a speed of 44 km/h for one hour and for next one hour car is moving with a speed of 34 Km/h. Find the average speed of car in km/h and meter/sec ?

Q.3. If a car is starting moving with a speed of 50 m/sec and if the acceleration is 4 m/sec. Find the speed after 10 sec and also find the distance after 10 sec?

Q.4. If a car is start moving from rest and get a speed of 40 m/sec. Find the distance after 10 sec?

Q.5. If a car is moving with a speed of 44 km/h and after applying break the speed decreases 18 km/h in 5 sec. Find the acceleration.

Q.6. If a ship cross a river in 20 sec with a average speed of 45 m/sec and return in 40 sec with same speed. Find the distance covered by ship and also find the Average Velocity of the ship.

science ncert solutions

7. A bullet of mass 20 gram hits a target with 50 m/sec and stop in 0.02 sec.

a). The distance covered by bullet before stoping.

b). Force exerted by target on bullet.

8. A bullet of mass 20 gram hits a target with 150 m/sec and stop in 0.04 sec.

a). The distance covered by bullet before stoping.

b). Force exerted by target on bullet.

9. If a car is moving with initial velocity of 90 km/h and sudden applying break and velocity decreases to 54 km/h in 10 sec. Find the acceleration of car.

10. If a car is moving with initial velocity of 108 km/h and sudden applying break and velocity decreases to 72 km/h in 20 sec. Find the acceleration of car.

11. A body of mass 80 Kg start moving from rest and covered a distance of 1600 mm. with a acceleration of 20 m/ sec^2. Find final velocity of body and also find the momentum of body.

12. A ball of mass 250 g moving with velocity of 20 m/sec and hits another ball of mass 4kg (rest). Find the combined velocity of balls after collision.

13. A car of mass 50 kg moving with velocity of 20 m/sec for 20 sec and hits tree wall. Find the distance covered by car and also find the impact force.

14. A ball of mass 300 g moving with velocity of 20 m/sec and hits another ball of mass 4kg (rest). Find the combined velocity of balls after collision.

15. A car of mass 80 kg moving with velocity of 20 m/sec for 20 sec and hits tree wall. Find the distance covered by car and also find the impact force.

16. If a car is moving with initial velocity of 150 km/h and sudden applying break and velocity decreases to 90 km/h in 10 sec. Find the acceleration of car.

Surface area and Volume (Part - 4)

2019-11-15T02:55:00.000-08:00

Surface area and Volume (Part - 4)

1. Om Shanti sweets stall was placing an order for making cardboard boxes for packing their sweets . Two sizes of boxes were required . The bigger of dimensions 15 cm * 20 cm * 5 cm and the smaller od dimensions 15 cm * 10 cm * 5 cm . For all the overlaps , 5 % of the total surface area required extra . If the cost of the cardboard is 4 Rs for 1000 cm^2 , find the cost of cardboard required for supplying 250 boxes of each kind .

Sol. For bigger box

Length of box l = 15 cm , breath b = 20 cm and height h = 5 cm

Total surface area of 1 bigger box = 2 ( l b + b h + h l )

Total surface area of 1 bigger box = 2 ( 15 * 20 + 20 * 5 + 5 * 15 )

Total surface area of 1 bigger box = 2 ( 300 + 100 + 75 )

Total surface area of 1 bigger box = 2 ( 475 )

Total surface area of 1 bigger box = 950 cm^2

Area of cardboard for overlap = 5 % of 950 cm^2 = 950 * ( 5 / 100 ) = 47.5 cm^2

Total area of cardboard for 1 bigger box = 950 + 47.5 = 997.5 cm^2

Therefore , the area of cardboard for 250 bigger boxes = 250 * 997.5 cm^2 = 249375 cm^2

For smaller box

Length of box l = 15 cm , breath b = 10 cm and height h = 5 cm

Total surface area of 1 bigger box = 2 ( l b + b h + h l )

Total surface area of 1 bigger box = 2 ( 15 * 10 + 10 * 5 + 5 * 15 )

Total surface area of 1 bigger box = 2 ( 150 + 50 + 75 )

Total surface area of 1 bigger box = 2 ( 275 )

Total surface area of 1 bigger box = 550 cm^2

Area of cardboard for overlap = 5 % of 550 cm^2 = 550 * ( 5 / 100 ) = 27.5 cm^2

Total area of cardboard for 1 bigger box = 550 + 27.5 = 577.5 cm^2

Therefore , the area of cardboard for 250 bigger boxes = 250 * 577.5 cm^2 = 144375 cm^2

S the area of cardboard for 500 boxes = 249375 + 144375 = 393750 cm^2

Total cost of cardboard at the rate of Rs 4 per 1000 cm^2 = ( 393750 * 4 ) / 1000 = Rs 1575

Hence , the total cost of cardboard for 500 boxes is Rs 1575 .

maths ncert solutions

2. Om Shanti sweets stall was placing an order for making cardboard boxes for packing their sweets . Two sizes of boxes were required . The bigger of dimensions 20 cm * 20 cm * 5 cm and the smaller od dimensions 10 cm * 12 cm * 5 cm . For all the overlaps , 5 % of the total surface area required extra . If the cost of the cardboard is 4 Rs for 1000 cm^2 , find the cost of cardboard required for supplying 250 boxes of each kind .

Sol. For bigger box

Length of box l = 20 cm , breath b = 20 cm and height h = 5 cm

Total surface area of 1 bigger box = 2 ( l b + b h + h l )

Total surface area of 1 bigger box = 2 ( 20 * 20 + 20 * 5 + 5 * 20 )

Total surface area of 1 bigger box = 2 ( 400 + 100 + 100 )

Total surface area of 1 bigger box = 2 ( 600 )

Total surface area of 1 bigger box = 1200 cm^2

Area of cardboard for overlap = 5 % of 1200 cm^2 = 1200 * ( 5 / 100 ) = 60 cm^2

Total area of cardboard for 1 bigger box = 1200 + 60 = 1260 cm^2

Therefore , the area of cardboard for 250 bigger boxes = 250 * 1260 cm^2 = 315000 cm^2

For smaller box

Length of box l = 10 cm , breath b = 12 cm and height h = 5 cm

Total surface area of 1 bigger box = 2 ( l b + b h + h l )

Total surface area of 1 bigger box = 2 ( 10 * 12 + 12 * 5 + 5 * 10 )

Total surface area of 1 bigger box = 2 ( 120 + 60 + 50 )

Total surface area of 1 bigger box = 2 ( 230 )

Total surface area of 1 bigger box = 460 cm^2

Area of cardboard for overlap = 5 % of 460 cm^2 = 460 * ( 5 / 100 ) = 23 cm^2

Total area of cardboard for 1 bigger box = 460 + 23 = 483 cm^2

Therefore , the area of cardboard for 250 bigger boxes = 250 * 483 cm^2 = 120750 cm^2

S the area of cardboard for 500 boxes = 315000 + 120750 = 435750 cm^2

Total cost of cardboard at the rate of Rs 4 per 1000 cm^2 = ( 435750 * 4 ) / 1000 = Rs 1743

Hence , the total cost of cardboard for 500 boxes is Rs 1743 .

Why do we fall ill (Part – 3)

2019-11-14T01:27:00.002-08:00

Chronic Diseases and Poor Health

We now that acute and chronic diseases have different effect our health , any disease that cause the poor functioning of any part of our body will affect our health . So the proper functioning of all the parts of our body is necessary for good health . When we are talking about acute disease , which is over a very soon , will not have time to cause major effects on general health and on other hand chronic diseases will not do so .

For example :- When we are fall ill due to cough and cold , we are getting better within the weak in most of cases . We never fell more bad effect on our health like we do not have weakness , mostly we do not have breathing problems . We do not losing weight .

If we have chromic disease , in such diseases we feel weakness , losing weight . These type of disease take a long time to recover , some times it takes a few years to recover and after the few year recovery , we are not able to do things like before means effect of disease will going on for a very long time .

For example :- When we are falling ill due to acute disease , we take leave of few days form our work . In other hand if we are falling ill due to chromic disease we are not able to say any thing about when we are back to our work . Some time it takes a few years to recover but we have no idea about the time taken for recovery .

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Causes of Disease

When we are thinking about the cause of disease , we think about the different levels of cause of disease . In general when any body is suffering from loose motion we can say that the cause of disease is infection due to virus . In the case of babies we think this happens by the unclean drinking water but most of babies do not drink unclean water but they have loose motions . We will find the answer of this.

One reason might be that the baby are not healthy that’s why the is getting loose motions and in other hand the healthy baby of same age does not have loose motions .

When a baby does not get proper nourished . so the lack of nourishment becomes the second level cause of disease . It is seen in the house which is poor .

It is also possible that the baby has some genetic difference that makes it more likely to suffer from loose motion . Without the virus , the genetic difference and the poor nourishment alone would not lead to loose motion .

The poorness of the family also lead to the third level of cause of disease because they have not proper environment to protect their children form virus . It is not necessary that all the diseases have single cause , it may be a multiple cause disease .

Assignment on Surface area and Volume

2019-11-13T03:04:00.000-08:00

Assignment on Surface area and Volume

1. A plastic box of 15 m long 12 m wide and 600 cm deep is to be made . It is to be open at top . Ignoring the thickness of plastic sheet , determine

a). The area of sheet required for making the box.

b). The cost of sheet for it , if a sheet measuring 1 m^2 costs Rs 10 .

2. A plastic box of 10 m long 10 m wide and 650 cm deep is to be made . It is to be open at top . Ignoring the thickness of plastic sheet , determine

a). The area of sheet required for making the box.

b). The cost of sheet for it , if a sheet measuring 1 m^2 costs Rs 15 .

3. The length, breath and height of a room are 4000 cm, 3000 cm and 3000 cm respectively . Find the cost of white washing the walls of the room and the ceiling at the rate of 10 Rs per m^2 .

4. The length, breath and height of a room are 2000 cm, 1500 cm and 1500 cm respectively . Find the cost of white washing the walls of the room and the ceiling at the rate of 10 Rs per m^2 .

5. The floor of a rectangular hall has a perimeter 200 m . If the cost of painting the four walls at the rate of 10 Rs per m^2 is 10000 , find the height of the hall . [ Hint :- Area of the four walls = Lateral surface area ]

maths-ncert-soltutions

6. The floor of a rectangular hall has a perimeter 100 m . If the cost of painting the four walls at the rate of 10 Rs per m^2 is 5000 , find the height of the hall . [ Hint :- Area of the four walls = Lateral surface area ]

7. The floor of a rectangular hall has a perimeter 150 m . If the cost of painting the four walls at the rate of 10 Rs per m^2 is 8000 , find the height of the hall . [ Hint :- Area of the four walls = Lateral surface area ]

8. The paint in a certain container is sufficient to paint an area equal to 8.88 m^2 . How many bricks of dimensions 20 cm * 10 cm * 7 cm can be painted out of this container ?

9. The paint in a certain container is sufficient to paint an area equal to 7 m^2 . How many bricks of dimensions 18 cm * 10 cm * 5 cm can be painted out of this container ?

10. The paint in a certain container is sufficient to paint an area equal to 5 m^2 . How many bricks of dimensions 12 cm * 10 cm * 4 cm can be painted out of this container ?

Assignment - 6 (Motion , Force and Law of Motion)

2019-11-12T00:34:00.000-08:00

Motion , Force and Law of Motion

Q.1. If a car is moving with a speed of 80 km/h for one hour and for next one hour car is moving with a speed of 60 Km/h. Find the average speed of car in km/h and meter/sec ?

Q.2. If a car is moving with a speed of 72 km/h for one hour and for next one hour car is moving with a speed of 54 Km/h. Find the average speed of car in km/h and meter/sec ?

Q.3. If a car is starting moving with a speed of 40 m/sec and if the acceleration is 4 m/sec. Find the speed after 10 sec and also find the distance after 10 sec?

Q.4. If a car is start moving from rest and get a speed of 30 m/sec. Find the distance after 10 sec?

Q.5. If a car is moving with a speed of 36 km/h and after applying break the speed decreases 18 km/h in 5 sec. Find the acceleration.

Q.6. If a ship cross a river in 40 sec with a average speed of 36 m/sec and return in 40 sec with same speed. Find the distance covered by ship and also find the Average Velocity of the ship.

7. A bullet of mass 10 gram hits a target with 40 m/sec and stop in 0.02 sec.

a). The distance covered by bullet before stoping.

b). Force exerted by target on bullet.

8. A bullet of mass 12 gram hits a target with 120 m/sec and stop in 0.04 sec.

a). The distance covered by bullet before stopping.

b). Force exerted by target on bullet.

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9. If a car is moving with initial velocity of 90 km/h and sudden applying break and velocity decreases to 54 km/h in 10 sec. Find the acceleration of car.

10. If a car is moving with initial velocity of 126 km/h and sudden applying break and velocity decreases to 72 km/h in 20 sec. Find the acceleration of car.

11. A body of mass 75 Kg start moving from rest and covered a distance of 1600 mm. with a acceleration of 20 m/ sec^2. Find final velocity of body and also find the momentum of body.

12. A ball of mass 300 g moving with velocity of 20 m/sec and hits another ball of mass 4 kg (rest). Find the combined velocity of balls after collision.

13. A car of mass 60 kg moving with velocity of 20 m/sec for 20 sec and hits tree wall. Find the distance covered by car and also find the impact force.

14. A ball of mass 300 g moving with velocity of 20 m/sec and hits another ball of mass 4 kg (rest). Find the combined velocity of balls after collision.

15. A car of mass 70 kg moving with velocity of 20 m/sec for 20 sec and hits tree wall. Find the distance covered by car and also find the impact force.

16. If a car is moving with initial velocity of 140 km/h and sudden applying break and velocity decreases to 90 km/h in 10 sec. Find the acceleration of car.

Surface area and Volume (Part - 3)

2019-11-11T02:57:00.000-08:00

Surface area and Volume (Part - 3)

1. A cubical box has each edge 9 cm and another cuboidal box is 10 cm long , 10 cm wide and 8 cm high .

a). Which box has the greater lateral surface area and by how much ?

b). Which box has the smaller total surface area and by how much ?

Sol. a). Side of cubical box l = 9 cm

Curved surface area of cubical box = 4 l^2 cm^2 = 4 ( 9 )^2 cm^2 = 4 * 81 cm^2 = 324 cm^2

Length of cuboidal box l = 9 cm^2 breadth = 10 cm^2 and height = 8 cm^2

Curved surface area of cuboidal box = 2 ( l + b ) h cm^2 = 2 * ( 9 + 10 ) * 8 cm^2

Curved surface area of cuboidal box = 2 * 19 * 8 cm^2 = 304 cm^2

Difference between surface area of two boxes = 324 cm^2 – 304 cm^2 = 20 cm^2

Hence , the curved surface area of cubical box is more than cuboidal box by 20 cm^2 .

b). Side of cubical box l = 9 cm

Total surface area of cubical box = 6 l^2 cm^2 = 6 ( 9 )^2 cm^2 = 6 * 81 cm^2 = 486 cm^2

Length of cuboidal box l = 9 cm^2 breadth = 10 cm^2 and height = 8 cm^2

Total surface area of cuboidal box = 2 ( l b + b h + h l ) cm^2 = 2 * ( 9 * 10 + 10 * 8 + 8 * 9 ) cm^2

Total surface area of cuboidal box = 2 ( 90 + 80 + 72 ) cm^2

Total surface area of cuboidal box = 2 ( 242 ) cm^2 = 484 cm^2

Difference between total surface area of two boxes = 486 cm^2 – 484 cm^2 = 2 cm^2

Hence , the curved surface area of cubical box is more than cuboidal box by 2 cm^2 .

2. A small indoor green house is made of entirely of glass panes held together with tape . It is 28 cm long , 25 cm wide and 25 cm high .

a). What is the area of glass ?

b). How much tape is needed for all the 12 edges ?

sol. a). Length of green house l = 28 , breath b = 25 cm and height = 25 cm

Total surface area of green house = 2 ( l b + b h + h l ) cm^2 = 2 * ( 28 * 25 + 25 * 25 + 25 * 28 ) cm^2

Total surface area of green house = 2 * ( 700 + 625 + 700 ) cm^2 = 2 * 2025 cm^2

Total surface area of green house = 4050 cm^2

Hence , the area of glass for greenhouse is 4050 cm^2

b). Length of green house l = 28 cm , breath = 25 cm and height h = 25 cm

Length of tape for all 12 edges = 4 ( l + b + h ) cm = 4 ( 28 + 25 + 25 ) cm

Length of tape for all 12 edges = 4 * 78 cm = 312 cm

Hence , for all 12 edges , 312 cm tape is required .

3. A cubical box has each edge 9 cm and another cuboidal box is 8 cm long , 10 cm wide and 8 cm high .

a). Which box has the greater lateral surface area and by how much ?

b). Which box has the smaller total surface area and by how much ?

Sol. a). Side of cubical box l = 9 cm

Curved surface area of cubical box = 4 l^2 cm^2 = 4 ( 9 )^2 cm^2 = 4 * 81 cm^2 = 324 cm^2

Length of cuboidal box l = 8 cm^2 breadth = 10 cm^2 and height = 8 cm^2

Curved surface area of cuboidal box = 2 ( l + b ) h cm^2 = 2 * ( 8 + 10 ) * 8 cm^2

Curved surface area of cuboidal box = 2 * 18 * 8 cm^2 = 288 cm^2

Difference between surface area of two boxes = 324 cm^2 – 288 cm^2 = 36 cm^2

Hence , the curved surface area of cubical box is more than cuboidal box by 36 cm^2 .

b). Side of cubical box l = 9 cm

Total surface area of cubical box = 6 l^2 cm^2 = 6 ( 9 )^2 cm^2 = 6 * 81 cm^2 = 486 cm^2

Length of cuboidal box l = 8 cm^2 breadth = 10 cm^2 and height = 8 cm^2

Total surface area of cuboidal box = 2 ( l b + b h + h l ) cm^2 = 2 * ( 8 * 10 + 10 * 8 + 8 * 8 ) cm^2

Total surface area of cuboidal box = 2 ( 80 + 80 + 64 ) cm^2

Total surface area of cuboidal box = 2 ( 224 ) cm^2 = 448 cm^2

Difference between total surface area of two boxes = 486 cm^2 – 448 cm^2 = 38 cm^2

Hence , the curved surface area of cubical box is more than cuboidal box by 38 cm^2 .

maths ncert solutions

4. A small indoor green house is made of entirely of glass panes held together with tape . It is 25 cm long , 20 cm wide and 20 cm high .

a). What is the area of glass ?

b). How much tape is needed for all the 12 edges ?

sol. a). Length of green house l = 25 , breath b = 20 cm and height = 20 cm

Total surface area of green house = 2 ( l b + b h + h l ) cm^2 = 2 * ( 25 * 20 + 20 * 20 + 20 * 25 ) cm^2

Total surface area of green house = 2 * ( 500 + 400 + 500 ) cm^2 = 2 * 1400 cm^2

Total surface area of green house = 2800 cm^2

Hence , the area of glass for greenhouse is 2800 cm^2

b). Length of green house l = 25 cm , breath = 20 cm and height h = 20 cm

Length of tape for all 12 edges = 4 ( l + b + h ) cm = 4 ( 25 + 20 + 20 ) cm

Length of tape for all 12 edges = 4 * 65 cm = 260 cm

Hence , for all 12 edges , 260 cm tape is required .

Assignment - 5 (Based on Work and Energy)

2019-11-09T03:31:00.000-08:00

Assignment - 5 (Based on Work and Energy)

1. A object of mass 50 kg at a height. If the potential energy of object is 800 J. Find the height of object? (Take g = 10 m/sec^2 )

2. A object of mass 60 kg at a height. If the potential energy of object is 600 J. Find the height of object? (Take g = 10 m/sec^2 )

3. Two army men, each weight have 35 Kg, climbs on a rope for height 5 m. First army man takes 5 sec and second army man takes 10 sec. Find the power expended by army man ?

4. Two army men, each weight have 90 Kg, climbs on a rope for height 6 m. First army man takes 7 sec and second army man takes 14 sec. Find the power expended by army man ?

5. A army woman of mass 108 kg runs on staircase 36 steps in 10 sec. If the height of each staircase is 12 cm. Find her power? (Take g = 10 m/sec^2 )

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6. A army woman of mass 60 kg runs on staircase 30 steps in 10 sec. If the height of each staircase is 10 cm. Find her power? (Take g = 10 m/sec^2 )

7. A bulb of 7 w is used for 9 hours per day. Find units consumed in a day ?

8. A fan of 100 w is used for 9 hours per day. Find units consumed in a week ?

9. A television of 200 w is used for 9 hours per day. Find units consumed in a

month ?

10. A fan of 100 W and a bulb of 7 W is used for 9 hours per day. Find units consumed in a month ?

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11. A object of mass 25 kg at a height. If the potential energy of object is 800 J. Find the height of object? (Take g = 10 m/sec^2 )

12. A object of mass 15 kg at a height. If the potential energy of object is 600 J. Find the height of object? (Take g = 10 m/sec^2 )

13. Two army men, each weight have 60 Kg, climbs on a rope for height 5 m. First army man takes 5 sec and second army man takes 10 sec. Find the power expended by army man ?

14. Two army men, each weight have 120 Kg, climbs on a rope for height 6 m. First army man takes 7 sec and second army man takes 14 sec. Find the power expended by army man ?

15. A army woman of mass 81 kg runs on staircase 36 steps in 10 sec. If the height of each staircase is 12 cm. Find her power? (Take g = 10 m/sec^2 )

16. A army woman of mass 90 kg runs on staircase 30 steps in 10 sec. If the height of each staircase is 10 cm. Find her power? (Take g = 10 m/sec^2 )

17. A bulb of 7 w is used for 11 hours per day. Find units consumed in a day ?

18. A fan of 100 w is used for 11 hours per day. Find units consumed in a week ?

19. A television of 200 w is used for 11 hours per day. Find units consumed in a month ?

20. A fan of 100 W and a bulb of 7 W is used for 11 hours per day. Find units consumed in a month ?

Surface Area and Volume (Part - 2)

2019-11-08T01:37:00.000-08:00

Surface Area and Volume (Part - 2)

1. The floor of a rectangular hall has a perimeter 200 m . If the cost of painting the four walls at the rate of 10 Rs per m^2 is 12000 , find the height of the hall . [ Hint :- Area of the four walls = Lateral surface area ]

Sol. The length of hall = l m , breath = b m and height = h m

Perimeter of floor = 2 ( l + b )

According to question

2 ( l + b ) = 200 m ….(1)

Area of four walls of hall = 2 ( l + b ) h

So the cost of painting four walls at the rate of Rs 10 per m^2 = 2 ( l + b ) h * 10

= 200 h * 10

= Rs 2000 h

According to question ,

2000 h = 12000

h = 12000 / 2000

h = 6 m

Hence , the height of the hall is 6 m.

2. The floor of a rectangular hall has a perimeter 100 m . If the cost of painting the four walls at the rate of 10 Rs per m^2 is 6000 , find the height of the hall . [ Hint :- Area of the four walls = Lateral surface area ]

Sol. The length of hall = l m , breath = b m and height = h m

Perimeter of floor = 2 ( l + b )

According to question

2 ( l + b ) = 100 m ….(1)

Area of four walls of hall = 2 ( l + b ) h

So the cost of painting four walls at the rate of Rs 10 per m^2 = 2 ( l + b ) h * 10

= 100 h * 10

= Rs 1000 h

According to question ,

1000 h = 6000

h = 6000 / 1000

h = 6 m

Hence , the height of the hall is 6 m.

3. The floor of a rectangular hall has a perimeter 150 m . If the cost of painting the four walls at the rate of 10 Rs per m^2 is 9000 , find the height of the hall . [ Hint :- Area of the four walls = Lateral surface area ]

Sol. The length of hall = l m , breath = b m and height = h m

Perimeter of floor = 2 ( l + b )

According to question

2 ( l + b ) = 150 m ….(1)

Area of four walls of hall = 2 ( l + b ) h

So the cost of painting four walls at the rate of Rs 10 per m^2 = 2 ( l + b ) h * 10

= 150 h * 10

= Rs 1500 h

According to question ,

1500 h = 9000

h = 9000 / 1500

h = 6 m

Hence , the height of the hall is 6 m.

maths ncert solutions

4. The paint in a certain container is sufficient to paint an area equal to 8.88 m^2 . How many bricks of dimensions 22 cm * 10 cm * 7 cm can be painted out of this container ?

Sol. Length of the brick l = 22 cm , breath b = 10 cm and height h = 7 cm

Total surface area of brick = 2 ( l b + b h + h l )

Total surface area of brick = 2 ( 22 * 10 + 10 * 7 + 7 * 22 )

Total surface area of brick = 2 ( 220 + 70 + 154 )

Total surface area of brick = 2 * 444

Total surface area of brick = 888 cm^2 = 0.0888 m^2

Number of bricks to be painted = ( Total paint available ) / ( Paint for one brick )

Number of bricks to be painted = 8.88 m^2 / 0.0888 m^2 = 100

Hence , the paint of this container can paint 100 bricks.

5. The paint in a certain container is sufficient to paint an area equal to 7 m^2 . How many bricks of dimensions 20 cm * 10 cm * 5 cm can be painted out of this container ?

Sol. Length of the brick l = 20 cm , breath b = 10 cm and height h = 5 cm

Total surface area of brick = 2 ( l b + b h + h l )

Total surface area of brick = 2 ( 20 * 10 + 10 * 5 + 5 * 20 )

Total surface area of brick = 2 ( 200 + 50 + 100 )

Total surface area of brick = 2 * 350

Total surface area of brick = 700 cm^2 = 0.0700 m^2

Number of bricks to be painted = ( Total paint available ) / ( Paint for one brick )

Number of bricks to be painted = 7 m^2 / 0.0700 m^2 = 100

Hence , the paint of this container can paint 100 bricks.

6. The paint in a certain container is sufficient to paint an area equal to 5 m^2 . How many bricks of dimensions 15 cm * 10 cm * 4 cm can be painted out of this container ?

Sol. Length of the brick l = 15 cm , breath b = 10 cm and height h = 4 cm

Total surface area of brick = 2 ( l b + b h + h l )

Total surface area of brick = 2 ( 15 * 10 + 10 * 4 + 4 * 15 )

Total surface area of brick = 2 ( 150 + 40 + 60 )

Total surface area of brick = 2 * 250

Total surface area of brick = 500 cm^2 = 0.0500 m^2

Number of bricks to be painted = ( Total paint available ) / ( Paint for one brick )

Number of bricks to be painted = 5 m^2 / 0.0500 m^2 = 100

Hence , the paint of this container can paint 100 bricks.

Why do we fall ill ( Part – 2 )

2019-11-07T01:56:00.002-08:00

Why do we fall ill ( Part – 2 )

Healthy and disease free

We can say in simple language if we feel uncomfortable in any way , then generally it means disease . Disease means when we cause a specific or particular cause of discomfort . It does not mean the actual cause of discomfort . For example some is saying for viral fever without the knowing , the cause of fever.

We can also see the poor health without a particular disease . If we have no diseases , it does not mean that we are healthy . For example :- For a dancer its very important that he / she can stretch their body parts easily . If any dancer is not able to do this that person is not saying healthy . In other hand for singer or musicians it is necessary to control its breath , so that he / she can take their notes .

If we are not able to do this then we are not in Good Health without having a particular disease . When we are taking about disease we are taking about individual and if we are taking about Health , it means we are taking about societies and communities .

Disease and Its Cause

We know that there are many tissues in our body , it was briefly described in earlier chapter tissue . These tissues make a physical system or organ system that perform the various functions in our body . In the organ system has specific organs as its part that are performing a specific function in our body . the digestive system has stomach and intestines and it helps in digesting the food we are eating from outside . The musculoskeletal system which is made of bones and muscles , holds the body part together and make us able to move .

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In this if we are taking about disease it means that the functioning of particular organ or any responsible tissues are not performing their function properly . That means system of body changes . These changes give in rise of symptoms and sign of the disease . Symptoms are feeling of disease that seems us something is Wrong in our body . These symptoms may be fever , headache , cough , loss motion etc . It indicates that there are any disease in our body but it is not indicates about the particular disease .

For example :- If we are having a headache it means many of diseases can be possible , headache can be due to stress in mind , it also due to migraine , it also due to the decreasing eye sight .

Sign or disease are indicating to the diseases but it can not be said surely . For saying about a particular disease , the physicians take a step of lab tests that give the pin point result of any disease and then they are able to give treatment for that disease .

Surface area and Volumes

2019-11-06T01:18:00.001-08:00

Surface area and Volumes

Cube

Lateral Surface Area = 4 a^2

Total Surface Area = 6 a^2

Volume = a^3

Cuboid

Lateral Surface Area = 2 ( l h + b h ) = 2 ( l + b ) h

Total Surface Area = 2 ( l b + b h + h I )

Volume = l * b * h

Cylinder

Curved Surface Area = 2 ∏ r h

Total Surface Area = 2 ∏ r h + ∏ r^2

Volume = ∏ r^2 h

Cone

Curved Surface Area = ∏ r l

Total Surface Area = ∏ r l + ∏ r^2 = ∏ r ( r + l )

Volume = 1/3 ∏ r^2 h

Hemisphere

Curved Surface Area = 2 ∏ r^2

Total Surface Area = 2 ∏ r l + ∏ r^2 = 3 ∏ r^2

Volume = 2/3 ∏ r^3

Sphere

Surface Area = 4 ∏ r^2

Volume = 4/3 ∏ r^3

maths ncert solutions

1. A plastic box of 15 m long 12.5 m wide and 650 cm deep is to be made . It is to be open at top . Ignoring the thickness of plastic sheet , determine

a). The area of sheet required for making the box.

b). The cost of sheet for it , if a sheet measuring 1 m^2 costs Rs 10 .

Sol. Given l = 15 m b = 12.5 m h = 650 cm = 6.5 m

Area of sheet required for making plastic box = Total surface area of box – Area of top of box

Area of sheet required for making plastic box = 2 ( l b + b h + h l ) – l b

Area of sheet required for making plastic box = 2 ( 15 * 12.5 + 12.5 * 6.5 + 6.5 * 15 ) – 15 * 12.5

Area of sheet required for making plastic box = 2 ( 187.5 + 81.25 + 97.5 ) – 187.5

Area of sheet required for making plastic box = 2 * 366.25 – 187.5

Area of sheet required for making plastic box = 732.5 – 187.5

Area of sheet required for making plastic box = 545 m^2

Hence, area of sheet required for making plastic box is 545 m^2 .

b). Total cost of sheet = 10 * 545 = 5450 Rs

2. A plastic box of 10 m long 12.5 m wide and 650 cm deep is to be made . It is to be open at top . Ignoring the thickness of plastic sheet , determine

a). The area of sheet required for making the box.

b). The cost of sheet for it , if a sheet measuring 1 m^2 costs Rs 15 .

Sol. Given l = 10 m b = 12.5 m h = 650 cm = 6.5 m

Area of sheet required for making plastic box = Total surface area of box – Area of top of box

Area of sheet required for making plastic box = 2 ( l b + b h + h l ) – l b

Area of sheet required for making plastic box = 2 ( 10 * 12.5 + 12.5 * 6.5 + 6.5 * 10 ) – 10 * 12.5

Area of sheet required for making plastic box = 2 ( 125 + 81.25 + 65 ) – 125

Area of sheet required for making plastic box = 2 * 271.25 – 125

Area of sheet required for making plastic box = 542.5 – 125

Area of sheet required for making plastic box = 417.5 m^2

Hence, area of sheet required for making plastic box is 417.5 m^2 .

b). Total cost of sheet = 15 * 417.5 = 6262.50 Rs

3. The length, breath and height of a room are 5000 cm, 4000 cm and 3000 cm respectively . Find the cost of white washing the walls of the room and the ceiling at the rate of 10 Rs per m^2 .

Sol. Given l = 5000 cm = 50 m b = 4000 cm = 40 m h = 3000 cm = 30 m

Area of four walls and ceiling = Total surface area of room – area of floor

Area of four walls and ceiling = 2 ( l b + b h + h l ) – l b

Area of four walls and ceiling = 2 ( 50 * 40 + 40 * 30 + 30 * 50 ) – 50 * 40

Area of four walls and ceiling = 2 ( 2000 + 1200 + 1500 ) – 2000

Area of four walls and ceiling = 2 * 4700 – 2000

Area of four walls and ceiling = 9400 – 2000

Area of four walls and ceiling = 7400 m^2

Therefore, Area of four walls and ceiling is 7400 m^2 .

Hence, the cost of white washing the walls of the room and the ceiling = 10 * 7400 = Rs 74,000/-

4. The length, breath and height of a room are 2500 cm, 2000 cm and 1500 cm respectively . Find the cost of white washing the walls of the room and the ceiling at the rate of 10 Rs per m^2 .

Sol. Given l = 2500 cm = 25 m b = 2000 cm = 20 m h = 1500 cm = 15 m

Area of four walls and ceiling = Total surface area of room – area of floor

Area of four walls and ceiling = 2 ( l b + b h + h l ) – l b

Area of four walls and ceiling = 2 ( 25 * 20 + 20 * 15 + 15 * 25 ) – 25 * 20

Area of four walls and ceiling = 2 ( 500 + 300 + 375 ) – 500

Area of four walls and ceiling = 2 * 1175 – 500

Area of four walls and ceiling = 2350 – 500

Area of four walls and ceiling = 1850 m^2

Therefore, Area of four walls and ceiling is 1850 m^2 .

Hence, the cost of white washing the walls of the room and the ceiling = 10 * 1850 = Rs 18,500 /-

Why do we fall ill (Part - 1)

2019-11-05T02:20:00.000-08:00

Why do we fall ill (Part - 1)

We know that health and disease are complex issues interconnected with human body . When we are taking about health and disease we find that it is very complicated to understand these. We know that cells are the basic unit of our body . These cells are made up of various things like as protein , carbohydrates , fats , liquids and other things . The functioning of our body is done by the help of cells . Cells moves one place to another place . Some cells do not move , there are repair going on . in our body , there are various organs and tissues that are performing various activities in our body like as heart is doing an activity of pumping blood , lungs is doing an activity of breathing , kidney of our body is filtering urine and brain is control room of whole body means brain controlling activities of our body .

All the activities of our body are interconnected . If any of them is not doing there activity it means it effects other functions also .

For example :- If kidney is not filtering urine in such case , it became poisonous and the brain is not able think properly .

For performing these activities energy and raw materials are needed from outside the body . So food is needed for the proper functioning of body . It leads to proper functioning of cells and tissues .

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Health and its failure

The meaning of word health is very complicated to understand . When we say the health is not good what does it means . The word health indicates the idea of ‘Being well’ . It can define as proper functioning of our body parts . If any old citizen is able to do their daily activities like as walking , going market , taking with others then we say that “ That old citizen is healthy “. Same in other hand If any old citizen is not able to do their daily activities like as walking , going market , taking with others then we say that “ That old citizen is not healthy “ or “poor health”. Health is a state of being well enough to function well physically , mentally and socially .

Personal and community issues both matter foe health

Health is a state of being well enough to function well physically , mentally and socially . It can not be something that each one of can achieve entirely on our own . The health is depend on environment for example if air in our surroundings is not fresh we are not able to breath properly . In modern century we are living in cities , town , village in these our health is also depend on social environment . If there are waste collected near our house or our street , that cause poor health due to various disease . We need food for proper health . So good economic condition is needed for good health .

We are living in society , if we are living in our society with piece , happiness and without fear . It cause the good individual health .

Assignment – 5 ( Heron’s Formula )

2019-11-04T03:46:00.000-08:00

Assignment – 5 ( Heron’s Formula )

1. A board in a shape of triangle having equal dimensions ‘a’. Find the area of board using Heron’s formula. If its perimeter is 180 cm. What will be the area of board ?

2. A board in a shape of triangle having equal dimensions ‘a’. Find the area of board using Heron’s formula. If its perimeter is 120 cm. What will be the area of board ?

3. The triangle board having the side walls of 122 cm , 22 cm and 120 cm . The rent of advertisement on board is 6000 rs per cm^2 per year . A company hired this walls for their advertisement for 3 months how much this company have to pay .

4. The triangle board having the side walls of 16 cm , 18 cm and 18 cm . The rent of advertisement on board is 6000 rs per cm^2 per year . A company hired this walls for their advertisement for 3 months how much this company have to pay .

5. The sides of a park is 15 cm , 11 cm and 6 cm. Find the area of this park.

6. The sides of a park is 30 cm , 22 cm and 12 cm. Find the area of this park.

7. Find the area of triangle two sides of which 9 cm and 5 cm and the perimeter is 21 cm.

8. Find the area of triangle two sides of which 8 cm and 6 cm and the perimeter is 22 cm.

9. Side of a triangle are in the ratio of 18 : 10 : 14 and its perimeter is 210 cm . Find its area .

maths ncert solutions

10. Side of a triangle are in the ratio of 9 : 5 : 7 and its perimeter is 210 cm . Find its area .

11. An isosceles triangle has perimeter 25 cm and each of the equal side is 8 cm . Find the area of triangle.

12. An isosceles triangle has perimeter 25 cm and each of the equal side is 9 cm . Find the area of triangle.

13. A field in the shape of quadrilateral ABCD having angle B is 90 where AB = 3 m BC = 4 m CD = 8 m DA = 9 m

14. A field in the shape of quadrilateral ABCD having angle B is 90 where AB = 8 m BC = 6 m CD = 7 m DA = 9 m

15. A field in the shape of quadrilateral ABCD where AB = 3 m BC = 4 m AC = 5 cm CD = 8 m DA = 9 m

16. A field in the shape of quadrilateral ABCD where AB = 8 m BC = 6 m AC = 10 cm CD = 7 m DA = 9 m

17. A triangle and a parallelogram have the same base and the same area . If the sides of triangles are 13 cm , 14 cm and 15 cm and the parallelogram stands on the base 14 cm. Find the height of parallelogram.

18. A triangle and a parallelogram have the same base and the same area . If the sides of triangles are 6 cm , 8 cm and 10 cm and the parallelogram stands on the base 6 cm. Find the height of parallelogram.

Assignment - 4 (Work and Energy)

2019-11-02T06:40:00.000-07:00

Assignment - 4 (Work and Energy)

1. A object of mass 40 kg at a height. If the potential energy of object is 800 J. Find the height of object? (Take g = 10 m/sec^2 )

2. A object of mass 40 kg at a height. If the potential energy of object is 600 J. Find the height of object? (Take g = 10 m/sec^2 )

3. Two army men, each weight have 60 Kg, climbs on a rope for height 5 m. First army man takes 5 sec and second army man takes 10 sec. Find the power expended by army man ?

4. Two army men, each weight have 100 Kg, climbs on a rope for height 6 m. First army man takes 7 sec and second army man takes 14 sec. Find the power expended by army man ?

5. A army woman of mass 72 kg runs on staircase 36 steps in 10 sec. If the height of each staircase is 12 cm. Find her power? (Take g = 10 m/sec^2 )

6. A army woman of mass 50 kg runs on staircase 30 steps in 10 sec. If the height of each staircase is 10 cm. Find her power? (Take g = 10 m/sec^2 )

7. A bulb of 7 w is used for 10 hours per day. Find units consumed in a day ?

8. A fan of 100 w is used for 10 hours per day. Find units consumed in a week ?

9. A television of 200 w is used for 10 hours per day. Find units consumed in a

month ?

10. A fan of 100 W and a bulb of 7 W is used for 10 hours per day. Find units consumed in a month ?

science ncert solutions

11. A object of mass 20 kg at a height. If the potential energy of object is 800 J. Find the height of object? (Take g = 10 m/sec^2 )

12. A object of mass 20 kg at a height. If the potential energy of object is 600 J. Find the height of object? (Take g = 10 m/sec^2 )

13. Two army men, each weight have 50 Kg, climbs on a rope for height 5 m. First army man takes 5 sec and second army man takes 10 sec. Find the power expended by army man ?

14. Two army men, each weight have 140 Kg, climbs on a rope for height 6 m. First army man takes 7 sec and second army man takes 14 sec. Find the power expended by army man ?

15. A army woman of mass 90 kg runs on staircase 36 steps in 10 sec. If the height of each staircase is 12 cm. Find her power? (Take g = 10 m/sec^2 )

16. A army woman of mass 81 kg runs on staircase 30 steps in 10 sec. If the height of each staircase is 10 cm. Find her power? (Take g = 10 m/sec^2 )

17. A bulb of 7 w is used for 12 hours per day. Find units consumed in a day ?

18. A fan of 100 w is used for 12 hours per day. Find units consumed in a week ?

19. A television of 200 w is used for 12 hours per day. Find units consumed in a month ?

20. A fan of 100 W and a bulb of 7 W is used for 12 hours per day. Find units consumed in a month ?

Heron’s Formula ( Part – 6 )

2019-11-01T02:46:00.000-07:00

Heron’s Formula ( Part – 6 )

1. A floral design on a floor is made up of 16 tiles which are triangular , the sides of triangle being 18 cm , 56 cm and 70 cm . Find the cost of polishing the tiles at the rate of 50 p per cm^2 ?

Sol. The sides of triangles are a = 18 cm b = 56 cm c = 70 cm

The semi perimeter of triangle S = ( a + b + c ) / 2

The semi perimeter of triangle S = ( 18 + 56 + 70 ) / 2

The semi perimeter of triangle S = 144 / 2

The semi perimeter of triangle S = 72 cm

Using Heron’s Formula , area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

area of triangle = √ 72 ( 72 – 18 ) ( 72 – 56 ) ( 72 – 70 )

area of triangle = √ 72 * 54 * 16 * 2

area of triangle = √ 2 * 2 * 2 * 3 * 3 * 2 * 3 * 3 * 3 * 2 * 2 * 2 * 2 * 2

area of triangle = 2 * 2 * 2 * 2 * 3 * 3 √ 6

area of triangle = 144 √ 6

area of triangle = 352.73 cm^2

Area of 16 tiles = 16 * 352.73

Area of 16 tiles = 5643.62 cm^2

The cost of polishing the tiles = 0.50 * 5643.62

The cost of polishing the tiles = Rs 2821.81 approx Rs 2822

maths ncert solutions

2. A floral design on a floor is made up of 10 tiles which are triangular , the sides of triangle being 18 cm , 56 cm and 70 cm . Find the cost of polishing the tiles at the rate of 50 p per cm^2 ?

Sol. The sides of triangles are a = 18 cm b = 56 cm c = 70 cm

The semi perimeter of triangle S = ( a + b + c ) / 2

The semi perimeter of triangle S = ( 18 + 56 + 70 ) / 2

The semi perimeter of triangle S = 144 / 2

The semi perimeter of triangle S = 72 cm

Using Heron’s Formula , area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

area of triangle = √ 72 ( 72 – 18 ) ( 72 – 56 ) ( 72 – 70 )

area of triangle = √ 72 * 54 * 16 * 2

area of triangle = √ 2 * 2 * 2 * 3 * 3 * 2 * 3 * 3 * 3 * 2 * 2 * 2 * 2 * 2

area of triangle = 2 * 2 * 2 * 2 * 3 * 3 √ 6

area of triangle = 144 √ 6

area of triangle = 352.73 cm^2

Area of 16 tiles = 10 * 352.73

Area of 16 tiles = 3527.30 cm^2

The cost of polishing the tiles = 0.50 * 3527.30

The cost of polishing the tiles = Rs 1763.65 approx Rs 1764

3. A field in the shape of a trapezium whose parallel sides are 50 m and 20 m . The non parallel sides are 28 m and 26 m . Find the area of the field .

Sol. Draw CF || AD and CG || AB

In quadrilateral ADCF ,

CF || AD { By construction }

CD || AF { ABCD is a trapezium }

Therefore , BF = AB – AF = 50 – 20 = 30 cm

The sides of triangle are a = 26 cm b = 28 cm c = 30 cm

The semi perimeter of triangle S = ( a + b + c ) / 2

The semi perimeter of triangle S = ( 26 + 28 + 30 ) / 2

The semi perimeter of triangle S = 84 / 2

The semi perimeter of triangle S = 42 cm

Using Heron’s Formula , area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

area of triangle = √ 42 ( 42 – 26 ) ( 42 – 28 ) ( 42 – 30 )

area of triangle = √ 42 * 16 * 14 * 12

area of triangle = √ 2 * 3 * 7 * 2 * 2 * 2 * 2 * 2 * 7 * 2 * 2 * 3

area of triangle = 2 * 2 * 2 * 2 * 3 * 7

area of triangle = 336 cm^2

Area of triangle BCF = ½ * BF * CG

½ * 30 * CG = 336

CG = ( 336 * 2 ) / 30

CG = 22.4 cm

So Area of trapezium = ½ * ( AB + CD ) * CG

Area of trapezium = ½ * ( 50 + 20 ) * 22.4

Area of trapezium = ½ * 70 * 22.4

Area of trapezium = 35 * 22.4

Area of trapezium = 784 cm^2

Natural Resources ( Part – 6 )

2019-10-31T08:11:00.000-07:00

Natural Resources ( Part – 6 )

The Carbon Cycle

As we know that Carbon is found in different elements in elementary form like as diamond and graphite . It is found in carbon – di – oxide in our atmosphere . In various minerals it is found in carbonate and carbonate salts . All life forms are based on carbon containing molecules like proteins , carbohydrates , fats , nucleic acid . Carbon is also incorporated in life cycle in the process of photosynthesis which is formed in the presence of sunlight. This process converts carbon dioxide into glucose molecules from the atmosphere or dissolved in water . These glucose molecules are converted into other molecules or used to provide energy .

The utilisation of glucose is to provide energy to living things involves the process of respiration in which oxygen may be used or may not be used to convert back glucose into carbon dioxide . This carbon dioxide goes back to the atmosphere . The amount of carbon dioxide in the atmosphere increases due to the combustion of heating , cooking , transportation etc. Due to the burning of fossil fuel , the amount of carbon dioxide in our atmosphere increases .

The greenhouse effect

We know that heat is trapped inside the glass so the temperature inside the glass is more than surrounding . So the temperature is maintained during winters . this is known as Greenhouse . Some gases escapes from earth surface that increases the average temperature of the atmosphere in world wide . This is known as greenhouse effect . Carbon dioxide gas is one of the green house effect gas . An increase in the carbon dioxide gas cause more heat in atmosphere and leads to global warming .

The oxygen cycle

We know that oxygen is very important for our survival on earth . In our atmosphere , 21 % of our atmosphere is oxygen gas . It also find in crust form in our atmosphere in the form of carbon dioxide .

It is also found in our atmosphere in the form of oxides of metals . It is also a essential component of carbohydrates , fats , nucleic acid and proteins . When we are talking about oxygen cycle , it means maintaining the amount of oxygen in atmosphere . The oxygen of atmosphere is used for combustion , respiration and in the formation of oxide of nitrogen . But it is restored only by the process of photosynthesis . We know that oxygen is most important for respiration but in the process of nitrogen fixing by bacteria does not take place in the presence of oxygen .

Ozone Layer

The elemental oxygen is normally found in our atmosphere in the form of diatomic molecules . In the upper the molecules contains three molecules of oxygen that is known as Ozone . In compare to diatomic molecules and ozone , ozone is poisonous . We are very lucky that ozone is not near to earths surface . But in other hand ozone protects us from harmful radiations by the sun .

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Now days a very common problem is Ozone layer depletion . This is causes due to the more and more use of different man made things like refrigerator , A.C. . These equipments releases chlorine and fluorine free radical that react with ozone and make a hole in ozone layer . Many people are working towards stopping all further damage to the ozone layer .

Heron’s Formula ( Part – 5 )

2019-10-30T05:20:00.000-07:00

Heron’s Formula ( Part – 5 )

1. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours each piece measuring 10 cm , 25 cm and 25 cm . How much cloth of each colour is required for the umbrella ?

Sol. The sides of triangle are

a = 10 cm b = 25 cm c = 25 cm

Semi perimeter of triangle ( S ) = ( a + b + c ) / 2

S = ( 10 + 25 + 25 ) / 2

S = 60 / 2

S = 30 cm

Using Heron’s Formula , Area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

Area of triangle = √ 30 ( 30 – 10 ) ( 30 – 25 ) ( 30 – 25 )

Area of triangle = √ 30 * 20 * 5 * 5

Area of triangle = √ 2 * 3 * 5 * 2 * 2 * 5 * 5 * 5

Area of triangle = 2 * 5 * 5 √ 2 * 3

Area of triangle = 50 √ 6

Area of triangle = 50 * 2.45

Area of triangle = 122.47 cm^2

Area of 10 triangular pieces of cloths = 10 * 122.47 = 1224.70 cm^2

The area of cloths of each other colour = 1224.70 / 2 = 612.37 cm^2

2. A kite in the shape of square with a diagonal 16 cm and isosceles triangle of base 4 cm and side 3 cm each is to be made of three different shades . How much paper of each shade has been used in it ?

Sol. In the triangle ,

Base = 16 cm height = 8 cm

Area of triangle = ½ * base * height

Area of triangle = ½ * 16 * 8

Area of triangle = 8 * 8

Area of triangle = 64 cm^2

Kite is in shape of square , so area in both sides is same

Area of kite of square shape = 2 * 64 = 128 cm^2

And area of isosceles triangle a = 3 cm b = 3 cm c = 4 cm

Semi perimeter of triangle ( S ) = ( a + b + c ) / 2

S = ( 3 + 3 + 4 ) / 2

S = 10 / 2

S = 5 cm

Using Heron’s Formula , Area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

Area of triangle = √ 5 ( 5 – 3 ) ( 5 – 3 ) ( 5 – 4 )

Area of triangle = √ 5 * 2 * 2 * 1

Area of triangle = 2 √ 5

Area of triangle = 2 * 2.24

Area of triangle = 4.48 cm^2

Total area of the kite of shape square with isosceles triangle = 128 + 4.48

Total area of the kite of shape square with isosceles triangle = 132.48 cm^2

maths ncert solutions

3. An umbrella is made by stitching 12 triangular pieces of cloth of two different colours each piece measuring 10 cm , 25 cm and 25 cm . How much cloth of each colour is required for the umbrella ?

Sol. The sides of triangle are

a = 10 cm b = 25 cm c = 25 cm

Semi perimeter of triangle ( S ) = ( a + b + c ) / 2

S = ( 10 + 25 + 25 ) / 2

S = 60 / 2

S = 30 cm

Using Heron’s Formula , Area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

Area of triangle = √ 30 ( 30 – 10 ) ( 30 – 25 ) ( 30 – 25 )

Area of triangle = √ 30 * 20 * 5 * 5

Area of triangle = √ 2 * 3 * 5 * 2 * 2 * 5 * 5 * 5

Area of triangle = 2 * 5 * 5 √ 2 * 3

Area of triangle = 50 √ 6

Area of triangle = 50 * 2.45

Area of triangle = 122.47 cm^2

Area of 10 triangular pieces of cloths = 12 * 122.47 = 1469.64 cm^2

The area of cloths of each other colour = 1469.64 / 2 = 734.82 cm^2

4. A kite in the shape of square with a diagonal 48 cm and isosceles triangle of base 12 cm and side 9 cm each is to be made of three different shades . How much paper of each shade has been used in it ?

Sol. In the triangle ,

Base = 48 cm height = 24 cm

Area of triangle = ½ * base * height

Area of triangle = ½ * 48 * 24

Area of triangle = 24 * 24

Area of triangle = 576 cm^2

Kite is in shape of square , so area in both sides is same

Area of kite of square shape = 2 * 576 = 1152 cm^2

And area of isosceles triangle a = 9 cm b = 9 cm c = 12 cm

Semi perimeter of triangle ( S ) = ( a + b + c ) / 2

S = ( 9 + 9 + 12 ) / 2

S = 30 / 2

S = 15 cm

Using Heron’s Formula , Area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

Area of triangle = √ 15 ( 15 – 9 ) ( 15 – 9 ) ( 15 – 12 )

Area of triangle = √ 15 * 6 * 6 * 3

Area of triangle = √ 3 * 5 * 6 * 6 * 3

Area of triangle = 3 * 6 √ 5

Area of triangle = 18 * 2.24

Area of triangle = 40.32 cm^2

Total area of the kite of shape square with isosceles triangle = 1152 + 40.32

Total area of the kite of shape square with isosceles triangle = 1192.32 cm^2

Natural Resources ( Part – 5 )

2019-10-29T06:01:00.003-07:00

Natural Resources ( Part – 5 )

Biogeochemical Cycle

A constant interaction between biotic and abiotic components of the biosphere makes it dynamic , these components transfer heat and energy to different components.

The water Cycle

We know that the water evaporates from the water bodies and condence this water and leads to rain . We also know that the water from sea and oceans also evaporates but these water bodies does not became dry . We know that the water of ocean and sea water evaporates and condence and leads to rain . This rain falls on the land some amount of this water seeps by soil and remaining water flow back to the water bodies . The process of evaporating water from water bodies and goes back to the water bodies , this complete process is known as the water cycle . The water seeps by the soil finds its way though its surface springs , known as underground water . This underground water is used by us for many process by tube wells and ponds . This water is used for irrigation , life process and full filling the need of water ( For drinking , Cooking ) . When the water flows on the rocks , or the land , then many nutrients and minerals are dissolved in the water and goes to the sea and rives by the flow of water this leads to marine organisms .

The Nitrogen Cycle

Science ncert solutions

We know that makes a 78 % of our atmosphere and nitrogen is also a part of many molecules essential for our life proteins , some vitamins , DNA and RNA . Nitrogen is also a main part of alkaloids and urea too. Nitrogen is main part for all life cycle and life is so simple if we use nitrogen present in our atmosphere . Some nitrogen fixing bacteria converts these hidden nitrogen into nitrites and these bacteria are present freely in atmosphere . These nitrogen fixing bacteria present in the roots of the plants . Due to the lighting , the high pressure and temperature creates the nitrogen into nitrogen oxide.

These nitrogen oxide dissolved in water to give nitric and nitrous acids and falls on the land along with rain and these are utilized by various life forms .

Plants take these nitrates and nitrites and convert them into amino acid . This amino acid is used to make proteins . Some other complex compounds are made by nitrogen . These nitrogen and complex compound are taken by the animals . When the plants or animals die some other type of bacteria convertes back these nitrogens into nitrides and nitrites . A other type of bacteria converts the nitrides and nitrites into nitrogen elements . This whole process is known as nitrogen cycle in which nitrogen passes from its elemental form in the atmosphere into simple in the soil and water . Which get converts in various nitrogen molecules in life cycles and converts back into the elementary nitrogen .

Heron's Formula ( Part - 4 )

2019-10-28T08:19:00.000-07:00

Heron's Formula ( Part - 4 )

1. A rhombus shaped field have green grass for 9 cows to graze . If the each side of the rhombus is 15 m and its longer diagonal is 24 m . How much area of grass field will each cow be getting ?

Sol. Join the diagonal of quadrilateral ABCD

The sides of triangle ABC are

a = 15 m b = 15 m c = 24 m

so , semi perimeter of triangle

S = ( a + b + c ) / 2

S = ( 15 + 15 + 24 ) / 2

S = 54 / 2

S = 27 m

Area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

Area of triangle = √ 27 ( 27 – 15 ) ( 27 – 15 ) ( 27 – 24 )

Area of triangle = √ 3 * 3 * 3 * 12 * 12 * 3

Area of triangle = √ 3 * 3 * 3 * 2 * 2 * 3 * 2 * 2 * 3 * 3

Area of triangle = 2 * 2 * 3 * 3 * 3

Area of triangle = 108 m ^ 2

Area of quadrilateral ABCD = 2 * Area of triangle

Area of quadrilateral ABCD = 2 * 108

Area of quadrilateral ABCD = 216 m^2

The area grazed by cow = total area / number of cows

The area grazed by cow = 216 / 9

The area grazed by cow = 24 m^2

maths ncert solution

2. A rhombus shaped field have green grass for 18 cows to graze . If the each side of the rhombus is 15 m and its longer diagonal is 24 m . How much area of grass field will each cow be getting ?

Sol. Join the diagonal of quadrilateral ABCD

The sides of triangle ABC are

a = 15 m b = 15 m c = 24 m

so , semi perimeter of triangle

S = ( a + b + c ) / 2

S = ( 15 + 15 + 24 ) / 2

S = 54 / 2

S = 27 m

Area of triangle = √ s ( s – a ) ( s – b ) ( s – c )

Area of triangle = √ 27 ( 27 – 15 ) ( 27 – 15 ) ( 27 – 24 )

Area of triangle = √ 3 * 3 * 3 * 12 * 12 * 3

Area of triangle = √ 3 * 3 * 3 * 2 * 2 * 3 * 2 * 2 * 3 * 3

Area of triangle = 2 * 2 * 3 * 3 * 3

Area of triangle = 108 m ^ 2

Area of quadrilateral ABCD = 2 * Area of triangle

Area of quadrilateral ABCD = 2 * 108

Area of quadrilateral ABCD = 216 m^2

The area grazed by cow = total area / number of cows

The area grazed by cow = 216 / 18

The area grazed by cow = 12 m^2