POV: You are a mob boss who needs some vital inside information from the rival gang so that you can get an upper hand in the on going gang war. You were a known face in the underworld so it wasn't easy for you to just go and do you tasks. You are always scrutinized by everyone. Rival mafia leaders try to decode your body language and attempt to get clues as to what your next move would be, and how could they counteract to it. You asked your stylist to dress you up in a kind of dress you would never be caught dead wearing. You usually wore suits to radiate the aura of authority and superiority but subtly. It was a skimpy red dress and you put on a wig different from your preferred hair style and a pair of contact lenses different from your original eye color. You then wore your signature red lipstick. When you looked at your reflection in your mirror.....you weren't comfortable wearing dresses like these but you had to suck it up and go and do your mission. You reached the club and saw him immediately.....the next victim of your seduction, the rival mob boss. You went to him and sat on his lap and wrapped your legs around his waist as he enjoyed the attention his ugly ass was getting from a gorgeous woman. You spoked in a whispery low tone, too different from your bold loud speech that you usually did. "Tell me what these hands can do better than destroying empires left and right" you purred in his ear. "You know me so well babygirl, but instead of telling you I will show you." he sad, desperately caught in your charms as you felt something turgid poking your thigh. 'ugh' you grimaced internally but successfully hid it from him. He carried you to the private section of the exclusive club as you made out with him, his hand reached to your back to grab the zipper of your dress in an attempt to take it off. 'Its ok, its just a few seconds' you motivated yourself as his slimy hands travelled to find the hook of your bra. In a second, he became unconscious and landed in your arms. You called your gang men, and had him in your car as you took out a napkin laced with something. An antidote. An antidote to ricin that was laced in your lipstick and made him unconscious. You wiped it around your lips and put the trusty red ricin lipstick away at a different place in one of the many innumerable pockets in your dress so it doesn't get exchanged with your regular one. You pulled out your regular one and walked out the club to a different car to not raise any suspicions. For those who don't know, ricin is an anesthetic drug used for make patients unconscious before surgery. Did you like this pov? should I post something like this on more slowed down songs on this channel?
@zoya koradia You should be a writer I really love the pov if you post more like this specially on Maria's I will subscribe u for sure u are so amazing and u should be a k drama writer 💕
Wow...this became....the male version....this hits different 😶 Why this version giving me so much of those LGBTQ+ Anime scenes....vibes. ...especially gay love scenes. .... No being homophobic but...those are eternal......so unusual and unconditional love.....😃😃😃😃
Height of cylinder, h = 14 cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that the formula to find the curved surface area of a cylinder is 2πrh. So 2πrh =88 cm2 (r is the radius of the base of the cylinder) 2×(22/7)×r×14 = 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm. 2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22/7 Solution: Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface area of tank = 2πr(r+h) unit square = [2×(22/7)×0.7(0.7+1)] = 7.48 square meters Therefore, 7.48 square meters of the sheet are required. 3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its  (i) inner curved surface area, (ii) outer curved surface area (iii) total surface area (Assume π=22/7) Solution: Let r1 and r2 be the inner and outer radii of the cylindrical pipe r1 = 4/2 cm = 2 cm r2 = 4.4/2 cm = 2.2 cm Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm (i) curved surface area of outer surface of pipe = 2πr1h = 2×(22/7)×2×77 cm2 = 968 cm2 (ii) curved surface area of outer surface of pipe = 2πr2h = 2×(22/7)×2.2×77 cm2 = (22×22×2.2) cm2 = 1064.8 cm2 (iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe. = 2r1h+2r2h+2π(r12-r22) = 9668+1064.8+2×(22/7)×(2.22-22) = 2031.8+5.28 = 2038.08 cm2 Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2. 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7) Solution: A roller is shaped like a cylinder. Let h be the height of the roller and r be the radius. h = Length of roller = 120 cm Radius of the circular end of roller = r = (84/2) cm = 42 cm Now, CSA of roller = 2πrh = 2×(22/7)×42×120 = 31680 cm2 Area of field = 500×CSA of roller = (500×31680) cm2 = 15840000 cm2 = 1584 m2. Therefore, the area of the playground is 1584 m2. 5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2. (Assume π = 22/7) Solution: Let h be the height of a cylindrical pillar and r be the radius. Given: Height cylindrical pillar = h = 3.5 m Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25 cm = 0.25 m CSA of pillar = 2πrh = 2×(22/7)×0.25×3.5 = 5.5 m2 Cost of painting 1 m2 area = Rs. 12.50 Cost of painting 5.5 m2 area = Rs (5.5×12.50) = Rs. 68.75 Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2 is Rs 68.75. 6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7) Solution: Let h be the height of the circular cylinder and r be the radius. The radius of the base of the cylinder, r = 0.7 m CSA of cylinder = 2πrh CSA of cylinder = 4.4 m2 Equating both equations, we have 2×(22/7)×0.7×h = 4.4 Or h = 1 Therefore, the height of the cylinder is 1 m. 7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. (Assume π = 22/7) Solution: Inner radius of circular well, r = 3.5/2 m = 1.75 m Depth of circular well, say h = 10 m (i) Inner curved surface area = 2πrh = (2×(22/7 )×1.75×10) = 110 m2 Therefore, the inner curved surface area of the circular well is 110 m2. (ii) Cost of plastering 1 m2 area = Rs. 40 Cost of plastering 110 m2 area = Rs (110×40) = Rs. 4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400. 8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = 22/7) Solution: Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder = 2×(22/7)×0.025×28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2. 9. Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. (ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7) Solution: Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of the cylindrical tank is 2πrh = 2×(22/7)×2.1×4.5 m2 = (44×0.3×4.5) m2 = 59.4 m2 Therefore, the CSA of the tank is 59.4 m2. (ii) Total surface area of tank = 2πr(r+h) = 2×(22/7)×2.1×(2.1+4.5) = 44×0.3×6.6 = 87.12 m2 Now, let us assume that S m2 steel sheet was used in making the tank. S(1 -1/12) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was actually used while making such a tank. 10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22/7)  Solution: Say h = height of the frame of the lampshade, which has a cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use the curved surface area formula to find the cloth required for covering the lampshade, which is 2πrh = (2×(22/7)×10×35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required to cover the lampshade. 11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7) Solution: The radius of the circular end of the cylindrical penholder, r = 3cm Height of penholder, h = 10.5cm Surface area of a penholder = CSA of pen holder + Area of base of penholder = 2πrh+πr2 = 2×(22/7)×3×10.5+(22/7)×32= 1584/7 Therefore, the area of cardboard sheet used by one competitor is 1584/7 cm2 So, the area of cardboard sheets used by 35 competitors = 35×1584/7 = 7920 cm2 Therefore, 7920 cm2 cardboard sheet will be needed for the competition. NCERT Solutions for Class 9 Maths Chapter 13helps to find the surface area of various geometrical objects in a simplified and easy way. NCERT Solutions for Class 9 Maths Exercise 13.2 explains how to find the surface area of the cylinder and helps to solve the problems based on it. Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 These NCERT Solutions help you understand and solve all questions of Chapter 13.It helps to secure the best score in Maths board exams.It contains all the important questions from the examination point of view.Good explanations help students to remember the concept effectively.    NCERT Solutions For Class 9 Maths Chapter 13 Surface Area And Volume Exercises NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.1NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.4NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.5NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.7NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.8NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.9 NCERT Solutions For Class 9 Maths Chapters Chapter 1 Number SystemsChapter 2 PolynomialChapter 3 Coordinate GeometryChapter 4 Linear Equation In Two VariablesChapter 5 Introduction To Euclids GeometryChapter 6 Lines And AnglesChapter 7 Geometry Of TrianglesChapter 8 QuadrilateralsChapter 9 Areas Of Parallelograms And TriangleChapter 10 CirclesChapter 11 ConstructionChapter 12 Herons FormulaChapter 13 Surface Area And VolumeChapter 14 StatisticsChapter 15 Introduction To Probability Join BYJU'S Learning Program
[10/12, 6:52 PM] Anshuman: Q1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’. (a) Suma is swimming in a pond. (b) A donkey is carrying a load on its back. (c) A wind mill is lifting water from a well. (d) A green plant is carrying out photosynthesis. (e) An engine is pulling a train. (f) Food grains are getting dried in the sun. (g) A sailboat is moving due to wind energy. Work is done whenever the given conditions are satisfied: (i) A force acts on a body. (ii) There is a displacement of the body. (a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, work is done by Seema while swimming. (b) While carrying a load, the donkey has to apply a force in the upward direction. But, displacement of the load is in the forward direction. Since, displacement is perpendicular to force, the work done is zero. (c) A wind mill works against the gravitational force to lift water. Hence, work is done by the wind mill in lifting water from the well. (d) In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero. (e) An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train. (f) Food grains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in the Sun. (g) Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat. Q2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object? Since the body returns to a point which is on the same horizontal line through the point of projection, no displacement has taken place against the force of gravity, therefore, no work is done by the force due to gravity. Q3. A battery lights a bulb. Describe the energy changes involved in the process. Within the electric cell of the battery the chemical energy changes into electrical energy. The electric energy on flowing through the filament of the bulb, first changes into heat energy and then into the light energy. [10/12, 6:53 PM] Anshuman: Q5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer. The work done is zero. This is because the gravitational force and displacement are perpendicular to each other. Q6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why? It does not violate the law Of conservation of energy. Whatever, is the decrease in PE due to loss of height, same is the increase in the KE due to increase in velocity of the body. Q7. What are the various energy transformations that occur when you are riding a bicycle? The chemical energy of the food changes into heat and then to muscular energy. On paddling, the muscular energy changes into mechanical energy Q8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going? Energy transfer does not take place as no displacement takes place in the direction of applied force. The energy spent is used to overcome inertia of rest of the rock.
OHH MYY GAAWWD. how is this so perfect? I--- 😶
this sounds beautiful 🥺❤️
REAL OR SLOWED?🥵💞
The line--- " aakhon se chule , haato se chuna mana hai " is just❤❤
What does this line mean?
Actually not Indian sohh--
@@ahhhhh__ it means touch me with your eyes touching with hands is not allowed
Why is this so underrated 🤧❤️
Yaa😊❤
i really dont know
Why doesn't this have more views- istg-
I just love this more than the original one💜✨ why is this soo under rated😏
No Only "OLD IS GOLD" Old version is more than good, nice, excellent, outstanding and also my favourite song.😊❤❤❤❤❤❤❤❤❤❤❤💖💞
0:53 was lit 🔥🤧🥵
The Voice! OMG! Just fall in love with the voice!!
Kis ki voice he ye???
@@intro106 neha kakkar ki
@@Miss_powerful neha kakkar ka to dusra he, ye ksi male singer ki voice he
@@intro106 ha male voice himesh reshamiya ka hai 😊
@@Miss_powerful wo to naak se gaata he uski nai he 😂
OMG, how is this so perfect, I cant even-✋🏻😭
Pure nostalgia 💫✨
it's more sensational 👀
Aashiq banaya apne 😩🔥
This song give me shivers down my spine 😍
Here before the song get famous 😩✨
I am not an indian but I think this is the power of indian songs
Where are you from 🤔
this is so goooood I---
uffff this soooonggg 🖤😭
3:31 fav line🐱
Me😊😊❤❤❤
This gives me vibes of dancing in front of the mafia's enemy to get the secret code from him🙂🖤🖤🖤
that's ma girl ~~
Effect of ff 😂💥
@@Kunmuns_gallery exactly😂😂
Imagine after all this tempting dancing that he doesn't give you 😂
@@evlin-up9pm 🤭😂😂
I m not the only one imagining Jimin here.....
I am imagining the one in your profile 👀
@@Akira-zz8jn hmmm bohot tejj ho rhe ho🌚
I'm imaging the worse in my profile
@@Zozoee35 🌚
omg the edits we could make w this song 😍😍😍
This song take me to the good vibes world 😘😘😘
Wonderful and beautiful Creative sense of the word🤍✨✨
Creating fake scenario of being a protagonist in amocie where i have to seduce the don to get the pass of the vault
Pls- 💀
Lol 😂😂
Same girl same!!!!! 😈😈
💀
@@shriyasinha5943 i found my twin-
0:44 time stamp for me
😊😊😊😊😊😊🥰🥰🥰🥰🥰🥰💝💝💝💝😉😉😉😉
Urvashi looks like Gigi Hadid in that picture lol
Yepp
True af😂
100%true
She loves JAIN malik
So true
@@kailashdevi6718 smj a gayi 😂😂😂😂😂😂
Wow...just loved it plssss do lut gaye- jubin song plsss
AASHIQ BANAYA APNEE...!!🥵💞
Yawrrr kamal ho gya hand free ky sath to itna Acha laga waawaa
Ishh😍this background picture😍
Night time is the best to listen the song 😌✨ it's blessed my ears so slow😌✨❤️😍......
2:15 Timestamp For Myself
I love this....by the way do you do punjabi songs? If you do could you please do G.O.A.T. by Diljit Dosanjh? He is my fav singer 💕🥺
I like this song in this voice and version
From the another to versions
Just say🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
Just wow 😲 so beautiful song I love this song forever ❤️😍
One of the most popular wedding song...❤😊
3:33
Aayee🥵👏♥️
Amar Khub Valo Lage Songta ❤
1:50 2:34 Damn Fire 🔥☠️
2:15
Uff this song.... 🥵🔥🔥🔥
Why am I finding this now❤️🔥🔥🔥🔥😭
Ishhhh 😭🔥
2:37 and 2:48 killed it 😶
haha!!!😂😂😂
3:10 this line 🔥
POV: You are a mob boss who needs some vital inside information from the rival gang so that you can get an upper hand in the on going gang war. You were a known face in the underworld so it wasn't easy for you to just go and do you tasks. You are always scrutinized by everyone. Rival mafia leaders try to decode your body language and attempt to get clues as to what your next move would be, and how could they counteract to it.
You asked your stylist to dress you up in a kind of dress you would never be caught dead wearing. You usually wore suits to radiate the aura of authority and superiority but subtly. It was a skimpy red dress and you put on a wig different from your preferred hair style and a pair of contact lenses different from your original eye color. You then wore your signature red lipstick. When you looked at your reflection in your mirror.....you weren't comfortable wearing dresses like these but you had to suck it up and go and do your mission.
You reached the club and saw him immediately.....the next victim of your seduction, the rival mob boss. You went to him and sat on his lap and wrapped your legs around his waist as he enjoyed the attention his ugly ass was getting from a gorgeous woman. You spoked in a whispery low tone, too different from your bold loud speech that you usually did.
"Tell me what these hands can do better than destroying empires left and right" you purred in his ear. "You know me so well babygirl, but instead of telling you I will show you." he sad, desperately caught in your charms as you felt something turgid poking your thigh. 'ugh' you grimaced internally but successfully hid it from him. He carried you to the private section of the exclusive club as you made out with him, his hand reached to your back to grab the zipper of your dress in an attempt to take it off. 'Its ok, its just a few seconds' you motivated yourself as his slimy hands travelled to find the hook of your bra.
In a second, he became unconscious and landed in your arms. You called your gang men, and had him in your car as you took out a napkin laced with something. An antidote. An antidote to ricin that was laced in your lipstick and made him unconscious. You wiped it around your lips and put the trusty red ricin lipstick away at a different place in one of the many innumerable pockets in your dress so it doesn't get exchanged with your regular one. You pulled out your regular one and walked out the club to a different car to not raise any suspicions.
For those who don't know, ricin is an anesthetic drug used for make patients unconscious before surgery.
Did you like this pov? should I post something like this on more slowed down songs on this channel?
LMAO YES 🙌
Damnnnnn
Woahh
Get a life
@zoya koradia
You should be a writer I really love the pov if you post more like this specially on Maria's I will subscribe u for sure u are so amazing and u should be a k drama writer 💕
Omg ….. it’s just ❤️❤️❤️❤️
🔥🔥🔥🔥
Sukoon dilane wala song 😊🔥
Neha Kakkar voice- 🔥
Urvashi Rautela beauty -🥵 uffff 🥰 this song _______________🔥🥵
Aye👀♥️👅
It seems as a drug to me superb ♥ ♥
This music a masterpiece 😍😍
O my godddd mera khoya howa song mil gya
jan jan keh ke jan tune mere le li what a line🥰🥰🥰
Voice is sooo lovely 🥰🥰😘😘😘
Damn Broo
Armies are literally everywhere ⬅➡⬆⬇↗↘↙↖
Ikr! 🙃😂💜
Yo 💦🌚
💜
Yeah 💜
FRRRR
Deep voice- just wow😗😗
Graceful song ❤ I love it😊
0:53 UfffSSS🥵💞
Nice song 🤩👍
Uff this level❤️❤️❤️
Ankhon se chu le chuna hatho se mana h 💥💘
Deewane ruk ja tera humse saamna h this line hit me harder than my father's belt 😫❤😊
Your father beat you
Emraan hasmi ❤
Ess mata nosto song... 💞💞🥰🥰
OpSs 😚😗
2:18is just.....uffffff
0:54 😮
Wow...this became....the male version....this hits different 😶
Why this version giving me so much of those LGBTQ+ Anime scenes....vibes. ...especially gay love scenes. ....
No being homophobic but...those are eternal......so unusual and unconditional love.....😃😃😃😃
Army!💜
@@crazhyformyself3176 💜
Same sis I also imagine the same thing and it's hit different
Same here! 😅
❤️❤️👍
Super
This music make me crazy 😉😉😉😉😉
2:4 TOP NOTCH
Height of cylinder, h = 14 cm
Let the diameter of the cylinder be d
Curved surface area of cylinder = 88 cm2
We know that the formula to find the curved surface area of a cylinder is 2πrh.
So 2πrh =88 cm2 (r is the radius of the base of the cylinder)
2×(22/7)×r×14 = 88 cm2
2r = 2 cm
d =2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22/7
Solution:
Let h be the height and r be the radius of a cylindrical tank.
Height of cylindrical tank, h = 1m
Radius = half of diameter = (140/2) cm = 70cm = 0.7m
Area of sheet required = Total surface area of tank = 2πr(r+h) unit square
= [2×(22/7)×0.7(0.7+1)]
= 7.48 square meters
Therefore, 7.48 square meters of the sheet are required.
3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its

(i) inner curved surface area,
(ii) outer curved surface area
(iii) total surface area
(Assume π=22/7)
Solution:
Let r1 and r2 be the inner and outer radii of the cylindrical pipe
r1 = 4/2 cm = 2 cm
r2 = 4.4/2 cm = 2.2 cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm
(i) curved surface area of outer surface of pipe = 2πr1h
= 2×(22/7)×2×77 cm2
= 968 cm2
(ii) curved surface area of outer surface of pipe = 2πr2h
= 2×(22/7)×2.2×77 cm2
= (22×22×2.2) cm2
= 1064.8 cm2
(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.
= 2r1h+2r2h+2π(r12-r22)
= 9668+1064.8+2×(22/7)×(2.22-22)
= 2031.8+5.28
= 2038.08 cm2
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to
move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7)
Solution:
A roller is shaped like a cylinder.
Let h be the height of the roller and r be the radius.
h = Length of roller = 120 cm
Radius of the circular end of roller = r = (84/2) cm = 42 cm
Now, CSA of roller = 2πrh
= 2×(22/7)×42×120
= 31680 cm2
Area of field = 500×CSA of roller
= (500×31680) cm2
= 15840000 cm2
= 1584 m2.
Therefore, the area of the playground is 1584 m2.
5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2.
(Assume π = 22/7)
Solution:
Let h be the height of a cylindrical pillar and r be the radius.
Given:
Height cylindrical pillar = h = 3.5 m
Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25 cm = 0.25 m
CSA of pillar = 2πrh
= 2×(22/7)×0.25×3.5
= 5.5 m2
Cost of painting 1 m2 area = Rs. 12.50
Cost of painting 5.5 m2 area = Rs (5.5×12.50)
= Rs. 68.75
Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2 is Rs 68.75.
6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)
Solution:
Let h be the height of the circular cylinder and r be the radius.
The radius of the base of the cylinder, r = 0.7 m
CSA of cylinder = 2πrh
CSA of cylinder = 4.4 m2
Equating both equations, we have
2×(22/7)×0.7×h = 4.4
Or h = 1
Therefore, the height of the cylinder is 1 m.
7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2.
(Assume π = 22/7)
Solution:
Inner radius of circular well, r = 3.5/2 m = 1.75 m
Depth of circular well, say h = 10 m
(i) Inner curved surface area = 2πrh
= (2×(22/7 )×1.75×10)
= 110 m2
Therefore, the inner curved surface area of the circular well is 110 m2.
(ii) Cost of plastering 1 m2 area = Rs. 40
Cost of plastering 110 m2 area = Rs (110×40)
= Rs. 4400
Therefore, the cost of plastering the curved surface of the well is Rs. 4400.
8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find
the total radiating surface in the system. (Assume π = 22/7)
Solution:
Height of cylindrical pipe = Length of cylindrical pipe = 28m
Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m
Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder
= 2×(22/7)×0.025×28 m2
= 4.4m2
The area of the radiating surface of the system is 4.4m2.
9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in
diameter and 4.5m high.
(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)
Solution:
Height of cylindrical tank, h = 4.5m
Radius of the circular end , r = (4.2/2)m = 2.1m
(i) the lateral or curved surface area of the cylindrical tank is 2πrh
= 2×(22/7)×2.1×4.5 m2
= (44×0.3×4.5) m2
= 59.4 m2
Therefore, the CSA of the tank is 59.4 m2.
(ii) Total surface area of tank = 2πr(r+h)
= 2×(22/7)×2.1×(2.1+4.5)
= 44×0.3×6.6
= 87.12 m2
Now, let us assume that S m2 steel sheet was used in making the tank.
S(1 -1/12) = 87.12 m2
This implies, S = 95.04 m2
Therefore, 95.04m2 steel was actually used while making such a tank.
10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.
The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22/7)

Solution:
Say h = height of the frame of the lampshade, which has a cylindrical shape
r = radius
Total height is h = (2.5+30+2.5) cm = 35cm and
r = (20/2) cm = 10cm
Use the curved surface area formula to find the cloth required for covering the lampshade, which is 2πrh
= (2×(22/7)×10×35) cm2
= 2200 cm2
Hence, 2200 cm2 cloth is required to cover the lampshade.
11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)
Solution:
The radius of the circular end of the cylindrical penholder, r = 3cm
Height of penholder, h = 10.5cm
Surface area of a penholder = CSA of pen holder + Area of base of penholder
= 2πrh+πr2
= 2×(22/7)×3×10.5+(22/7)×32= 1584/7
Therefore, the area of cardboard sheet used by one competitor is 1584/7 cm2
So, the area of cardboard sheets used by 35 competitors = 35×1584/7 = 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be needed for the competition.
NCERT Solutions for Class 9 Maths Chapter 13helps to find the surface area of various geometrical objects in a simplified and easy way. NCERT Solutions for Class 9 Maths Exercise 13.2 explains how to find the surface area of the cylinder and helps to solve the problems based on it.
Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2
These NCERT Solutions help you understand and solve all questions of Chapter 13.It helps to secure the best score in Maths board exams.It contains all the important questions from the examination point of view.Good explanations help students to remember the concept effectively.



NCERT Solutions For Class 9 Maths Chapter 13 Surface Area And Volume Exercises
NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.1NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.4NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.5NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.7NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.8NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.9
NCERT Solutions For Class 9 Maths Chapters
Chapter 1 Number SystemsChapter 2 PolynomialChapter 3 Coordinate GeometryChapter 4 Linear Equation In Two VariablesChapter 5 Introduction To Euclids GeometryChapter 6 Lines And AnglesChapter 7 Geometry Of TrianglesChapter 8 QuadrilateralsChapter 9 Areas Of Parallelograms And TriangleChapter 10 CirclesChapter 11 ConstructionChapter 12 Herons FormulaChapter 13 Surface Area And VolumeChapter 14 StatisticsChapter 15 Introduction To Probability
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You are actually chu****
@@baka7454🗿
bhai sambhar ke kya kya ho rha hai tujhe .. calm down .reduce maths tension 😢😂😅
2:22
Exactly what I wanted😉
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0:30 timestamp
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this song vibes are -----------------------------------------------song
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This song has a major wattpad mafia vibes-
The line: " Hoonto se padh le vo jo honto pe lika hai"
Is the sexist one❤❤❤❤😅
Love it🥵❤️❤️❤️❤️❤️
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[10/12, 6:52 PM] Anshuman: Q1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
(a) Suma is swimming in a pond.
(b) A donkey is carrying a load on its back.
(c) A wind mill is lifting water from a well.
(d) A green plant is carrying out photosynthesis.
(e) An engine is pulling a train.
(f) Food grains are getting dried in the sun.
(g) A sailboat is moving due to wind energy.
Work is done whenever the given conditions are satisfied:
(i) A force acts on a body.
(ii) There is a displacement of the body.
(a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, work is done by Seema while swimming.
(b) While carrying a load, the donkey has to apply a force in the upward direction. But, displacement of the load is in the forward direction. Since, displacement is perpendicular to force, the work done is zero.
(c) A wind mill works against the gravitational force to lift water. Hence, work is done by the wind mill in lifting water from the well.
(d) In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.
(e) An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.
(f) Food grains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in the Sun.
(g) Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.
Q2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Since the body returns to a point which is on the same horizontal line through the point of projection, no displacement has taken place against the force of gravity, therefore, no work is done by the force due to gravity.
Q3. A battery lights a bulb. Describe the energy changes involved in the process.
Within the electric cell of the battery the chemical energy changes into electrical energy. The electric
energy on flowing through the filament of the bulb, first changes into heat energy and then into the light energy.
[10/12, 6:53 PM] Anshuman: Q5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
The work done is zero. This is because the gravitational force and displacement are perpendicular to each other.
Q6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
It does not violate the law Of conservation of energy. Whatever, is the decrease in PE due to loss of height, same is the increase in the KE due to increase in velocity of the body.
Q7. What are the various energy transformations that occur when you are riding a bicycle?
The chemical energy of the food changes into heat and then to muscular energy. On paddling, the muscular energy changes into mechanical energy
Q8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Energy transfer does not take place as no displacement takes place in the direction of applied force. The energy spent is used to overcome inertia of rest of the rock.
I think this song is drugs for me to think about her my dopamine level 🥵
Ishq❤🎉
Wawaaaaaa