Check out other songs on my channel too. And agar kisi gaane ka slowed reverb chahiye to comment karo, I'll try my best to upload them as soon as possible . ❤ Kaise hua X Timro Pratiksha ( Full Version) : ruclips.net/video/Wj1Ny-FY8Dc/видео.htmlsi=4DSDJ72U0jh2pTSJ ruclips.net/video/Wj1Ny-FY8Dc/видео.htmlsi=4DSDJ72U0jh2pTSJ
Bro I gonna fall in freaking love with this I never liked to hear original mercy but bro I am gonna die on your version i can't express what aura you maked. Lit bro.
Sounds like you actually know what " slowed " means, I have been searching for the slowed version of this song but they were still too fast, this is perfect 😩🤌
[31/10, 7:07 PM] Anshuman: Question 1. “ On the whole, the small society of Rameswaram was very rigid in terms of the segregation of different social groups,” says the author. Which social groups does he mention? Were these groups easily identifiable (for example, by the way they dressed)? Were they aware only of their differences or did they also naturally share friendships and experiences? (Think of the bedtime stories in Kalam’s house; of who his friends were ; and of what used to take place in the pond near his house.) The author speaks both of people who were very aware of the differences among them and those who tried to bridge these differences. Can you identify such people in the text? Narrate two incidents that show how differences can be created, and also how they can be resolved. How can people change their attitudes? Answer: The author talks about the people who belong to various castes and follow various religious preachings. Yes, these groups were easily identifiable. Their dressing, traditions, culture and rituals were different. They did share their personal experiences and friendships. Lakshmana Sastry summoned the teacher who separated the author and his friend in the class and told him that he should not spread the poison of social inequality and communal intolerance in the minds of innocent children. The school teacher encouraged communal differences and Lakshmana Sastry and Sivasubramania Iyer discouraged this malpractice. The influential people can do both the things. A teacher has the ability to bridge communal differences and can play with sentiments of the innocent and ignorant people. This is what the new teacher did. But the Science teacher Sivasubramania Iyer changed his wife’s attitude and showed her the right path. Question 2. Why did Abdul Kalam want to leave Rameswaram? What did his father say to this? What do you think his words mean? Why do you think he spoke those words? Answer: Abdul Kalam wanted to leave Rameswaram to study at the district headquarters in Ramanathapuram. His father said that he knew he had to go away to grow. He gave the example of a seagull and said that a seagull flies across the sun alone and without a nest. He spoke these words because he intended to hone his skills. He knew the harsh reality of life that children may have to live far from their parents to make their career and earn their livelihood. So he showed his wisdom and intelligence in uttering these words [31/10, 7:36 PM] Anshuman: Hn
Let AB and CD be two equal cords (i.e., AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E. It is now to be proven that the line segments AE = DE and CE = BE Construction Steps: Step 1: From the center of the circle, draw a perpendicular to AB i.e., OM ⊥ AB Step 2: Similarly, draw ON ⊥ CD. Step 3: Join OE. Now, the diagram is as follows-  Proof: From the diagram, it is seen that OM bisects AB and so, OM ⊥ AB Similarly, ON bisects CD and so, ON ⊥ CD It is known that AB = CD. So, AM = ND - (i) and MB = CN - (ii) Now, triangles ΔOME and ΔONE are similar by RHS congruency since ∠OME = ∠ONE (They are perpendiculars) OE = OE (It is the common side) OM = ON (AB and CD are equal and so, they are equidistant from the centre) ∴ ΔOME ≅ ΔONE ME = EN (by CPCT) - (iii) Now, from equations (i) and (ii) we get, AM+ME = ND+EN So, AE = ED Now from equations (ii) and (iii) we get, MB-ME = CN-EN So, EB = CE (Hence proved). 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. Solution: From the question we know the following: (i) AB and CD are 2 chords which are intersecting at point E. (ii) PQ is the diameter of the circle. (iii) AB = CD. Now, we will have to prove that ∠BEQ = ∠CEQ For this, the following construction has to be done: Construction: Draw two perpendiculars are drawn as OM ⊥ AB and ON ⊥ D. Now, join OE. The constructed diagram will look as follows:  Now, consider the triangles ΔOEM and ΔOEN. Here, (i) OM = ON [Since the equal chords are always equidistant from the centre] (ii) OE = OE [It is the common side] (iii) ∠OME = ∠ONE [These are the perpendiculars] So, by RHS congruency criterion, ΔOEM ≅ ΔOEN. Hence, by CPCT rule, ∠MEO = ∠NEO ∴ ∠BEQ = ∠CEQ (Hence proved). 4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).  Solution: The given image is as follows:  First, draw a line segment from O to AD such that OM ⊥ AD. So, now OM is bisecting AD since OM ⊥ AD. Therefore, AM = MD - (i) Also, since OM ⊥ BC, OM bisects BC. Therefore, BM = MC - (ii) From equation (i) and equation (ii), AM-BM = MD-MC ∴ AB = CD 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip? Solution:  Let the positions of Reshma, Salma and Mandip be represented as A, B and C, respectively. From the question, we know that AB = BC = 6cm. So, the radius of the circle i.e. OA = 5cm Now, draw a perpendicular BM ⊥ AC. Since AB = BC, ABC can be considered as an isosceles triangle. M is mid-point of AC. BM is the perpendicular bisector of AC, and thus it passes through the centre of the circle. Now, let AM = y and OM = x So, BM will be = (5-x). By applying Pythagorean theorem in ΔOAM we get, OA2 = OM2 +AM2 ⇒ 52 = x2 +y2 - (i) Again, by applying Pythagorean theorem in ΔAMB, AB2 = BM2 +AM2 ⇒ 62 = (5-x)2+y2 - (ii) Subtracting equation (i) from equation (ii), we get 36-25 = (5-x)2 +y2 -x2-y2 Now, solving this equation we get the value of x as x = 7/5 Substituting the value of x in equation (i), we get y2 +(49/25) = 25 ⇒ y2 = 25 - (49/25) Solving it we get the value of y as y = 24/5 Thus, AC = 2×AM = 2×y = 2×(24/5) m AC = 9.6 m So, the distance between Reshma and Mandip is 9.6 m. 6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. Solution: First, draw a diagram according to the given statements. The diagram will look as follows.  Here, the positions of Ankur, Syed and David are represented as A, B and C, respectively. Since they are sitting at equal distances, the triangle ABC will form an equilateral triangle. AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O. Also, O is the centroid of the ΔABC. OA is the radius of the triangle. OA = 2/3 AD Let the side of a triangle a metres then BD = a/2 m. Applying Pythagoras theorem in ΔABD, AB2 = BD2+AD2 ⇒ AD2 = AB2 -BD2 ⇒ AD2 = a2 -(a/2)2 ⇒ AD2 = 3a2/4 ⇒ AD = √3a/2 OA = 2/3 AD 20 m = 2/3 × √3a/2 a = 20√3 m So, the length of the string of the toy is 20√3 m. This exercise has six questions based on the topic - Equal Chords and Their Distances from the Centre. There are two theorems mentioned based on it; Theorem 1: Equal chords of a circle are equidistant from the centre. Theorem 2: Chords equidistant from the centre of a circle are equal in length. For all exercise questions for Class 9 Maths, Chapter 10, reach us at BYJU’S. Solve previous year questions papers along with solutions, to get an idea of the question pattern. You can also get notes and books here, along with tips and tricks of learning for the 9th standard syllabus of Maths and Science. The questions in Exercise 10.4 have long answers with construction and proofs. NCERT solutions provide answers to all the problems of Maths and Science subject for all the classes.
![Tareefan - Badshah [Slowed+Reverb]](/img/1.gif)
Check out other songs on my channel too. And agar kisi gaane ka slowed reverb chahiye to comment karo, I'll try my best to upload them as soon as possible . ❤
Kaise hua X Timro Pratiksha ( Full Version) : ruclips.net/video/Wj1Ny-FY8Dc/видео.htmlsi=4DSDJ72U0jh2pTSJ ruclips.net/video/Wj1Ny-FY8Dc/видео.htmlsi=4DSDJ72U0jh2pTSJ
It's 5:30 in the morning and I am listening to this masterpiece ❤
Exams?
Instead of studying
same vro xD
@@notantprateek ya
Ha toh?
High lvl remixxx🥵🔥😌
Thank you ❤️
Just realize, this is 10 times x better than the real one 🍷🗿
Thnx bro ❤
DUDE U MADE A MASTERPIECE!! I CANT STOP LISTENING TO THIS
❤❤
It's night 3.42 and I am solving physics questions and I listen this masterpiece ❤🔥😜
Boards ?
@@shinchanslowed No neet aspirant ❣️👨🏻⚕️
@@MyselfJivan_k all the best for your exams bro ❤
@@shinchanslowed 🙏🏻❣️
Masterpiece....❤😚, btw its 11:55 and I'm listening this instead of sleeping...😁
Bro I gonna fall in freaking love with this I never liked to hear original mercy but bro I am gonna die on your version i can't express what aura you maked. Lit bro.
Thank you so much ❤
Ambulance bheju?
@@farhanbaig8639 nahi bhaiii itni mehnat mat kar.
@@poojaspills yeah your face don't deserve it tho
@@farhanbaig8639 🌚😹😹😹
Feels energetic after listening to this masterpiece 🔥🔥 💖
*Man 😩 u just made a whole masterpiece .....like tf .....this one is literally a chef kiss 😩🙌*
Thnx ❤
Masterpiece by Badshah ❤️
Omgggg its soo freaking good😳😳😳💗
Last one Baadshah is really superb bro💯💯💯
Fabulous beat uuuuuf
🤯
Beat...😌🖤
❤️❤️❤️
When u r kpop stan and badshah fan too 😈
Really you Korean fan I mean kpop fan you are 💜ᗷTS⟭⟬💜 army
Padshah ☠️
Lit broo!! That's one me!! When u heard badshah rap nd suga's rap on loop🔥🔪
Real🏳🌈
Yesss❤💜
Badshah bhai + slowed songs = drugs ^_^
Bhai katai zher remix 5star remix
High afff!!!❤️
Bruhhhh vroo u killed it!! This song is now more smooth.. dang! Thankss alot for this..
Thnx ❤️
Superb masterpiece ❤️🤌😭
Vibe 🔥💥
❤❤
2 am💗🤤
Laar kyu tapak rhi bhai
It's amazing i can't explain this ...😌❣️
Zoodddd yrr🔥🔥
Goosebumps 🥵🥵👍
That's dope🥶
Next level remix mza agya ....
Haye ❤️🔥🔥🥵
Better that original 🔥👀
I m in love with slowed songs
Arey yaar 🥵🔥🔥🔥
It's masterpiece
This is crazy🛋
Hindi song in slowed and reverb 🥵😬❣
[13/01, 7:20 PM] ABHIJIT: We know that,
Speed=
time
Distance
Time =5s
Distance =3625 m
Total distance to be travelled by sound =2×3625, as it is the distance taken by sonar wave to transmit and receive =7250 m
So, Speed =
5
7250
=1450m/s
[13/01, 7:21 PM] ABHIJIT: Ultrasound is passed through one end of the metal and a detector is installed at the other end. When the metal is perfect, the ultrasonic waves travel as expected and hence are detected on the other end. For a metal with a defect, the waves are reflected back and hence don't reach the other end. When no detection is received, it can be concluded that the metal has a defect.
[13/01, 7:23 PM] ABHIJIT: The outer ear is called ‘pinna’. It collects the sound from the surroundings. The collected sound passes through the auditory canal. At the end of the auditory canal there is a thin membrane called the ear drum or tympanic membrane. When a compression of the medium reaches the eardrum the pressure on the outside of the membrane increases and forces the eardrum inward. Similarly, the eardrum moves outward when a rarefaction reaches it. In this way the eardrum vibrates. The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in the middle ear. The middle ear transmits the amplified pressure variations received from the sound wave to the inner ear. In the inner ear, the pressure variations are turned into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.
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Vibess!!✨🤧
Vibes 💀😳❤️
Master piece 💪
Oh god😳😩💜❤️
Damn😩
Thnx ❤
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Thank you ❤
@@shinchanslowed most welcome 🥰
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It's 4:00 I am listing to refresh my mind to start study (board student👩🎓?
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Luv it❤❤
Listening while solving Equations.....👀💗🌸
Same 💀
Sounds like you actually know what " slowed " means, I have been searching for the slowed version of this song but they were still too fast, this is perfect 😩🤌
Thank you ❤️
It's 1 pm and I am still playing song
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Aag lagi diii✨❤️🥺
Song is love and remix is literally good ❤️
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baap fadu remeix
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[31/10, 7:07 PM] Anshuman: Question 1.
“ On the whole, the small society of Rameswaram was very rigid in terms of the segregation of different social groups,” says the author.
Which social groups does he mention? Were these groups easily identifiable (for example, by the way they dressed)?
Were they aware only of their differences or did they also naturally share friendships and experiences? (Think of the bedtime stories in Kalam’s house; of who his friends were ; and of what used to take place in the pond near his house.)
The author speaks both of people who were very aware of the differences among them and those who tried to bridge these differences. Can you identify such people in the text?
Narrate two incidents that show how differences can be created, and also how they can be resolved. How can people change their attitudes?
Answer:
The author talks about the people who belong to various castes and follow various religious preachings. Yes, these groups were easily identifiable. Their dressing, traditions, culture and rituals were different.
They did share their personal experiences and friendships. Lakshmana Sastry summoned the teacher who separated the author and his friend in the class and told him that he should not spread the poison of social inequality and communal intolerance in the minds of innocent children.
The school teacher encouraged communal differences and Lakshmana Sastry and Sivasubramania Iyer discouraged this malpractice.
The influential people can do both the things. A teacher has the ability to bridge communal differences and can play with sentiments of the innocent and ignorant people. This is what the new teacher did. But the Science teacher Sivasubramania Iyer changed his wife’s attitude and showed her the right path.
Question 2.
Why did Abdul Kalam want to leave Rameswaram?
What did his father say to this?
What do you think his words mean? Why do you think he spoke those words?
Answer:
Abdul Kalam wanted to leave Rameswaram to study at the district headquarters in Ramanathapuram.
His father said that he knew he had to go away to grow. He gave the example of a seagull and said that a seagull flies across the sun alone and without a nest.
He spoke these words because he intended to hone his skills. He knew the harsh reality of life that children may have to live far from their parents to make their career and earn their livelihood. So he showed his wisdom and intelligence in uttering these words
[31/10, 7:36 PM] Anshuman: Hn
I am sitting in the toilet and i am listening to this masterpiece
I don't know how to react on this comment 💀
Bsdk
😂😂🤣🤣🤣
Wow ❤️
Literally no one:
That username gives me nostalgia lol
This is #vibesoverload
❤️❤️🥴
@@shinchanslowed ❤
Yaar konsi app Mai se ye masterpiece bnaa te ho yar bta do love it❣️💕
FL Studio Mobile
@@shinchanslowed and aap ye songs download Kahan se krte ho?
@@shinchanslowed btw thank you so much for replying me tbh koi reply nhi karta but apke pass big heart hai apne subscribers ke liye😉❤️
@@LoffiOp y2mate , pagalworld , mr.jatt
@@LoffiOp thank you so much ❤️😊
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I am listening 🎧 this song 4.20 am fucking osm vibe
Let AB and CD be two equal cords (i.e., AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E.
It is now to be proven that the line segments AE = DE and CE = BE
Construction Steps:
Step 1: From the center of the circle, draw a perpendicular to AB i.e., OM ⊥ AB
Step 2: Similarly, draw ON ⊥ CD.
Step 3: Join OE.
Now, the diagram is as follows-

Proof:
From the diagram, it is seen that OM bisects AB and so, OM ⊥ AB
Similarly, ON bisects CD and so, ON ⊥ CD
It is known that AB = CD. So,
AM = ND - (i)
and MB = CN - (ii)
Now, triangles ΔOME and ΔONE are similar by RHS congruency since
∠OME = ∠ONE (They are perpendiculars)
OE = OE (It is the common side)
OM = ON (AB and CD are equal and so, they are equidistant from the centre)
∴ ΔOME ≅ ΔONE
ME = EN (by CPCT) - (iii)
Now, from equations (i) and (ii) we get,
AM+ME = ND+EN
So, AE = ED
Now from equations (ii) and (iii) we get,
MB-ME = CN-EN
So, EB = CE (Hence proved).
3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
From the question we know the following:
(i) AB and CD are 2 chords which are intersecting at point E.
(ii) PQ is the diameter of the circle.
(iii) AB = CD.
Now, we will have to prove that ∠BEQ = ∠CEQ
For this, the following construction has to be done:
Construction:
Draw two perpendiculars are drawn as OM ⊥ AB and ON ⊥ D. Now, join OE. The constructed diagram will look as follows:

Now, consider the triangles ΔOEM and ΔOEN.
Here,
(i) OM = ON [Since the equal chords are always equidistant from the centre]
(ii) OE = OE [It is the common side]
(iii) ∠OME = ∠ONE [These are the perpendiculars]
So, by RHS congruency criterion, ΔOEM ≅ ΔOEN.
Hence, by CPCT rule, ∠MEO = ∠NEO
∴ ∠BEQ = ∠CEQ (Hence proved).
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Solution:
The given image is as follows:

First, draw a line segment from O to AD such that OM ⊥ AD.
So, now OM is bisecting AD since OM ⊥ AD.
Therefore, AM = MD - (i)
Also, since OM ⊥ BC, OM bisects BC.
Therefore, BM = MC - (ii)
From equation (i) and equation (ii),
AM-BM = MD-MC
∴ AB = CD
5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Solution:

Let the positions of Reshma, Salma and Mandip be represented as A, B and C, respectively.
From the question, we know that AB = BC = 6cm.
So, the radius of the circle i.e. OA = 5cm
Now, draw a perpendicular BM ⊥ AC.
Since AB = BC, ABC can be considered as an isosceles triangle. M is mid-point of AC. BM is the perpendicular bisector of AC, and thus it passes through the centre of the circle.
Now,
let AM = y and
OM = x
So, BM will be = (5-x).
By applying Pythagorean theorem in ΔOAM we get,
OA2 = OM2 +AM2
⇒ 52 = x2 +y2 - (i)
Again, by applying Pythagorean theorem in ΔAMB,
AB2 = BM2 +AM2
⇒ 62 = (5-x)2+y2 - (ii)
Subtracting equation (i) from equation (ii), we get
36-25 = (5-x)2 +y2 -x2-y2
Now, solving this equation we get the value of x as
x = 7/5
Substituting the value of x in equation (i), we get
y2 +(49/25) = 25
⇒ y2 = 25 - (49/25)
Solving it we get the value of y as
y = 24/5
Thus,
AC = 2×AM
= 2×y
= 2×(24/5) m
AC = 9.6 m
So, the distance between Reshma and Mandip is 9.6 m.
6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Solution:
First, draw a diagram according to the given statements. The diagram will look as follows.

Here, the positions of Ankur, Syed and David are represented as A, B and C, respectively. Since they are sitting at equal distances, the triangle ABC will form an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB2 = BD2+AD2
⇒ AD2 = AB2 -BD2
⇒ AD2 = a2 -(a/2)2
⇒ AD2 = 3a2/4
⇒ AD = √3a/2
OA = 2/3 AD
20 m = 2/3 × √3a/2
a = 20√3 m
So, the length of the string of the toy is 20√3 m.
This exercise has six questions based on the topic - Equal Chords and Their Distances from the Centre. There are two theorems mentioned based on it;
Theorem 1: Equal chords of a circle are equidistant from the centre.
Theorem 2: Chords equidistant from the centre of a circle are equal in length. For all exercise questions for Class 9 Maths, Chapter 10, reach us at BYJU’S. Solve previous year questions papers along with solutions, to get an idea of the question pattern. You can also get notes and books here, along with tips and tricks of learning for the 9th standard syllabus of Maths and Science. The questions in Exercise 10.4 have long answers with construction and proofs. NCERT solutions provide answers to all the problems of Maths and Science subject for all the classes.
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