Love Mera Hit Hit || Slowed + Reverb || Neeraj shridhar || Slowed ~ Vibezzz

#srk #sharukhkhan #deepikapadukone
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Thanks for watching 🎧❤️
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Instagram 👇🏽
armann._fit?igs...
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Song Credit 🎧
Song : Love Mera Hit Hit
Artist : Sharukh Khan, Deepeka Padukone
Singers : Neeraj shridhar, Tulsi Kumar
Lyrics : Ashish Pandit
Music Director : Pritam Chakraborty
Labels : T-Series
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Subscribe for more peace ❤️🎧
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Disclaimer: This Following audio/video is strictly meant for promotional purposes. We Do not Wish to make any commercial use of this & intended to Showcase the Creativity of the artist involved.
The original copyright (s) is (are) Solely owned by the Companies/ Original-Artist(s) / Record-label(s). All the contents are intended to Showcase the Creativity of the artist involved and is Strictly done for a promotional purposes.
*Disclaimer: As per 3rd Section of fair use guidelines borrowing small bit of material from an original work is more likely to be considered fair use. Copyright Disclaimer Under Section 107 of the Copyright Act 1976, allowance is made for fair use.

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[@Fun_of_boy_]

2024 mein bhi sunne ka maza hi kuch or with slowed+reverb

[@hritam9576]

Fall in love with this reverb song 😍❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

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Maza hi a gya

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Don't shy to say that we are from 9xm generation 😅 i used to listen this song when I reads in 9th n 10th

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2024 me kon kon sun raha h

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Srk power 😊

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0:23

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Golden vibe ✨❤️😍

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favorite song 🤫🤫🤫🤫👿👿👿👿👿👿👿👿👿👿👿👿👿

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Love you ❤❤❤❤

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Osam song and bor

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My favourite srk ❤❤

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kon kon RCB ka reel dekh ke ye gana sunne aya hai 😂

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Fav From Chiildhood

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Classy song❤❤

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Song hit hit❤❤

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❤

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I love it❤❤

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Nice song ❤❤❤

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My favourite song ❣️🥰💖💕💗💝💞

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Mera bhi favourite song hai ji❤

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Hii

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आप कहा से हो

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My best song ❣️❣️

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What a song 🎵😘 love it

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Nyc song

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BEST İOVE VALENTINE SONG😍😍😍

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Song lovely 😍🌹

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hmm

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Bullet raja ( south movi 9

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0:01

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🌹🌹I love it❤️❤️

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when listen this song then i will enjoyed " 😃🗿

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Kon kon Mera trha 2024 mai sun rha Hy

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Wow fadu song bro

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Amazing

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Pagal kar diya

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Kya ese songs or copyright claim ni ata

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Aata hai

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To चैनल monetiz नी होगा

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get over yourself ,,,,,,,,,you should muscular ,,,,,,,go gym ,,,,,,dont chase them

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I don’t cl😮😅c😅c😊

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Islam ki @@@

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Bhai Tu marketing team se hain kya ?

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The Volume (V) of each cube is = 64 cm3 This implies that a3 = 64 cm3 ∴ a = 4 cm Now, the side of the cube = a = 4 cm Also, the length and breadth of the resulting cuboid will be 4 cm each, while its height will be 8 cm. So, the surface area of the cuboid = 2(lb+bh+lh) = 2(8×4+4×4+4×8) cm2 = 2(32+16+32) cm2 = (2×80) cm2 = 160 cm2 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Answer: The diagram is as follows:  Now, the given parameters are The diameter of the hemisphere = D = 14 cm The radius of the hemisphere = r = 7 cm Also, the height of the cylinder = h = (13-7) = 6 cm And the radius of the hollow hemisphere = 7 cm Now, the inner surface area of the vessel = CSA of the cylindrical part + CSA of the hemispherical part (2πrh+2πr2) cm2 = 2πr(h+r) cm2 2×(22/7)×7(6+7) cm2 = 572 cm2 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. Answer: The diagram is as follows:  Given that the radius of the cone and the hemisphere (r) = 3.5 cm or 7/2 cm The total height of the toy is given as 15.5 cm. So, the height of the cone (h) = 15.5-3.5 = 12 cm  ∴ The curved surface area of the cone = πrl (22/7)×(7/2)×(25/2) = 275/2 cm2 Also, the curved surface area of the hemisphere = 2πr2 2×(22/7)×(7/2)2 = 77 cm2 Now, the total surface area of the toy = CSA of the cone + CSA of the hemisphere = (275/2)+77 cm2 = (275+154)/2 cm2 = 429/2 cm2 = 214.5cm2 So, the total surface area (TSA) of the toy is 214.5cm2 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. Answer: It is given that each side of the cube is 7 cm. So, the radius will be 7/2 cm.  We know, The total surface area of solid (TSA) = surface area of the cubical block + CSA of the hemisphere - Area of the base of the hemisphere ∴ TSA of solid = 6×(side)2+2πr2-πr2 = 6×(side)2+πr2 = 6×(7)2+(22/7)×(7/2)×(7/2) = (6×49)+(77/2) = 294+38.5 = 332.5 cm2 So, the surface area of the solid is 332.5 cm2 5. A hemispherical depression is cut out from one face of a cubical wooden block, such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. Answer: The diagram is as follows:  Now, the diameter of the hemisphere = Edge of the cube = l So, the radius of the hemisphere = l/2 ∴ The total surface area of solid = surface area of the cube + CSA of the hemisphere - Area of the base of the hemisphere TSA of the remaining solid = 6 (edge)2+2πr2-πr2 = 6l2 + πr2 = 6l2+π(l/2)2 = 6l2+πl2/4 = l2/4(24+π) sq. units 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm, and the diameter of the capsule is 5 mm. Find its surface area.  Answer: Two hemispheres and one cylinder are shown in the figure given below.  Here, the diameter of the capsule = 5 mm ∴ Radius = 5/2 = 2.5 mm Now, the length of the capsule = 14 mm So, the length of the cylinder = 14-(2.5+2.5) = 9 mm ∴ The surface area of a hemisphere = 2πr2 = 2×(22/7)×2.5×2.5 = 275/7 mm2 Now, the surface area of the cylinder = 2πrh = 2×(22/7)×2.5×9 (22/7)×45 = 990/7 mm2 Thus, the required surface area of the medicine capsule will be = 2×surface area of hemisphere + surface area of the cylinder = (2×275/7) × 990/7 = (550/7) + (990/7) = 1540/7 = 220 mm2 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m, respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.) Answer: It is known that a tent is a combination of a cylinder and a cone.  From the question, we know that Diameter = 4 m Slant height of the cone (l) = 2.8 m Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m Height of the cylinder (h) = 2.1 m So, the required surface area of the tent = surface area of the cone + surface area of the cylinder = πrl+2πrh = πr(l+2h) = (22/7)×2(2.8+2×2.1) = (44/7)(2.8+4.2) = (44/7)×7 = 44 m2 ∴ The cost of the canvas of the tent at the rate of ₹500 per m2 will be = Surface area × cost per m2 44×500 = ₹22000 So, Rs. 22,000 will be the total cost of the canvas. 8. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. Answer: The diagram for the question is as follows:  From the question, we know the following: The diameter of the cylinder = diameter of the conical cavity = 1.4 cm So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7 Also, the height of the cylinder = height of the conical cavity = 2.4 cm  Now, the TSA of the remaining solid = surface area of the conical cavity + TSA of the cylinder = πrl+(2πrh+πr2) = πr(l+2h+r) = (22/7)× 0.7(2.5+4.8+0.7) = 2.2×8 = 17.6 cm2 So, the total surface area of the remaining solid is 17.6 cm2 Exercise 13.1 of Class 10 Maths has problems

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0:03

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