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

Jaan ki yaad atak hai song sun kar❤❤❤

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

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

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

Maza hi a gya

2024 me kon kon sun raha h

Srk power 😊

Golden vibe ✨❤️😍

My favourite srk ❤❤

Love you ❤❤❤❤

🥵

Fav From Chiildhood

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favorite song 🤫🤫🤫🤫👿👿👿👿👿👿👿👿👿👿👿👿👿

👌👌👌👌👌

Song hit hit❤❤

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Osam song and bor

Nice song ❤❤❤

What a song 🎵😘 love it

Kon kon Mera trha 2024 mai sun rha Hy

Classy song❤❤

I love it❤❤

❤❤❤❤❤🎉🎉🎉🎉🎉

BEST İOVE VALENTINE SONG😍😍😍

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Tax ❤❤❤❤❤

My best song ❣️❣️

when listen this song then i will enjoyed " 😃🗿

My favourite song ❣️🥰💖💕💗💝💞

Nyc song

😊

Wow fadu song bro

🌹🌹I love it❤️❤️

Song lovely 😍🌹

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

🔥🔥🔥

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Amazing

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😟i💔

Gf kaha pe milta hai koi batai ga jara😅😢

I don’t cl😮😅c😅c😊

Kya ese songs or copyright claim ni ata

Pagal kar diya

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

Islam is true religion ,read quran guys

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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