A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th
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A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Last Answer : According to question find the radius of the sphere

A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Description : A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Last Answer : Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r2 = 8 8 r = 8 cm l2 = h2 + r2 = (9)2 + (8)2 l = = 12 cm C.S.A (cone) = = 301.71 cm2

In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th

Description : In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th

Last Answer : answer:

A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Last Answer : Volume of cone = Volume of sphere 1 / 3π(2.1)2 × 8.4 = 4 / 3 πr3 ⇒ r3 = (2.1)2 × 8.4 / 4 = (2.1)3 ⇒ r = 2.1 cm ∴ Radius of the sphere = 2.1 cm

A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

Last Answer : Volume of cone = Volume of sphere 1 / 3π(2.1)2 × 8.4 = 4 / 3 πr3 ⇒ r3 = (2.1)2 × 8.4 / 4 = (2.1)3 ⇒ r = 2.1 cm ∴ Radius of the sphere = 2.1 cm

How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3

How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th

Description : A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th

Last Answer : 92239223% Volume of cone = 1313πr2h = 13×227×1×7=22313×227×1×7=223 cu. cm Volume of cubical block = (10 × 5 × 2) cm3 = 100 cm3 ∴ Wastage of wood = (100−227)100×100(100−227)100×100 = 27832783% = 92239223%

A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th

Description : A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th

Last Answer : answer:

A cylinder, a cone and a sphere are of the same radius -Maths 9th

Description : A cylinder, a cone and a sphere are of the same radius -Maths 9th

Last Answer : Let r be the common radius of a cylinder, cone and a sphere. Then, height of the cylinder = Height of the cone = Height of the sphere = 2r Let 'I' be the slant height of the cone. Then l = root under( √r2 + h2) = root under( ... , S1 : S2 :S3 = 4 πr2 : √5 πr2 : 4 πr2 ∴ S1 : S2 : S3 = 4 : √5 : 4

A sphere, a cylinder and a cone respectively are of the same radius and same height. Find the ratio of their curved surfaces. -Maths 9th

Description : A sphere, a cylinder and a cone respectively are of the same radius and same height. Find the ratio of their curved surfaces. -Maths 9th

Last Answer : answer:

A spherical metal of radius 10 cm is melted and made into 1000 smaller spheres of equal sizes. In this process the surface area of the -Maths 9th

Description : A spherical metal of radius 10 cm is melted and made into 1000 smaller spheres of equal sizes. In this process the surface area of the -Maths 9th

Last Answer : Option (C) is correct. Solution: Let the radius of the small spheres be r' cm. Volume of metal remains the same in both cases. So, vol of the spherical metal of radius 10 cm = total ... Total Surface area of 1000 smaller spheres: 1000*4π12 = 4000π Hence, the surface area increased by 10 times.

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Description : Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Last Answer : Let 'r' be the radius of sphere Surface area of sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 154 = 4 x 22/7 x r2 ⇒ r 2 = 154 x 7/4 x 22 = 49/4 ⇒ r = 7/2 cm = 3.5 cm

The surface area of a sphere of radius 5 cm -Maths 9th

Description : The surface area of a sphere of radius 5 cm -Maths 9th

Last Answer : Radius of the sphere (r1) = 5 cm Radius of the base of cone (r2) = 4 cm Let r сm be the height of the cone. Surface area of sphere = 4 πr2 ⇒ 4 π(5)2 = 100 π cm2 Curved surface area of cone = πrl = 4 πl ... ∴ Volume of cone = 1/3 πr2h = 1/3 x 22/7 x 42 x 3 = 352/7 cm3 = 50.29 cm3 (Approximately)

Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Description : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165

A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Last Answer : Let radius of iron rod = r ∴∴ Height = 8r ∴∴ Volume of iron rod =π×(r)2×8r⇒8πr3=π×(r)2×8r⇒8πr3 ⇒⇒ Radius of spherical ball =r2=r2 Volume of spherical ball =43π(r2)3=43π(r2)3 Let n balls are casted ∴n×43π(r38)=8πr3∴n×43π(r38)=8πr3 ⇒n6=8⇒n=48

A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Last Answer : Let radius of cylindrical rod =r ⇒ height =4r Volume of cylindrical rod =πr2h =πr2(4r) =4πr3 Volume of spherical balls of radius r=34​πr3 No. of balls =34​πr34πr3​=

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.

Diameter of the base of a cone is 10.5 cm -Maths 9th

Description : Diameter of the base of a cone is 10.5 cm -Maths 9th

Last Answer : Radius of cone (r) = 10.5/2 cm Slant height of cone (l) = 10 cm Curved surface area of cone = πrl = 22/7 x 10.5/2 x 10 = 165 cm2

If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Description : If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Last Answer : 20.8 cm Let the edge of the single cube be ‘a’ cm. Then, total volume melted = Volume of cube formed ⇒ (6)3 + (8)3 + (10)3 = a3 ⇒ a3 = 216 + 512 + 1000 = 1728 ... Diagonal of the new cube = 3–√a=(3–√×12)3a=(3×12) cm = 20.8 cm (approx.)

If a wire of resistance R is melted and recast to half of its length, then the new resistance of the wire will be - (1) R/4 (2) R/2 (3) R (4) 2R

Description : If a wire of resistance R is melted and recast to half of its length, then the new resistance of the wire will be - (1) R/4 (2) R/2 (3) R (4) 2R

Last Answer : (1) R/4

If a wire of resistance R is melted and recast to half of its length, then the new resistance of the wire will be

Description : If a wire of resistance R is melted and recast to half of its length, then the new resistance of the wire will be

Last Answer : R/4

There are two identical cubes. Out of one cube, a sphere of maximum volume (VS) is cut off. Out of the second cube, a cone of maximum volume -Maths 9th

Description : There are two identical cubes. Out of one cube, a sphere of maximum volume (VS) is cut off. Out of the second cube, a cone of maximum volume -Maths 9th

Last Answer : answer:

A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be : -Maths 9th

Description : A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be : -Maths 9th

Last Answer : answer:

A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Description : A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Last Answer : answer:

A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

The circumference of the base of 9 m high wooden solid cone is 44 m. Find the slant height of the cone. -Maths 9th

Description : The circumference of the base of 9 m high wooden solid cone is 44 m. Find the slant height of the cone. -Maths 9th

Last Answer : Circumference of the base of a cone = 2πr

The circumference of the base of 9 m high wooden solid cone is 44 m. Find the slant height of the cone. -Maths 9th

Description : The circumference of the base of 9 m high wooden solid cone is 44 m. Find the slant height of the cone. -Maths 9th

Last Answer : Circumference of the base of a cone = 2πr

If the radius of a sphere is 2r, then its volume will be -Maths 9th

Description : If the radius of a sphere is 2r, then its volume will be -Maths 9th

Last Answer : As, r=2r Volume of sphere = 4​/3π(2r)^3 =32/3​πr^3

If the radius of a sphere is 2r, then its volume will be -Maths 9th

Description : If the radius of a sphere is 2r, then its volume will be -Maths 9th

Last Answer : (d) Given, radius of a sphere = 2r Volume of a sphere =4/3 π(Radius)3 = 4/3 π(2r)3 = 4/3 π 8r3 = (32 πr3)/3 cu units Hence the volume of a sphere is (32 πr3)/3 cu units.

The radius of sphere is 2r, then find its volume. -Maths 9th

Description : The radius of sphere is 2r, then find its volume. -Maths 9th

Last Answer : Volume of the sphere = 4/3.π.(2r)3 = 32/3πr3

A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

Description : A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

Last Answer : answer:

Find the volume of cone of radius r/2 and height ‘2h’. -Maths 9th

Description : Find the volume of cone of radius r/2 and height ‘2h’. -Maths 9th

Last Answer : Volume of cone = 1 / 3π × (r / 2)2 × 2h = 1 / 3π × r2 / 4 × 2h = 1 / 6 πr2h cu. units.

Find the volume of cone of radius r/2 and height ‘2h’. -Maths 9th

Description : Find the volume of cone of radius r/2 and height ‘2h’. -Maths 9th

Last Answer : Volume of cone = 1 / 3π × (r / 2)2 × 2h = 1 / 3π × r2 / 4 × 2h = 1 / 6 πr2h cu. units.

The total surface area of a cone whose radius is r/2 and slant height 2l is -Maths 9th

Description : The total surface area of a cone whose radius is r/2 and slant height 2l is -Maths 9th

Last Answer : Total surface area of cone = πr(r+l) Given, radius = r/2​ and slant height = 2l Therefore, new total surface area of cone = πr/2​(r​/2+2l) = π(r/4^2​+rl) = πr(l+r/4​)

The total surface area of a cone whose radius is r/2 and slant height 2l is -Maths 9th

Description : The total surface area of a cone whose radius is r/2 and slant height 2l is -Maths 9th

Last Answer : Radius (r)=r/2 & slant height=2l TSA (S)=PIE R (l+r) =22/7×r/2(2l+r/2) =11/7×r(2l+r/2)

The radius and slant height of a cone... -Maths 9th

Description : The radius and slant height of a cone... -Maths 9th

Last Answer : Let the radius of cone (r) = 4x cm and the slant height of the cone (l) = 7x cm Curved surface area of cone = πrl ∴ πrl = 792 cm2 ⇒ 22/7 x 4x x 7x = 792 ⇒ x2 = 792/22 x 4 = 9 ⇒ x = 3 cm ∴ Radius of the cone = 4 x 3 = 12 cm

The radius and height of a cone are in the ratio 3 : 4 -Maths 9th

Description : The radius and height of a cone are in the ratio 3 : 4 -Maths 9th

Last Answer : Let the radius ofthe cone (r) = 3x cm Height of the cone (h) = 4x cm Volume of the cone = 1/3 πr2h ⇒ 301.44 = 1/3 x 3.14 x (3x)2 .4x ⇒ x3 = 301.44/3.14 x 12 = 8 ⇒ x3 = 23 ⇒ x = 2 ... 4 x 2 = 8 cm Slant height of the cone (l) = root under (√r2 + h2 ) = root under (√62 + 82)= √100 = 10 cm

A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Description : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.

Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2

Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Description : Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Last Answer : Required ratio = 4πr2 / 4/3.πr3 = 3 x 4 x π x (1)2 / 4 x π x (1)3 = 3/1 (Since, r = 1) i.e., 3 : 1

If the radius of a sphere is doubled... -Maths 9th

Description : If the radius of a sphere is doubled... -Maths 9th

Last Answer : Surface area of sphere = 4πr2 When radius is doubled then new surface area = 4π(2r)2 = 4π x 4r2 = 4(4πr2 ) = 4 x original surface area. ∴​ Surface area becomes 4 tim es.

In Fig., a right circular cylinder just encloses a sphere of radius r. Find -Maths 9th

Description : In Fig., a right circular cylinder just encloses a sphere of radius r. Find -Maths 9th

Last Answer : (i) Surface areas S1 of the sphere = 4 πr2 (ii) We have Radius of the cylinder = r Height of the cylinder = h = 2r ∴ Curved surface area S2 of the cylinder ... 2 πrh = 2 πr x 2r = 4 πr2 (iii) S1/S2 = 4 πr2/4 πr2 = 1/1 ∴ S1 : S2 = 1 : 1