A cylinder is filled to 4/5 th of its volume. It is, then tilted so that the level of water coincides with one edge of its bottom and top edge -Maths 9th

A cylinder is filled to 4/5 th of its volume. It is, then tilted so that the level of water coincides with one edge of its bottom and top edge -Maths 9th
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A rigid square plate ABCD of unit side rotates in its own plane about the middle-point of CD until the new position of A coincides with -Maths 9th

Description : A rigid square plate ABCD of unit side rotates in its own plane about the middle-point of CD until the new position of A coincides with -Maths 9th

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If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

Calculate the edge of the cube if its volume is 1331 cm3 -Maths 9th

Description : Calculate the edge of the cube if its volume is 1331 cm3 -Maths 9th

Last Answer : Let the edge of the cube be a Volume of a cube = a×a×a 1331= a×a×a 11 =a ( as 1331 is the cube of 11 Therefore , Edge of a cube is 11 cm

Calculate the edge of the cube if its volume is 1331 cm3 -Maths 9th

Description : Calculate the edge of the cube if its volume is 1331 cm3 -Maths 9th

Last Answer : Let the edge of the cube be a Volume of a cube = a×a×a 1331= a×a×a 11 =a ( as 1331 is the cube of 11 Therefore , Edge of a cube is 11 cm

the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Description : the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Last Answer : T.S.A = 3*154 = 462 cm² C.S.A = 154 cm² C.S.A = 2πrh T.S.A = 2πr(r+h) Now, In T.S.A = 2πrr + 2πrh 462 = 2πrr + 2πrh 462 = 2*22/7*r*r + 154 462 - 154 = 2*22/7*r*r 308*7/2*22 = r*r 49 = r*r R = 7 cm ... 7*h 154/44 = h 7/2 =h H = 3.5 cm or 7/2 cm Now volume = πrrh = 22/7 * 7* 7 *7/2 = 11*49 = 539 cm³

Define :Volume and surface area of a hollow cylinder. -Maths 9th

Description : Define :Volume and surface area of a hollow cylinder. -Maths 9th

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A right circular solid cone of maximum possible volume is cut off from a solid metallic right circular cylinder of volume V. -Maths 9th

Description : A right circular solid cone of maximum possible volume is cut off from a solid metallic right circular cylinder of volume V. -Maths 9th

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The magnitude of the volume of a closed right circular cylinder of unit height divided by the magnitude of the total surface area of the -Maths 9th

Description : The magnitude of the volume of a closed right circular cylinder of unit height divided by the magnitude of the total surface area of the -Maths 9th

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If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Description : If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Last Answer : (a) 0Let h be the first term and k be the common ratio of a GP, then a = hkp - 1, b = hkq - 1, c = hkr - 1∴ (q - r) log a + (r - p) log b + (p - q) log c = log [hkp -1]q - r + log [hkq -1]r - p + log[hkr -1]p - ... r + r - p + p - q) (kp - 1)q - r (kq -1)r - p (kr -1)p - q = log(ho ko) = log 1 = 0.

he 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. -Maths 9th

Description : he 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. -Maths 9th

Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.

Neutral axis of a beam always coincides with a. Axis passing through bottom of beam b. Axis passing through height h/2 from bottom c. Axis passing through height h/3 from bottom d. Axis passing through centroid

Description : Neutral axis of a beam always coincides with a. Axis passing through bottom of beam b. Axis passing through height h/2 from bottom c. Axis passing through height h/3 from bottom d. Axis passing through centroid

Last Answer : d. Axis passing through centroid

A prism is completely filled with 3996 cubes that have edge lengths of 1/3 in.What is the volume of the prism?

Description : A prism is completely filled with 3996 cubes that have edge lengths of 1/3 in.What is the volume of the prism?

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If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th

Description : If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th

Last Answer : Let r and h be radius and height of the cyclinder, then C.S.A. = 2πrh Now, radius is doubled and height is halved. ∴ New radius = 2r and new height = h / 2 New C.S.A. = 2π × 2r × h / 2 = 2πrh .

If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th

Description : If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th

Last Answer : Let r and h be radius and height of the cyclinder, then C.S.A. = 2πrh Now, radius is doubled and height is halved. ∴ New radius = 2r and new height = h / 2 New C.S.A. = 2π × 2r × h / 2 = 2πrh .

In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th

Description : In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th

Last Answer : The curved surface area will remain same. So, there is no change in the curved surface area of cylinder . Hence the curved surface area will remain same.

In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th

Description : In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th

Last Answer : The curved surface area will remain same. So, there is no change in the curved surface area of cylinder . Hence the curved surface area will remain same.

A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. -Maths 9th

Description : A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. -Maths 9th

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Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Description : Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Last Answer : ThenP (A) = P (B) = P (C) = \(rac{2}{4}\) = \(rac{1}{2}\) andP (A ∩ B) = P ({HH}) = \(rac{1}{4}\), P (A ∩ C) = P ({HT}) = \(rac{1}{4}\)P ( ... C)Thus condition (i) is satisfied, i.e., the events are pairwise independent. But condition (ii) is not satisfied and so the three events are not independent

The depth of the centre of pressure on a vertical rectangular gate 8 m wide and 6 m high, when the water surface coincides with the top of the gate, is (A) 2.4 m (B) 3.0 m (C) 4.0 m (D) 5.0 m

Description : The depth of the centre of pressure on a vertical rectangular gate 8 m wide and 6 m high, when the water surface coincides with the top of the gate, is (A) 2.4 m (B) 3.0 m (C) 4.0 m (D) 5.0 m

Last Answer : Answer: Option B

Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Description : Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Last Answer : Edge of cubical tank = 1.5 m ∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m² Area of floor = (1.5)² = 2.25 m² ∴ Total surface area = 9 + 2.25 = 11.25 m² Edge of square tile = 25 m = 0.25 m² ∴ Area of 1 tile = (0.25)2 = .0625 m²

The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Description : The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Last Answer : Total surface area of a cube = 726 cm2 6 × (side)2 = 726 (side)2 = 121 side = 11 cm Hence, the length of the edge of cube is 11 cm.

The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Description : The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Last Answer : Total surface area of a cube = 726 cm2 6 × (side)2 = 726 (side)2 = 121 side = 11 cm Hence, the length of the edge of cube is 11 cm.

Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Last Answer : Edge of cube (a) = 10 cm (i) ∴ Lateral surface area = 4a² = 4 x (10)² = 4 x 100 cm²= 400 cm² (ii) Total surface area = 6a² = 6 x(10)² cm² = 6 x 100 = 600 cm²

A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high -Maths 9th

Description : A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high -Maths 9th

Last Answer : From the question statement, we have Edge of a cube = 10cm Length, l = 12.5 cm Breadth, b = 10cm Height, h = 8 cm (i) Find the lateral surface area for both the figures Lateral surface ... . Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2

an edge of cube is increased by 10%. find the percentage by which the surface area of cube has increased -Maths 9th

Description : an edge of cube is increased by 10%. find the percentage by which the surface area of cube has increased -Maths 9th

Last Answer : each side of cube=l incresed by 10%=L+L/10 =11L/10 surface area of the new cube= 6l^2 =6x 11L/10 x 11L/10 =726/100L^2 % increase= 726/100L^2 - 6L^2 / 6L^2 x 100 =126L^2/100 / 6L^2 x 100 =126L^2/600L^2 x 100 =126L^2/ 6L^2 =126/6 =21 %

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

an edge of cube is increased by 10%. find the percentage by which the surface area of cube has increased -Maths 9th

Description : an edge of cube is increased by 10%. find the percentage by which the surface area of cube has increased -Maths 9th

Last Answer : each side of cube=l incresed by 10%=L+L/10 =11L/10 surface area of the new cube= 6l^2 =6x 11L/10 x 11L/10 =726/100L^2 % increase= 726/100L^2 - 6L^2 / 6L^2 x 100 =126L^2/100 / 6L^2 x 100 =126L^2/600L^2 x 100 =126L^2/ 6L^2 =126/6 =21 %

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

A cubical box has each edge 10 cm -Maths 9th

Description : A cubical box has each edge 10 cm -Maths 9th

Last Answer : (i) Lateral surface area of cubical box = 4a2 = 4 x 102 = 400 cm2 Lateral surface area of cuboidal box = 2h (l + b) = 2 x 8(12.5 +10) = 16 X 22.5 = 360 cm2 Thus, lateral surface area of ... x 305cm2 = 610 cm2 Thus, total surface area of cuboidal box is greater by (610 - 600) cm2 = 10 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.)

A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Description : A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Last Answer : Answer We have, r=0.7m, h=8m ∴ Total surface area = 2πr2+2πrh=2πr(r+h)=2×722​×0.7×8.7m2 Required cost = Rs. {2×722​×0.7×8.7×10}=Rs.382.80

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​=

The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

Description : The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

Last Answer : answer:

In hyperboloid cooling towers the material is filled at 1. Tower’s bottom 2. Tower’s top 3. Near tower 4. None of above

Description : In hyperboloid cooling towers the material is filled at 1. Tower’s bottom 2. Tower’s top 3. Near tower 4. None of above

Last Answer : Tower’s top

A piece of metal of specific gravity 13.6 is placed in mercury of specific gravity 13.6, what fraction of it volume is under mercury? (A) The metal piece will simply float over the mercury (B) The metal piece will be immersed in mercury by half (C) Whole of the metal piece will be immersed with its top surface just at mercury level (D) Metal piece will sink to the bottom

Description : A piece of metal of specific gravity 13.6 is placed in mercury of specific gravity 13.6, what fraction of it volume is under mercury? (A) The metal piece will simply float over the mercury (B ... will be immersed with its top surface just at mercury level (D) Metal piece will sink to the bottom

Last Answer : Answer: Option C

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.

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

The curved surface area of a cylinder is 154 cm2. -Maths 9th

Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th

Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3

The curved surface area of a cylinder is 154 cm2. -Maths 9th

Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th

Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3

What is the number of surfaces of a right circular cylinder ? -Maths 9th

Description : What is the number of surfaces of a right circular cylinder ? -Maths 9th

Last Answer : Number of surfaces of right circular cylinder are three.

A cone, a hemisphere and a cylinder -Maths 9th

Description : A cone, a hemisphere and a cylinder -Maths 9th

Last Answer : V1 (volume of cone) = 1/3 πr2r V2 (volume of hemisphere) = 2/3 πr3 V3 (volume of cylinder) = πr2 .r V1: V2: V3 = 1/3 πr3 : 2/3πr3 : πr3 = 1/3 : 2/3 : 1 V 1: V 2: V 3 = 1 : 2 : 3

If the radius of a cylinder is doubled -Maths 9th

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

Last Answer : Volume of cylinder = πr2h New volume = πR2H = π.(2r)2.(h/2) = 2( πr2h) ∴ New volume = 2 x original volume.

A cylinder and a cone have equal -Maths 9th

Description : A cylinder and a cone have equal -Maths 9th

Last Answer : Curved surface area of cylinder/Curved surface area of cone = 2πrh/πrl = 2πrh/πr root under√(r2 + h2) 8/5 = 2h/root under√(r2 + h2) ⇒ 64/25 = 4h2/r2 + h2 ⇒ 64r2 + 64h2 = 100 h2 ⇒ 64r2 = 100h2 - 64h2 ⇒ 64r2 = 36h2 ⇒ r2/h2 = 36/64 = 9/16 ⇒ r/h = 3/4 ∴ r : h = 3 : 4

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

If the lateral surface of a cylinder is 94.2 -Maths 9th

Description : If the lateral surface of a cylinder is 94.2 -Maths 9th

Last Answer : Height of the cylinder (h) = 5 cm Let r сm be the radius of the base Lateral surface area of cylinder = 94.2 cm2 ⇒ 2 πrh = 94.2 cm2 2 x 3.14 x r x 5 = 94.2 ⇒ r = 94.2/2 x 3.14 x 5 = 94.2 ... Thus, radius of the base of cylinder = 3 cm. (ii) Volume of cylinder = πr2h = 3.14 x 32 x 5 = 141.3 cm3

A sphere and a right circular cylinder -Maths 9th

Description : A sphere and a right circular cylinder -Maths 9th

Last Answer : Let the radius of sphere and cylinder be r and h be the height of cylinder. Then according to the question. Volume of sphere = Volume of cylinder ⇒ 4/3πr3 = πr2h ⇒ r = 3/4.h Diameter of the cylinder = ... x 100 = h/2 x 1/h x 100 = 50% Thus, the diameter of the cylinder exceeds its height by 50%.