A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th
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1 Answer

Answer :

Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.
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The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th

Description : The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th

Last Answer : Let length, breadth, and height of the rectangular hall be l, b, and h respectively. Area of four walls = 2lh+2bh = 2(l+b)h Perimeter of the floor of hall = 2(l+b) = 250 m Area of four walls = 2( ... of paining the walls is Rs. 15000. 15000 = 2500h Or h = 6 Therefore, the height of the hall is 6 m.

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Last Answer : Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of cylindrical tank is 2πrh = 2 (22/7) 2.1 4.5 m2 = (44 0.3 ... ) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was used in actual while making such a tank.

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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

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Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Description : Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

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It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

Description : It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

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Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

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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

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Description : Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th

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Last Answer : Sides of the triangle are a=50m,b=65m,c=65m Area of triangle, by Heron's formula =s(s−a)(s−b)(s−c)​where, s=2a+b+c​s=250+65+65​s=90 Area of triangle = 90(40)(25)(25)​Area of triangle = 1500m2 Cost of laying grass = Area ×7 Cost of laying grass =1500×7 Cost of laying grass = Rs 10500

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Description : Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

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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

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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

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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

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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

A cone of height 24 cm has a curved surface -Maths 9th

Description : A cone of height 24 cm has a curved surface -Maths 9th

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A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Description : A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Last Answer : answer:

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

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A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : NEED ANSWER

A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : Solution of the question

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

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A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

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Description : It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th

Last Answer : Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface are of tank = 2πr( ... [2 (22/7) 0.7(0.7+1)] = 7.48 Therefore, 7.48 square meters of the sheet are required.

From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

Description : From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

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The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Last Answer : The roller is in the form of a cylinder of diameter = 84 cm ⇒ Radius of the roller(r) = 842 cm = 42 cm Length of the roller (h) = 120 cm Curved surface area of the ... roller = 31680 cm2 = 3168010000m2 ∴ Area of the playground levelled in 500 revolutions = 500 x 3168010000m2 = 1584m2

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Last Answer : A roller is shaped like a cylinder. Let h be the height of the roller and r be the radius. h = Length of roller = 120 cm Radius of the circular end of roller = r = (84/2) cm = 42 cm Now, CSA of roller = 2πrh = ... , we have 2 (22/7) 0.7 h = 4.4 Or h = 1 Therefore, the height of the cylinder is 1 m.

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

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A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

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A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

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A person donates cylindrical bowls of diameter 7 cm -Maths 9th

Description : A person donates cylindrical bowls of diameter 7 cm -Maths 9th

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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.

If three cylinders of radius r and height h are placed vertically such that the curved surface of each cylinder touches the curved surfaces -Maths 9th

Description : If three cylinders of radius r and height h are placed vertically such that the curved surface of each cylinder touches the curved surfaces -Maths 9th

Last Answer : hr2 (3-√−π2)(3−π2) The bases of the three cylinders when placed as given are as shown in the figure : Let the radius of the base of each cylinder = r cm. We are required to find the volume of air. ... ∠C = 60º) = 3 x 60o360o πr2=πr2260o360o πr2=πr22 ∴ Required volume = (3-√r2−π2r2)h=(3-√−π2)r2h.

If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

Description : If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

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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

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Description : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Last Answer : Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of ... 22/7) (5.25)2 = 173.25 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th

Description : The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th

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The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th

Description : The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th

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A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

Description : A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

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The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

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Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th

Description : Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th

Last Answer : Radius of cone, r = 24/2 m = 12m Slant height, l = 21 m Formula: Total Surface area of the cone = πr(l+r) Total Surface area of the cone = (22/7)×12×(21+12) m2 = 1244.57m2