Mensuration Archives - Noon Academy

A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :
(i) the total surface area of the can in contact with water when the sphere is in it.
(ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

(i)Given radius of the cylinder, r = 3.5 cm Diameter of the sphere = height of the cylinder = 3.5×2 = 7 cm So radius of sphere, r = 7/2 = 3.5 cm Height of cylinder, h = 7 cm Total surface area of...

Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboid pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given speed of water flow = 15 km/h Diameter of pipe = 14 cm So radius of pipe, r = 14/2 = 7 cm = 0.07 m Dimensions of cuboidal pond = 50 m × 44 m Height of water in pond = 21 cm = 0.21 m...

The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate (i) the radius of the solid sphere. (ii) the number of cones recast. (Use π = 3.14).

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Given surface area of the solid metallic sphere = 1256 cm2 4R2 = 1256 4×3.14×R2 = 1256 R2 = 1256/4×3.14 R2 = 100 R = 10 Hence the radius of solid sphere is 10 cm. (ii)Volume of the...

The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained? (2007)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 616 cm2 4R2 = 616 4×(22/7)R2 = 616 R2 = 616×7/4×22 R2 = 49 R = 7 Volume of the solid metallic sphere = (4/3)R3 = (4/3)×73 = (1372/3) cm3 Diameter of...

A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel. (2003)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cone, h = 11 cm Radius of the cone, r = 2.5 cm Volume of the cone = (1/3)r2h = (1/3)×2.52×11 = (11/3)×6.25 cm3 When lead shots are dropped into vessel, (2/5) of water...

A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones. (2016)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of metallic cones, r = 2 cm Height of cone, h = 3 cm Volume of cone = (1/3)r2h = (1/3)×22×3 = 4 cm3 Radius of the solid sphere, R = 6 cm Volume of the solid sphere = (4/3)R3 =...

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained. (2005)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the metallic sphere, R = 10.5 cm Volume of the sphere = (4/3)R3 = (4/3)×10.53 = 1543.5 cm3 Radius of cone, r = 3.5 cm Height of the cone, h = 3 cm Volume of the cone =...

A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the metal cylinder, r = 14 cm Height of the metal cylinder, h = 21 cm Radius of the sphere, R = 3.5 cm Volume of the metal cylinder = r2h = (22/7)×142×21 = 22×2×14×21 =...

Find the number of metallic circular discs with 1.5 cm base diameter and height 0.2 cm to be melted to form a circular cylinder of height 10 cm and diameter 4.5 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the circular cylinder, h = 10 cm Diameter of circular cylinder = 4.5 cm So radius, r = 4.5/2 = 2.25 cm Volume of circular cylinder = r2h = ×2.252×10 = 50.625 cm3 Base...

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Edge of the cube, a = 44 cm Volume of cube = a3 = 443 = 85184 cm3 Diameter of shot = 4 cm So radius of shot, r = 4/2 = 2 cm Volume of a shot = (4/3)r3 = (4/3)×(22/7)×23 = 704/21 cm3 Number...

How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given dimensions of the cuboidal solid = 9 cm× 11 cm× 12 cm Volume of the cuboidal solid = 9×11×12 = 1188 cm3 Diameter of shot = 3 cm So radius of shot, r = 3/2 = 1.5 cm Volume of shot =...

A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Radius of the solid circular cylinder, r = 14 cm Height, h = 12 cm Volume of the cylinder = r2h = ×142×12 = ×196×12 = 2352 = 2352×22/7 = 7392 cm3 Edge of the cube, a = 2 cm Volume of cube...

There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given radius of the glass jar, R = 8 cm Diameter of the sphere = 12 cm Radius of the sphere, r = 12/2 = 6 cm When the sphere is removed from the jar, volume of water decreases. Let h be...

A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given internal diameter of cylindrical can = 21 cm Radius of the cylindrical can, R = 21/2 cm Diameter of sphere = 10.5 cm Radius of the sphere, r = 10.5/2 = 21/4 cm Let the rise in water...

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone. (2002)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal diameter of hollow sphere = 4 cm Internal radius, r = 4/2 = 2 cm External diameter = 8 cm External radius, R = 8/2 = 4 cm Volume of the hollow sphere, V = (4/3)(R3-r3) V =...

A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is ½ cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone correct to one decimal place.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal radius of the tube, r = 3 cm Thickness of the tube = ½ cm = 0.5 cm External radius of tube = 3+0.5 = 3.5 cm Height of the tube, h = 21 cm Volume of the tube = (R2-r2)h =...

A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm and height is 72 cm, find the uniform thickness of the cylinder.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the sphere, r = 6 cm Volume of the sphere = (4/3)r3 = (4/3)×63 = 288 cm3 Let r be the internal radius of the hollow cylinder. External radius of the hollow cylinder, R = 4...

A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given inner diameter of the pipe = 6 cm So inner radius, r = 6/2 = 3 cm Outer diameter = 10 cm Outer radius, R = 10/2 = 5 cm Let h be the height of the pipe. Volume of pipe = (R2-r2)h =...

Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: For same material, density will be same. Density = mass/Volume Mass of the smaller sphere, m1 = 1 kg Mass of the bigger sphere, m2 = 7 kg The spheres are melted to form a new sphere. So...

A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 2.5 mm = 0.25 cm Radius of the cylinder, r = 12 cm Volume of the cylinder = r2h = ×122×0.25 = ×144×0.25 = 36 cm3 Let R be the radius of the sphere. Volume...

Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of each sphere, r = 2 cm Volume of a sphere = (4/3)r3 = (4/3)×23 = (4/3)×8 = (32/3) cm3 Volume of 8 spheres = 8×(32/3) = (256/3) cm3 Let R be radius of new sphere. Volume of...

The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 9 cm Diameter of the cylinder = 40 cm Radius of the cylinder, r = 40/2= 20 cm Volume of the cylinder = r2h = ×202×9 = ×400×9 = 3600 cm3 Height of the...

A rectangular water tank of base 11 m x 6 m contains water upto a height of 5 m. if the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given dimensions of the cylindrical vessel = 22 m × 20 m Let the rainfall be x cm. Volume of water = (22×20×x)m3 Diameter of the cylindrical base = 2 m So radius of the cylindrical base =...

A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the hemisphere, r = 8 cm Volume of the hemisphere, V = (2/3)r3 = (2/3)×83 = (1024/3) cm3 Radius of cone, R = 6 cm Since hemisphere is melted and recasted into a cone, the...

The radius of a sphere is 9 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire in metres.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Radius of the sphere, r = 9 cm Volume of the sphere, V = (4/3)r3 = (4/3)××93 = 12××81 = 972 cm3 Diameter of the wire = 2 mm So radius of the wire = 2/2 = 1 mm = 0.1 cm Since the sphere is...

The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the metallic sphere = 6 cm Radius of the sphere, r = 6/2 = 3 cm Volume of the sphere, V = (4/3)r3 = (4/3)××33 = 4××9 = 36 cm3 Length of the wire, h = 36 m = 3600 cm Since...

The adjoining figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1 : 200, find
(i) the total surface area of the solid in π m².
(ii) the volume of the solid in π litres.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 8 cm Radius of the cylinder, r = 3 cm Radius of hemisphere , r = 3 cm Scale = 1:200 Hence actual radius, r = 200×3 = 600 Actual height, h = 200×8 = 1600...

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given height of the cylinder, H = 13 cm Radius of the cylinder, r = 5 cm Radius of the hemisphere, r = 5 cm Height of the cone, h = 12 cm Radius of the cone, r = 5 cm Slant height of the cone, l =...

A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take π = 3.14)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given height of the cylinder, H = 10 cm Height of the cone, h = 6 cm Common diameter = 3.5 cm Common radius, r = 3.5/2 = 1.75 cm Volume of the solid = Volume of the cone + Volume of the...

The adjoining figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and the cone are each of 4 cm. Find the volume of the solid.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given common radius, r = 7 cm Height of the cone, h = 4 cm Height of the cylinder, H = 4 cm Volume of the solid = Volume of the cone + Volume of the cylinder + Volume of the hemisphere =...

A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket. (Use π = 3.14).

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given diameter of the cylinder = 6 cm Radius of the cylinder, r = 6/2 = 3 cm Height of the cylinder, H = 12 cm Slant height of the cone, l = 5 cm Radius of the cone, r = 3 cm Height of the...

A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let the radius of the dome be r. Internal diameter = 2r Given internal diameter is equal to total height. Total height of the building = 2r Height of the hemispherical area = r So height...

A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m³.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Let the radius of the hemisphere be r. Inner diameter = 2r Given greatest height equal to inner diameter. So total height of the hall = 2r Height of the hemispherical part = r Height of cylindrical...

A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is 2/3 of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given radius of the cone, r = 3.5 cm Radius of hemisphere, r = 3.5 cm = 7/2 cm Volume of hemisphere = (2/3)r3 = (2/3)×(22/7)×(7/2)3 = (2/3)×(22/7)×(7/2)×(7/2)×(7/2) = (22/3)×(7/2)×(7/2) =...

The adjoining figure shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. Find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the hemisphere, r = 5 cm Radius of cone, r = 5 cm Height of the cone, h = 7 cm Volume of the solid = Volume of the hemisphere + Volume of the cone = (2/3)r3 + (1/3)r2h =...

The adjoining figure shows a wooden toy rocket which is in the shape of a circular cone mounted on a circular cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted green and the cylindrical portion red, find the area of the rocket painted with each of these colours. Also, find the volume of the wood in the rocket. Use π = 3.14 and give answers correct to 2 decimal places.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; (i) Given height of the rocket = 26 cm Height of the cone, H = 6 cm Height of the cylinder, h = 26-6 = 20 cm Diameter of the cone = 5 cm Radius of the cone, R = 5/2 = 2.5 cm Diameter of...

From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given height of the cylinder, H = 30 cm Radius of the cylinder, r = 7 cm Height of cone, h = 24 cm Radius of cone, r = 7 cm Slant height of the cone, l = √(h2+r2) l = √(242+72) l =...

An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

solution; Given height of the tent above the ground = 85 m Height of the cylindrical part, H = 50 m height of the cone, h = 85-50 h = 35 m Diameter of the base, d = 168 m Radius of the base of...

A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas used to make the tent.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given diameter of the cylindrical part of tent, d = 24 m Radius, r = d/2 = 24/2 = 12 m Height of the cylindrical part, H = 11 m Since vertex of cone is 16 m above the ground, height of cone, h =...

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given radius of the cone, r = 3.5 cm Radius of hemisphere, r = 3.5 cm Total height of the toy = 15.5 cm Height of the cone = 15.5 – 3.5 = 12 cm Slant height of the cone, l = √(h2+r2) l = √(122+3.52)...

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in the given figure). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 10 cm Radius of the cylinder, r = 3.5 cm Radius of the hemisphere = 3.5 cm Total surface area of the article = curved surface area of the cylinder +...

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given edge of the cube, a = 7 cm Diameter of the hemisphere, d = 7 cm Radius, r = d/2 = 7/2 = 3.5 cm Surface area of the hemisphere = 2r2 = 2×(22/7)×3.52 = 44×12.25/7 = 539/7 = 77 cm2 Surface area...

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of the wood in the entire stand, correct to 2 decimal places.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Dimensions of the cuboid = 15 cm× 10 cm × 3.5 cm Volume of the cuboid = 15×10×3.5 = 525 cm3 Radius of each depression, r = 0.5 cm Depth, h = 1.4 cm Volume of conical depression = (1/3)r2h...

16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of the water filled in the box.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given dimensions of the box = 16 cm ×8 cm ×8 cm So volume of the box = lbh = 16×8×8 = 1024 cm3 Radius of the glass sphere, r = 2 cm Volume of the sphere = (4/3)r3 = (4/3)×(22/7)×23 =...

A cone of maximum volume is curved out of a block of wood of size 20 cm x 10 cm x 10 cm. Find the volume of the remaining wood.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given dimensions of the block of wood = 20 cm × 10 cm× 10 cm Volume of the block of wood = 20×10×10 = 2000 cm3 [Volume = lbh] Diameter of the cone, d = 10 cm Radius of the cone , r = d/2 = 10/2 = 5...

From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Given edge of the cube, a = 14 cm Radius of the cone, r = 14/2 = 7 cm Height of the cone, h = 14 cm Volume of the cube = a3 = 143 = 14×14×14 = 2744 cm3 Volume of the cone = (1/3)r2h =...

The given figure shows a solid trophy made of shining glass. If one cubic centimetre of glass costs Rs 0.75, find the cost of the glass for making the trophy

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given side of the cube, a = 28 cm Radius of the cylinder, r = 28/2 = 14 cm Height of the cylinder, h = 28 cm Volume of the cube = a3 = 283 = 28×28×28 = 21952 cm3 Volume of the cylinder =...

Write whether the following statements are true or false. Justify your answer :
(i) The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
(ii) The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals the volume of a hemisphere of radius r.
(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1 : 2 : 3.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Let the radius of sphere be r. Then height of the cylinder, h = 2r Radius of cylinder = r Volume of cylinder = r2h = ×r2×2r = 2r3 Volume of sphere = (4/3)r3 = (2/3)× 2r3 = (2/3)× Volume...

The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 1256 cm2 4r2 = 1256 4×3.14×r2 = 1256 r2 = 1256×/3.14×4 r2 = 100 r = 10 cm Total surface area of the hemisphere = 3r2 = 3×3.14×102 = 3×3.14×100 = 942 cm2...

The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its capacity. Find the volume of water pumped into the tank.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal diameter of the hemispherical tank, d = 14 m So radius, r = 14/2 = 7 m Volume of the tank = (2/3)r3 = (2/3)×(22/7)×(7)3 = 718.667 m3 = 718.67 m3 (approx) = 718.67 kilolitre...

A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the hemispherical bowl, r = 3.5 cm = 7/2 cm Volume of the hemisphere = (2/3)r3 = (2/3)×(22/7)×(7/2)3 = 11×49/6 = 539/6 Hence the volume of the hemispherical bowl is...

Find the volume of a sphere whose surface area is 154 cm².

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 154 cm2 4r2 = 154 4×(22/7)×r2 = 154 r2 = (154×7)/(4×22) r2 = 49/4 r = 7/2 Volume of the sphere = (4/3)r3 = (4/3)×(22/7)×(7/2)3 = 539/3 = 179.666 = 179.67...

A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given side of the cube, a = 4 cm Volume of the cube = a3 = 43 = 4×4×4 = 64 cm3 Diameter of the sphere = 4 cm So radius of the sphere, r = d/2 = 4/2 = 2 cm Volume of the sphere = (4/3)r3 =...

(a) If the ratio of the radii of two sphere is 3 : 7, find :
(i) the ratio of their volumes.
(ii) the ratio of their surface areas.
(b) If the ratio of the volumes of the two sphere is 125 : 64, find the ratio of their surface areas.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Let the radii of two spheres be r1 and r2. Given ratio of their radii = 3:7 Volume of sphere = (4/3)r3 Ratio of the volumes = (4/3)r13/(4/3)r23 = r13/ r23 = 33/73 = 27/343 Hence the...

A sphere and a cube have the same surface area. Show that the ratio of the volume of the sphere to that of the cube is √6 :√π

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let r be the radius of the sphere and a be the side of the cube. Surface area of sphere = 4r2 Surface area of cube = 6a2 Given sphere and cube has same surface area. 4r2 = 6a2 r2/a2 = 6/4...

The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the spherical balloon, r = 7 cm Radius of the spherical balloon after air is pumped, R = 14 cm Surface area of the sphere = 4r2 Ratio of surface areas of the balloons =...

A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given inner diameter of the brass bowl, d = 10.5 cm Radius, r = d/2 = 10.5/2 = 5.25 cm Curved surface area of the bowl = 2r2 = 2×(22/7)×5.252 = 173.25 cm2 Rate of tin plating = Rs.16 per...

Find:
(i) the curved surface area.
(ii) the total surface area of a hemisphere of radius 21 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given radius of the hemisphere, r = 21 cm Curved surface area of the hemisphere = 2r2 = 2×(22/7)×212 = 2×22×3×21 = 2772 cm2 Hence the curved surface area of the hemisphere is 2772 cm2....

Find the diameter of a sphere whose surface area is 154 cm2 .

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 154 cm2 Surface area of the sphere = 4r2 4×(22/7)×r2 = 154 r2 = 154 ×7/(22×4) = 49/4 r = √49/2 Diameter = 2×r = 2×√49/2 = √49 = 7 Hence the diameter of...

Find the surface area of a sphere of diameter:
(i) 21 cm
(ii) 3.5 cm

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given diameter of the sphere, d = 21 cm Radius, r = d/2 = 21/2 = 10.5 Surface area of the sphere = 4r2 = 4×(22/7)×10.52 = 1386 cm2 Hence the surface area of the sphere is 1386 cm2....

Find the volume of a sphere of radius :
(i) 0.63 m
(ii) 11.2 cm

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given radius of the sphere, r = 0.63 m Volume of the sphere, V = (4/3)r3 = (4/3)×(22/7)×0.633 = 1.047 m3 = 1.05 m3 (approx) Hence the volume of the sphere is 1.05 m3. (ii) Given radius...

Find the surface area of a sphere of radius : (i) 14 cm (ii) 10.5 cm

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: (i) Given radius of the sphere, r = 14 cm Surface area of the sphere = 4r2 = 4×(22/7)×142 = 4×22×14×2 = 2464 cm3 Hence the surface area of the sphere is 2464 cm2. (ii) Given radius of the...

A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find
(i) the radius of the cone.
(ii) the (lateral) surface area of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

(i)Given radius of the semi circular lamina, r = 35 cm A cone is formed by folding it. So the slant height of the cone, l = 35 cm Let r1 be radius of cone. Semicircular perimeter of lamina becomes...

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

So the height of the resulting cone, h = 8 cm Radius, r = 6 cm Slant height, l = 10 cm Volume of the cone, V = (1/3)r2h V = (1/3)×3.14×62×8 V = (1/3)×3.14×36×8 V = 3.14×12×8 V = 301.44 cm3 Hence the...

The volume of a right circular cone is 9856 cm3 and the area of its base is 616 cm2 . Find
(i) the slant height of the cone.
(ii) total surface area of the cone.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Given base area of the cone = 616 cm2 r2 = 616 (22/7)×r2 = 616 r2 = 616×7/22 r2 = 196 r = 14 Given volume of the cone = 9856 cm3 (1/3)r2h = 9856 (1/3)×(22/7)×142 ×h = 9856 h =...

The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given perimeter of the base of a cone = 44 cm 2r = 44 2×22/7×r = 44 r = 44×7/(2×22) r = 7 cm Slant height, l = 25 height, h = √(l2-r2) h = √(252-72) h = √(625-49) h = √576 h = 24 cm Volume...

The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given perimeter of the base of a cone = 44 cm 2r = 44 2×22/7×r = 44 r = 44×7/(2×22) r = 7 cm Slant height, l = 25 height, h = √(l2-r2) h = √(252-72) h = √(625-49) h = √576 h = 24 cm Volume...

Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of Rs 80 per metre.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the conical tent, d = 20 m radius, r = d/2 = 20/2 = 10 m Slant height, l = 42 m Curved surface area of the conical tent = rl = (22/7)×10×42 = 22×10×6 = 1320 m2 So the...

(a) The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.
(b) The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.
(c) The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find: (i) the ratio of their volumes.
(ii) the ratio of their lateral surface areas.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (a) Let r1 and r2 be the radius of the given cones and h be their height. Ratio of radii, r1:r2 = 3:4 Volume of cone, V1 = (1/3)r12h Volume of cone, V2 = (1/3)r22h V1 /V1 = (1/3)r12h/...

A Jocker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cone, h = 24 cm Radius, r = 7 cm We know that l2 = h2+r2 l2 = 242+72 l2 = 576+49 l2 = 625 l = √625 l = 25 Curved surface area = rl = (22/7)×7×25 = 22×25 = 550 cm2 So...

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100 m2.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given slant height of conical tomb, l = 25 m Base diameter, d = 14 m So radius, r = 14/2 = 7 m Curved surface area = rl = (22/7)×7×25 = 550 m2 Hence the curved surface area of the cone is...

The height of a cone is 15 cm. If its volume is 1570 cm2 , find the radius of the base. (Use π = 3.14)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of a cone, h = 15 cm Volume of the cone = 1570 cm3 (1/3)r2h = 1570 (1/3)3.14 ×r2×15 = 1570 5 ×3.14×r2 = 1570 r2 = 1570/5×3.14 = 314/3.14 = 100 r = 10 Hence the radius of the...

If the volume of a right circular cone of height 9 cm is 48π cm3 , find the diameter of its base.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Given height of a cone, h = 9 cm Volume of the cone = 48 (1/3)r2h = 48 (1/3)r2×9 = 48 3r2 = 48 r2 = 48/3 = 16 r = 4 So diameter = 2×radius = 2×4 = 8 cm Hence the diameter of the cone is 8...

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter, d = 3.5 m So radius, r = 3.5/2 = 1.75 Depth, h = 12 m Volume of the cone = (1/3)r2h =(1/3)×(22/7)×1.752×12 = (22/7)× 1.752×4 = 38.5 m3 = 38.5 kilolitres [1 kilolitre = 1m3]...

Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius, r = 7 cm Slant height, l = 25 cm We know that l2 = h2+r2 Height of the conical vessel, h = √(l2-r2) = √(252-72) = √(625-49) = √576 = 24 cm Volume of the cone = (1/3)r2h...

Find the volume of the right circular cone with
(i) radius 6 cm and height 7 cm
(ii) radius 3.5 cm and height 12 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Given radius, r = 6 cm Height, h = 7 cm Volume of the cone = (1/3)r2h =(1/3)×(22/7)×62×7 = 22×12 = 264 cm3 Hence the volume of the cone is 264 cm3. (ii) Given radius, r = 3.5 cm Height,...

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find , (i)radius of the base (ii)total surface area of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given curved surface area of the cone = 308 cm2 Slant height of the cone, l = 14 cm rl = 308 (22/7)×r×14 = 308 r = 308×7/(22×14) = 7 Hence the radius of the cone is 7 cm. (ii)Total...

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

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the cone = 10.5 cm Radius, r = d/2 = 10.5/2 = 5.25 cm Slant height of the cone, l = 10 cm Curved surface area of the cone = rl = (22/7)×5.25×10 = 165 cm2 Hence the curved...

Write whether the following statements are true or false. Justify your answer.
(i) If the radius of a right circular cone is halved and its height is doubled, the volume will remain unchanged.
(ii) A cylinder and a right circular cone are having the same base radius and same height. The volume of the cylinder is three times the volume of the cone.
(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Volume of cone = (1/3)r2h If radius is halved and height is doubled, then volume = (1/3)(r/2)2 2h = (1/3)r2h/2 If the radius of a right circular cone is halved and its height is...

A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Length of the can, l = 5 cm Width, b = 4 cm Height, h = 15 cm Volume of the can = lbh = 5×4×15 = 300 cm3 (ii) Diameter of the cylinder, d = 7 cm Radius, r = d/2 = 7/2 = 3.5 cm Height,...

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given length of the pencil, h = 14 cm Diameter of the pencil = 7 mm radius, R = 7/2 mm = 7/20 cm Diameter of the graphite = 1 mm Radius of graphite, r = ½ mm = 1/20cm Volume of graphite =...

The given figure shows a metal pipe 77 cm long. The inner diameter of a cross-section is 4 cm and the outer one is 4.4 cm. Find its
(i) inner curved surface area
(ii) outer curved surface area
(iii) total surface area.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the metal pipe = 77 cm Inner diameter = 4 cm Inner radius, r = d/2 = 4/2 = 2 cm Outer diameter = 4.4 cm Outer radius, R = d/2 = 4.4/2 = 2.2 cm (i)Inner curved surface area...

A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal diameter of the tube = 11.2 cm Internal radius, r = d/2 = 11.2/2 = 5.6 cm Length of the tube, h = 21 cm Thickness = 0.4 cm Outer radius, R= 5.6+0.4 = 6 cm Volume of the...

Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let r1 and r2 be the radius of the two cylinders and h1 and h2 be their heights. Given ratio of the diameter = 3:4 Then the ratio of radius r1:r2 = 3:4 Given volume of both jars are same....

The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm2.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given the ratio of curved surface area and the total surface area = 1:2 Total surface area = 616 cm2 Curved surface area = 616/2 = 308 cm2 2rh = 308 rh = 308/2 = 308×7/2×22 rh = 49 …(i)...

(i) The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm2 . Find the height and the volume of the cylinder.
(ii) The total surface area of a cylinder is 352 cm2 . If its height is 10 cm, then find the diameter of the base.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Let r be the radius and h be the height of the cylinder. Given the sum of radius and height of the cylinder, r+h = 37 cm Total surface area of the cylinder = 1628 cm2 2r(r+h) = 1628...

The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let the radius of the base of a right circular cylinder be r and height be h. Volume of the cylinder, V1 = r2h The radius of the base of a right circular cylinder is halved and the height...

Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the cylinder, r = 3.5 cm Height of the cylinder, h = 7.5 cm Total surface area = 2r(r+h) Curved surface area = 2rh Ratio of Total surface area to curved surface area =...

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Answer correct to the nearest. 100 words. Solution: Height of the barrel of a pen, h = 7 cm Diameter, d = 5mm = 0.5 cm Radius, r = d/2 = 0.5/2 = 0.25 cm Volume of the barrel of pen, V = r2h =...

A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm2. Find (i) the height of the cylinder correct to one decimal place. (ii) the volume of the cylinder correct to one decimal place. (Take π = 3.14)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given diameter of the cylinder, d = 20 cm Radius, r = d/2 = 20/2 = 10 cm Curved surface area = 1000 cm2 2rh = 1000 2×3.14×10×h = 1000 62.8h = 1000 h = 1000/62.8 = 15.9 cm Hence the...

The area of the curved surface of a cylinder is 4400 cm2, and the circumference of its base is 110 cm. Find
(i) the height of the cylinder.
(ii) the volume of the cylinder.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given curved surface area of a cylinder = 4400 cm2 Circumference of its base = 110 cm 2r = 110 r = 110/2 = (110×7)/2×22 = 17.5 cm (i) Curved surface area of a cylinder, 2rh = 4400...

A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m3.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the pole, h = 7 m Diameter of the pole, d = 20 cm radius, r = d/2 = 20/2 = 10 cm = 0.1m Volume of the pole = r2h = (22/7)×0.12×7 = 0.22 m3 Weight of wood per m3 = 225 kg...

If the volume of a cylinder of height 7 cm is 448 π cm3, find its lateral surface area and total surface area.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given Height of the cylinder, h = 7 cm Volume of the cylinder, V = 448 cm3 r2h = 448 ×r2×7 = 448 r2 = 448/7 = 64 r = 8 Lateral surface area = 2rh = 2××8×7 = 112 cm2 Total surface area =...

A road roller (in the shape of a cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the road roller, d = 0.7 m Radius, r = d/2 = 0.7/2 = 0.35 m Width, h = 1.2 m Curved surface area of the road roller = 2rh =2 ×(22/7)×0.35×1.2 = 2.64 m2 Area of the play...

(i) How many cubic metres of soil must be dug out to make a well 20 metres deep and 2 metres in diameter?
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 50 per m2, find the cost of plastering.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given diameter of the well, d = 2 m Radius, r = d/2 = 2/2 = 1 m Depth of the well, h = 20 m Volume of the well, V = r2h = (22/7)×12×20 = 62.85 m3 Hence the amount of soil dug out to...

In the given figure, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 16 cm When rectangular foil of length 22 cm is folded to form cylinder, the base circumference of the cylinder is 22cm 2r = 22 r = 22/2 = 3.5 cm Volume of...

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto a height of 12 cm, find how many litres of milk is needed to serve 1600 students.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the cylindrical glass, d = 7 cm Radius, r = d/2 = 7/2 = 3.5 cm Height of the cylindrical glass, h = 12 cm Volume of the cylindrical glass, V = r2h = (22/7)×3.52×12 = 462...

An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2m. Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area,
(ii) its capacity in litres.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the cylinder, d = 35 cm radius, r = d/2 = 35/2 = 17.5 cm Height of the cylinder, h = 1.2 m = 120 cm (i)Outer lateral surface area = 2rh = 2×(22/7)×17.5×120 = 13200 cm2...

Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of π.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the cylinder, r = 5 cm Height of the cylinder, h = 10 cm Total surface area = 2r(r+h) = 2×5(5+10) = 2×5×15 = 150 cm2 Hence the total surface area of the solid cylinder is...