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NCERT Solutions For Class 10th Maths Chapter 13 : Surface Areas And Volumes

CBSE NCERT Solutions For Class 10th Maths Chapter 13 : Surface Areas And Volumes. NCERT Solutins For Class 10 Mathematics. Exercise 13.1, Exercise 13.2, Exercise 13.3, Exercise 13.4, Exercise 13.5


NCERT Solutions for Class X Maths Chapter 13 Surface Areas And Volumes – Mathematics CBSE

Exercise 13.1 (NCERT)

Question 1: 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution: Side of cube

10 surface area volume exercise solution

Length of new cuboid = 8 cm, height = 4 cm, width = 4 cm

Surface Area can be calculated as follows:

10 surface area volume exercise solution

Alternate Method:

Surface area of cube = 6 x side2

When two cubes are joined end to end, then out of 12 surfaces; two surfaces are lost due to joint. Thus, we need to take surface area of 10 surfaces and hence surface area can be given as follows:

= 10 x side 2

= 10 x 42 = 160 cm2

 

Question 2: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

10 surface area volume exercise solution

Solution: Radius = 7 cm

Height of cylindrical portion = 13 – 7 = 6 cm

Curved surface are of cylindrical portion can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of hemispherical portion can be calculated as follows:

10 surface area volume exercise solution

Total surface are = 308 + 264 = 572 sq cm

Question 3: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

10 surface area volume exercise solution

Solution: Radius of cone = 3.5 cm, height of cone = 15.5 – 3.5 = 12 cm

Slant height of cone can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of cone can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of hemispherical portion can be calculated as follows:

10 surface area volume exercise solution

Hence, total surface area = 137.5 + 77 = 214.5 sq cm

Question 4: A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

10 surface area volume exercise solution

Solution: The greatest diameter = side of the cube = 7 cm

Surface Area of Solid = Surface Area of Cube – Surface Area of Base of Hemisphere + Curved Surface Area of hemisphere

Surface Area of Cube = 6 x Side2

= 6 x 7 x 7 = 294 sq cm

Surface Area of Base of Hemisphere

10 surface area volume exercise solution

Curved Surface Area of Hemisphere = 2 x 38.5 = 77 sq cm

Total Surface Area = 294 – 38.5 + 77 = 332.5 sq cm

Question 5: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution: This question can be solved like previous question. Here the curved surface of the hemisphere is a depression, unlike a projection in the previous question

Total Surface Area

10 surface area volume exercise solution

Question 6: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

10 surface area volume exercise solution

Solution: Height of Cylinder = 14 – 5 = 9 cm, radius = 2.5 cm

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Curved Surface Area of two Hemispheres

10 surface area volume exercise solution

Total Surface Area

10 surface area volume exercise solution

Question 7: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

10 surface area volume exercise solution

Solution: Radius of cylinder = 2 m, height = 2.1 m and slant height of conical top = 2.8 m

Curved Surface Area of cylindrical portion

10 surface area volume exercise solution

Curved Surface Area of conical portion

10 surface area volume exercise solution

Total CSA

10 surface area volume exercise solution

Cost of canvas = Rate x Surface Area

= 500 x 44 = Rs. 22000

Question 8: From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Solution: Radius = 0.7 cm and height = 2.4 cm

Total Surface Area of Structure = Curved Surface Area of Cylinder + Area of top of cylinder + Curved Surface Area of Cone

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Area of top

10 surface area volume exercise solution

Slant height of cone can be calculated as follows:

10 surface area volume exercise solution

Curved Surface Area of Cone

10 surface area volume exercise solution

Hence, remaining surface area of structure

10 surface area volume exercise solution

Question 9: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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.

10 surface area volume exercise solution

Solution: Radius = 3.5 cm, height = 10 cm

Total Surface Area of Structure = CSA of Cylinder + CSA of two hemispheres

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Surface Area of Sphere

10 surface area volume exercise solution

Total Surface Area

10 surface area volume exercise solution

Surface Area Volume

 

Exercise 13.2 (NCERT)

Question 1: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Solution: radius = 1 cm, height = 1 cm

Volume of hemisphere

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Question 2: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution: Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Question 3: A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution: Length of cylinder = 5 – 2.8 = 2.2 cm, radius = 1.4 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of two hemispheres

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Volume of syrup = 30% of total volume

10 surface area volume exercise solution

Volume of syrup in 45 gulabjamuns = 45 x 7.515 = 338.184 cm3

Question 4: 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 depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution: Dimensions of cuboid = 15 cm x 10 cm x 3.5 cm, radius of cone = 0.5 cm, depth of cone = 1.4 cm

Volume of cuboid = length x width x height

= 15 x 10 x 3.5 = 525 cm3

Volume of cone

10 surface area volume exercise solution

Volume of wood = Volume of cuboid – 6 x volume of cone

10 surface area volume exercise solution

Question 5: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution: radius of cone = 5 cm, height of cone = 8 cm, radius of sphere = 0.5 cm

Volume of cone

10 surface area volume exercise solution

Volume of lead shot

10 surface area volume exercise solution

Number of lead shots

10 surface area volume exercise solution

Question 6: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass.

Solution: radius of bigger cylinder = 12 cm, height of bigger cylinder = 220 cm

Radius of smaller cylinder = 8 cm, height of smaller cylinder = 60 cm

Volume of bigger cylinder

10 surface area volume exercise solution

Volume of smaller cylinder

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Mass = Density x volume

10 surface area volume exercise solution

Question 7: A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution: Radius of cone = 60 cm, height of cone = 120 cm

Radius of hemisphere = 60 cm

Radius of cylinder = 60 cm, height of cylinder = 180 cm

Volume of cone

10 surface area volume exercise solution

Volume of hemisphere

10 surface area volume exercise solution

Volume of solid

10 surface area volume exercise solution

Volume of cylinder

10 surface area volume exercise solution

Volume of water left in the cylinder

10 surface area volume exercise solution

Question 8: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution: Radius of cylinder = 1 cm, height of cylinder = 8 cm, radius of sphere = 8.5 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of sphere

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Exercise 13.3 (NCERT)

Question 1: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution: Radius of sphere = 4.2 cm, radius of cylinder = 6 cm

Volume of sphere

10 surface area volume exercise solution

Volume of cylinder

10 surface area volume exercise solution

Since volume of cylinder = Volume of sphere

Hence, height of cylinder

Question 2: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution: Radii of spheres = 6 cm, 8 cm, 10 cm

Volume of sphere

10 surface area volume exercise solution

Total volume of three spheres

10 surface area volume exercise solution

Hence radius of biggest sphere

10 surface area volume exercise solution

Question 3: A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution: Radius of well = 3.5 m, depth of well = 20 m

Dimensions of rectangular platform = 22 m x 14 m

Volume of earth dug out

10 surface area volume exercise solution

Area of top of platform = Area of Rectangle – Area of Circle

(because circular portion of mouth of well is open)

10 surface area volume exercise solution

Height = Volume/Area

10 surface area volume exercise solution

Question 4: A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution: Radius of well = 1.5 m, depth of well = 14 m, width of embankment = 4 m

Radius of circular embankment = 4 + 1.5 = 5.5 m

Volume of earth dug out

10 surface area volume exercise solution

Area of top of platform = (Area of bigger circle – Area of smaller circle)

Height = Volume/Area

10 surface area volume exercise solution

Question 5: A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Solution: Radius of cylinder = 6 cm, height of cylinder = 15 cm

Radius of cone = 3 cm, height of cone = 12 cm

Radius of hemispherical top on ice cream = 3 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Volume of hemisphere

10 surface area volume exercise solution

Volume of ice cream

10 surface area volume exercise solution

Hence, number of ice creams = Volume of cylinder/Volume of ice cream

10 surface area volume exercise solution

Question 6: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution: Radius of coin = 0.875 cm, height = 0.2 cm

Dimensions of cuboid = 5.5 cm x 10 cm x 3.5 cm

Volume of coin

10 surface area volume exercise solution

Volume of cuboid = 5.5 x 10 x 3.5 = 192.5 cm3

Number of coins

10 surface area volume exercise solution

Question 7: A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution: Radius of cylinder = 18 cm, height = 32 cm

Height of cone = 24 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone = Volume of cylinder

Volume of cone

10 surface area volume exercise solution

Hence, radius of cone can be calculated as follows:

10 surface area volume exercise solution

Now, slant height of conical heap can be calculated as follows:

10 surface area volume exercise solution

Question 8: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution: Depth = 1.5 m, width = 6 m, height of standing water = 0.08 m

In 30 minutes, length of water column = 5 km = 5000 m

Volume of water in 30 minutes = 1.5 x 6 x 5000 = 45000 cubic m

Area = Volume/Height

10 surface area volume exercise solution

Question 9: A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution: Radius of pipe = 10 cm = 0.1 m, length = 3000 m/h

Radius of tank = 5 m, depth = 2 m

Volume of water in 1 hr through pipe

10 surface area volume exercise solution

Volume of tank

10 surface area volume exercise solution

Time taken to fill the tank = Volume of tank/Volume of water in 1 hr

10 surface area volume exercise solution

Exercise 13.4 (NCERT)

Question 1: A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution: We have R = 2, r = 1 cm and h = 14 cm

Volume of frustum

10 surface area volume frustum exercise solution

Question 2: The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution: Slant height l = 4 cm, perimeters = 18 cm and 6 cm

Radii can be calculated as follows:

10 surface area volume frustum exercise solution

Curved surface area of frustum

10 surface area volume frustum exercise solution

Question 3: A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

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Solution: R = 10 cm, r = 4 cm, slant height = 15 cm

Curved surface area of frustum

10 surface area volume frustum exercise solution

Area of upper base

10 surface area volume frustum exercise solution

Hence, total surface area

10 surface area volume frustum exercise solution

Question 4: A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.

Solution: Height of frustum = 16 cm, R = 20 cm, r = 8 cm

Volume of frustum

10 surface area volume frustum exercise solution

Cost of milk @ Rs. 20 per 1000 cubic cm

10.44992 x 20 = Rs. 208.99

For calculating surface area, we need to find slant height which can be calculated as follows:

10 surface area volume frustum exercise solution

Surface area of frustum

10 surface area volume frustum exercise solution

Cost of metal sheet @ Rs. 8 per 100 sq cm = 19.5936 x 8 = Rs. 156.75

Question 5: A metallic right circular cone 20 cm high and whose vertical angle is 60 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution: Volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.

10 surface area volume frustum exercise solution

In the given figure; AO = 20 cm and hence height of frustum LO = 10 cm

In triangle AOC we have angle CAO = 30 (halft of vertical angle of cone BAC)

Therefore;

10 surface area volume frustum exercise solution

Using similarity cirteria in triangles AOC and ALM it can be shown that LM = 10/√3 (because LM bisects the cone through its height)

Similarly, LO = 10 cm

Volume of frustum can be calculated as follows:

10 surface area volume frustum exercise solution

Volume of cylinder is given as follows:

10 surface area volume frustum exercise solution

Exercise 13.5 (NCERT Book)

Question 1: A copper wire, 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Solution: For copper wire: Diameter = 3 mm = 0.3 cm

For cylinder; length h = 12 cm, d = 10 cm

Density of copper = 8.88 gm cm3

Curved surface area of cylinder can be calculated as follows:

10 surface area volume exercise solution

Length of wire can be calculated as follows:

10 surface area volume exercise solution

Now, volume of wire can be calculated as follows:

10 surface area volume exercise solution

Mass can be calculated as follows:

Mass = Density × Volume

= 8.88 × 88.7364

= 788 g (approx)

Question 2: A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

10 surface area volume exercise solution

Solution: In triangle ABC;

AC2 = AB2 + BC2

Or, AC2 = 32 + 42

Or, AC2 = 9 + 16 = 25

Or, AC = 5 cm

In triangle ABC and triangle BDC;

10 surface area volume exercise solution

So, we get following equations:

In triangle BDC;

DC2 = BC2 – BD2

Or, DC2 = 42 – 2.42

Or, DC2 = 16 – 5.76 = 10.24

Or, DC = 3.2

From above calculations, we get following measurements for the double cone formed:

Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm

Volume of cone

Curved surface area of cone

Lower Cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm

Volume of cone

10 surface area volume exercise solution

Curved surface area of cone

10 surface area volume exercise solution

Total volume = 19.29216 + 10.85184 = 30.144 cm3

Total surface area = 30.144 + 22.608 = 52.752 cm2

Question 3: A cistern measuring 150 cm x 120 cm x 110 cm has 129600 cm3 water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Solution: Volume of cistern = length x width x depth

= 150 x 120 x 110

= 1980000 cm3

Vacant space = Volume of cistern – Volume of water

= 1980000 – 129600 = 1850400 cm3

Volume of brick = length x width x height

= 22.5 x 7.5 x 6.5

Since the brick absorbs one seventeenth its volume hence remaining volume will be equal to 16/17 the volume of brick

Remaining volume

10 surface area volume exercise solution

Number of bricks = Remaining volume of cistern/remaining volume of brick

10 surface area volume exercise solution

Question 5: An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

10 surface area volume exercise solution

Solution: Curved surface area of cylinder

10 surface area volume exercise solution

Slant height of frustum can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of frustum

10 surface area volume exercise solution

Total curved surface area

10 surface area volume exercise solution

Exercise 13.1 (NCERT)

Question 1: 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution: Side of cube

10 surface area volume exercise solution

Length of new cuboid = 8 cm, height = 4 cm, width = 4 cm

Surface Area can be calculated as follows:

10 surface area volume exercise solution

Alternate Method:

Surface area of cube = 6 x side2

When two cubes are joined end to end, then out of 12 surfaces; two surfaces are lost due to joint. Thus, we need to take surface area of 10 surfaces and hence surface area can be given as follows:

= 10 x side 2

= 10 x 42 = 160 cm2

 

Question 2: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

10 surface area volume exercise solution

Solution: Radius = 7 cm

Height of cylindrical portion = 13 – 7 = 6 cm

Curved surface are of cylindrical portion can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of hemispherical portion can be calculated as follows:

10 surface area volume exercise solution

Total surface are = 308 + 264 = 572 sq cm

Question 3: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

10 surface area volume exercise solution

Solution: Radius of cone = 3.5 cm, height of cone = 15.5 – 3.5 = 12 cm

Slant height of cone can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of cone can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of hemispherical portion can be calculated as follows:

10 surface area volume exercise solution

Hence, total surface area = 137.5 + 77 = 214.5 sq cm

Question 4: A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

10 surface area volume exercise solution

Solution: The greatest diameter = side of the cube = 7 cm

Surface Area of Solid = Surface Area of Cube – Surface Area of Base of Hemisphere + Curved Surface Area of hemisphere

Surface Area of Cube = 6 x Side2

= 6 x 7 x 7 = 294 sq cm

Surface Area of Base of Hemisphere

10 surface area volume exercise solution

Curved Surface Area of Hemisphere = 2 x 38.5 = 77 sq cm

Total Surface Area = 294 – 38.5 + 77 = 332.5 sq cm

Question 5: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution: This question can be solved like previous question. Here the curved surface of the hemisphere is a depression, unlike a projection in the previous question

Total Surface Area

10 surface area volume exercise solution

Question 6: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

10 surface area volume exercise solution

Solution: Height of Cylinder = 14 – 5 = 9 cm, radius = 2.5 cm

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Curved Surface Area of two Hemispheres

10 surface area volume exercise solution

Total Surface Area

10 surface area volume exercise solution

Question 7: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

10 surface area volume exercise solution

Solution: Radius of cylinder = 2 m, height = 2.1 m and slant height of conical top = 2.8 m

Curved Surface Area of cylindrical portion

10 surface area volume exercise solution

Curved Surface Area of conical portion

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Total CSA

10 surface area volume exercise solution

Cost of canvas = Rate x Surface Area

= 500 x 44 = Rs. 22000

Question 8: From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Solution: Radius = 0.7 cm and height = 2.4 cm

Total Surface Area of Structure = Curved Surface Area of Cylinder + Area of top of cylinder + Curved Surface Area of Cone

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Area of top

10 surface area volume exercise solution

Slant height of cone can be calculated as follows:

10 surface area volume exercise solution

Curved Surface Area of Cone

10 surface area volume exercise solution

Hence, remaining surface area of structure

10 surface area volume exercise solution

Question 9: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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.

10 surface area volume exercise solution

Solution: Radius = 3.5 cm, height = 10 cm

Total Surface Area of Structure = CSA of Cylinder + CSA of two hemispheres

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Surface Area of Sphere

10 surface area volume exercise solution

Total Surface Area

10 surface area volume exercise solution

Surface Area Volume

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Exercise 13.2 (NCERT)

Question 1: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Solution: radius = 1 cm, height = 1 cm

Volume of hemisphere

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Question 2: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution: Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Question 3: A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution: Length of cylinder = 5 – 2.8 = 2.2 cm, radius = 1.4 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of two hemispheres

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Volume of syrup = 30% of total volume

10 surface area volume exercise solution

Volume of syrup in 45 gulabjamuns = 45 x 7.515 = 338.184 cm3

Question 4: 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 depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution: Dimensions of cuboid = 15 cm x 10 cm x 3.5 cm, radius of cone = 0.5 cm, depth of cone = 1.4 cm

Volume of cuboid = length x width x height

= 15 x 10 x 3.5 = 525 cm3

Volume of cone

10 surface area volume exercise solution

Volume of wood = Volume of cuboid – 6 x volume of cone

10 surface area volume exercise solution

Question 5: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution: radius of cone = 5 cm, height of cone = 8 cm, radius of sphere = 0.5 cm

Volume of cone

10 surface area volume exercise solution

Volume of lead shot

10 surface area volume exercise solution

Number of lead shots

10 surface area volume exercise solution

Question 6: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass.

Solution: radius of bigger cylinder = 12 cm, height of bigger cylinder = 220 cm

Radius of smaller cylinder = 8 cm, height of smaller cylinder = 60 cm

Volume of bigger cylinder

10 surface area volume exercise solution

Volume of smaller cylinder

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Mass = Density x volume

10 surface area volume exercise solution

Question 7: A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution: Radius of cone = 60 cm, height of cone = 120 cm

Radius of hemisphere = 60 cm

Radius of cylinder = 60 cm, height of cylinder = 180 cm

Volume of cone

10 surface area volume exercise solution

Volume of hemisphere

10 surface area volume exercise solution

Volume of solid

10 surface area volume exercise solution

Volume of cylinder

10 surface area volume exercise solution

Volume of water left in the cylinder

10 surface area volume exercise solution

Question 8: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution: Radius of cylinder = 1 cm, height of cylinder = 8 cm, radius of sphere = 8.5 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of sphere

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Exercise 13.3 (NCERT)

Question 1: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution: Radius of sphere = 4.2 cm, radius of cylinder = 6 cm

Volume of sphere

10 surface area volume exercise solution

Volume of cylinder

10 surface area volume exercise solution

Since volume of cylinder = Volume of sphere

Hence, height of cylinder

Question 2: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution: Radii of spheres = 6 cm, 8 cm, 10 cm

Volume of sphere

10 surface area volume exercise solution

Total volume of three spheres

10 surface area volume exercise solution

Hence radius of biggest sphere

ALSO READ:  NCERT Solutions for Class 8th Science Chapter 7 : Conservation of Plants and Animals

10 surface area volume exercise solution

Question 3: A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution: Radius of well = 3.5 m, depth of well = 20 m

Dimensions of rectangular platform = 22 m x 14 m

Volume of earth dug out

10 surface area volume exercise solution

Area of top of platform = Area of Rectangle – Area of Circle

(because circular portion of mouth of well is open)

10 surface area volume exercise solution

Height = Volume/Area

10 surface area volume exercise solution

Question 4: A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution: Radius of well = 1.5 m, depth of well = 14 m, width of embankment = 4 m

Radius of circular embankment = 4 + 1.5 = 5.5 m

Volume of earth dug out

10 surface area volume exercise solution

Area of top of platform = (Area of bigger circle – Area of smaller circle)

10 surface area volume exercise solution

Height = Volume/Area

10 surface area volume exercise solution

Question 5: A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Solution: Radius of cylinder = 6 cm, height of cylinder = 15 cm

Radius of cone = 3 cm, height of cone = 12 cm

Radius of hemispherical top on ice cream = 3 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Volume of hemisphere

10 surface area volume exercise solution

Volume of ice cream

10 surface area volume exercise solution

Hence, number of ice creams = Volume of cylinder/Volume of ice cream

10 surface area volume exercise solution 10 surface area volume exercise solution

Question 6: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution: Radius of coin = 0.875 cm, height = 0.2 cm

Dimensions of cuboid = 5.5 cm x 10 cm x 3.5 cm

Volume of coin

10 surface area volume exercise solution

Volume of cuboid = 5.5 x 10 x 3.5 = 192.5 cm3

Number of coins

10 surface area volume exercise solution

Question 7: A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution: Radius of cylinder = 18 cm, height = 32 cm

Height of cone = 24 cm

Volume of cylinder

Volume of cone = Volume of cylinder

Volume of cone

10 surface area volume exercise solution

Hence, radius of cone can be calculated as follows:

10 surface area volume exercise solution

Now, slant height of conical heap can be calculated as follows:

10 surface area volume exercise solution

Question 8: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution: Depth = 1.5 m, width = 6 m, height of standing water = 0.08 m

In 30 minutes, length of water column = 5 km = 5000 m

Volume of water in 30 minutes = 1.5 x 6 x 5000 = 45000 cubic m

Area = Volume/Height

10 surface area volume exercise solution

Question 9: A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution: Radius of pipe = 10 cm = 0.1 m, length = 3000 m/h

Radius of tank = 5 m, depth = 2 m

Volume of water in 1 hr through pipe

10 surface area volume exercise solution

Volume of tank

10 surface area volume exercise solution

Time taken to fill the tank = Volume of tank/Volume of water in 1 hr

10 surface area volume exercise solution

Exercise 13.4 (NCERT)

Question 1: A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution: We have R = 2, r = 1 cm and h = 14 cm

Volume of frustum

10 surface area volume frustum exercise solution

Question 2: The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution: Slant height l = 4 cm, perimeters = 18 cm and 6 cm

Radii can be calculated as follows:

10 surface area volume frustum exercise solution

Curved surface area of frustum

10 surface area volume frustum exercise solution

Question 3: A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution: R = 10 cm, r = 4 cm, slant height = 15 cm

Curved surface area of frustum

10 surface area volume frustum exercise solution

Area of upper base

10 surface area volume frustum exercise solution

Hence, total surface area

10 surface area volume frustum exercise solution

Question 4: A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.

Solution: Height of frustum = 16 cm, R = 20 cm, r = 8 cm

Volume of frustum

10 surface area volume frustum exercise solution

Cost of milk @ Rs. 20 per 1000 cubic cm

10.44992 x 20 = Rs. 208.99

For calculating surface area, we need to find slant height which can be calculated as follows:

10 surface area volume frustum exercise solution

Surface area of frustum

10 surface area volume frustum exercise solution

Cost of metal sheet @ Rs. 8 per 100 sq cm = 19.5936 x 8 = Rs. 156.75

Question 5: A metallic right circular cone 20 cm high and whose vertical angle is 60 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution: Volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.

10 surface area volume frustum exercise solution

In the given figure; AO = 20 cm and hence height of frustum LO = 10 cm

In triangle AOC we have angle CAO = 30 (halft of vertical angle of cone BAC)

Therefore;

10 surface area volume frustum exercise solution

Using similarity cirteria in triangles AOC and ALM it can be shown that LM = 10/√3 (because LM bisects the cone through its height)

Similarly, LO = 10 cm

Volume of frustum can be calculated as follows:

10 surface area volume frustum exercise solution

Volume of cylinder is given as follows:

10 surface area volume frustum exercise solution

Exercise 13.5 (NCERT Book)

Question 1: A copper wire, 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Solution: For copper wire: Diameter = 3 mm = 0.3 cm

For cylinder; length h = 12 cm, d = 10 cm

Density of copper = 8.88 gm cm3

Curved surface area of cylinder can be calculated as follows:

10 surface area volume exercise solution

Length of wire can be calculated as follows:

10 surface area volume exercise solution

Now, volume of wire can be calculated as follows:

10 surface area volume exercise solution

Mass can be calculated as follows:

Mass = Density × Volume

= 8.88 × 88.7364

= 788 g (approx)

Question 2: A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

10 surface area volume exercise solution

Solution: In triangle ABC;

AC2 = AB2 + BC2

Or, AC2 = 32 + 42

Or, AC2 = 9 + 16 = 25

Or, AC = 5 cm

In triangle ABC and triangle BDC;

10 surface area volume exercise solution

So, we get following equations:

10 surface area volume exercise solution

In triangle BDC;

DC2 = BC2 – BD2

Or, DC2 = 42 – 2.42

Or, DC2 = 16 – 5.76 = 10.24

Or, DC = 3.2

10 surface area volume exercise solution

From above calculations, we get following measurements for the double cone formed:

Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm

Volume of cone

10 surface area volume exercise solution

Curved surface area of cone

10 surface area volume exercise solution

Lower Cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm

Volume of cone

10 surface area volume exercise solution

Curved surface area of cone

10 surface area volume exercise solution

Total volume = 19.29216 + 10.85184 = 30.144 cm3

Total surface area = 30.144 + 22.608 = 52.752 cm2

Question 3: A cistern measuring 150 cm x 120 cm x 110 cm has 129600 cm3 water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Solution: Volume of cistern = length x width x depth

= 150 x 120 x 110

= 1980000 cm3

Vacant space = Volume of cistern – Volume of water

= 1980000 – 129600 = 1850400 cm3

Volume of brick = length x width x height

= 22.5 x 7.5 x 6.5

Since the brick absorbs one seventeenth its volume hence remaining volume will be equal to 16/17 the volume of brick

Remaining volume

10 surface area volume exercise solution

Number of bricks = Remaining volume of cistern/remaining volume of brick

10 surface area volume exercise solution

Question 5: An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Solution: Curved surface area of cylinder

Slant height of frustum can be calculated as follows:

Curved surface area of frustum

Total curved surface area

10 surface area volume exercise solution

CBSE NCERT Solutions For Class 10th Maths Chapter 13 : Surface Areas And Volumes. NCERT Solutins For Class 10 Mathematics. Exercise 13.1, Exercise 13.2, Exercise 13.3, Exercise 13.4, Exercise 13.5


NCERT Solutions for Class X Maths Chapter 13 Surface Areas And Volumes – Mathematics CBSE

Exercise 13.1 (NCERT)

Question 1: 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution: Side of cube

10 surface area volume exercise solution

Length of new cuboid = 8 cm, height = 4 cm, width = 4 cm

Surface Area can be calculated as follows:

10 surface area volume exercise solution

Alternate Method:

Surface area of cube = 6 x side2

When two cubes are joined end to end, then out of 12 surfaces; two surfaces are lost due to joint. Thus, we need to take surface area of 10 surfaces and hence surface area can be given as follows:

= 10 x side 2

= 10 x 42 = 160 cm2

 

Question 2: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

10 surface area volume exercise solution

Solution: Radius = 7 cm

Height of cylindrical portion = 13 – 7 = 6 cm

Curved surface are of cylindrical portion can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of hemispherical portion can be calculated as follows:

10 surface area volume exercise solution

Total surface are = 308 + 264 = 572 sq cm

Question 3: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

10 surface area volume exercise solution

Solution: Radius of cone = 3.5 cm, height of cone = 15.5 – 3.5 = 12 cm

Slant height of cone can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of cone can be calculated as follows:

10 surface area volume exercise solution

Curved surface area of hemispherical portion can be calculated as follows:

10 surface area volume exercise solution

Hence, total surface area = 137.5 + 77 = 214.5 sq cm

Question 4: A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

10 surface area volume exercise solution

Solution: The greatest diameter = side of the cube = 7 cm

Surface Area of Solid = Surface Area of Cube – Surface Area of Base of Hemisphere + Curved Surface Area of hemisphere

Surface Area of Cube = 6 x Side2

= 6 x 7 x 7 = 294 sq cm

Surface Area of Base of Hemisphere

10 surface area volume exercise solution

Curved Surface Area of Hemisphere = 2 x 38.5 = 77 sq cm

Total Surface Area = 294 – 38.5 + 77 = 332.5 sq cm

Question 5: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution: This question can be solved like previous question. Here the curved surface of the hemisphere is a depression, unlike a projection in the previous question

Total Surface Area

10 surface area volume exercise solution

Question 6: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

10 surface area volume exercise solution

Solution: Height of Cylinder = 14 – 5 = 9 cm, radius = 2.5 cm

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Curved Surface Area of two Hemispheres

10 surface area volume exercise solution

Total Surface Area

10 surface area volume exercise solution

Question 7: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

10 surface area volume exercise solution

Solution: Radius of cylinder = 2 m, height = 2.1 m and slant height of conical top = 2.8 m

Curved Surface Area of cylindrical portion

10 surface area volume exercise solution

Curved Surface Area of conical portion

10 surface area volume exercise solution

Total CSA

10 surface area volume exercise solution

Cost of canvas = Rate x Surface Area

= 500 x 44 = Rs. 22000

Question 8: From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Solution: Radius = 0.7 cm and height = 2.4 cm

Total Surface Area of Structure = Curved Surface Area of Cylinder + Area of top of cylinder + Curved Surface Area of Cone

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Area of top

10 surface area volume exercise solution

Slant height of cone can be calculated as follows:

10 surface area volume exercise solution

Curved Surface Area of Cone

10 surface area volume exercise solution

Hence, remaining surface area of structure

10 surface area volume exercise solution

Question 9: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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.

10 surface area volume exercise solution

Solution: Radius = 3.5 cm, height = 10 cm

Total Surface Area of Structure = CSA of Cylinder + CSA of two hemispheres

Curved Surface Area of Cylinder

10 surface area volume exercise solution

Surface Area of Sphere

10 surface area volume exercise solution

Total Surface Area

10 surface area volume exercise solution

Surface Area Volume

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Exercise 13.2 (NCERT)

Question 1: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Solution: radius = 1 cm, height = 1 cm

Volume of hemisphere

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Question 2: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

ALSO READ:  NCERT Solutions for Class 10th Science Chapter 3 : Metals and Non-metals

Solution: Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Question 3: A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution: Length of cylinder = 5 – 2.8 = 2.2 cm, radius = 1.4 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of two hemispheres

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Volume of syrup = 30% of total volume

10 surface area volume exercise solution

Volume of syrup in 45 gulabjamuns = 45 x 7.515 = 338.184 cm3

Question 4: 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 depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution: Dimensions of cuboid = 15 cm x 10 cm x 3.5 cm, radius of cone = 0.5 cm, depth of cone = 1.4 cm

Volume of cuboid = length x width x height

= 15 x 10 x 3.5 = 525 cm3

Volume of cone

10 surface area volume exercise solution

Volume of wood = Volume of cuboid – 6 x volume of cone

10 surface area volume exercise solution

Question 5: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution: radius of cone = 5 cm, height of cone = 8 cm, radius of sphere = 0.5 cm

Volume of cone

10 surface area volume exercise solution

Volume of lead shot

10 surface area volume exercise solution

Number of lead shots

10 surface area volume exercise solution

Question 6: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass.

Solution: radius of bigger cylinder = 12 cm, height of bigger cylinder = 220 cm

Radius of smaller cylinder = 8 cm, height of smaller cylinder = 60 cm

Volume of bigger cylinder

10 surface area volume exercise solution

Volume of smaller cylinder

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Mass = Density x volume

10 surface area volume exercise solution

Question 7: A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution: Radius of cone = 60 cm, height of cone = 120 cm

Radius of hemisphere = 60 cm

Radius of cylinder = 60 cm, height of cylinder = 180 cm

Volume of cone

10 surface area volume exercise solution

Volume of hemisphere

10 surface area volume exercise solution

Volume of solid

10 surface area volume exercise solution

Volume of cylinder

10 surface area volume exercise solution

Volume of water left in the cylinder

10 surface area volume exercise solution

Question 8: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution: Radius of cylinder = 1 cm, height of cylinder = 8 cm, radius of sphere = 8.5 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of sphere

10 surface area volume exercise solution

Total volume

10 surface area volume exercise solution

Exercise 13.3 (NCERT)

Question 1: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution: Radius of sphere = 4.2 cm, radius of cylinder = 6 cm

Volume of sphere

10 surface area volume exercise solution

Volume of cylinder

10 surface area volume exercise solution

Since volume of cylinder = Volume of sphere

Hence, height of cylinder

Question 2: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution: Radii of spheres = 6 cm, 8 cm, 10 cm

Volume of sphere

10 surface area volume exercise solution

Total volume of three spheres

10 surface area volume exercise solution

Hence radius of biggest sphere

10 surface area volume exercise solution

Question 3: A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution: Radius of well = 3.5 m, depth of well = 20 m

Dimensions of rectangular platform = 22 m x 14 m

Volume of earth dug out

10 surface area volume exercise solution

Area of top of platform = Area of Rectangle – Area of Circle

(because circular portion of mouth of well is open)

10 surface area volume exercise solution

Height = Volume/Area

10 surface area volume exercise solution

Question 4: A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution: Radius of well = 1.5 m, depth of well = 14 m, width of embankment = 4 m

Radius of circular embankment = 4 + 1.5 = 5.5 m

Volume of earth dug out

10 surface area volume exercise solution

Area of top of platform = (Area of bigger circle – Area of smaller circle)

10 surface area volume exercise solution

Height = Volume/Area

10 surface area volume exercise solution

Question 5: A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Solution: Radius of cylinder = 6 cm, height of cylinder = 15 cm

Radius of cone = 3 cm, height of cone = 12 cm

Radius of hemispherical top on ice cream = 3 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone

10 surface area volume exercise solution

Volume of hemisphere

10 surface area volume exercise solution

Volume of ice cream

10 surface area volume exercise solution

Hence, number of ice creams = Volume of cylinder/Volume of ice cream

10 surface area volume exercise solution 10 surface area volume exercise solution

Question 6: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution: Radius of coin = 0.875 cm, height = 0.2 cm

Dimensions of cuboid = 5.5 cm x 10 cm x 3.5 cm

Volume of coin

10 surface area volume exercise solution

Volume of cuboid = 5.5 x 10 x 3.5 = 192.5 cm3

Number of coins

10 surface area volume exercise solution

Question 7: A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution: Radius of cylinder = 18 cm, height = 32 cm

Height of cone = 24 cm

Volume of cylinder

10 surface area volume exercise solution

Volume of cone = Volume of cylinder

Volume of cone

10 surface area volume exercise solution

Hence, radius of cone can be calculated as follows:

10 surface area volume exercise solution

Now, slant height of conical heap can be calculated as follows:

10 surface area volume exercise solution

Question 8: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution: Depth = 1.5 m, width = 6 m, height of standing water = 0.08 m

In 30 minutes, length of water column = 5 km = 5000 m

Volume of water in 30 minutes = 1.5 x 6 x 5000 = 45000 cubic m

Area = Volume/Height

10 surface area volume exercise solution

Question 9: A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution: Radius of pipe = 10 cm = 0.1 m, length = 3000 m/h

Radius of tank = 5 m, depth = 2 m

Volume of water in 1 hr through pipe

10 surface area volume exercise solution

Volume of tank

10 surface area volume exercise solution

Time taken to fill the tank = Volume of tank/Volume of water in 1 hr

10 surface area volume exercise solution

Exercise 13.4 (NCERT)

Question 1: A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution: We have R = 2, r = 1 cm and h = 14 cm

Volume of frustum

Question 2: The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution: Slant height l = 4 cm, perimeters = 18 cm and 6 cm

Radii can be calculated as follows:

10 surface area volume frustum exercise solution

Curved surface area of frustum

10 surface area volume frustum exercise solution

Question 3: A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution: R = 10 cm, r = 4 cm, slant height = 15 cm

Curved surface area of frustum

10 surface area volume frustum exercise solution

Area of upper base

10 surface area volume frustum exercise solution

Hence, total surface area

10 surface area volume frustum exercise solution

Question 4: A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.

Solution: Height of frustum = 16 cm, R = 20 cm, r = 8 cm

Volume of frustum

10 surface area volume frustum exercise solution

Cost of milk @ Rs. 20 per 1000 cubic cm

10.44992 x 20 = Rs. 208.99

For calculating surface area, we need to find slant height which can be calculated as follows:

10 surface area volume frustum exercise solution

Surface area of frustum

10 surface area volume frustum exercise solution

Cost of metal sheet @ Rs. 8 per 100 sq cm = 19.5936 x 8 = Rs. 156.75

Question 5: A metallic right circular cone 20 cm high and whose vertical angle is 60 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution: Volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.

10 surface area volume frustum exercise solution

In the given figure; AO = 20 cm and hence height of frustum LO = 10 cm

In triangle AOC we have angle CAO = 30 (halft of vertical angle of cone BAC)

Therefore;

10 surface area volume frustum exercise solution

Using similarity cirteria in triangles AOC and ALM it can be shown that LM = 10/√3 (because LM bisects the cone through its height)

Similarly, LO = 10 cm

Volume of frustum can be calculated as follows:

10 surface area volume frustum exercise solution

Volume of cylinder is given as follows:

10 surface area volume frustum exercise solution

Exercise 13.5 (NCERT Book)

Question 1: A copper wire, 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Solution: For copper wire: Diameter = 3 mm = 0.3 cm

For cylinder; length h = 12 cm, d = 10 cm

Density of copper = 8.88 gm cm3

Curved surface area of cylinder can be calculated as follows:

10 surface area volume exercise solution

Length of wire can be calculated as follows:

10 surface area volume exercise solution

Now, volume of wire can be calculated as follows:

10 surface area volume exercise solution

Mass can be calculated as follows:

Mass = Density × Volume

= 8.88 × 88.7364

= 788 g (approx)

Question 2: A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

10 surface area volume exercise solution

Solution: In triangle ABC;

AC2 = AB2 + BC2

Or, AC2 = 32 + 42

Or, AC2 = 9 + 16 = 25

Or, AC = 5 cm

In triangle ABC and triangle BDC;

10 surface area volume exercise solution

So, we get following equations:

10 surface area volume exercise solution

In triangle BDC;

DC2 = BC2 – BD2

Or, DC2 = 42 – 2.42

Or, DC2 = 16 – 5.76 = 10.24

Or, DC = 3.2

10 surface area volume exercise solution

From above calculations, we get following measurements for the double cone formed:

Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm

Volume of cone

10 surface area volume exercise solution

Curved surface area of cone

10 surface area volume exercise solution

Lower Cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm

Volume of cone

10 surface area volume exercise solution

Curved surface area of cone

10 surface area volume exercise solution

Total volume = 19.29216 + 10.85184 = 30.144 cm3

Total surface area = 30.144 + 22.608 = 52.752 cm2

Question 3: A cistern measuring 150 cm x 120 cm x 110 cm has 129600 cm3 water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Solution: Volume of cistern = length x width x depth

= 150 x 120 x 110

= 1980000 cm3

Vacant space = Volume of cistern – Volume of water

= 1980000 – 129600 = 1850400 cm3

Volume of brick = length x width x height

= 22.5 x 7.5 x 6.5

Since the brick absorbs one seventeenth its volume hence remaining volume will be equal to 16/17 the volume of brick

Remaining volume

10 surface area volume exercise solution

Number of bricks = Remaining volume of cistern/remaining volume of brick

10 surface area volume exercise solution

Question 5: An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

10 surface area volume exercise solution

Solution: Curved surface area of cylinder

Slant height of frustum can be calculated as follows:

Curved surface area of frustum

Total curved surface area

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