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This chapter is an important part of the mathematics syllabus and deals with the calculation of surface areas and volumes of various geometric shapes such as cubes, cylinders, cones, spheres, etc.
The chapter provides students with the fundamental concepts of surface areas and volumes, such as the formulae for calculating the surface area and volume of different objects. Additionally, the chapter also covers various practical applications of surface areas and volumes, making it an essential topic for students to learn.
The NCERT Solutions provided here have been designed to help students understand the concepts covered in this chapter in a clear and concise manner. These solutions have been prepared by experienced teachers and subject matter experts, keeping in mind the latest CBSE guidelines and syllabus.
The solutions provide step-by-step explanations to all the questions in the NCERT textbook, ensuring that students understand the concepts thoroughly. Additionally, the solutions also provide tips and tricks to solve problems quickly and accurately.
We hope that these NCERT Solutions will help students in their preparation for the Class 10 board exams and also help them build a strong foundation for higher studies in mathematics.

Download PDF of NCERT Solutions for Class 10th Maths Chapter 13 Surface Areas And Volumes

Answers of Maths NCERT Solutions for Class 10th Maths Chapter 13 Surface Areas And Volumes

Unit 13

Surface Areas and Volumes

Exercise 13.1

Question 1:

2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboids.

Answer:

Given that,

Volume of cubes = 64 cm3 (Edge)3 = 64

Edge = 4 cm

If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.

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*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.

Answer:oF1DjRR5mha or56r0 gG9jyvPgqNs1Fa5Wn5wc52BVYUhSWgogOurHM1NnXoW44Gya6YJmL0ryjQMz9zSpUl9nTIOA1cNUnZHqJgG6d8U9XDAA4Nnt5jDkFJy7ccBTl 7QnCq4

It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm).

Height of hemispherical part = Radius = 7 cm Height of cylindrical part

Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part
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*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.lVn1ZG

Answer:

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It can be observed that the radius of the conical part and the hemispherical part is same (i.e., 3.5 cm).

Height of hemispherical part = Radius (r) = 3.5 = cm 

Height of conical part (h) = 15.5 −3.5 = 12 cm
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Total surface area of toy = CSA of conical part + CSA of hemispherical part
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*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. 

lVn1ZG

Answer:

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From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7cm.

Radius (r) of hemispherical part =  = 3.5cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

− Area of base of hemispherical part

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Question 5:

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

Answer:

Diameter of hemisphere = Edge of cube = l
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Radius of hemisphere =

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

− Area of base of hemispherical part

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Question 6:

A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
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Answer:

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It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part

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Length of cylindrical part (h) = Length of the entire capsule − 2 × r

= 14 − 5 = 9 cm 

Surface area of capsule = 2×CSA of hemispherical part + CSA of cylindrical part

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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.)

lVn1ZG

Answer:

Given that,

Height (h) of the cylindrical part = 2.1 m Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m Slant height (l) of conical part = 2.8 m     

Area of canvas used = CSA of conical part + CSA of cylindrical part

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Cost of 1 m2 canvas = Rs 500

Cost of 44 m2 canvas = 44 × 500 = 22000

Therefore, it will cost Rs 22000 for making such a tent.

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.

Answer:
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Given that,

Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm Diameter of the cylindrical part = 1.4 cm 

Therefore, radius (r) of the cylindrical part = 0.7 cm

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Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

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SAMOABUoNXBIRTFbHq5bvlH05jDZlf4nY2Y5qea6WXHGApTbFXh6qgP7QSRoN vcNgb8awSZTlTp6K21XHhzVYippc 02cgsVWkpV0bR2f8UlkdbhtLEW1X95glipuHJj3jTtSQ

The total surface area of the remaining solid to the nearest cm2 is 18 cm2.

*Question 9:

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

Given that,

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm Height of cylindrical part (h) = 10 cm 

Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part

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Exercise 13.2

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.

Answer:

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Given that,

Height (h) of conical part = Radius(r) of conical part = 1 cm Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm 

Volume of solid = Volume of conical part + Volume of hemispherical part

=13r2h+23r3

=13121+2313

=3+2π3=π

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 aluminum 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.) 

lVn1ZG

Answer:

From the figure, it can be observed that Height (h1) of each conical part = 2 cm

Height (h2) of cylindrical part = 12 − 2 × Height of conical part 

= 12 − 2 ×2 = 8 cm 

Radius (r) of cylindrical part = Radius of conical part

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Volume of air present in the model = Volume of cylinder + 2 × Volume of cones

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*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 (see the given figure).
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Answer:

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It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part

Length of each hemispherical part = Radius of hemispherical part = 1.4 cm Length (h) of cylindrical part = 5 − 2 × Length of hemispherical part 

= 5 − 2 × 1.4 = 2.2 cm 

Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part
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Volume of 45 gulab jamuns

Volume of sugar syrup = 30% of volume

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 cuboids 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 (see the following figure).
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Answer:

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Depth (h) of each conical depression = 1.4 cm 

Radius (r) of each conical depression = 0.5 cm

Volume of wood = Volume of cuboid − 4 × Volume of cones
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*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.

Answer:

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Height (h) of conical vessel = 8 cm Radius (r1) of conical vessel = 5 cm 

Radius (r2) of lead shots = 0.5 cm

Let n number of lead shots were dropped in the vessel. 

Volume of water spilled = Volume of dropped lead shots

Hence, the number of lead shots dropped in the vessel is 100.
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*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 8 g mass. [Use π = 3.14]

Answer:

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From the figure, it can be observed that Height (h1) of larger cylinder = 220 cm

Radius (r1) of larger cylinder

Height (h2) of smaller cylinder = 60 cm Radius (r2) of smaller cylinder = 8 cm 

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Mass of   iron= 8 g

Mass of 111532.8iron = 111532.8 × 8 = 892262.4 g = 892.262 kg

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.

Answer:
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Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm Height (h2) of conical part of solid = 120 cm 

Height (h1) of cylinder = 180 cm

Radius (r) of cylinder = 60 cm

Volume of water left = Volume of cylinder − Volume of solid

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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.

Answer:

Height (h) of cylindrical part = 8 cm

Radius (r2) of cylindrical part

Radius (r1) spherical part

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Volume of vessel = Volume of sphere + Volume of cylinder
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Hence, she is wrong.

Exercise 13.3

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.

Answer:

Radius (r1) of hemisphere = 4.2 cm Radius (r2) of cylinder = 6 cm 

Let the height of the cylinder be h.

The object formed by recasting the hemisphere will be the same in volume. 

Volume of sphere = Volume of cylinder

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Hence, the height of the cylinder so formed will be 2.74 cm.

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.

Answer:

Radius (r1) of 1st sphere = 6 cm Radius (r2) of 2nd sphere = 8 cm Radius (r3) of 3rd sphere = 10 cm 

Let the radius of the resulting sphere be r.

The object formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres.

Volume of 3 spheres = Volume of resulting sphere
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Therefore, the radius of the sphere so formed will be 12 cm.

Question 3:

A 20 m deep well with a 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. lVn1ZG

Answer:

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The shape of the well will be cylindrical. Depth (h) of well = 20 m

Radius (r) of circular end of well

Area of platform = Length × Breadth = 22 × 14 m2 Let height of the platform = H 

Volume of soil dug from the well will be equal to the volume of soil scattered on the platform.

Volume of soil from well = Volume of soil used to make such platform

Therefore, the height of such platform will be 2.5 m.

*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.

Answer:

The shape of the well will be cylindrical. Depth (h1) of well = 14 m
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Radius (r1) of the circular end of well =

Width of embankment = 4 m

From the figure, it can be observed that our embankment 
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will be in a cylindrical shape having outer radius (r2) as

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and inner radius (r1) as

Let the height of embankment be h2.

Volume of soil dug from well = Volume of earth used to form embankment

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Therefore, the height of the embankment will be 1.125 m.

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.

Answer:

Height (h1) of cylindrical container = 15 cm

Radius (r1) of circular end of container =

Radius (r2) of circular end of ice-cream cone

Height (h2) of conical part of ice-cream cone = 12 cm

Let n ice-cream cones be filled with ice-cream of the container.

Volume of ice-cream in cylinder = n × (Volume of 1 ice-cream cone + Volume of hemispherical shape on the top)
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Therefore, 10 ice-cream cones can be filled with the ice-cream in the container.

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?

lVn1ZG

Answer:

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Coins are cylindrical in shape.

Height (h1) of cylindrical coins = 2 mm = 0.2 cm

Radius (r) of circular end of coins

Let n coins be melted to form the required cuboids. Volume of n coins = Volume of cuboids

Therefore, the number of coins melted to form such a cuboid is 400.4LoQcZ91btZUfj4tCcfTT1DJU1 LXm4qXuBo8lSA6Ped 5Dk0E5csNHACYlpmkGUzUtsutosRZhi4ZXq rum mOBmw94EeINdUqhp3D64hATV8

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.

Answer:

fa1BfJmyaqWy11Jbt3g006zBJz8Szv4SiFtqgS5B6TaFAzLb4pwUNbCgmE54Gd5Y9kxkP20WD75S N1WN EtUNSMPs2NEz345UcDSBszN5gAc 7Fw9
svvZGR CNeQdwJm3PI69bo8tUfbfWJWmzMbjkQy1nHxZyBe5ro5o7Wj1kXDGO2VJ3dQzC J3tqT YlSI87Zgv2iKbPDqGpNktMJV5enyksEqMdgc Y0B RyIE 8Fkq1s1ziQgA

Height (h1) of cylindrical bucket = 32 cm Radius (r1) of circular end of bucket = 18 cm Height (h2) of conical heap = 24 cm 

Let the radius of the circular end of conical heap be r2.

The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.

Volume of sand in the cylindrical bucket = Volume of sand in conical heap
qqS9Xp6sHFkdVmkrWOnMNf64bvTk3 IzK2Yki3uM68ZOB34ThZoiOxhD2SM65t3sT1VYyRQ61AUieOr rybFgBSF9BomACV TLCsTQcGS1BAJ03vUicP7QxdZiFQe9 P1P5C eIMOF6YkxKOzg0QH nkaMUXxci0yih7crzZBSr6IqxUfJOz2yNy IashUoDcnZJyqFk6J0neCwLDGL3pxWBiai ktFMVAQUqm16nuvOlVRkQZO p5ohfRsXpKGxxIsNO6V dKacogOKO9Be4

wcQcYwaMFsnut3dkQnTwlF541rTxgYZ3 5QmFLcUpbt9Y AyGXsjkpolOks1 AlKQ4IsDgb THFXB91rUIkLQ 3yeY5WjupeDLhU3lVtXtGj7MLnC Njp0ZiajA9KE8A0gPLnA

Slant height

Therefore, the radius and slant height of the conical heap are 36 cm and  respectively.

*Question 8:

Water in 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?

Answer:

hxre1nTizbgJRVc1w KBdfTAiB PiwJtABRD4L1Tfor wCg5bcTfOe8m5jOGlBC0jdrromHeodcXKtc4

Consider an area of cross-section of canal as ABCD. Area of cross-section = 6 × 1.5 = 9 m2

Speed of water

Volume of water that flows in 1 minute from canal

Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m3   

PO6m WEPnkeGLH2LjUuvzjD9ax1ouY08iaXURXapvjk1tBBbMSs3FF K cegEXlnvtivhz7hqjs2UuaKlU TZaSYGZLYSwRqauQWeZhGz0cAOuGt3 JeCDSvlEWR7QPJZph8OKw

Let the irrigated area be A. Volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal.

Vol. of water flowing in 30 minutes from canal = Vol. of water irrigating the reqd. area
lkxBT7GSzAS4GeywVCMoszrGVJqkmWwZPHUKzQSwwaHvSdjPPjxpeB6XLM2WCGyj3FfJKZVs6OsE7vJAYxdfYMW29FlcnLgGqNONCevWFtj4s74828MwL qFe dl yWCSwQ2i8E

A = 562500 m2

Therefore, area irrigated in 30 minutes is 562500 m2.

Question 9:

A farmer connects a pipe of internal diameter 20 cm form 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?

Answer:
Fs7r6dXy6mCP5Fcwv1 PSLYyl8uMGZClPDnBmUits6272IzQDELJANIbyU Zizu8LIzXdgEV4lnuo8oipQMVVrLkOmbNH9crl7ku7MUpjqqdO6LvwmTCOpYcPrEJwbDTAMl3RE

Consider an area of cross-section of pipe as shown in the figure.

Radius (r1) of circular end of pipe

Area of cross-section

Speed of water

Volume of water that flows in 1 minute from pipe

Volume of water that flows in t minutes from pipe = t × 0.5π m3

wfe I4co DQOLXiawgOJ EkNfUfT83u5YEGcY2ciX1 wA4HzLoB2uzWkgHu3 MGAUirnCo9pheGUqsdlSEPbyye6tpDE9UwJqzUXbQw9WTtXmDkQoBaGoaxc8V dDEUCLXEThG4

Radius (r2) of circular end of cylindrical tank

Depth (h2) of cylindrical tank = 2 m

Let the tank be filled completely in t minutes.

Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.

Volume of water that flows in t minutes from pipe = Volume of water in tank t × 0.5π = π ×(r2)2 ×h

t × 0.5 = 52 ×2

t = 100

Therefore, the cylindrical tank will be filled in 100 minutes.

Exercise 13.4

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. lVn1ZG

Answer:

THs1cUoElffBNSmQG8YXhxNBKmQtOeWxSB3clMdwu2QtcRP36BJ66O9GsGIvamMH9D4mAV

Radius (r1) of upper base of glass

Radius (r2) of lower base of glass

Capacity of glass = Volume of frustum of cone

YRCDvQhKYD5N1C3S2jO0jWxfitNbcXodD9zwK1WGQcawquinjLRzzDBOM2sK5iMw0HVdwGwqG4APV6 scLxOZunFqDx7Hg2P1FkH7ddDuWx 6 w7H8txR3UW2YBJ NKB9t5UVPU

Therefore, the capacity of the glass is.

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.

Answer:

P3PMc0E JxTsGuJSvk1EJjJxkpvAL2Sohi3IaBN

Perimeter of upper circular end of frustum = 2πr1=187axGqOSaru5UdXRCSiTtYe9RZ3NhpqBjsiJz91yLcIRFX6 8Q0S8m3mHXrTSpP6Mb6 MbdLVxTSRPNA

Perimeter of lower end of frustum = 2πr2= 6g KXvE9mADXXMb8piP6izfLf1E2eYomiRhSDKUnU5js5a9ZIEUkFJZLC 0OpZjUUgP be6O6R FigZiDNVzdkdHoOBpT2r6gW5knB58cj9XftlZoxCBUiS13 x6PcWVQf 5FCn8

Slant height (l) of frustum = 4 cm CSA of frustum = π(r1 + r2) l
N zjRRYMolbcz0ZeZJ1T1xdvyr 27IYXiAw1p0MjejbBLnUV j5fMKkK wFs1ZMZnuiULNbSSu6sIBeNC1Gohmkn

Therefore, the curved surface area of the frustum is 48 cm2.

*Question 3:

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure given below). 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 use for making it. A e359tYlgPFmal3z7B UAVpEDYgABGGecTK2JqW8RrMNXdc3TUOdPYU8c1NMtsnAClCmV5gU1acOp Y 7eUPHpaaY1yENv4mQ4fjIlGw6J2K8hD

Answer:
x3ux uXdCQtPBP1BYAG5Siq brJ4wVVvzgmX a55drMrM2KmT0m3a4ZE3GkL77Fq02e7JE7QCt yDBgDoucXP6fl0s2R KIfN89v3SRUbecB2JuzskzcupcTJ8gH1Z3Tdg a2eI

Radius (r2) at upper circular end = 4 cm

Radius (r1) at lower circular end = 10 cm Slant height (l) of frustum = 15 cm 

Area of material used for making the fez = CSA of frustum + Area of upper circular end

heLHAWDzwXlGDPQ71U2waLi rjAp

= π (10 + 4) 15 + π (4)

= π (14) 15 + 16 π 

tYwTYxcU8kr9HDxq6GVpfSnlbJ

Therefore, the area of material used for making it is.

*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. [Take π = 3.14]

Answer:

lcV5TpSYxnCCwb6TBhWpka5X cBf6BaI5R7c yBrvl8Z60LlIYSYyfaTVAD8zjY1Tm FpuaU7v24D NPftVThEZ SjStbBnjh qyDD5no7MLZLZZxDE Z2PwXOAKiiIja4cXrEY

Radius (r1) of upper end of container = 20 cm

Radius (r2) of lower end of container = 8 cm Height (h) of container = 16 cm 

Slant height (l) of frustum

lJ68czmfGChwyOmKUZpb6SxalBVzCsmDNwS3F4fsDnGwBqXe0wwdv6mHSvMoyEblhLKNnlTgmxFoTsbhSvWM5b5Wj7jbGglJ5 V2yZnuFW2x4IC2

Capacity of container = Volume of frustum
8RQEfM tc tmYsekHsrrhtKCR3Sry4I4mitpjv l4TPsiLAIV7YOQ3K2L9nFvOBFg2oa8f2vUfbMjlm9eQxF4L1KbNGOVQdmrkIUfeuNmcbdHESMzOYI yn2GaR58Th1ZX7LbME

Cost of 1 litre milk = Rs 20

Cost of 10.45 litre milk = 10.45 × 20

= Rs 209

Area of metal sheet used to make the container

heLHAWDzwXlGDPQ71U2waLi rjAp

= π (20 + 8) 20 + π (8)

= 560 π + 64 π = 624 π cm2  

Cost of 100 cm2 metal sheet = Rs 8

C0nbxM1bzzulhNTp0tMIjCkXkRw2825szYn Z8Y8EM8gUzTgeNCoYnYaFjIuVOgNjSsTblz0ybnRfLZYEip5

Therefore, the cost of the milk which can completely fill the container is

Rs 209 and the cost of metal sheet used to make the container is 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 is drawn into a wire of diameter cm, find the length of the wire.lVn1ZG

Answer:

u2l qB4

In ΔAEG,

In ΔABD,

9suog2ZfFyOwYsviZTm8z8upv4cMh4nPMeyq94WgYaz hobWQ5inovO5WiYjXSuu3vP2bRmg4746tyslaC vnBN0Z C5Jpq9AXcyCYRn5vJ50rZdYXn6ewKD50ED yN4yisrFT8

Radius (r1) of upper end of frustum

Radius (r2) of lower end of container

Height (h) of container = 10 cm

Volume of frustum
S6FoKdt4L1aSmefiTV2JKo7qY 8SWIMlHZydC9Q0FiA6jBdN2FXg38gBIkIM6Ks4QvpnbY IzwpaOGoB94zmhKYoPFYUOzS rFo2fjbdeuJhHDYvU3HiE79RW5qqq0kwWlBAzQ

Radius (r) of wire

Let the length of wire be l.

Volume of wire = Area of cross-section × Length

= (πr2) (l)
KgpFYbfzaEixa8xtV hOnY 4Yfj zzOgtWmFYc4JjQaKuRkrCimDDjuhaVdPJGlnhwrmf Yff re5O5ZrjiORvDCuGSK66h4McT6UNPL

Volume of frustum = Volume of wire

220009=π1322×l

220009×32×32=l

220009×32×32×722=l

l=7964.44m

Exercise 13.5

*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.

Answer:sAzh B5LwNtedslDSbouVg3nD7ZDSvo0DYeQIeMXdgSMCsLB9uPFmuZb4gFgirp t0R3SYX96Pc 8LuYyDZCMwFpHKH45 GdKl UdfXOKdMygR6fsv6emWgCY38CQYuojYuQyik

It can be observed that 1 round of wire will cover 3 mm height of cylinder.

ImhIFyeox2ROKfGUDwUEKp9

Length of wire required in 1 round = Circumference of base of cylinder

= 2πr = 2π × 5 = 10π 

Length of wire in 40 rounds = 40 × 10π

7mKb1eMEgZAxV qCwngbOG4FIVLGWUBXytRBANNiffkxIVkQ MxtUJvB2mJN BnUCLnWZRIFLaQNHireWS038z6FrJXJI6C TrHrheiCpsD8WCQcv4jYcgjjA0ZQ7zeFF6wQpY

= 1257.14 cm = 12.57 m 

Radius of wire (Equation alignment needed)

Volume of wire = Area of cross-section of wire × Length of wire

= π(0.15)2 × 1257.14 

= 88.898 cm3

Mass = Volume × Density

= 88.898 × 8.88= 789.41 gm 

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. (Choose value of π as found appropriate.)

Answer:

l0VhavQ71uUzaCm4hB6Uyi DUUEUFILpmeo5mvFpPOzkrobGO5z s9xKLlokrfiDWW2GrcOUm4xWuS WkJxRhrThvO

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

Hypotenuse yTdWJkbAiodj BJuS4zYtp5cA8KpoRjsL2snfHygNQdxtHNqXRtNaY0YTbstv43oXHHG3MmyYDRw7x1 Wh7UwJQfCJVEPEbDtEXh1m5WDuypZ0Y8aKyIzfaNluBlpbT8DmeJR35nvsjWHgyCLT0w4bhhwuWxaJLt3Y1 z98BI5qWtp4tB Ig1LhI

=5 cm
IuEAy0hQmFsOOIK5nCT 1 TKB3Oe8PR7RnlRkZ5O2IuNteHUqsT6 61ribhaogQDtXjSulu3Cc7iVGkRuwlzhRtK EKmaHmkU0doVpKf01OgTfjs2Jq6t0gpiJVp8sU7UlVuceE

Area of ΔABC

zv5bUWrTZzIrx8oIDiNIJYUhPpDLuJzkZAmYbl74unrup2XdUa7 DNApf5Wwqru8Zji6Dt P Dwbxbvr3sQxmUYA8qIpk Ru GiMtr15jkOm 6HZY0U0SGe Rjxggf sd2qXZA

Volume of double cone = Volume of cone 1 + Volume of cone 2
aEd2VD7NLSE4CJVbeO ZdJEzh32A7PgdGC4Tj8wM7bVRIJ3nbo

= 30.14 cm3

Surface area of double cone = Surface area of cone 1 + Surface area of cone 2

= πrl1 + πrl2  
M

= 52.75 cm2

Question 3:

A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of 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 × 7.5 cm × 6.5 cm?

Answer:

Volume of cistern = 150 × 120 × 110

= 1980000 cm3

Volume to be filled in cistern = 1980000 – 129600 

= 1850400 cm3

Let n numbers of porous bricks were placed in the cistern. Volume of n bricks = n × 22.5 × 7.5 × 6.5 

= 1096.875n

As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these bricks
PCCV7B9t8VAkTAN3Qz3xjfhmH XLNm2yRBsvu6bmYnBcpFTTEJ36pJXK3FN

n = 1792.41

Therefore, 1792 bricks were placed in the cistern.

Question 4:

In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer:

Area of the valley = 7280 km2

If there was a rainfall of 10 cm in the valley then amount of rainfall in the valley = Area of the valley × 10 cm

Amount of rainfall in the valley = 7280 km2 × 10 cm

U3BM471g0qzktq0aCxj7SNEEwefS0H9L4UyzAlURPJj5Eep2biNflkYDSQ7YIe9Z4u7nxTW40B1TLiSk09SETPddiP2OBusElVxNsIPaMIJJH6xG DdKwcNzj7NQe4jGiLgmdhI

Length of each river,

Breadth of each river, b = 75 m Depth of each river, h = 3 m Volume of each river = l × b × h 

= 1072000 × 75 × 3 m

= 2.412 × 108 m3

Volume of three such rivers = 3 × Volume of each river

= 3 × 2.412 × 108 m3

= 7.236 × 108 m3

Thus, the total rainfall is approximately same as the volume of the three rivers.

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 the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given figure).

Answer:8PSpNnaNQN5uUG4EIWaB

HMm3b2qszs4j3SdPcGt3c5OjVGrtYZ6fXfm8rYMQ47Tko2wRk7WyZnku FrccZdUxzbrczSoBn4z1u12jRSSScBjwtemx6cztcretey

Radius (r1) of upper circular end of frustum part

Radius (r2) of lower circular end of frustum part = Radius of circular end of cylindrical part

Height (h1) of frustum part = 22 − 10 = 12 cm Height (h2) of cylindrical part = 10 cm

Slant height (l) of frustum part

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

7FNRWSgtNztus7 d5yCXG9IFcVu3Giebh26yCptB8Dz7n mRmcb6eUMUDdbcHOMJDpq5WhGyNDYsQqWklLLMpeFZf2CkyQg16i pr3fBH4qekfg7qQdY0Hrj4v3 F

Question 6:

Derive the formula for the curved surface area and total surface area of the frustum of cone.

Answer:

k2fpDR2s1PdhXpnbV0fY3X0tTrksBG7xytdHl7plgikZdJQXXDOnSoQq9fxzB 4TQgdWF8Yr hAt3G5VerqnBnwyTeifTDpJSMR4MF BNztG O ce0vZ yaarOIgGvAh7CgJd0g

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r1 and r2 be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

In ΔABG and ΔADF, DF||BG

∴ ΔABG ∼ ΔADF

I TnULivxmnYGQda8wDvoSytZW49Rhi5COvr90JJLU7Bq6RGoE4V1DL158ptzABv8rb2FJU2C8

CSA of frustum DECB = CSA of cone ABC − CSA cone ADE

IZZwuZEVEHkC8AMIfFQ6TJMkg0u 3JSPDUAIkrCl7NJI2v0r zrKwFiP6ypPqpbOhjUl4Ql c2NmhvCIl3 X4Nxcvc0donBzvEPGtb BXPfWuLxRc Pd7yuofpdYHMGuv35Xy 0
dKEYy8dVgkq4 qPmZiB7BHukbLWfFR9eENcILlQXy9oDhSslgBlvd5GhW IArHJR33SoEddW1GFa72XGXMrPwJr5dcNWfRn1FCCeGCzN9Hjr4coSFgyy7oqjSO8tP nW8 JHijM

CSA of frustum
weYdZoZXLV2EKnKiNwpqT1v4gtgmSqAPPieyodfKjTCAmlr3sD65XJ REqQPxFIquBjN

Question 7:

Derive the formula for the volume of the frustum of a cone.

Answer:

EENB4wOqITYec9EUuNJEuGntBAqUBKeVPwrnWs9 9WNlxnRxQHQVD0fase19c4dSUeEFQoEk0wDNQX4jUsWzQzmfGWaoocFbvzT0 vfy4EI k1ETfmAqgNn3ImFZ1s0MPoSRpQE

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base.

Let r1 and r2 be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

In ΔABG and ΔADF, DF||BG

∴ ΔABG ∼ ΔADF

G70w3MLTjYEUCocy45YhNyeFJdTluoXsbvlUjibLJUVUE2Qgz QpJYBp rbtEQT0FTd6FAypiV2s4QQkLeDZYC5kRkNlvr3DylhSx rII 4vX2DoLIrUNN1Da9oMHzy8ncXC8FA

DnRWTKo78GqhPlgeUyyN1HY

Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE

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ytk FyFm94MsfP1uew7fe7qohq2 Xt00pQ2qUKJe r2aLR4Qsk e

Oz3lU3yf7ymXkzVl51JPm2xlsUTK1G2gnVA2B4YT4upkIRZPoF02Vne3vcnXk ZVw NgK2GVE1NL9gRZMKn rFfGnOlYPCvXH

Chapter 13, Surface Areas and Volumes, of class 10 Maths, is one of the most important chapters. The weightage of this chapter in the final exam is around 12 to 13 marks. On average, there are 4 questions asked from this chapter, based on surface areas and volumes. The distribution of marks with respect to questions are 3+3+3+4, where marks could vary depending upon the question. Topics covered in Chapter 13, Surface Areas and Volumes are;

  • The surface area of a combination of Solids
  • The volume of a combination of solids
  • Conversion of solid from one shape to another
  • Frustum of a cone

List of Exercises in class 10 Maths Chapter 13:

Exercise 13.1 Solutions 9 Question ( 7 long, 2 short)

Exercise 13.2 Solutions 8 Question ( 7 long, 1 short)

Exercise 13.3 Solutions 9 Question ( 9 long)

Exercise 13.4 Solutions 5 Question ( 5 long)

Exercise 13.5 Solutions 7 Question ( 7 long)

NCERT Solutions for Class 10 Maths Chapter 13- Surface Areas and Volumes. These NCERT solutions of chapter Surface Areas and Volumes are made available for the students who want to excel in the board exam. Let us discuss here, the importance of chapter in your academic and real life. You will face many real-life scenarios where the fundamentals of surface areas and volumes of real objects have to be configured. Some of the examples are rectangular boxes, gas cylinders, footballs, etc. These have 3D shapes which have both surface area and volume. And to calculate these quantities, we have to learn the formulas based on the dimensions of the object. Therefore, this solution of class 10, will help you learn about the formulas of surface areas and volumes of such 3D shapes such as Sphere, Cylinder, Cone, Cuboid, and combination of any two solids. With the help of these solved questions, you can not only solve the questions present in your academics but also real-life ones.

Key Features of NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes

  • The solutions are the great source for the students who want to get the unsolved questions present in each exercise of 10th Standard Maths book for the 13th chapter.
  • These works as reference material for the student.
  • It serves a great help for the revision of chapter 13, of all types of questions, asked in the exam.
  • Student’s can attain really good marks from surface area and volume chapter, with the help of solutions.
  • It is completely based on the syllabus prescribed by Central Board of Secondary Education(CBSE) guidelines.

Conclusion

The NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes are an essential resource for students looking to develop a strong understanding of calculating surface areas and volumes of different geometric shapes. The chapter covers various topics, such as the formulae for calculating the surface area and volume of objects such as cubes, cylinders, cones, and spheres.
Experienced teachers and subject matter experts have carefully designed these solutions to provide step-by-step explanations for all the NCERT textbook questions. The solutions also offer helpful tips and tricks to help students solve problems quickly and accurately.
Using these solutions, students can strengthen their understanding of the subject, improve their problem-solving skills, and prepare effectively for the Class 10 board exams. These solutions are an excellent resource for students who wish to excel in mathematics and build a strong foundation for higher studies.
Overall, the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes are indispensable for students looking to score well in mathematics and enhance their analytical and mathematical abilities.

How can we score full marks in Class test of NCERT Solutions for Class 10 Maths Chapter 13?

By using the NCERT Solutions for Class 10 Maths Chapter 13 provided in SWC’s website makes you reach full marks in class tests as well as in board exams. These solutions are highly imperative for easy and quick revision during the class tests and exams. Meanwhile, this serves as the best study material for learners.

How NCERT Solutions for Class 10 Maths Chapter 13 helpful in board exams?

NCERT Solutions for Class 10 Maths Chapter 13 implements answers with detailed descriptions as per term limit specified by the SWC for self-evaluation. Practising these questions will ensure that students have a good preparation for all sorts of questions that can be devised in the finals.

What are the main concepts covered in NCERT Solutions for Class 10 Maths Chapter 13?

The main topics of NCERT Solutions for Class 10 Maths Chapter 13 are the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another and frustum of a cone.


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