NCERT Solutions Class 10 Maths Chapter 13
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes
Class 10 is a turning point in the life of every student. Whether a student wants to pursue subjects of his choice or attend the college of his dreams, a lot of it depends on their performance in the Class 10 board exam. Chapter 13 is one of those chapters that often confuses students, and they end up losing marks in the examination. Practising questions related to Chapter 13 regularly can help students have a better understanding of the chapter. Referring to NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks will enable students to solve practice questions given in their NCERT books with precision, and master the topic. Solutions are there not just for Mathematics, but for all other subjects as well so that students don’t have to look elsewhere for any assistance.
NCERT Solutions Class 10 Mathematics Chapter 13 are curated by subject matter experts. These have answers to every question given at the end of NCERT Class 10 Chapter 13 textbook.
When it comes to solving questions from NCERT Class 10 Mathematics Chapter 13, students are expected to understand theoretical concepts like what surface area & volume is, along with other practical elements like calculating them. Extramarks’ NCERT Solutions for Class 10 Mathematics Chapter 13 are as per the latest syllabus of Class 10 CBSE pattern. . Thus, making it an ideal solution for their CBSE Class 10 Mathematics board exam preparations.
NCERT Solutions for Class 10 Maths
Mathematics is one of those subjects in Class 10 that helps to raise the overall percentage of students. To increase the chances of scoring full marks in Class 10 Mathematics, students need not only practise but do so with precision. A good place to begin for Class 10 Mathematics would be with NCERT Solutions by Extramarks. Every chapter-wise solution comes with solved answers to textbook questions and in-depth explanations of the same. All the chapters are listed below: :
- Chapter 1 – Real Numbers
- Chapter 2 – Polynomials
- Chapter 3 – Pair of Linear Equations in Two Variables
- Chapter 4 – Quadratic Equations
- Chapter 5 – Arithmetic Progressions
- Chapter 6 – Triangles
- Chapter 7 – Coordinate Geometry
- Chapter 8 – Introduction to Trigonometry
- Chapter 9 – Some Applications of Trigonometry
- Chapter 10 – Circles
- Chapter 11 – Constructions
- Chapter 12 – Areas Related to Circles
- Chapter 13 – Surface Areas and Volumes
- Chapter 14 – Statistics
- Chapter 15 – Probability
Chapter 13 – Surface Area and Volumes
Chapter 13 – Surface Area and Volumes of Class 10 Mathematics is a lesson loaded with formulas. It teaches the students how to calculate areas and volumes of different shapes. The chapter comprises 5 exercises 13.1 – 13.5. NCERT Solutions for Class 10 Mathematics Chapter 13 has solved answers to every question in exercise 13.1 to 13.5.
Just like any other introduction from any other chapter, in Chapter 13 of Class 10 Mathematics, you would be required to recall what you studied in Class 9 about cubes, cylinders, circles, etc. This is why at Extramarks we never recommend memorising anything. We always suggest understanding a particular concept. The sole reason is that, by memorising things,you are not able to retain them in your mind for a long time. But if you understand a concept, chances are you will remember it forever. That’s the cardinal rule in Mathematics.
13.2 Surface Area of a Combination of Solids
Some items are a combination of two or more solid shapes. At the end of this chapter, students will also learn how to calculate the surface area and volumes of a combination of solids using formulas.
13.3 Volume of a Combination of Solids
For calculating the volume of shapes that are a combination of two or more solid shapes, students must find out the volume area of individual shapes first.
13.4 Conversion of a Solid From One Form to Another
|Total surface area of sphere = curved surface area of sphere||4 π r2|
|Total surface area of cone||πr(r+l)|
|Curved surface area of cone||πrl|
|Total surface area of cuboid||2(lb+bh+hl)|
|Total surface area of cylinder||2 πr(h+r)|
Volumes of 3D objects:
|Volume of sphere||4/3 πr³|
|Volume of hemisphere||2πr3/3|
|Volume of cone||(⅓)πr2h|
|Volume of cube||s3|
|Volume of cuboid||lbh|
|Volume of cylinder||πr2h|
Even if a solid is converted from one form to another, then the volume remains the same. Let us understand this with an example:
If there is water in the cuboid shape (which has dimensions of 20 * 22) which is transferred into a cylinder that has a height of 3.5 m and 2 m then what is the height of the water level in the cuboid shape if the water (once transferred into the cylinder) fills the cylinder to the brim?
Solution – From the theorem on volumes, we know that the volume of the water in the cylinder and the volume of the water in the cuboid would be the same. If the water fills the cylinder to the brim, then:
Volume of water in cylinder = π * r² * h = 22/7 * 1 * 3.5 = 11.
Volume of water in cuboid = l * b * h = 20 * 22 * h.
Since 20 * 22 * h = 3.5 * π, we get h = 11/(20 * 22) = 2.5 cm.
Q1. 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: It is known that the coins are cylindrical in shape.
So, height (h1) of the cylinder = 2 mm = 0.2 cm
Radius (r) of circular end of coins = 1.75/2 = 0.875 cm
Now, the number of coins to be melted to form the required cuboids be “n”
So, Volume of n coins = Volume of cuboids
n × π × r2 × h1 = l × b × h
n×π×(0.875)2×0.2 = 5.5×10×3.5
Or, n = 400
Q2. 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: Radius (r1) of the upper base = 4/2 = 2 cm
Radius (r2) of lower the base = 2/2 = 1 cm
Height = 14 cm
Now, Capacity of glass = Volume of frustum of cone
So, Capacity of glass = (⅓)×π×h(r12+r22+r1r2)
= (⅓)×π×(14)(22+12+ (2)(1))
∴ The capacity of the glass = 102×(⅔) cm3
Key Features of NCERT Solutions for Class 9 Maths Chapter 13
- NCERT Solutions for Class 10 Mathematics Chapter 13 is a study material to help Class 10 students solve the exercise questions given in NCERT Chapter 13 with accuracy.
- NCERT Solutions Class 10 Mathematics Chapter 13 is prepared by experienced faculty as per the latest CBSE Class 10 syllabus.
- A student doesn’t require any aid from teachers or parents to understand the answers given NCERT Solutions Class 10 Mathematics Chapter 13. The language used in it is uncomplicated and designed in a way that everything becomes self-explanatory.
- All numerical problems are solved step-wise with appropriate diagrams and explanations as and when required.
Q.1 Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Q.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.
Q.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.
Q.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.
Q.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.
Q.6 A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the following figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Q.7 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.
Q.8 A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the following 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.
Q.9 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.
Q.10 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.)
Q.11 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 following figure)
Q.12 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 (see the following figure).
Q.13 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.
Q.14 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. (Use π = 3.14)
Q.15 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.
Q.16 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.
Q.17 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.
Q.18 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.
Q.19 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.
Q.20 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.
Q.21 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.
Q.22 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?
Q.23 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.
Q.24 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?
Q.25 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?
Q.26 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.
Q.27 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.
Q.28 A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the following figure). 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.
Q.30 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.
Q.31 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.
Q.32 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.)
Q.33 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?
Q.34 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.
Q.35 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 following figure).
Q.36 Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Q.37 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Q.38 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 â‚¹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)
FAQs (Frequently Asked Questions)
1. What are the topics covered in NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks?
The topics covered in NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks include:
– Surface area of a combination of solids
– Volume of a combination of solids
– Conversion of solid from one shape to another
– Frustum of a cone
2. From where can I access NCERT Solutions Class 10 Mathematics Chapter 13?
You can access NCERT Solutions Class 10 Mathematics Chapter 13 from the Extramarks website or app.