NCERT Solutions for Exercise 13.3 Class 10 Maths Chapter 13-Surface Area and Volumes

NCERT Solutions for Exercise 13.3 Class 10 Maths Chapter 13 - Surface Area and Volumes

Edited By Ramraj Saini | Updated on Nov 27, 2023 05:43 PM IST

NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3

NCERT Solutions for Exercise 13.3 Class 10 Maths Chapter 13 Surface Area and Volumes are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 Maths ex 13.3 concentrate on solids conversion. For example, a cylinder-shaped candle can be melted and poured into a cubical container. The candle has been reshaped into a different shape. The volume of the candle remains constant, which is interesting to note. Students can determine that an object's volume remains constant even when it is converted from one solid shape to another.

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NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3
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Surface Areas and Volumes Class 10 Maths Chapter 13 Exercise: 13.3
Q9 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? Answer: Area of the cross-section of pipe is Speed of water is given to be = 3 km/hr Thus, the volume of water flowing through a pipe in 1 min. is Now let us assume that the tank will be completely filled after t minutes. Then we write : Hence the time required for filling the tank completely in 100 minutes.
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Benefits of NCERT Solutions for Exercise 13.3 Class 10 Maths Chapter 13
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NCERT Solutions for Exercise 13.3 Class 10 Maths Chapter 13 - Surface Area and Volumes

With the help of this exercise students understand that when Solids are remoulded and melted to change their shape, but their volume remains the same. With this understanding, students will be able to create equations. If students are asked to find the height of a shape in a question, they can simply equate the two equations where the volume is the same. They will be able to locate the missing dimension in this manner. 10th class Maths exercise 13.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Surface Areas and Volumes Class 10 Maths Chapter 13 Exercise: 13.3

Q1 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:

Let us assume the height of the cylinder to be h.

Since the material is melted and recast thus its volume will remain the same.

So, Volume of sphere = Volume of obtained cylinder.

$\frac{4}{3}\pi r^3_s\ =\ \pi r^2_c h$

$\frac{4}{3}\pi \times 4.2^3\ =\ \pi \times 6^2 \times h$

$h\ =\ \frac{4}{3}\times \frac{4.2\times4.2\times 4.2}{36}$

$h\ =\ 2.74\ cm$

Hence the height of the cylinder is 2.74 cm.

Q2 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:

According to the question, small spheres are melted and converted into a bigger sphere. Thus the sum of their volume is equal to the volume of the bigger sphere.

The volume of 3 small spheres = Volume of bigger sphere

Let us assume the radius of the bigger sphere is r.

$\frac{4}{3}\pi \left ( r^3_1\ +\ r^3_2\ +\ r^3_3 \right )\ =\ \frac{4}{3}\pi r^3$

$r^3_1\ +\ r^3_2\ +\ r^3_3 \ =\ r^3$

$r^3\ =\ 6^3\ +\ 8^3\ +\ 10^3$

$r\ =\ 12\ cm$

Hence the radius of the sphere obtained is 12 cm.

Q3 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.

Answer:

According to the question, the volume of soil dug will be equal to the volume of the platform created.

Thus we can write :

The volume of soil dug = Volume of platform

$\\\Rightarrow \pi r^2h\ =\ Area\times h\\\\\Rightarrow \pi \times \left ( \frac{7}{2} \right )^2\times 20\ =\ 22\times 14 \times h\\\\\Rightarrow h\ =\ \frac{5}{2}\ m$

Thus the height of the platform created is 2.5 m.

Q4 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:

According to the question, the volume is conserved here :

The volume of soil dug out = Volume of the embankment made.

Let the height of the embankment is h.

$\\\pi r^2_1h'\ =\ \pi\left ( r^2_2\ -\ r^2_1 \right )h\\\\\pi \left ( \frac{3}{2} \right )^2\times 14\ =\ \pi\left ( \left ( \frac{11}{2} \right )^2\ -\ \left ( \frac{3}{2} \right )^2 \right )h\\\\h\ =\ \frac{9}{8}\ m$

Hence the height of the embankment made is 1.125 m.

Q5 A container shaped like a right circular cylinder having a diameter of 12 cm and a 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:

Let the number of cones that can be filled with ice cream be n.

Then we can write :

The volume of a cylinder containing ice cream = n ( volume of 1 ice cream cone )

$\\\pi r^2_{cy}h_{cy}\ =\ n\left ( \frac{1}{3}\pi r^2_{co}h_{co}\ +\ \frac{2}{3} \pi r^3 \right )\\\\\pi\times 6^2\times 15\ =\ n\left ( \frac{1}{3}\times \pi \times 3^2 \times 12\ +\ \frac{2}{3} \pi \times 3^3 \right )\\\\n\ =\ \frac{36\times 15}{54}\\\\n=\ 10$

Hence the number of cones that can be filled is 10.

Q6 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?

Answer:

Let us assume the number of coins that need to be melted be n.

Then we can write :

The volume of n coins = Volume of cuboid formed.

$n\left ( \pi r^2 h \right )\ =\ lbh'$

$n\left ( \pi \times \left ( \frac{1.75}{2} \right )^2\times 0.2 \right )\ =\ 5.5\times 10\times 3.5$

$n\ =\ 400$

Thus the required number of coins is 400.

Q7 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:

According to question volume will remain constant thus we can write :

The volume of bucket = Volume of heap formed.

$\pi r^2_1h_1\ =\ \frac{1}{3}\pi r^2_2 h_2$

Let the radius of heap be r.

$\pi\times 18^2 \times 32\ =\ \frac{1}{3}\times \pi \times r^2\times 24$

$r\ =\ 18\times 2\ =\ 36\ cm$

And thus the slant height will be

$l\ =\ \sqrt{r^2\ +\ h^2}$

$=\ \sqrt{36^2\ +\ 24^2}$

$=\ 12\sqrt{13}\ cm$

Hence the radius of heap made is 36 cm and its slant height is .

Q8 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?

Answer:

Speed of water is: 10 Km/hr

And the volume of water flow in 1 minute is :

$=\ 9\times \frac{10000}{60}\ =\ 1500\ m^3$

Thus the volume of water flow in 30 minutes will be : $=\ 1500\ \times 30\ =\ 45000\ m^3$

Let us assume irrigated area be A. Now we can equation the expression of volumes as the volume will remain the same.

$45000\ =\ \frac{A\times 8}{100}$

$A\ =\ 562500\ m^2$

Thus the irrigated area is $562500 \:m^2$ .

Q9 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?
Answer:
Area of the cross-section of pipe is $=\ \pi r^2$
$=\ \pi (0.1)^2\ =\ 0.01 \pi\ m^2$
Speed of water is given to be = 3 km/hr
Thus, the volume of water flowing through a pipe in 1 min. is
$=\ \frac{3000}{60}\times 0.01 \pi$
$=\ 0.5 \pi\ m^3$
Now let us assume that the tank will be completely filled after t minutes.
Then we write :
$t\times 0.5 \pi\ =\ \pi r^2h$
$t\times 0.5 \ =\ 5^2\times 2$
$t\ =\ 100$
Hence the time required for filling the tank completely in 100 minutes.

Benefits of NCERT Solutions for Exercise 13.3 Class 10 Maths Chapter 13

Exercise 13.3 Class 10 Maths is based on Conversion of Solid from One Shape to Another teaching us about shape conversion due to melting and recasting.
Class 10 Maths chapter 13 exercise 13.3 also covers the important concepts of exercise 13.2.
Class 10 Maths chapter 13 exercise 13.3 also contains concepts related to time value for objects to be filled or empty.

Key Features For Class 10 Maths Chapter 13 Exercise 13.3

Step-by-Step Solutions: Clear and detailed explanations for each problem in Ex 13.3 class 10, ensuring students understand the methodology.
Conceptual Understanding: Explanation of surface area and volume concepts involved in the exercises, aiding students in grasping the underlying principles are discussed in class 10 maths ex 13.3.
Variety of Problems: In class 10 ex 13.3, diverse types of problems, offering a broad spectrum of challenges to reinforce understanding.
Visual Aids, Diagrams, and Examples: Use of visual representations, diagrams, and real-life examples to simplify complex concepts and aid visual learners.
Application-Based Problems: In 10th class maths exercise 13.3 answers, application-oriented problems that relate statistics to real-world scenarios, fostering practical understanding and problem-solving skills.

Also, See:

NCERT Solutions for Class 10 Maths Chapter 13- Surface Area and Volume