NCERT Solutions for class 10 Maths Chapter 13

NCERT Solutions for class 10 Maths Chapter 13 Surface areas and volumes all exercises (प्रश्नावली 13.1 to प्रश्नावली 13.5) for UP Board 2020-21 and CBSE Board as well in Hindi and English medium free PDF form to download.

Download NCERT Solutions Offline Apps 2020-2021 based on updated NCERT Solutions and latest NCERT Books 2020-21. Previous year CBSE Exam question, important questions for practice are also given below.

NCERT Solutions for class 10 Maths Chapter 13

Class:	10
Subject:	Maths – गणित
Chapter 13:	Surface areas and volumes

10th Maths Chapter 13 Solutions

CBSE NCERT Solutions for class 10 Maths Chapter 13 Surface areas and volumes exercises 13.1 to 13.5 are given below to free download in PDF or view online without downloading. All the solutions are updated for new academic session 2020-21 for UP Board, MP Board, Gujrat, CBSE board, etc., who are using latest NCERT Books for course.

Previous Years Questions

ONE MARK QUESTIONS
1. Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere? [CBSE 2017]
THREE MARKS QUESTIONS
1. A metallic solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. How many cones will be made? [CBSE 2017]
FOUR MARKS QUESTIONS
1. In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If the tank is filled completely then what will be the height of standing water used for irrigation the park. Write your views on recycling of water. [CBSE 2017]
2. The height of a cone is 30 cm. From its topside a small cone is cut by a plane parallel to its base. If volume of smaller cone is 1/27 of the given cone, then at what height it is cut from its base? [CBSE 2017]
3. The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts. [CBSE 2017]

Objective of Surface areas and volumes

1. To identify situations where there is a need of finding surface area and where there is a need of finding volume of a solid figure and to find the surface areas & volumes of cuboids, cubes, cylinders, cones spheres and hemispheres, using their respective formulae. To solve some problems related to daily life situations involving surface areas and volumes of above solid figures.
2. In this chapter, we will deal with problems such as finding the area of sheet to covers a solid body, area of an object which is combination of two objects, finding the number of one object, required for creating another different object, finding the cost of ploughing a given field at a given rate, finding the cost of constructing a water tank with a particular capacity, finding area & volume of frustum (Frustum is a Latin word meaning ‘piece cut off’ and its plural form is ‘Frusta’) and the conversion of one solid into another solid on the basis of volume and so on.
For solving above type of problems, we need to find the perimeters and areas of simple closed plane figures (figure which lie in a plane) and surface areas and volumes of solid figures (figures which do not lie wholly in a plane). You are already familiar with the concepts of perimeters, areas, surface areas and volumes. In this chapter, we will study these in details.

Historical Facts!

Archimedes of Syracuse, Sicily, is remembered as the greatest Greek mathematician of the ancient era. He contributed significantly in geometry regarding the areas of plane figures and the areas and volumes of solid figures. He proved that the volume of a sphere is equal to two-third the volume of a circumscribed cylinder.

Important Questions on Class 10 Maths Chapter 13

व्यास 7 m वाला 20 m गहरा एक कुआँ खोदा जाता है और खोदने से निकली हुई मिट्टी को समान रूप से फैलाकर 22 m × 14 m वाला एक चबूतरा बनाया गया है। इस चबूतरे की ऊँचाई ज्ञात कीजिए।

दिया है,
कुँए की त्रिज्या (r) = 7/2 m
कुँए की गहराई = 20 m
चबूतरे की लंबाई = 22 m
चबूतरे की चौड़ाई = 14 m
माना चबूतरे की ऊँचाई = H
प्रश्नानुसार, कुँए से निकली हुई मिट्टी का आयतन = चबूतरे की मिट्टी का आयतन
πr^2 h = 22 × 14 × H
⇒ 22/7 × (7/2)^2 × 20 = 22 × 14 × H
⇒ 11 × 7 × 10 = 22 × 14 × H
⇒ (11 × 7 × 10)/(22 × 14) = H
⇒ H = 2.5 m
अतः, चबूतरे की ऊँचाई 2.5 m है।

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.

Radius of metallic sphere (r1) = 4.2 cm,
Radius of cylinder (r2) = 4.2 cm
Let, the height of cylinder = h
According to question, volume of sphere = volume of cylinder
⇒ 4/3 πr1^3 = πr2^2 h
⇒4/3 π(4.2)^3 = π(6)^2 h
⇒ h = 4/3 × (4.2 × 4.2 × 4.2)/36
= 1.4 × 1.4 × 1.4 = 2.74 cm
Hence, the height of cylinder is 2.74 cm.

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?

Width of canal = 6 m,
Depth of canal = 1.5 m
Speed of water = 10 km/h = 1000/60 m/min
Volume of water in 1 minute = 6 × 1.5 ×1000/60 = 1500 m^3
Therefore, the volume of water in 30 minutes
= 30 × 1500 = 45000 m^3
Let the area of irrigated field = A m^2
Therefore,
Volume of water from canal in 30 minutes = volume of water in the field
⇒ 45000 = (A × 8)/100
⇒ A = (45000 × 100)/8 = 562500 m^2
Hence, canal irrigate 562500 m^2 area in 30 minutes.

दो घनों, जिसमें से प्रत्येक का आयतन 64 cm^3 है, के संलग्न फलकों को मिलाकर एक ठोस बनाया जाता है। इससे प्राप्त घनाभ का पृष्ठीय क्षेत्रफल ज्ञात कीजिए।

दिया है,
घन का आयतन = 64 cm^3
(भुजा) 3 = 64 cm^3
भुजा = 4 cm
यदि घनों के संलग्न फलकों को मिलाया जाता है तो प्राप्त घनाभ की भुजाएँ 4 cm, 4 cm, 8 cm होंगी।
घनाभ का पृष्ठीय क्षेत्रफल
= 2(lb + bh + hl)
= 2(4 × 4 + 4 × 8 + 4 × 8) cm^2
= 2(16 + 32 + 32) cm^2
= 2(80) cm^2
= 160 cm^2
अतः, इसप्रकार इससे प्राप्त घनाभ का पृष्ठीय क्षेत्रफल 160 cm^2 है।

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.

Height of conical part (h) = radius of conical part (r) = 1 cm
Radius of conical part (r) = radius of hemispherical part (r) = 1 cm
Volume of Solid = Volume of conical part + Volume of hemispherical part
= 1/3 πr^2 h + 2/3 πr^3
=1/3 π〖.1〗^2.1 + 2/3 π〖.1〗^3 = π〖cm〗^3