NCERT Solutions of class 10 maths Exercise 13.3-Surface areas and Volumes
NCERT Solutions of class 10 maths Exercise 13.3 of the chapter 13-Surface areas and Volumes-Here NCERT Solutions of the exercise 13.3 are the Solutions of the chapter ‘Surface areas and Volumes’ of the NCERT maths coursebook of class 10 endorsed by CBSE a reputed Board of India for school education.All questions of exercise 13.3 are solved by an expert of maths in a step-by-step way, so everyone would comprehend the profundity of the idea of maths.
There are a variety of questions in chapter- 13 based on surface areas and volumes of compound figures made of cubes, cuboids, cylinders, cones, and spheres. In exercise 13.3, the questions based on evaluation of the volumes, height, and radius of solids when one solid is melted and recast into another solid. In this exercise for the evaluation of physical quantities like height, radius, surface areas, and volume, you need to keep in mind the following points.
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
- In the case when one metallic object is melted and recast into another type of article then the volume of both will be the same.
- In the case when the earth is taken out from a well to be dug out and something is made by the earth taken out from the well then the volume of the well is equivalent to the thing to be made.
- In a question, if you are given the speed of the water (distance/unit time)then the volume of water streamed in per unit time =depth ×height ×speed.
NCERT Solutions Class 10 Science from chapter 1 to 16
The questions of exercise 13.3 are very important for your board examination of 2020-21. We likewise have highlighted the questions which are important and most important, so that you could pay attention to those questions.
NCERT Solutions of class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
NCERT Solutions of class 10 maths Exercise 13.3 -Surface areas and Volumes
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
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.(Imp)
Ans.We are given the radius of the sphere = 4.2 cm
Radius of the cylinder = 6 cm
Let the height of cylinder = h
The volume of sphere = Volume of cylinder
Volume of sphere = and Volume of cylinder is =
h = 2.74
Hence the height of the cylinder = 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.(Imp)
Ans. We are given the radii of three metalic spheres 6 cm, 8 cm and 10 cm respectively
Sum of the volumes of metalic spheres = Volume of a single solid sphere
The volume of sphere =
Let the radius of the single solid sphere = r
r ³ = 6³ + 8³ + 10³
r³ = 216 + 512+1000
r³ = 1728
Hence the radius of single solid sphere is 12 cm
Q3.A 20 m deep well with a diameter of 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.(Very Imp)
Ans. We are given the depth of the well, h = 20 m, radius, r of the well = 7/2= 3.5 cm
Let the height of the platform made by the earth taken out from the well =H
The shape of well is the shape of a cylinder
The volume of well = The volume of the earth taken out from the well= Volume of the platform = πr²h =22/7 ×3.5²×20 =22×0.5×3.5×20 = 770 m³
The volume of plateform = Length ×Breadth × Height = 22 × 14 × H
Therefore
22 ×14 H = 770
H = 2.5 m
Hence the height of the plateform = 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.(Most Imp)
Ans.Let the height of the embankment= H
Radius of well = 3/2 = 1.5 m, height, h =14 m
The volume of well = The volume of the earth taken out from the well= Volume of the embankment
Volume of well =πr²h =π(1.5)²14
Outer radius of embankment =1.5 +4 = 5.5 m
Inner radius of embankment = 1.5 m
Volume of embankment =π(5.5)²H – π(1.5)²H=πH(5.5²- 1.5²)=πH(30.25 -2.25)=28πH
28πH = π(1.5)²14
28H =2.25×14
H = 2.25/2 =1.125
Hence the height of embankment is 1.125 m
Q5. A container shaped like 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 hemisperical shape on the top .Find the number of such cones which can be filled with ice-cream.(Most Imp)
Ans.
We are given diameter of right circular cylinder= 12 cm,so radius,r = 12/2 = 6 cm
Height of the cylinder ,h =15 cm
Volume of cylinder = πr²h = π(6)²15 =540π cm³
The volume of ice-cream = Volume of cone + Volume of hemisphere
The height,h of the cone =12 cm and radius of cone and hemisphere,r= 6/2= 3 cm
Volume of cone =
= 3π ×12 = 36π
Volume of hemisphere =
= 2π× 9 = 18π
∴ The volume of ice-cream = 36π + 18π =54 π cm³
Number of ice-cream cones to be filled
Hence the number of cones to be filled are 10
Q6.How many silver coins, 1.75 cm in diameter and of thickness 2 mm,must be meted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm ?(Imp)
Ans.Let number of coins to be formed = n
Radius of coins,r = 1.75/2 = 0.875 cm and thickness,h=2 mm = 0.2 cm
Volume of coin = πr²h
=4.4 × 0.125×0.875=0.48125 cm³
Volume of cuboid = Length× Breadth ×Height =5.5 cm × 10 cm × 3.5 cm =192.5 cm³
Number of coins to be melted
= 400
Hence the number of coins to be melted are 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 conical heap is 24 cm,find the radius and slant height of the heap. (Very Imp)
Ans. The volume of the bucket = Volume of conical heap
We are given the height of the cylindrical bucket,h = 32 cm and radius of base, = 18 cm
Volume of cylndrical bucket = πr²h = π(18)²32
The height of conical heap =24 cm
Volume of conical heap =
The volume of the bucket = Volume of conical heap
r² = (18 ×18× 4)
r²=9 ×2×9×2×4
r = 9×2×2= 36 cm
Hence radius of the conical heap = 36 cm
Slant height of the conical heap,l =√(r² +h²) = √(36² + 24²)=√(1296 +576)=√(1872)=12√13 cm
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?(Most Imp)
Ans.Let area to be irrigated A in time 30 minutes=30 ×60=1800 seconds
Standing water in the area= 8 cm= o.08 m
We are given width of canal is =6 m and depth=1.5 m
The speed of water =10 km/h = (10×1000)/(60×60) =25/9 m/s
Volume of water flowed in the area in per second =Width ×Depth ×Speed of water=6×1.5×25/9=2×0.5×25=25 m³/s
In 30 minutes (1800 s) the water flowed in the area =25 ×1800=45000 m³
The volume of standing water in 30 minutes = A×h =0.08 A
0.08 A = 45000
A = 45000/0.08= 562500 m²
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?(Most Imp)
Ans. We are given the diameter of the pipe = 20 cm = 0.20 m, so its radius =.20/2=0.10 m
The speed of water in the pipe = 3 km/h = 3000/(60×60) =3000/3600=(5/6) m/s
The volume of water flowed per second through the pipe=πr²h=π(0.10)²×5/6=0.05π/6 m³/s
The diameter of cylindrical tank = 10 m,so its radius =10/2= 5m
Depth(h) of the tank = 2 m
Volume of the tank =πr²h =π(5²)×2= 50π m³
Time taken by the pipe in filling the tank =(Volume of tank)/(Volume of water flowed per second inside the pipe) = 50π /(0.05π/6) =300/0.05= 6000 seconds=100 minutes= 1h 40 minutes
Science and Maths NCERT Solutions for Class 9,10,11 and 12
NCERT Solutions of class 9 maths
Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |
Chapter 2-Polynomial | Chapter 10-Circles |
Chapter 3- Coordinate Geometry | Chapter 11-Construction |
Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |
Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |
Chapter 6-Lines and Angles | Chapter 14-Statistics |
Chapter 7-Triangles | Chapter 15-Probability |
Chapter 8- Quadrilateral |
NCERT Solutions of class 9 science
CBSE Class 9-Question paper of science 2020 with solutions
CBSE Class 9-Sample paper of science
CBSE Class 9-Unsolved question paper of science 2019
NCERT Solutions of class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
NCERT solutions of class 11 maths
Chapter 1-Sets | Chapter 9-Sequences and Series |
Chapter 2- Relations and functions | Chapter 10- Straight Lines |
Chapter 3- Trigonometry | Chapter 11-Conic Sections |
Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
Chapter 8- Binomial Theorem | Chapter 16- Probability |
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
NCERT solutions of class 12 maths
Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
Chapter 7- Integrals | |
Chapter 8-Application of Integrals |
50 important science questions for cbse 10 class
Chemical properties of acid and base
Chemistry viva voce questions & answers
Chemistry practical based questions
Solutions of previous years science & maths question papers
Solutions of class 10 Science question paper 2019 CBSE
Solutions of class 10 CBSE maths question paper 2020
CBSE Class 10 science question paper 2020 SET -3 solutions
Download e-book of Science NCERT 634 questions with solutions for Polytechnic and NDA entrance exams