NCERT Maths Specific Questions Of Mensuration for 9 and 10 grade
Mensuration is the part of mathematics that was born in the lap of geometry in which we study physical quantity like area, volume, perimeter and another related quantity of all kinds of geometrical figures. The questions of mensuration obligatorily are a piece of the maths study material of each class and the questions of mensuration are likewise asked in each competitive exam additionally so this post of Future Study Point is helpful for everyone. Mensuration is ordered into two sections (i) Two-dimensional geometrical figure (ii) Three-dimensional geometrical figure, here we will study specific questions of mensuration for which students commonly required assistance from his mentor or educator. These are the specific questions of mensuration that are posed in the CBSE board exam and in the competitive exams. We trust you will get support from the way of solutions for illuminating the questions. Questions are clarified through a step-by-step way by a specialist in maths so that everyone could comprehend the way of solution.
This post consists of a few examples with the solutions and some of the unsolved questions which are taken from the NCERT and different maths CBSE guides.
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
NCERT Maths Specific Questions Of Mensuration for 9 and 10 grade
Example(1)-Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Ans.
We have to determine total cost of cardboard required to supply 250 boxes of each type
The cardboard required to prepare bigger box = TSA of box + 5% of TSA of the box
TSA of box = 2(lb + bh + hl) = 2(25×20 + 20 × 5 + 5×25)
⇒2×(500 + 100 + 125) = 2 ×725 =14 50 sq.cm
Additional cardboard is required to buid the box = 5 % of 1450 =5/100 ×(1450)=72.5
Cardboard required to build bigger box = 1450 + 72.5 =1522 .5 sq.cm
Cardboard required for 250 bigger boxes = 250× 1522.5 = 380625
Therefore 380625 sq.cm cardboard is required to prepare 250 bigger boxes
TSA of small box of the dimension = 2(15×12 + 12×5 + 15× 5) = 2 ×(180 + 60 +75) = 2×315= 630 sq.cm
Additional cardboard is required to build smaller box = 5% of 630 = 5/100 ×(630) = 31.5 sq.cm
Cardboard required to build this box = 630 + 31.5 = 661.5 sq.cm
Cardboard required for 250 smaller boxes = 250× 661.5 = 165375
Therefore 165375 sq.cm cardboard is required to prepare 250 boxes
Cardboard required to prepare 250 boxes of both kinds = 380625 + 165375 = 546000 sq.cm
The cost of 1000 sq.cm cardboard is = Rs 4
Cost of 546000 sq.cm cardboard will be
= (4/1000)×546000 = 2184
Therefore the cost of cardboard required to prepare 250 boxes of each type is Rs 2184
Example 2- A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm andlength is 1 m.
Answer.
The length of road roller(h) = 1 m
The diameter of road roller = 84 cm
Its radius = 42 cm
The area occupied by it on the road in one revolution
= The curved surface area of the road
= 2πrh
= 2×(22/7) ×0.42×1 = 2.64
Therefore area occupied by it on the road in one revolution =2.64 sq.m
In 750 revolution the whole of the road is covered
In 750 revolution the roller will cover the area
=2.64 × 750=1980
Therefore the area of road = 1980 sq.m
Example 3. A cubical block of the side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of solid.
Ans. The greatest diameter of the hemisphere surmounted on one of the face of a cube is the side of the cube
The side of the cube = 7 cm
The surface area of solid =TSA of cube + CSA of hemisphere –area of the circular base of the hemisphere
⇒ 6a² + 2πr² – πr²= 6a² + πr²
a = 7 cm, r = 7/2 = 3.5 cm
⇒ 6 × a² + π ×r²
= 6×7²+(22/7)×3.5²
=6×49 +22×0.5×3.5
= 294 + 38.5
= 332.5
The surface area of solid = 332.5 sq.cm
Q4.A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends(see fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Ans. The length of the entire capsule = 14 mm
The diameter of the capsule = 5 mm
Radius of the capsule = 2.5 mm
The length of the cylindrical part of the capsule = 14 –(2.5 + 2.5) = 9 mm
The surface area of capsule = The curved surface area of its cylindrical part + curved surface area of its two hemispherical ends
⇒ 2πrh + 2×2πr²
=2πr(h +2r)
=2(22/7)× 2.5 (9+2×2.5)
=220
Therefore 220 sq.mm is the entire surface area of the capsule
Q5.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 a length 5 cm and a diameter of 2.8 cm.
Ans.
The length of gulab jamun = 5 cm
The length of its cylindrical part (h) = 5 –(1.4 + 1.4) =5 –2.8=2.2
Therefore the length(h) of cylindrical part is 2.2 cm
Diameter of gulam jamun = 2.8 cm
Radius of gulab jamun = 2.8/2 = 1.4 cm
Volume of gulab jamun(V) = Volume of cylinderical part + Volume of two hemispherical part
V = πr²h +2×(2/3)πr³
=(22/7)×1.4×1.4×2.2+(4/3)×(22/7)×1.4×1.4×1.4
=13.552+(34.496/3) + (4/3)×(22/7)×1.4×1.4×1.4
==13.552+(34.496/3)
=(40.656+34.496)/3 =75.152/3
The volume of 45 gulab jamun = 45×(75.152/3) =1127.28
45 gulab jamun contain sugar syrup =30 % of 1127.28
= .30 ×1127.28 = 338.184 ≈ 338
Therefore 338 cubic.cm sugar syrup contained in 45 gulab jamun.
Q6.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 figure)
Ans. The volume of wood in the pen stand = Volume of the wooden cuboid – Volume of 4 depressions
The volume of the wooden cuboid = l × b × h
l = 15 cm, b = 10 cm, h = 3.5 cm
⇒The volume of the wooden cuboid = 15 × 10 × 3.5 = 525 cubic.cm
The volume of 4 conical depression of each of radius(r) 0.5 cm and height(h) 1.4 cm
Since volume of the cone = (1/3)πr²h
The volume of 4 conical depression = 4×(1/3)×(22/7)×0.5×0.5×1.4
= 1.47
∴ The volume of wood in the pen stand = 525 – 1.47 =523.53
Therefore 523.53 cubic.cm is the volume of wood in the pen stand
Solve the following important questions generally asked in the CBSE board exams and competitive entrance exams.
Q1. The water is flowing within a circular pipe of diameter 1.4 cm at the speed of 1.8 km/h, how much time it will take to fill a cuboidal tank of 30000 liters.
Q2. A solid metallic sphere of the diameter of 16 cm is melted and recast into spherical bullets, how many bullets can be formed of the diameter 4 mm.
Q3. A solid metallic cylinder of the diameter 40 cm and height 50 cm is melted and recast into a wire of 4 mm diameter, find the length of wire so formed.
Q4. What will be the ratio of radius and height of a circular cylinder if its volume and radius is equal to the volume and radius of a sphere?
Q5. If the ratio of the volume of two-sphere is 27: 64, find the ratio between their surface areas.
Q6. The level of water in a cylindrical vessel of the 9 cm is raised by 12 cm when a certain ball is dropped inside it and submerged completely, finds the diameter of the ball.
We will be posting similar questions continually for your help, please subscribe to our website for the update of the information
NCERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions for class 9 maths
NCERT Solutions for class 9 science
NCERT Solutions for 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 for Class 10 Science
NCERT Solutions for 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 for Class 11 Physics
chapter 3-Motion in a Straight Line
NCERT Solutions for Class 11 Chemistry
Chapter 1-Some basic concepts of chemistry
NCERT Solutions for Class 11 Biology
NCERT solutions for 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 |
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
