RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

Other Exercises

Question 1.
The height of a cone is 15 cm. If its volume is 500π cm3, then find the radius of its base.
Solution:
Volume of cone = 500π cm3
and height (h) = 15 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q1.1

Question 2.
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
Solution:
Volume of a cone = 48π cm3
Height (h) = 9 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q2.1

Question 3.
If the height and slant height of a cone are 21 cm and 28 cm respectively. Find its volume.
Solution:
Height of a cone (h) = 21 cm
and slant height (l) = 28 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q3.1

Question 4.
The height of a conical vessel is 3.5 cm. If its capacity is 3.3 litres of milk. Find the diameter of its base.
Solution:
Capacity of conical vessel = 3.3 litres
Volume = 3.3 m3
= 3.3 x 1000 = 3300 cm2
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q4.1

Question 5.
If the radius and slant height of a cone are in the ratio 7 : 13 and its curved surface area is 286 cm2, find its radius.
Solution:
Two ratio in radius and slant height of a cone = 7 : 13
Let radius (r) = 7x
and slant height (1) = 13x
Curved surface area = πrl
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q5.1

Question 6.
Find the area of canvas required for a conical tent of height 24 m and base radius 7 m.
Solution:
Base radius of the closed cone (r) = 7 cm
and vertical height (h) = 24 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q6.1

Question 7.
Find the area of metal sheet required in making a closed hollow cone of base radius 7 cm and height 24 cm. making a closed hollow cone of base radius 7 cm and height 24 cm.
Solution:
Base radius of the closed cone (r) = 7 cm
and vertical height (h) = 24 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q7.1

Question 8.
Find the length of cloth used in making a conical pandal of height 100 m and base radius 240 m, if the cloth is 100π m wide.
Solution:
Height of conical pandal (A) = 100 m
Base radius (r) = 240 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q8.1

Hope given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2

Other Exercises

Question 1.
Find the volume of a right circular cone with:
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
(iii) height 21 cm and slant height 28 cm. [NCERT]
Solution:
(i) Radius of a cone (r) = 6 cm
and height (h) = 7 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q1.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q1.2

Question 2.
Find the capacity in litres of a conical vessel with:
(i) radius 7 cm, slant height 25 cm,
(ii) height 12 cm, slant height 13 cm. [NCERT]
Solution:
(i) Radius of the conical vessel (r) = 7 cm
Slant height (h) = 25 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q2.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q2.2

Question 3.
Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.
Solution:
Ratio in the heights of two cones =1:3
and ratio in their radii = 3: 1
Let radius of first cone (r1) = x
and of second cone (r2) = 3x
and height of first cone (h1) = 3y
and of second cone (h2)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q3.1

Question 4.
The radius and the height of a right circular cone are in the ratio 5 : 12. If its volume is 314 cubic metre, find the slant height and the radius (Use π = 3.14).
Solution:
Ratio in the radius and height of a cone = 5 : 12
Volume = 314 cm3
Let radius (r) = 5x
and height (h) = 12x
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q4.1

Question 5.
The radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use π = 3.14).
Solution:
Ratio in the radius and height of a right circular cone = 5 : 12
Volume = 2512 cm3
Let radius (r) = 5x
Then height (h) = 12x
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q5.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q5.2

Question 6.
The ratio of volumes of two cones is 4 : 5 and the ratio of the radii of their bases is 2:3. Find the ratio of their vertical heights.
Solution:
Ratio in volumes of two cones = 4:5
and ratio in radii = 2:3
Let radius of the first cone (r1) = 2x
Then radius of second cone (r2) = 3x
Let h1, h2 be their heights respectively
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q6.1

Question 7.
Ratio between their vertical heights = 9:5 7. A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1.
Solution:
Let r be the radius and h be the height of a cylinder and a cone, then
Volume of cylinder = πr2h
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q7.1

Question 8.
If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?
Solution:
Let r be the radius and h be the height of the cone, then
Volume = \(\frac { 1 }{ 3 }\) πr2h
By halving the radius and same height the
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q8.1

Question 9.
A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π = 3.14). [NCERT]
Solution:
Diameter of conical heap of wheat = 9 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q9.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q9.2

Question 10.
Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm.
Solution:
Diameter of the base of solid cone = 14 cm
and vertical height (h) = 51 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q10.1

Question 11.
A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area.
Solution:
Length of sides of a right angled triangle are 6.3 cm and 10 cm
By turning around the longer side, a cone is formed in which radius (r) = 6.3 cm
and height (h) = 10 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q11.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q11.2

Question 12.
Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.
Solution:
Side of cube = 14 cm ,
Radius of the largest cone that can be fitted
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q12.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q12.2

Question 13.
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find:
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone. [NCERT]
Solution:
Volume of a right circular cone = 9856 cm3
Diameter of the base = 28 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q13.1

Question 14.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? [NCERT]
Solution:
Diameter of the top of the conical pit = 3.5 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q14.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q14.2

Question 15.
Monica has a piece of Canvas whose area is 551 m2. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m2. Find the volume of the tent that can be made with it. [NCERT]
Solution:
Area of Canvas = 551 m2
Area of wastage = 1 m2
Actual area = 551 – 1 = 550 m2
Base radius of the conical tent = 7 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q15.1
Let l be the slant height and h be the vertical
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q15.2

Hope given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1

RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1

Other Exercises

Question 1.
What do you understand by the word ‘statistics’ in (i) singular form, (ii) plural forms.
Solution:
The word ‘statistics’ is used in both its singular as well as its plural senses.
In singular sense, statistics piay be defined as the science of collection, presentation, analysis and interpretation of numerical data.
In plural sense, statistics means numerical facts or observations collected with definite purpose.
For example, income and expenditure of persons in a particular locality, number of males and females in a particular town are statistics.

Question 2.
Describe some fundamental characteristics of statistics.
Solution:
Statistics in plural sense have the following characteristics:
(i) A single observation does not form statistics. Statistics are a sum total of observations.
(ii) Statistics are expressed quantitatively and not qualitatively.
(iii) Statistics are collected with a definite purpose.
(iv) Statistics in an experiment are comparable and can be classified into various groups.

Question 3.
What are (i) primary data? (ii) secondary data? Which of the two – the primary or the secondary data – is more reliable and why?
Solution:
(i) Primary data : When an investigator collects the data himself with a definite plan Or design in his mind, it is called primary data.
(ii) Secondary data : Data which are not originally collected rather obtained from the published or unpublished, sources are called secondary data.
Primary data are reliable and relevent because they are original in character and are collected by some individuals or by some institutions or by research bodies.

Question 4.
Why do we group data?
Solution:
When the number of observations is large, then arranging data in ascending or descending order is tedius job and it does not tell us much except minimum or maximum(s) of data. So, to make it easily understandable and clear, we tabulate data in the form of a table.

Question 5.
Explain the meaning of the following terms:
(i) variate
(ii) class intervals
(iii) class size
(iv) class mark
(v) frequency
(vi) class limits
(vii) true class limits.
Solution:
(i) Variate : The observations of a data are called variate.
(ii) Class intervals : When the presentations of data in classes, or groups, then groups are called classes or class intervals.
(iii) Class size : The difference between upper limit and lower limit is called class size.
(iv) Class mark : The mean of lower limit and upper limit is called class mark or mid value. Therefore class mark
= \(\frac { lower limit + upper limit }{ 2 }\)
(v) Frequency : The number of times an observation occurs in the given data, is called frequency of that observation.
(vi) Class limits : Every class has two limits : lower limit and upper limit.
(vii) True class limits : Whenever inclusive method is used, it is necessary tp make an adjustment to determine the correct class intervals, and to have continuity. If a-b is a class in inclusive method, then in order to change it into exclusive method, it becomes
a – \(\frac { h }{ 2 }\) – b + \(\frac { h }{ 2 }\) where h = \(\frac { 1 }{ 2 }\) [lower limit of a class – upper limit of previous class]

Question 6.
The ages of ten students of a group are given below. The ages have been recorded in years and months.
8-6, 9-0, 8-4, 9-3, 7-8, 8-11, 8-7, 9-2, 7¬10, 8-8
(i) What is the lowest age?
(ii) What is the highest age?
(iii) Determine the range?
Solution:
From the given data
(i) Lowest age is 7 years 8 months
(ii) The highest age is 9 years, 3 months
(iii) Range = Highest term – Lowest term
= 9 years 3 months – 7 years 8 months
= 1 years 7 months

Question 7.
The monthly pocket money of six friends is given below:
₹45, ₹30, ₹40, ₹50, ₹25, ₹45
(i) What is the highest pocket money?
(ii) What is the lowest pocket money?
(iii) What is the range?
(iv) Arrange the amounts of pocket money in ascending order.
Solution:
From the given data
(i) Highest pocket money = ₹50
(ii) Lowest pocket money = ₹25
(iii) Range = Highest term – Lowest term = ₹50-₹25 = ₹25
(iv) In ascending order : ₹25, ₹30, ₹40, ₹45, ₹45, ₹50

Question 8.
Write the class-size in each of the following:
(i) 0-4,5-9,10-14
(ii) 10-19, 20-29, 30-39
(iii) 100-120, 120-140, 160-180
(iv) 0-0.25, 0.25-0.50, 0.50-0.75
(v) 5-5,01, 5.01-5.02, 5.02-5.03
Solution:
(i) In 0-4, 5-9, 10-14
0-4 means from 0 to 4, similarly 5-9 means 5 to 9 and 10-14 means 10 to 14
class-size is 5
(ii) In 10-19, 20-29, 30-39
Here 10-19, means 10 to 19, 20-29 means 20 to 29 and 30-39 means 30 to 39
Class-size = 10
(iii) 100-120, 120-140, 160-180
Here 100-120, 120-140, 160-180
Then 120 – 100 = 20, 140 – 120 = 20, 180 – 160 = 20 is class-size = 20
(iv) 0-0.25, 0.25-0.50, 0.50-0.75
Here 0.25 – 0 = 0.25, 0.50 – 0.25 = 0.25 and 0.75 – 0.50 = 0.25
∴ Class-size = 0.25
(v) 5-5.01, 5.01-5.02, 5.02-5.03
Here 5.01 – 5 = 0.01, 5.02 – 5.01 = 0.01 and 5.03 – 5.02 = 0.01
∴ Class-size = 0.01

Question 9.
The final marks in mathematics of 30 students are as follows:
53, 61, 48, 60, 78, 68, 55, 100, 67, 90
75, 88, 77, 37, 84, 58, 60, 48, 62, 56
44, 58, 52, 64, 98, 59, 70, 39, 50, 60
(i) Arrange these marks in the ascending order,30 to 39 one group, 40 to 49 second group, etc.
Now answer the following:
(ii) What is the highest score?
(iii) What is the lowest score?
(iv) What is the range?
(v) If 40 is the pass mark how many have failed?
(vi) How many have scored 75 or more?
(vii) Which observations between 50 and 60 have not actually appeared?
(viii)How many have scored less than 50?
Solution:
(i) Arranging in ascending order:
37, 39, 44, 48, 48, 50, 52, 53, 55, 56, 58, 58, 59, 60, 60, 60, 61, 62, 64, 67, 68, 70, 75, 77, 78, 84, 88, 90, 98, 100
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 9.1
(ii) Highest score =100
(iii) Lowest score = 37
(iv) Range =  Highest score – Lowest score = 100 – 37 = 63
(v) If 40 is marks, then the number of students who failed = 2
(vi) No. of students who scored 75 or more = 8
(vii) Between 50 and 60, the scores which are missing 51, 54, 57
(viii) Number of students who scored less then 50 = 2 + 3 = 5

Question 10.
The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight.
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were bom on that day?
(vi) How many babies weigh below 2.5 kg?
(vii) How many babies weigh more than 2.8 kg?
(viii)How many babies weigh 2.8 kg?
Solution:
(i) Weights in descending order are
3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.4, 2.3, 2.2, 2.1
(ii) Highest weight = 3.1
(iii) Lowest weight = 2.1
(iv) Range = Highest weight – Lowest wight = 3.1 – 2.1 = 1.0
(v) No. of babies who bom = 15
(vi) No. of babies whose weights are below 2.5 kg = 4
(vii) No. of babies whose weight are more than 2.8 kg = 4.
(viii) No. of babies whose weight is 2.8 kg = 2

Question 11.
The number of runs scored by a cricket player in 25 innings are as follows:
26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47.
(i) Rearrange these runs in ascending order.
(ii) Determine the player, is highest score.
(ii) How many times did the player not score a run?
(iv) How many centuries did he score?
(v) How many times did he score more than 50 runs?
Solution:
(i) Arranging in ascending order
0, 0, 0, 15, 26, 34, 35, 39, 42, 42, 47, 48, 48, 53, 54, 64, 64, 67, 71, 82, 84, 94, 105, 124, 139
(ii) Highest score is 139
(iii) 3 times when his score is 0
(iv) No. of times, he made century = 3
(v) No. of times his score more then 50 runs = 12

Question 12.
The class size of a distribution is 25 and the first class-interval is 200-224. There are seven class-intervals.
(i) Write the class-intervals.
(ii) Write the class-marks of each interval
Solution:
Class size = 25
First class is 200-224
∴ Number of class = 7
∴ Class interval will be
(i) 200-224, 225-249, 250-274, 275-299, 300-324, 325-349, 350-374
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 12.1
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 12.2

Question 13.
Write the class size and class limits in each of the following:
(i) 104, 114, 124, 134, 144, 154 and 164
(ii) 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97 and 102
(iii) 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5
Solution:
(i) 104, 114, 124, 134, 144, 154 and 164
Here class size = 114 – 104 = 10
Here in first class, 104 – \(\frac { 10 }{ 2 }\)
= 104 – 5 = 99 and 104 + 5 = 109
∴ Class will be 99-109
In this way other classes will be 109-119, 119-129, 129-139, 139-149, 149-159, 159¬169
(ii) In 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, 102
Here class size = 52 – 47 = 5
∴ In first class, 47 – \(\frac { 5 }{ 2 }\) and 47 + \(\frac { 5 }{ 2 }\)
= 44.5-49.5
In this way other classes will be = 49.5-54.5, 54.5-59.5, 59.5-64.5, 64.5-69.5,69.5-74.5, 74.5-79.5, 79.5-84.5, 84.5-89.5, 89.5-94.5,94.5-99.5, 99.5-104.5 070 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5
Here class interval (size) = 17.5 – 12.5 = 5
In the class = 12.5 – \(\frac { 5 }{ 2 }\) , 12.5 + \(\frac { 5 }{ 2 }\)
⇒ 12.5 – 2.5, 12.5 + 2.5
⇒ 10, 15
∴ First class will be 10-15
In this way other classes 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50

Question 14.
Following data gives the number of children in 41 families:
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.
Represent it in the form of a frequency distribution.
Solution:
Frequency distribution table is given below:
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 14.1

Question 15.
The marks scored by 40 students of class IX in mathematics are given below:
81, 55, 68, 79, 85, 43, 29, 68, 54, 73, 47, 35, 72, 64, 95, 44, 50, 77, 64, 35, 79, 52, 45, 54, 70, 83, 62, 64, 72, 92, 84, 76, 63, 43, 54, 38, 73, 68, 52, 54.
Prepare a frequency distribution with class size of 10 marks.
Solution:
Maximum marks = 95,
minimum marks = 29,
range = 95 – 29 = 66
Frequency distribution is given below
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 15.1

Question 16.
The heights (in cm) of 30 students of class IX are given below:155, 158, 154, 158, 160, 148, 149, 150, 153, 159, 161, 148, 157, 153, 157, 162, 159, 151, 154, 156, 152, 156, 160, 152, 147, 155, 163, 155, 157, 153.
Prepare a frequency distribution table with 160-164 as one of the class intervals.
Solution:
Maximum height = 163 cm,
minimum height = 147 cm,
range 163 – 147 = 16
Frequency distribution of the given data is given below
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 16.1

Question 17.
The monthly wages of 30 workers in a factory are given below:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878,840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.
Solution:
Highest wage = ₹890,
lower wage = ₹804,
range = 890 – 804 = 86
Frequency distribution table is given below
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 17.1

Question 18.
The daily maximum temperatures (in degree Celsius) recorded in a certain city during the month of November are as follows:
25.8, 24.5, 25.6, 20.7, 21.8, 20.5, 20.6, 20.9, 22.3, 22.7, 23.1, 22.8, 22.9, 21.7, 21.3, 20.5, 20.9, 23.1, 22.4, 21.5, 22.7, 22.8, 22.0, 23.9, 24.7, 22.8, 23.8, 24.6, 23.9, 21.1
Represent them as a frequency distribution table with class size 1°C.
Solution:
Maximum temperature = 25.8
Minimum temperature = 20.5
Range = 25.8 – 20.5 = 5.3
Frequency distribution table is given below:
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 18.1

Question 19.
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking* one of the class intervals as 210-230 (230 not included):
220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.
Solution:
Maximum wages = ₹320
Minimum wages = ₹210
Range = 320 – 210 = 110
Required frequency table is given below:
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 19.1

Question 20.
The blood groups of 30 students of class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Find out which is the most common and which is the rarest blood group among these students. (NCERT)
Solution:
Frequency distribution table is given below:
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 20.1

Question 21.
Three coins were tossed 30 times. Each time the number of heads occuring was noted down as follow:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above. (NCERT)
Solution:
Frequency distribution table is given below:
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 21.1

Question 22.
Thirty children were asked about the number of hours they watched T.V. programmes in the previous week. The results were found as follows:
1 6 2 35 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week? (NCERT)
Solution:
Highest number of hours = 17
Lowest number of hours = 1
(i) The frequency table is given below:
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 22.1
(ii) No. of children watching T.V. for 15 or more hours a week = 2

Question 23.
The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows:
-12.5, -10.8, -18.6, -8.4, -10.8, -4.2, -4.8, -6.7, -13.2, -11.8, -2.3, 1.2, 2.6, 0, 2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, -2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, -5.8, -8.9, -14.6, -12.3, -11.5,-7.8,-2.9.
Represent them as frequency distribution table taking -19.9 to -15 as the first class interval.
Solution:
Maximum temperature = -18.6
Minimum temperature = 3.4
Range = 3.4 – (-18.6) = 3.4 + 18.6 = 22
The required frequency distribution table is given below:
RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 23.1
Hope given RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1

Other Exercises

Question 1.
Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.
Solution:
Radius of the base of the cone = 21 cm
and slant height (l) = 60 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 1.1

Question 2.
The radius of a cone is 5 cm and vertical height is 12 cm. Find the area of the curved surface.
Solution:
Radius of the base of a cone = 5 cm
Vertical height (h) = 12 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 2.1

Question 3.
The radius of a cone is 7 cm and area of curved surface is 176 cm2. Find the slant height.
Solution:
Curved surface area of a cone = 176 cm2
and radius (r) = 1 cm2
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 3.1

Question 4.
The height of a cone is 21 cm. Find the area of the base if the slant height is 28 cm.
Solution:
Height of the cone (h) = 21 cm
Slant height (l) = 28 cm
∴ l2 = r2 + h2
⇒ r2 = l2– h2 = (28)2 – (21 )2
⇒ 784 – 441 = 343 …(i)
Now area of base = πr2
= \(\frac { 22 }{ 7 }\) x 343 [From (i)]
= 22 x 49 = 1078 cm2

Question 5.
Find the total surface area of a right circular cone with radius 6 cm and height 8 cm.
Solution:
Radius of base of cone (r) = 6 cm
and height (h) = 8 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 5.1

Question 6.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm. [NCERT]
Solution:
Radius of base of a cone (r) = 5.25 cm
and slant height (l) = 10 cm
Curved surface area = πrl
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 6.1

Question 7.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. [NCERT]
Solution:
Slant height of a cone (l) = 21 m
and diameter of its base = 24 m
∴ Radius (r) = \(\frac { 24 }{ 2 }\) = 12 m
Now total surface area = πr(l + r)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 7.1

Question 8.
The area of the curved surface of a cone is 60π cm2. If the slant height of the cone be 8 cm, find the radius of the base.
Solution:
Curved surface area of a cone = 6071 cm2
Slant height (l) = 8 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 8.1

Question 9.
The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. What is the slant height ? (Use π = 22/7).
Solution:
Surface area of a cone = 4070 cm2
Diameter of its base = 70 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 9.1

Question 10.
The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm2, find its radius. (Use π = 22/7)
Solution:
Curved surface area of a cone = 792 cm2
Ratio in radius and slant height = 4:7
Let radius = 4x
Then slant height = 7x
∴ Curved surface area πrl = 792
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 10.1

Question 11.
A Joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. [NCERT]
Solution:
Radius of the base of a conical cap (r) = 7 cm
and height (h) = 24 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 11.1

Question 12.
Find the ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4 : 3.
Solution:
Let diameters of each cone = d
Then radius (r) = \(\frac { d }{ 2 }\)
Ratio in their slant heights = 4 : 3
Let slant height of first cone = 4x
and height of second cone = 3x
Now curved surface area of the first cone = 2πrh
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 12.1

Question 13.
There are two cones. The curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.
Solution:
In two cones, curved surface of the first cone = 2 x curved surface of the second cone
Slant height of the second cone = 2 x slant height of first cone
Let r1 and r2 be the radii of the two cones
and let height of the first cone = h
Then height of second cone = 2h
∴ Curved surface of the first cone = 2πr1h
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 13.1

Question 14.
The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surfaces.
Solution:
Let diameter of one cone = d
and diameter of second cone = d
∴ Radius of the first cone (r) = \(\frac { d }{ 2 }\)
and of second cone (r2) = \(\frac { d }{ 2 }\)
Ratio in their slant heights = 5:4
Let slant height of the first cone = 5x
Then that of second cone = 4x
Now curved surface of the first cone = 2πrh
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 14.1

Question 15.
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone. [NCERT]
Solution:
Area of curved surface of a cone = 308 cm2
and slant height (l) = 14 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 15.1

Question 16.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2. [NCERT]
Solution:
Slant height of a cone (l) = 25 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 16.1

Question 17.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m2 canvas is Rs. 70, find the cost of the canvas required to make the tent. [NCERT]
Solution:
Height of conical tent (A) = 10 m
Radius of the base (r) = 24 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 17.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 17.2

Question 18.
The circumference of the base of a 10 m height conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 m. (Use π = 22/7)
Solution:
Circumference of the base of a conical tent = 44 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 18.1

Question 19.
What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assum that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14) [NCERT]
Solution:
Height of the conical tent (h) = 8 m
and radius of the base (r) = 6 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 19.1
Now curved surface area of the tent = πrl = 3.14 x 6 x 10 = 188.4 m
Width of tarpaulin used = 3 m
∴ Length = 188.4 , 3 = 62.8 m
Extra length required = 20 cm = 0.2 m
∴ Total length of tarpaulin = 62.8 + 0.2 = 63 m

Question 20.
A bus stop is barricated from the remaining part of the road, by using 50 hollow cones made of recycled card-board. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m2, what will be the cost of painting these cones. (Use π = 3.14 and \(\sqrt { 1.04 } \) = 1.02) [NCERT]
Solution:
Diameter of the base of tent = 40 cm
∴ Radius of the base of cone (r) = \(\frac { 40 }{ 2 }\)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 20.1

Question 21.
A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius of each is to the height of each as 3 : 4.
Solution:
Let radius of cylinder = r
and radius of cone = r
and let height of cylinder = h
and height of cone = h
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 21.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 21.2

Question 22.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.
Solution:
Diameter of the cylindrical portion = 24 m 24
∴ Radius (r) = \(\frac { 24 }{ 2 }\) = 12 m
Height of cylindrical portion = 11 m
and total height = 16 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 22.1
∴ Height of conical portion = 16 – 11 = 5 m
∴ Slant height of the conical portion (l)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 22.2

Question 23.
A circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.
Solution:
Diameter of the cylindrical tent = 105 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 23.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 23.2

 

Hope given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

Other Exercises

Mark correct alternative in each of the following:
Question 1.
In a cylinder, if radius is doubled and height is halved, curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four times
Solution:
Let radius of the first cylinder (r1) = r
and height (h1) = h
Surface area = 2πrh
If radius is doubled and height is halved
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q1.1
∴ Their surface area remain same (c)

Question 2.
Two cylindrical jars have their diameters in the ratio 3:1, but height 1:3. Then the ratio of their volumes is
(a) 1 : 4
(b) 1 : 3
(c) 3 : 1
(d) 2 : 5
Solution:
Sol. Ratio in the diameters of two cylinder = 3:1
and ratio in their heights = 1:3
Let radius of the first cylinder (r1) = 3x
and radius of second (r2) = x
and height of the first (h1) = y
and height of the second (h2) = 3y
Now volume of the first cylinder = πr2h
= π(3x)2 x y = 9πx2y
and of second cylinder = π(x2) (3y)
∴ Ratio between then = 9πx2y : 3πx2y
= 3 : 1 (c)

Question 3.
The number of surfaces in right cylinder is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
The number of surfaces of a right cylinder is three. (c)

Question 4.
Vertical cross-section of a right circular cylinder is always a
(a) square
(b) rectangle
(c) rhombus
(d) trapezium
Solution:
The vertical cross-section of a right circular cylinder is always a rectangle. (b)

Question 5.
If r is the radius and h is height of the cylinder the volume will be
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q5.1
Solution:
Volume of a cylinder = πr2h (b)

Question 6.
The number of surfaces of a hollow cylindrical object is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
The number of surfaces of a hollow cylindrical object is 4. (d)

Question 7.
If the radius of a cylinder is doubled and the height remains same, the volume will be
(a) doubled
(b) halved
(c) same
(d) four times
Solution:
If r be the radius and h be the height, then volume = πr2h
If radius is doubled and height remain same,
the volume will be
= π(2r)2h = π x 4r2h
= 4πr2h = 4 x Volume
The volume is four times (d)

Question 8.
If the height of a cylinder is doubled and radius remains the same, then volume will be
(a) doubled
(b) halved
(c) same
(d) four times
Solution:
If r be the radius and h be the height, then volume of a cylinder = πr2h
If height is doubled and radius remain same, then volume = πr2(2h) = 2πr2h
∴ Its doubled (a)

Question 9.
In a cylinder, if radius is halved and height is doubled, the volume will be
(a) same
(b) doubled
(c) halved
(d) four times
Solution:
Let r be radius and h be height, then Volume = πr2h
If radius is halved and height is doubled
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q9.1

Question 10.
If the diameter of the base of a closed right circular cylinder be equal to its height h, then its whole surface area is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q10.1
Solution:
Let diameter of the base of a cylinder (r) = h
Then its height (h) = h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q10.2

Question 11.
A right circular cylindrical tunnel of diameter 2 m and length 40 m is to be constructed from a sheet of iron. The area of the iron sheet required in m2, is
(a) 40π
(b) 80π
(c) 160π
(d) 200π
Solution:
Diameter of a cylindrical tunnel = 2 m
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1m
and length (h) = 40 m
Curved surfae area = 2πrh = 2 x π x 1 x 40 = 80π (b)

Question 12.
Two circular cylinders of equal volume have their heights in the ratio 1 : 2. Ratio of their radii is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q12.1
Solution:
Let r1 and h1 be the radius and height of the
first cylinder, then
Volume = πr12h1
Similarly r1 and h2 are the radius and height of the second cylinder
∴ Volume = πr2h2
But their volumes are equal,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q12.2

Question 13.
The radius of a wire is decreased to one- third. If volume remains the same, the length will become
(a) 3 times
(b) 6 times
(c) 9 times
(d) 27 times
Solution:
In the first case, r and h1, be the radius and height of the cylindrical wire
∴ Volume = πr2h1 …(i)
In second case, radius is decreased to one third
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q13.1
∴ In second case height is 9 times (c)

Question 14.
If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q14.1
Solution:
Let r be the radius and h be the height then volume = πr2h
If height is doubled and volume is same and let x be radius then πr2h = π(x)2 x 2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q14.2

Question 15.
The volume of a cylinder of radius r is 1/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q15.1
Solution:
Let r be the radius and h be the height, then volume = πr2h
This volume is \(\frac { 1 }{ 4 }\) of the volume of a rectangular box
∴ Volume of box = 4πr2h
Let side of base of box = x and height h,
then volume = x2h
∴ 4πr2h = x2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q15.2

Question 16.
The height ft of a cylinder equals the circumference of the cylinder. In terms of ft, what is the volume of the cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q16.1
Solution:
In a cylinder,
h = circumference of the cylinder
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q16.2

Question 17.
A cylinder with radius r and height ft is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?
(a) 2πr(r + h)
(b) πr(r + 2h)
(c) πr(2r + h)
(d) 2πr2 + h
Solution:
r is the radius of the base and ft is the height of a closed cylinder
Then total surface area = 2πr(r + h ) (a)

Question 18.
The height of sand in a cylindrical-shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q18.1
Solution:
Let h be the height and d be the diameter of a cylinder, then
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q18.2

Question 19.
Two steel sheets each of length a1 and breadth a2 are used to prepare the surfaces of two right circular cylinders – one having volume vand height a2 and other having volume v2 and height a1. Then,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.1
Solution:
Length of each sheet = a1
and breadth = a2
Volume of cylinder = πr2h
In first case,
v1 is volume and a2 is the height
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.3

Question 20.
The altitude of a circualr cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q20.1
Solution:
In first case,
Let r be the radius and h be the height of the cylinder. Then,
∴ Lateral surface area = 2πrh
In second case,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q20.2

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

Other Exercises

Question 1.
Write the number of surface of a right circular cylinder.
Solution:
Three, two circular and one curved.

Question 2.
Write the ratio of total surface area to the curved surface area of a cylinder of radius r and height h.
Solution:
∵ Radius = r
and height = h
∴ Curved surface area = 2πrh
and total surface area = 2πr(h + r)
∴ Ratio = 2πr(h + r) : 2πrh
= h + r : h

Question 3.
The ratio between the radius of the base and height of a cylinder is 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.
Solution:
Ratio in radius and height of the cylinder = 2 : 3
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Q3.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Q3.2

Question 4.
If the radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then find the ratio of their volumes.
Solution:
Ratio of radii of two cylinder = 2:3
Let radius of first cylinder (r1) = 2x
and second cylinder (r2) = 3x
and ratio in their heights = 5:3
Let height of first cylinder (h1) = 5y
and height of second (h2) = 3y
∴ Volume of the first cylinder =πr2h
= π x (2x)2 x 5y = 20πx2y
and volume of second cylinder = π(3x)2 x 3y = 27πx2y
Now ratio between then,
= 20πx2y: 21πx2y
= 20 : 27

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2

Other Exercises

Question 1.
A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much? [NCERT]
Solution:
In first case, in a rectangular container of soft drink, the length of base = 5 cm
and Width = 4 cm
Height = 15 cm
∴ Volume of soft drink = lbh = 5 x 4 x 15 = 300 cm3
and in second case, in a cylindrical container, diameter of base = 7 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q1.1
∴ The soft drink in second container is greater and how much greater = 385 cm – 380 cm2 = 85 cm2

Question 2.
The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars? [NCERT]
Solution:
Radius of each pillar (r) = 20 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q2.1

Question 3.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 gm. [NCERT]
Solution:
Inner diameter of a cylindrical wooden pipe = 24 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q3.1

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, find:
(i) radius of its base
(ii) volume of the cylinder [Use π = 3.14] [NCERT]
Solution:
Lateral surface area of a cylinder = 94.2 cm2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q4.1

Question 5.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it? [NCERT]
Solution:
The capacity of a closed cylindrical vessel = 15.4 l
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q5.1

Question 6.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients? [NCERT]
Solution:
Diameter of the cylindrical bowl = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\)cm
Level of soup in it = 4 cm
∴ Volume of soup in one bowl for one patient
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q6.1

Question 7.
A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.
Solution:
Width of hollow cylinder (A) = 63 cm
Girth = 440 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q7.1

Question 8.
The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is ₹ 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places.
Solution:
Rate of painting = 50 paise per dm2
Total cost = ₹198
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q8.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q8.2

Question 9.
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:3. Calculate the ratio of their volumes and the ratio of their curved surfaces.
Solution:
Ratio in radii of two cylinders = 2:3
and ratio in their heights = 5:3
Let radius of the first cylinder (r1) = 2x
and radius of second cylinder (r2) = 3x
and height of first cylinders (h1) = 5y
and height of second cylinder (h2) = 3y
(i) Now volume of the first cylinder = πr2h = π(2x)2 x 5y = 20πx22y
and volume of tlie second cylinder = π(3x)2 x 3y = π x 9×2 x 3y = 27πx2y
Now ratio in their volume
= 20πx2y : 21πx2y = 20 : 27
(ii) Curved surface area of first cylinder = 2πrh = 2π x 2x x 5y =20πxy
and curved surface area of second cylinder = 2π x 3x x = 1 8πxy
∴ Ratio in their curved surface area
= 20πxy : 18πxy = 10 : 9

Question 10.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm2.
Solution:
Ratio in curved surface area and total surface area of a cylinder =1:2
Total surface area = 616 cm2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q10.1

Question 11.
The curved surface area of a cylinder is 1320 cm2 and its base had diameter 21 cm. Find the height and the volume of the cylinder. [Use π = 22/7]
Solution:
Curved surface area of a cylinder = 1320 cm2
Diameter of the base = 21 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q11.1

Question 12.
The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm3.
Solution:
Ratio between radius and height of a cylinder = 2:3
Volume =1617 cm3
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q12.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q12.2

Question 13.
A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length of form a cylinder. Find the volume of the cylinder so formed.
Solution:
Length of sheet = 44 cm
Breadth = 20 cm
By rolling along length, the height of cylinder (h) = 20cm
and circumference of the base = 44cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q13.1

Question 14.
The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.
Solution:
Curved surface area of a pillar = 264 m2
and volume = 924 m3
Let r be the radius and It be height, then 2πrh = 264
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q14.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q14.2

Question 15.
Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.
Solution:
Volumes of two cylinders are equal Ratio in their height h1 :h2 = 1: 2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q15.1

Question 16.
The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the” area of the curved surface. Find the volume of the cylinder.
Solution:
Height of a right circular cylinder = 10.5 m
3 x sum of areas of two circular faces
= 2 x area of curved surface
Let r be that radius,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q16.1

Question 17.
How many cubic metres of earth must be dugout to sink a well 21 m deep and 6 m diameter? Find the cost of plastering the inner surface of the well at ₹9.50 per m2.
Solution:
Diameter of a well = 6 m
∴ Radius (r) = \(\frac { 6 }{ 2 }\) = 3 m
Depth (h) = 21 m
∴ Volume of earth dugout = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q17.1

Question 18.
The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m. Find the volume of the timber that can be obtained from the trunk.
Solution:
Circumference of a cylindrical trunk of a tree = 176 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q18.1

Question 19.
A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.
Solution:
Diameter of cylindrical container = 56 cm
∴ Radius (r) = \(\frac { 56 }{ 2 }\) = 28 cm
Dimensions of a rectangular solid are = 32 cm x 22 cm x 14 cm
∴ Volume of solid = lbh
= 32 x 22 x 14 = 9856 cm3
∴ Volume of water in the container = 9856 cm3
Let h be the level of water, then
πr2h = 9856
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q19.1

Question 20.
A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.
Solution:
Length of metallic tube = 25 cm
Inner diameter = 10.4 cm
∴ Radius (r) = \(\frac { 10.4 }{ 2 }\) = 5.2 cm
Thickness of metal = 8 mm
∴ Outer radius (R) = 5.2 + 0.8 = 6.0 cm
Volume of metal used = π(R2 – r2) x h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q20.1

Question 21.
From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.
Solution:
Inner radius of a tap = 0.75 cm
Speed of flow of water in it = 7 m/s
Time = 1 hour
∴ Length of flow of water (h)
= 7 x 60 x 60 m = 25200 m
∴ Volume of water = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q21.1

Question 22.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.
Solution:
Size of rectangular sheet = 30 cm x 18 cm
∴ Length of sheet = 30 cm
and breadth = 18 cm
By folding length wise,
Height = 18 cm
and circumference = 30 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q22.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q22.2

Question 23.
How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec?
Solution:
Area of the cross-section of the pipe = 5 cm2
Speed of water flow = 30 cm/sec
Period = 1 minute
∴ Flow of water in 1 minute = 30 x 60 cm = 1800 cm
Area of mouth of pipe = 5 cm2
∴ Volume = 1800 x 5 = 9000 cm3
Volume of water in litres = 9000 ml
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q23.1

Question 24.
Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of ₹3.60 per cubic metre. Find also the cost of cementing its inner curved surface at ₹2.50 per square metre.
Solution:
Depth of well (h) = 280 m
Diameter = 3 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q24.1

Question 25.
Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Weights of copper wire = 13.2 kg
Diamter = 4 mm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q25.1

Question 26.
A solid cylinder has a total surface area of 231 cm2. Its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.
Solution:
Surface area of solid cylinder = 231 cm2
and curved surface area = \(\frac { 2 }{ 3 }\) of 231 cm2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q26.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q26.2

Question 27.
A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of a well = 14 m
∴ Radius (r) = y = 7 m
Depth (h) = 8 m
∴ Volume of the earth dugout = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q27.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q27.2

Question 28.
The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.
Solution:
Length of cylindrical tube = 14 cm
Difference betveen the outer surface and inner surface = 88 cm2
and volume of the tube = 176 cm3
Let R and r be the outer and inner radius of the tube
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q28.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q28.2

Question 29.
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank. The radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?
Solution:
Internal diameter of the pipe = 2 cm
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1 cm
Speed of water flow = 6m per second Water in 30 minutes (h) = 6 x 60 x 30 m = 10800 m
Volume of water = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q29.1

Question 30.
A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled?
Solution:
Diameter of cylindrical tank = 1.4 m
∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) = 0.7 m
and height (h) = 2.1 m
∴ Volume of water in the tank = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q30.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q30.2

Question 31.
The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m2. Find the volume of the cylinder.
Solution:
Sum of radius and height of a cylinder = 37 m
Let r be the radius and h be the height, then r + h = 37m …(i)
Total surface area of a solid cylinder = 1628m3
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q31.1

Question 32.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 10 m 10
∴ Radius (r) = \(\frac { 10 }{ 2 }\) = 5 m
Depth (h) = 8.4 m
∴ Volume of earth dugout = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q32.1

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1

Other Exercises

Question 1.
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height. [NCERT]
Solution:
Curved surface area of a cylinder = 4.4 m2
Radius (r) = 0.7 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q1.1

Question 2.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. [NCERT]
Solution:
Diameter of the pipe = 5 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q2.1

Question 3.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of 12.50 per m2. [NCERT]
Solution:
Diameter of cylindrical pillar = 50 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q3.1

Question 4.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same? [NCERT]
Solution:
Height of cylinder (h) = 1 m = 100 cm
Diameter of box = 140 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q4.1

Question 5.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder open from both sides = 4620 cm2
Area of base of ring = 115.5 cm2
Height (h) = 7 cm
Let outer radius (R) = R
and inner radius = r
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q5.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q5.2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q5.3

Question 6.
Find the ratio between the total surface area of a cylinder to its curved surface area, given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius of the cylinder (r) = 3.5 cm
and height (h) = 7.5 cm
Total surface area = 2πr (h + r)
and curved surface area = 2πrh
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q6.1

Question 7.
A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of ₹3.50 per 1000 cm2.
Solution:
Radius of the base of a cylindrical vessel (r) = 70 cm
and height (h) = 1.4 m = 140 cm
Total surface area (excluding upper lid) on both sides = 2πrh x 2 + πr2 x 2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q7.1

Question 8.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:
(i) inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of ₹40 per m2. [NCERT]
Solution:
Inner diameter of a well = 3.5 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q8.1

Question 9.
The students of a Vidyalaya were asked to participate s a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? [NCERT]
Solution:
Radius of cylinderical pen holder (r) = 3 cm
Height (h) = 10.5 cm
∴ Surface area of the pen holder
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q9.1

Question 10.
The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a play¬ground, find the cost of levelling this ground at the rate of 50 paise per square metre.
Solution:
Diameter of a roller = 1.5 m
∴ Radius = \(\frac { 1.5 }{ 2 }\) = 0.75 m = 75 cm
and length (h) = 84 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q10.1

Question 11.
Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m. What will be the cost of cleaning them at the rate of ₹2.50 per square metre? [NCERT]
Solution:
Number of pillars = 20
Diameter of one pillar = 0.50 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q11.1

Question 12.
A solid cylinder has total surface area of 462 cm2. Its curved surface area is one- third of its total surface area. Find the radius and height of the cylinder.
Solution:
Total surface of solid cylinder = 462 cm2
Curved surface area = \(\frac { 1 }{ 3 }\) of total surface area
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q12.1

Question 13.
The total surface area of a hollow metal cylinder, open at both ends of external radius 8 cm and height 10 cm is 338 π cm2. Taking r to be inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.
Solution:
Total surface area of a hollow metal cylinder = 338π cm2
Let R be the outer radius, r be inner radius and h be the height of the cylinder of the cylinder
∴ 2πRh + 2πrh + 2πR2 – 2πr2 = 338π
R = 8 cm, h = 10 cm
⇒ 2πh (R + r) + 2π(R2 – r2) = 338π
⇒ Dividing by 2π , we get
⇒ h(R + r) + (R2 – r2) = 169
⇒ 10(8 + r) + (8 + r) (8 – r) = 169
⇒ 80 + 10r + 64 – r2 = 169
⇒ 10r – r2 + 144 – 169 = 0
⇒ r2 – 10r + 25 = 0
⇒ (r-5)2 = 0
⇒ r = 5
∴ Thickness of the metal = R – r = 8 – 5 = 3 cm

Question 14.
Find the lateral curved surface area of a cylinderical petrol storage tank that is 4.2m in diameter and 4.5 m high. How much steel was actually used, if \(\frac { 1 }{ 12 }\) of steel actually used was wasted in making the closed tank? [NCERT]
Solution:
Diameter of a cylinderical tank = 4.2 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q14.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q14.2

 

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS

Other Exercises

Mark correct alternative in each of the following:
Question 1.
The length of the longest rod that can be fitted in a cubical vessel of edge 10 cm long, is
(a) 10 cm
(b) 10\(\sqrt { 2 } \) cm
(c) 10\(\sqrt { 3 } \) cm
(d) 20 cm
Solution:
Edge of cuboid (a) = 10 cm
∴ Longest edge = \(\sqrt { 3 } \) a cm
= \(\sqrt { 3 } \) x 10 = 10\(\sqrt { 3 } \) cm (c)

Question 2.
Three equal cubes are placed adjacently in a row. The ratio of the total surface area of the resulting cuboid to that of the sum of the surface areas of three cubes, is
(a) 7 : 9
(b) 49 : 81
(c) 9 : 7
(d) 27 : 23
Solution:
Let a be the side of three equal cubes
∴ Surface area of 3 cubes
= 3 x 6a2 = 18a2
and length of so formed cuboid = 3a
Breadth = a
and height = a
∴ Surface area = 2(lb + bh + hl)
= 2[3a x a + a x a+a x 3a] = 2[3a2 + a2 + 3a2] = 2 x 7a2 = 14a2
∴ Ratio in the surface areas of cuboid and three cubes = 14a2 : 18a2= 7:9 (a)

Question 3.
If the length of a diagonal of a cube is 8 \(\sqrt { 3 } \) cm, then its surface area is
(a) 512 cm2
(b) 384 cm2
(c) 192 cm2
(d) 768 cm2
Solution:
Length of the diagonal of cube = 8 \(\sqrt { 3 } \) cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q3.1

Question 4.
If the volumes of two cubes are in the ratio 8:1, then the ratio of their edges is
(a) 8 : 1
(b) 2\(\sqrt { 2 } \) : 1
(c) 2 : 1
(d) none of these
Solution:
Let volume of first cube = 8x3
and of second cube = x3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q4.1

Question 5.
The volume of a cube whose surface area is 96 cm2, is
(a) 16\(\sqrt { 2 } \) cm3
(b) 32 cm3
(c) 64 cm3
(d) 216 cm3
Solution:
Surface area of a cube = 96 cm2
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q5.1

Question 6.
The length, width and height of a rectangular solid are in the ratio of 3 : 2 : 1. If the volume of the box is 48 cm3, the total surface area of the box is
(a) 27 cm2
(b) 32 cm2
(c) 44 cm2
(d) 88 cm2
Solution:
Ratio in the dimensions of a cuboid =3 : 2 : 1
Let length = 3x
Breadth = 2x
and height = x
Then volume = lbh = 3x x 2x x x = 6×3
∴ 6x3 = 48 ⇒ x3= \(\frac { 48 }{ 6 }\) = 8 = (2)3
∴ x = 2
∴ Length (l) = 3 x 2 = 6 cm
Breadth (b) = 2 x 2 = 4 cm
Height (h) = 1 x 2 = 2 cm
Now surface area = 2[lb + bh + hl]
= 2[6 x 4 + 4 x 2 + 2 x 6] cm2
= 2[24 + 8-+ 12] = 2 x 44 cm2
= 88 cm2 (d)

Question 7.
If the areas of the adjacent faces of a rectangular block are in the ratio 2:3:4 and its volume is 9000 cm3, then the length of the shortest edge is
(a) 30 cm
(b) 20 cm
(c) 15 cm
(d) 10 cm
Solution:
Ratio in the areas of three adjacent faces of a cuboid = 2 : 3 : 4
Volume = 9000 cm3
Let the area of faces be 2x, 3x, Ax and
Let a, b, and c be the dimensions of the cuboid, then
∴ 2x = ab, 3x = be, 4x = ca
∴ ab x be x ca = 2x x 3x x 4x
a2b2c2 = 24 x 3
But volume = abc = 9000 cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q7.1
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q7.2

Question 8.
If each edge of a cube, of volume V, is doubled, then the volume of the new cube is
(a) 2V
(b) 4V
(c) 6V
(d) 8V
Solution:
Let a be the edge of a cube whose Volume = V
∴ a3 = V
By doubling the edge, we get 2a
Then volume = (2a)3 = 8a3
∴ Volume of new cube = 8a3 = 8V (d)

Question 9.
If each edge of a cuboid of surface area S is doubled, then surface area of the new cuboid is
(a) 2S
(b) 4S
(c) 6S
(d) 8S
Solution:
Let each edge of a cube = a
Then surface area = 6a2
∴ S = 6a2
Now doubling the edge, we get
New edge of a new cube = 2a
∴ Surface area = 6(2a)2
= 6 x 4a2 = 24a2
= 4 x 6a2 = 4S (b)

Question 10.
The area of the floor of a room is 15 m2. If its height is 4 m, then the volume of the air contained in the room is
(a) 60 dm3
(b) 600 dm3
(c) 6000 dm3
(d) 60000 dm3
Solution:
Area of a floor of a room = 15 m2
Height (h) = 4 m
∴ Volume of air in the room = Floor area x Height
= 15 m2 x 4 m = 60 m3
= 60 x 10 x 10 x 10 dm2 = 60000 dm2 (d)

Question 11.
The cost of constructing a wall 8 m long, 4 m high and 20 cm thick at the rate of ₹25 per m3 is
(a) ₹16
(b) ₹80
(c) ₹160
(d) ₹320
Solution:
Length of wall (l) = 8 m
Breadth (b) = 20 cm = \(\frac { 1 }{ 5 }\) m
Height (h) = 4 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q11.1

Question 12.
10 cubic metres clay in uniformaly spread on a land of area 10 acres. The rise in the level of the ground is
(a) 1 cm
(b) 10 cm
(c) 100 cm
(d) 1000 cm
Solution:
Volume of clay = 10 m3
Area of land = 10 acres
= 10 x 100 = 1000 m2
∴ Rise of level by spreading the clay
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q12.1

Question 13.
Volume of a cuboid is 12 cm3. The volume (in cm3) of a cuboid whose sides are double of the above cuboid is
(a) 24
(b) 48
(c) 72
(d) 96
Solution:
Volume of cuboid = 12 cm3
By doubling the sides of the cuboid the
volume will be = 12 cm3 x 2 x 2 x 2
= 96 cm3 (d)

Question 14.
If the sum of all the edges of a cube is 36 cm, then the volume (in cm3) of that cube is
(a) 9
(b) 27
(c) 219
(d) 729
Solution:
Sum of all edges of a cube = 36 cm
No. of edge of a cube are 12
∴ Length of its one edge = \(\frac { 36 }{ 12 }\) = 3 cm
Then volume = (edge)3 = (3)3 cm3
= 27 cm3 (b)

Question 15.
The number of cubes of side 3 cm that can be cut from a cuboid of dimensions 9 cm x 9 cm x 6 cm, is
(a) 9
(b) 10
(c) 18
(d) 20
Solution:
Dimensions of a cuboid = 9 cm x 9 cm x 6 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q15.1

Question 16.
On a particular day, the rain fall recorded in a terrace 6 m long and 5 m broad is 15 cm. The quantity of water collected in the terrace is
(a) 300 litres
(b) 450 litres
(c) 3000 litres
(d) 4500 litres
Solution:
Dimension of a terrace = 6mx5m
Level of rain on it = 15 cm
∴ Volume of water collected on it
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q16.1

Question 17.
If A1, A2 and A3 denote the areas of three adjacent faces of a cuboid, then its volume is
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q17.1
Solution:
Let l, b, h be the dimensions of the cuboid
∴ A1= lb, A2 = bh, A3 = hl
∴ A1 A2 A3 = lb.bh.hl = l2b2h2
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q17.2

Question 18.
If l is the length of a diagonal of a cube of volume V, then
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q18.1
Solution:
Volume of a cube = V
and longest diagonal = l
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q18.2

Question 19.
If V is the volume of a cuboid of dimensions x, y, z and A is its surface area, then \(\frac { A }{ V }\)
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q19.1
Solution:
A is surface area, V is volume and x, y and z are the dimensions
Then V = xyz
A = 2[xy + yz + zx]
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q19.2

Question 20.
The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is 5\(\sqrt { 5 } \) cm. Its surface area is
(a) 361 cm2
(b) 125 cm2
(c) 236 cm2
(d) 486 cm2
Solution:
Let x, y, z be the dimensions of a cuboid,
then x + y + z = 19 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q20.1

Question 21.
If each edge of a cube is increased by 50%, the percentage increase in its surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 125%
Solution:
Let in first case, edge of a cube = a
Then surface area = 6a2
In second case, increase in side = 50%
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q21.1

Question 22.
A cube whose volume is 1/8 cubic centimeter is placed on top of a cube whose volume is. 1 cm3. The two ,cubes are then placed on top of a third cube whose volume is 8 cm3. The height of the stacked cubes is
(a) 3.5 cm
(b) 3 cm
(c) 7 cm
(d) none of these
Solution:
Volume of first cube = \(\frac { 1 }{ 2 }\) cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS Q22.1

Hope given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS

Other Exercises

Question 1.
If two cubes of side 6 cm are joined face to face, then find the volume of the resulting cuboid.
Solution:
Side of a cube = 6 cm
∴ By joining two such cubes, the length of so
formed cuboid (l) = 6 x 2 = 12 cm
Breadth (b) = 6 cm
Height (h) = 6 cm
∴ Volume = lbh = 12 x 6 x 6 cm3
= 432 cm3

Question 2.
Three cubes of metal whose edges are in the ratio 3 : 4 : 5, are melted down into a single cube whose diagonals is 12 \(\sqrt { 3 } \) cm. Find the edges of three cubes.
Solution:
Ratio in the sides of three cubes = 3 : 4 : 5
Let side of first cube = 3x
and side of second cube = 4x
and side of third cube = 5x
∴ Sum of volume of three cubes
= (3x)3 + (4x)3 + (5x)3
= 27x3 + 64x3 + 125x3 = 216x3
∴ Volume of the cube formed Jby melting these three cubes = 216x3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS Q2.1
∴ Side of first cube = 3x = 3 x 2 = 6 cm
Side of second cube = 4x = 4×2 = 8 cm
and side of third cube = 5.r = 5 x 2 = 10 cm

Question 3.
If the perimeter of each face of a cube is 32 cm, find its lateral surface area. Note that four faces which meet the base of a cube are called its lateral faces.
Solution:
Perimeter of each face of a cube = 32 cm
∴ Length of edge = \(\frac { 32 }{ 4 }\) = 8 cm
and lateral surface area of the cube = 4 x (side)2
= 4 x 8 x 8 = 256 cm2

Question 4.
Find the edge of a cube whose surface area is 432 m2.
Solution:
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS Q4.1

Question 5.
A cuboid has total surface area of 372 cm2 and its lateral surface area is 180 cm2, find the area of its base.
Solution:
Total surface area of a cuboid = 372 cm2
and lateral surface area = 180 cm2
∴ Area of base and roof = 372 – 180 = 192 cm2
and area of base = \(\frac { 192 }{ 2 }\) = 96 cm2

Question 6.
Three cubes of each side 4 cm are joined end to end. Find the surface area of the resulting cuboid.
Solution:
By joining three cubes of side 4 cm each, end is end, we get a cuboid
Length of cuboid = 4 x 3 = 12 cm
Breadth = 4 cm
and height = 4 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS Q6.1
∴ Surface area = 2(lb + bh + hl)
= 2[12 x4+4×4 + 4x 12] cm2
= 2[48 + 16 + 48] cm2
= 2 x 112 = 224 cm2

Question 7.
The surface area of a cuboid is 1300 cm2. If its breadth is 10 cm and height is 20 cm, find its length.
Solution:
Surface area of a cuboid = 1300 cm2
Breadth (b) = 10 cm
and height (h) = 20 cm
Let l be the length, then
= 2 (lb + bh + hl) = 1300
lb+ bh + hl = \(\frac { 1300 }{ 2 }\) = 650
l x 10 + 10 x 20 + 20 x l = 650
10l + 20l + 200 = 650
⇒ 30l = 650 – 200 = 450
⇒ l = \(\frac { 450 }{ 30 }\) = 15
∴ Length of cuboid = 15 cm

Hope given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

Other Exercises

Question 1.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? [NCERT]
Solution:
Length of water tank (l) = 6 m
Breadth (b) = 5 m
and depth (h) = 4.5 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q1.1
∴ Volume of water in it = lbh
= 6 x 5 x 4.5 m3 = 135 m3
Capacity of water in litres = 135 x 1000 litres (1 m3 = 1000 l)
= 135000 litres

Question 2.
A cubical vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? [NCERT]
Solution:
Length of vessal (l) = 10 m
Breadth (b) = 8 m
Volume = 380 m3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q2.1

Question 3.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m3. [NCERT]
Solution:
Length of pit (l) = 8m
Width (b) = 6 m
and depth (h) = 3 m
∴ Volume of earth digout = lbh
= 8 x 6 x 3 = 144 m3
Cost of digging the pit at the rate of ₹30 per m3
= 144 x 30 = ₹4320

Question 4.
If the areas of three adjacent faces of a cuboid are 8 cm2, 18 cm2 and 25 cm2. Find the volume of the cuboid.
Solution:
Let x, y, z be the three adjacent faces of the cuboid, then
x = 8 cm2, y = 18 cm2, z = 25 cm2
and let l, b, h are the dimensions of the cuboid, then
x = lb = 8 cm2
y = bh = 18 cm2
z = hl = 25 cm2
∴ Volume = lbh
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q4.1

Question 5.
The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.
Solution:
Let breadth of a room (b) = x
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q5.1

Question 6.
Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.
Solution:
Edge of first cube = 6 cm
Edge of second cube = 8 cm
and edge of third cube = 10 cm
∴ Volume of 3 cubes = (6)3 + (8)3 + (10)3 cm3
= 216 + 512 + 1000 cm3
= 1728 cm3
∴ Edge of so formed cube
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q6.1

Question 7.
Two cubes, each of volume 512 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each volume = 512 cm3
∴ Side (edge) = \(\sqrt [ 3 ]{ 512 }\)
=\(\sqrt [ 3 ]{ { 8 }^{ 3 } }\)  = 8 cm
Now by joining two cubes, then Length of so formed cuboid (l)
= 8 + 8 = 16 cm
Breadth (b) = 8 cm
and height (h) = 8 cm
∴ Surface area = 2(lb + bh + hl)
= 2[16 x 8 + 8 x 8 + 8 x 16] cm2
= 2[128 + 64 + 128] cm2
= 2 x 320 = 640 cm2

Question 8.
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.
Solution:
Edge of metal cube = 12 cm
∴ Its volume = (Edge)3 = (12)3 cm33
= 1728 cm3
It is melted and form 3 cubes
Edge of one smaller cube = 6 cm
and edge of second smaller cube = 8 cm
∴ Volume of two smaller cubes = (6)3 + (8)3 cm3
= 216 + 512 cm3 = 728 cm3
∴ Volume of third smaller cube = 1728 – 728 = 1000 cm3
∴ Edge of the third cube = \(\sqrt [ 3 ]{ 1000 }\)
= \(\sqrt [ 3 ]{ (10)3 }\)  cm = 10 cm

Question 9.
The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air?
Solution:
Length of cinema hall (l) = 100 m
Breadth (b) = 50 m
and height (h) = 18 m
∴ Volume of air in it = lbh
= 100 x 50 x 18 m= 90000 m3
Air required for one person = 150 m3
∴ Number of persons in the hall = \(\frac { 90000 }{ 150 }\) = 600 persons

Question 10.
Given that 1 cubic cm of marble weighs 0.25 kg, the weight of marble block 28 cm in width and 5 cm thick is 112 kg. Find the length of the block.
Solution:
Weight of 1 cm3 = 0.25 kg
Breadth of the block (b) = 28 cm
Thickness (h) = 5 cm
and total weight of the block = 112 kg
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q10.1

Question 11.
A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a liquid can be placed in it? Also, find the volume of the wood used in it.
Solution:
Outer length of the closed wooden box (l) = 25 cm
Breadth (b) = 18 cm
and height (h) = 15 cm
Width of wood = 2 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q11.1
∴ Inner length = 25 – 2×2 = 25- 4 = 21cm
Breadth =18- 2×2 = 18-4 = 14 cm
and height =15- 2×2 = 15- 4=11 cm
Now outer volume = 25 x 18 x 15 cm3 = 6750 cm3
and inner volume = 21 x 14 x 11 cm3 = 3234 cm3
(i) Inner volume = 3234 cm3
(ii) Volume of wood = 6750 – 3234 = 3516 cm3

Question 12.
The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box?
Solution:
External length of a closed wooden box (L) = 48 cm
Width (B) = 36 cm
and height (H) = 30 cm
Thickness of wood = 1.5 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q12.1
∴ Internal length (l) = 48 – 2 x 1.5 cm = 48 – 3 = 45 cm
Width (b) = 36 – 2 x 1.5 cm
= 36 – 3 = 33 cm
and height (h) = 30 – 2 x 1.5 cm
= 30 – 3 = 27 cm
Now volume of internal box
= lbh = 45 x 33 x 27 cm3
Volume of one bricks = 6 x 3 x 0.75 cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q12.2

Question 13.
A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.
Solution:
Edge of a cube = 9 cm
Volume of cube = (9)3 cm3
= 729 cm3
Now length of vessel (l) = 15 cm
and breadth (b) = 12 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q13.1

Question 14.
A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7 m deep and the earth taken out is spread evenly on the field. By how many metres is the level of the field raised? Give the answer to the second place of decimal.
Solution:
Length of a field (l) = 200 m
Breadth (b) = 150 m
Length of plot = 50 m
and breadth = 40 m
Depth of plot = 7 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q14.1
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q14.2

Question 15.
A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a comer of the field and the earth taken out is spread- over the remaining area of the field. Find out the extent to which the level of the field has been raised.
Solution:
Length of a field (L) = 18m
and width (B) = 15 m
Length of pit (l) = 7.5 m
Breadth (b) = 6 m
and depth (h) = 0.8 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q15.1
∴ Volume of earth dugout = lbh
= 7.5 x 6 x 0.8 m3
= 45 x 0.8 = \(\frac { 45 x 4 }{ 5 }\) = 36 m3
Total area of the field = L x B
= 18 x 15 = 270 m2
and area of pit = lb = 7.5 x 6 = 45 m2
∴ Remaining area of the field excluding pit
= 270 – 45 = 225 m2
Let by spreading the earth on the remaining part of the field, the height = h
= 225 x h = 36
⇒ h = \(\frac { 36 }{ 225 }\) = \(\frac { 4 }{ 25 }\)= 0.16 m = 16 cm
∴ Level of field raised = 16 cm

Question 16.
A village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20 m x 15m x 6 m. For how many days will the water of this tank last? [NCERT]
Solution:
Population of a village = 4000
Water required per head per day = 150 litres
∴ Total water required = 4000 x 150 litres = 600000 litres
Dimensions of a tank = 20mx 15mx6m
∴ Volume of tank = 20 x 15 x 6 m3 = 1800 m3
Capacity of water in litres = 1800 x 1000 litres (1 m3 = 1000 litres)
= 1800000 litres
The water will last for = \(\frac { 1800000 }{ 600000 }\) = 3 days

Question 17.
A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in the figure. If the edge of each cube is 3 cm, find the volume of the structure built by the child. [NCERT]
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q17.1
Solution:
No. of cubes at the given structure = 1+ 2 + 3+ 4 + 5 = 15
Edge of one cube = 3 cm
∴ Volume of one cube = (3)3 = 3 x 3 x 3 cm3 = 27 cm3
∴Volume of the structure = 27 x 15 cm3 = 405 cm3

Question 18.
A godown measures 40 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown. [NCERT]
Solution:
Length of godown (L) = 40 m
Breadth (B) = 25 m
and height (H) = 10 m
∴ Volume of godown = LBH
= 40 x 25 x 10 = 10000 m3
Dimension of one wooden crates = 1.5 m x 1.25 m x 0.5 m
∴Volume of one crate = 1.5 x 1.25 x 0.5 m3 = 0.9375 m3
∴ Number of crates to be stored in the
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q18.1

Question 19.
A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required? [NCERT]
Solution:
Length of wall (L) = 10 m = 1000 cm
Height (H) = 4 m = 400 cm
Thickness (B) = 24 cm = 24 cm
∴ Volume of wall = LBH = 1000 x 24 x 400 cm3 = 9600000 cm3
Dimensions of one brick = 24 cm x 12 cm x 8 cm = 2304 cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q19.1

Question 20.
If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q20.1
Solution:
a, b, c are the dimensions of a cuboid S is the surface area and V is the volume
∴ V = abc and S = 2(lb + bc + ca)
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q20.2

Question 21.
The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V= xyz.
Solution:
Let a, b, c are the dimensions of a cuboid then,
x = ab, y = bc, z = ca
and V = abc
Now L.H.S. = V2
= (abc)= a2b2c2
= ab.bc.ca = xyz = R.H.S.
Hence V2 = xyz

Question 22.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? [NCERT]
Solution:
Speed of water in a river = 2 km/hr
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q22.1

Question 23.
Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?
Solution:
Width of canal (b) = 30 dm = 3 m
Depth (h) = 12 dm = 1.2 m
Speed of water = 100 km/hr
Length of water flow in 30 minutes = \(\frac { 1 }{ 2 }\) hr
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q23.1

Question 24.
Half cubic metre of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold- sheet.
Solution:
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q24.1

Question 25.
How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 g, find the weight of the empty box in kg.
Solution:
External length of open box (L) = 36 cm
Breadth (B) = 25 cm
and Height (H) = 16.5 cm
Width of iron sheet used = 1.5 cm
∴ Inner length (l) = 36 – 1.5 x 2 = 36 – 3 = 33 cm
Breadth (b) = 25 – 2 x 1.5 = 25 – 3 = 22 cm
and Height (h) = 16.5 – 1.5 = 15 cm
∴ Volume of the iron used = Outer volume – Inner volume
= 36 x 25 x 16.5 – 33 x 22 x 15
= 14850 – 10890 = 3960 cm3
Weight of 1 cm3 = 15 g
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q25.1

Question 26.
A rectangular container, whose base is a square of side 5 cm, stands on a horizontal table, and holds water upto 1 cm from the top. When a cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.
Solution:
Base of the container = 5 cm x 5 cm
Level of water upto 1 cm from the top After placing a cube in it, the water rises to the top and 2 cubic cm of water overflows,
(i) ∴ Volume of water = 5 x 5 x 1 + 2 = 25 + 2 = 27 cm3
∴ Volume of cube = 27 cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q26.1

Question 27.
A rectangular tank is 80 m long and 25 m broad. Water-flows into it through a pipe whose cross-section is 25 cm2, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes.
Solution:
Length of tank (l) = 80 m
Breadth (b) = 25 m
Area of cross section of the month of pipe = 25 cm2
and speed of water-flow =16 km/h
∴ Volume of water is 45 minutes
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q27.1

Question 28.
Water in a rectangular reservoir having base 80 m by 60 m is 6.5 m deep. In what time can the water be emptied by a pipe of which the cross-section is a square of side 20 cm, if the water runs through the pipe at the rate of 15 km/hr.
Solution:
Length of reservoir (l) = 80 m
Breadth (b) = 60 m
and depth (h) = 6.5 m
∴ Volume of water in it = lbh = 80 x 60 x 6.5 m3 = 31200 m3
Area of cross-section of the month of pipe = 20 x 20 = 400 cm2
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q28.1

Hope given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.