Surface Areas & Volume
Mathematics Class Tenth
RD Sharma exercise 16.1 q11 to 15
Question (11) A well of diameter 2 m is dug 14 meter deep. The earth taken out of it is spread evenly all around it to form an embankment of height 40 cm. Find the width of the embankment.
Solution:
[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (11) (a) Using diameter find out the radius of well. (b) Using radius and height of well calculate the volume which will be equal to the volume of earth dug out. (c) The volume of earth used in embankment will be equal to the earth dug from well. (d) Using the volume of earth, height of embankment and radius of well calculate the width of embankment. ]
Given, depth of the well = height of well = 14 m
Diameter of well = 2 m
Therefore, radius of well = 2m/2 = 1 m
Height of embankment around the well = 40 cm
= 40/100 = 0.4 m
Let, width of embankment `=w`
Therefore, radius of embankment including well `=(w+1)\ m`
Now, we know that, volume of a cylinder `=pi\ r^2\ h`
Thus, volume of well = volume of earth `=pi\ (1\ m)^2xx14\ m`
`=pi\ xx14\ m^3`
Thus, volume of earth `=pixx14\ m^3`
The embankment formed is similar to the shape of a hollow cylinder
And, the volume of embankment = volume of earth dug from well
Now, we know that, volume of material of hollow cylinder = External volume – Internal volume
`=pi\ R^2\ h– pi\ r^2\ h`
`=pi\ h(R^2–r^2)`
Here, R = (w+1) = radius of embankment including well
And, r = 1 m = radius of well
And, h = height of embankment = 0.4 m
Thus, volume of material in embankment `=pi\ h(R^2–r^2)`
`=>pi\ 14m^2 = pi\ 0.4\ m[(w+1)m^2–(1\ m)^2]`
`=>(w+1)m^2–(1\ m)^2 = (pi\ 14m)/(pi\ 0.4\ m)`
`=>(w^2+1^2+2w–1^2)m^2=35`
`=>w^2+2w=35`
`=>w^2+2w–35=0`
`=>w^2+7w–5w–35=0`
`=>w(w+7)–5(w+7)=0`
`=>(w–5)(w+7)=0`
Now, if (w+7) = 0
Therefore, w = –7 m
And, if (w–5) = 0
Therefore, w = 5 m
Now, discarding negative value, `w = 5\ m`
Thus, width of the embankment = 5 m Answer
Question (12) Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.
Solution:
[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (12) (a) Find out the radius of cone which will be equal to the half of the side of the cube (b) Find out the height of the cone which also be equal to the side of the cube. (c) By taking radius and height of the cone, find its volume using the formula to find the volume of a cone.]
Given, Side of the cube = 9 cm
Since, largest right circular cone is to be cut out of cube,
Thus, Diameter of base of the cone = side of the cube = 9 cm
Thus, radius of cube (r) = 9/2 cm = 4.5 cm
And height of the cone (h) = side of the cube = 9 cm
Thus, volume of the largest right circular cone to be cut out from given cube = ?
We know that, Volume of a right circular cone `=1/3\ pi\ r^2\ h`
Thus, volume of given cone in question `=1/3xx22/7xx(9/2\ cm)^2xx9\ cm`
`=(11xx3xx9xx9)/(7xx2)\ cm^3`
`=2673/14\ cm^3`
= 190.92 cm3
Thus, volume of the given cone = 190.92 cm3 Answer
Question (13) A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (13) (a) Find the area of cylindrical bucket. (b) The area of cylindrical bucket will be equal to the sand filled in it and will be equal to the conical heap formed on the ground. (c) Using height find the radius of the conical heap using formula to find the volume of a cone. (d) Using radius and height, find the slant height of cone using Pythagoras theorem.]
Given, Height of the cylindrical bucket (h) = 32 cm
And radius of cylindrical bucket = (r) = 18 cm
Height of conical heap formed by the sand filled in bucket = 24 cm
Therefore, radius of cone and slant height of cone = ?
Calculation of volume of sand in the bucket
We know that, Volume of a circular cylinder `=pi\ r^2\ h`
Thus, volume of given bucket `=pi\ (18\ cm)^2xx32\ cm`
`=pi\ 324xx32\ cm^3`
Thus, volume of bucket = volume of sand `=10368\ pi\ cm^3`
Now, volume of sand = volume of cone formed using that sand
Calculation of radius of cone
Now, we know that, volume of cone `=1/3\ pi\ r^2\ h`
Or, volume of cone = volume of sand `=1/3\ pi\ r^2xx24\ cm`
`=>10368\ pi\ cm^3 = 8\ r^2\ pi`
`=>r^2 = (10368\ pi\ cm^3)/(8\ pi)`
`=>r^2 = 1296\ cm^2`
Thus, r = 36 cm = Radius of cone
Calculation of slant height (`l`) of the cone
Here, height = 24 cm, radius = 36 cm
Thus, slant height (`l`) = ?
Here, ABC is a right angle,
Thus, According to Pythagoras theorem, we know that
AC2 = AB2 + BC2
`=> l^2 = (24\ cm)^2 + (36\ cm)^2`
= 576 cm2 + 1296 cm2
= 1872 cm2
Thus, `l = sqrt(1872\ cm^2)`
Thus, slant height (`l`) = 43.266 cm = 43.27 cm
Thus, Radius of cone = 36 cm and slant height of cone = 43.27 cm Answer
Question (14) Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen? [Use `pi =22/7` ]
Solution:
[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (14) (a) Here height of the water is equal to 1 cm. (b) Find the volume of water by multiplying length and breadth of rectangular surface and height of water. (c) Using volume of water and radius of cylinder find the height of water in cylindrical vessel using the formula of volume of cylinder.]
Given, length of rectangular surface = 6 m
= 6 × 100 = 600 cm
And, breadth of rectangular surface = 4 m
= 4 × 100 = 400 cm
And height of water = 1 cm
Radius of cylindrical vessel = 20 cm
Thus, height of water in cylindrical vessel when water is transferred to it = ?
Calculation of volume of water
Since, water is spread evenly on a surface, thus length and breadth of water will be equal to the length and breadth of given rectangular surface.
Volume of water = length of rectangle × Breadth of rectangle × height of water
= 600 cm × 400 cm × 1 cm
= 240000 cm3
Calculation of height of water when it is placed in cylinder
Now, We know that, volume of cylinder `=pi\ r^2\ h`
Thus, volume of water = volume of cylinder `=pi\ (20\ cm)^2\ h`
`=>240000 cm^3 = 22/7xx20xx20\ cm^2xxh`
Thus, `h = (240000\ cm^3xx7)/(20xx20xx22\ cm^2)`
Or, h = 190.90 cm
Thus, height of water in cylinder = 190.90 cm = 191 cm (approximately) Answer
Question (15) A conical flask is full of water. The flask has base radius r and height h. The water is poured into a cylindrical flask of base radius mr. Find the height of water in the cylindrical flask.
Solution:
[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (15) (a) Find the volume of conical flask which will give the volume of water. (b) Using the volume of water and radius of cylindrical flask, find the height of water in cylindrical flask.]
Given, radius of conical flask = r
And height of conical flask = h
Radius of cylindrical flask = mr
Water full in conical flask is poured in cylindrical flask, then height of water in cylindrical flask = ?
Calculation of volume of water in conical flask
We know that, Volume of cone `=1/3\ pi\ r^2h`
Thus, volume of given conical flask `=1/3\ pi\ r^2h`
Since conical flask is full of water, thus volume of water = volume of conical flask
`=1/3\ pi\ r^2h`
Calculation of height of water in cylindrical flask
Let, height of water in conical flask `=h_w`
We know that, volume of cylinder `=pi\ r^2\ h`
Since, water in conical flask is poured into cylindrical flask.
Thus, volume of water in cylindrical flask = volume of water in conical flask
Thus volume of water `=pi\ (m\ r)^2\ h_w`
`=>1/3\ pi\ r^2h = pi\ (m\ r)^2\ h_w`
`=>h_w = (pi\ r^2\ h)/(3xxpi\ m^2\ r^2)`
`=>h_w = h/(3\ m^2)`
Thus, height of water in cylindrical flask `=h/(3\ m^2)` Answer
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