RD Chapter 16 Surface Areas and Volumes Ex 16.1 Solutions
Question - 31 : - A well with inner radius 4 m is dug up and 14 m deep. Earth taken out of it has spread evenly all around a width of 3 m it to form an embankment. Find the height of the embankment?
Answer - 31 : -
Given,
Inner radius of thewell = 4 m
Depth of the well = 14m
We know that,
Volume of the cylinder= ╧А r2h
= ╧А ├Ч 42┬а├Ч14 тАж. (i)
From the question,itтАЩs told that
The earth taken outfrom the well is evenly spread all around it to form an embankment
And, the width of theembankment = 3 m
So, the outer radiusof the well = 3 + 4 m = 7 m
We know that,
Volume of the hollowembankment = ╧А (R2┬атАУ r2) ├Ч h
= ╧А ├Ч (72┬атАУ42) ├Ч h тАжтАж (ii)
On equating both theequations (i) and (ii), we get
╧А ├Ч 42┬а├Ч14 = ╧А ├Ч (72┬атАУ 42) ├Ч h
h = 42┬а├Ч14 / (33)
h = 6.78 m
Therefore, the heightof the embankment so formed is 6.78 m.
Question - 32 : - A well of diameter 3 m is dug up to 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.
Answer - 32 : -
Given,
Diameter of the well =3 m
So, the radius of thewell = 3/2 m = 1.5 m
Depth of the well (h)= 14 m
Width of theembankment (thickness) = 4 m
So, the radius of theouter surface of the embankment = (4 + 1.5) m = 5.5 m
Let the height of theembankment be taken as h m
We know that theembankment is a hollow cylinder
Volume of theembankment = ╧А (R2┬атАУ r2) ├Ч h
= ╧А (5.52┬атАУ1.52) ├Чh┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а┬а тАж.. (i)
Volume of earth dugout = ╧А ├Ч r2┬а├Ч h
= ╧А ├Ч 22┬а├Ч14┬а┬а┬а тАж.. (ii)
On equating both (i)and (ii) we get,
╧А (5.52┬атАУ1.52) ├Ч h┬а= ╧А ├Ч (3/2)2┬а├Ч 14
(30.25 тАУ 2.25) x h = 9x 14/ 4
h = 9 x 14/ (4 x 28)
h = 9/8 m
Therefore, the heightof the embankment is 9/8 m
Question - 33 : - Find the volume largest right circular cone that can be cut out of a cube whose edge is 9 cm.
Answer - 33 : -
Given,
The side of the cube =9 cm
The largest cone thatcan be cut from cube will have the base diameter = side of the cube
2r = 9
r = 9/2 cm = 4.5 cm
And,
Height of cone = sideof cube
So, height of cone (h)= 9 cm
Thus,
Volume of the largestcone to fit in = 1/3 ╧А ├Ч r2┬а├Ч h
= 1/3 ╧А ├Ч 4.52┬а├Ч9
= 190.93 cm3
Therefore, the volumeof the largest cone to fit in the cube has a volume of 190.93 cm3
Question - 34 : - 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?
Answer - 34 : -
Given,
Length of therectangular surface = 6 m = 600 cm
Breadth of therectangular surface = 4 m = 400 cm
Height of theperceived rain = 1 cm
So,
Volume of therectangular surface = length * breadth * height
= 600*400*1 cm3
= 240000 cm3┬а┬атАжтАжтАжтАжтАж..(i)
Also given,
Radius of thecylindrical vessel = 20 cm
Let the height of thecylindrical vessel be taken as h cm
We know that,
Volume of thecylindrical vessel = ╧А ├Ч r2┬а├Ч h
= ╧А ├Ч 202┬а├Чh┬а┬атАжтАжтАж.. (ii)
As all the rain wateris transferred to the cylindrical vessel
We can equate both (i)and (ii) for equal volumes,
240000 = ╧А ├Ч 202┬а├Чh
h = 190.9 cm
Therefore, the heightof the cylindrical vessel nearly 191 cm.