RD Chapter 16 Surface Areas and Volumes Ex 16.1 Solutions
Question - 21 : - 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.
Answer - 21 : -
Given,
Height of thecylindrical bucket = 32 cm
Radius of thecylindrical bucket = 18 cm
Height of conical heap= 24 cm
We know that,
Volume of cylinder = π× r2 × h
And, volume of cone =1/3 π × r2 × h
Then, from thequestion
Volume of the conicalheap = Volume of the cylindrical bucket
1/3 π × r2 ×24 = π × 182 × 32
r2 =182 x 4
r = 18 x 2 = 36 cm
Now,
Slant height of theconical heap (l) is given by
l = √(h2 +r2)
l = √(242 +362) = √1872
l = 43.26 cm
Therefore, the radiusand slant height of the conical heap are 36 cm and 43.26 cm respectively.
Question - 22 : - A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed.
Answer - 22 : -
Let the number ofcones made be n
Given,
Radius of metallicsphere = 5.6 cm
Radius of the cone =2.8 cm
Height of the cone =3.2 cm
We know that,
Volume of a sphere =4/3 π × r3
So, V1 =4/3 π × 5.63
And,
Volume of cone = 1/3 π× r2 × h
V2 =1/3 π × 2.82 × 3.2
Thus, the number ofcones (n) = Volume of the sphere/ Volume of the cone
n = 4/3 π × 5.63 /(1/3 π × 2.82 × 3.2)
n = (4 x 5.63)/(2.82 × 3.2)
n = 28
Therefore, 28 suchcones can be formed.
Question - 23 : - A solid cuboid of iron with dimensions 53 cm x 40 cm x 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe.
Answer - 23 : -
Let the length of thepipe be h cm.
Then, Volume of cuboid= (53 x 40 x 15) cm3
Internal radius of thepipe = 7/2 cm = r
External radius of thepipe = 8/2 = 4 cm = R
So, the volume of ironin the pipe = (External Volume) – (Internal Volume)
= πR2h – πr2h
= πh(R2– r2)
= πh(R – r) (R + r)
= π(4 – 7/2) (4 + 7/2)x h
= π(1/2) (15/2) x h
Then from the questionit’s understood that,
The volume of iron inthe pipe = volume of iron in cuboid
π(1/2) (15/2) x h = 53x 40 x 15
h = (53 x 40 x 15 x7/22 x 2/15 x 2) cm
h = 2698 cm
Therefore, the lengthof the pipe is 2698 cm.
Question - 24 : - The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder.
Answer - 24 : -
Given,
Internal diameter ofhollow spherical shell = 6 cm
So, the internalradius of hollow spherical shell = 6/2 = 3 cm = r
External diameter ofhollow spherical shell = 10 cm
So, the externaldiameter of hollow spherical shell = 10/2 = 5 cm = R
Diameter of thecylinder = 14 cm
So, the radius ofcylinder = 14/2 = 7 cm
Let the height ofcylinder be taken as h cm
Then, according to thequestion we have
Volume of cylinder =Volume of spherical shell
π × r2 ×h = 4/3 π × (R3 – r3)
π × 72 ×h = 4/3 π × (53 – 33)
h = 4/3 x 2
h = 8/3 cm
Therefore, the heightof the cylinder = 8/3 cm
Question - 25 : - A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone?
Answer - 25 : -
Given,
Internal diameter ofhollow sphere = 4 cm
So, the internalradius of hollow sphere = 2 cm
External diameter ofhollow sphere = 8 cm
So, the externalradius of hollow sphere = 4 cm
We know that,
Volume of the hollowsphere 4/3 π × (43 – 23) … (i)
Also given,
Diameter of the cone =8 cm
So, the radius of thecone = 4 cm
Let the height of thecone be x cm
Volume of the cone 1/3π × 42 ×h ….. (ii)
As the volume of thehollow sphere and cone are equal. We can equate equations (i) and (ii)
So, we get
4/3 π × (43 –23) = 1/3 π × 42 × h
4 x (64 – 8) = 16 x h
h = 14
Therefore, the heightof the cone so obtained will have a height of 14 cm
Question - 26 : - A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Answer - 26 : -
Given,
The internal radius ofhollow sphere = 2 cm
The external radius ofhollow sphere = 4 cm
We know that,
Volume of the hollowsphere 4/3 π × (43 – 23) … (i)
Also given,
The base radius of thecone = 4 cm
Let the height of thecone be x cm
Volume of the cone 1/3π × 42 ×h ….. (ii)
As the volume of thehollow sphere and cone are equal. We can equate equations (i) and (ii)
So, we get
4/3 π × (43 –23) = 1/3 π × 42 × h
4 x (64 – 8) = 16 x h
h = 14
Now,
Slant height of thecone (l) is given by
l = √(h2 +r2)
l = √(142 +42) = √212
l = 14.56 cm
Therefore, the heightand slant height of the conical heap are 14 cm and 14.56 cm respectively.
Question - 27 : - A spherical ball of radius 3 cm is melted and recast into three sphericalballs. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter ofthe third ball.
Answer - 27 : -
Given,
Radius of thespherical ball = 3 cm
We know that,
The volume of thesphere = 4/3 πr3
So, it’s volume (V) =4/3 πr3
That the ball ismelted and recast into 3 spherical balls.
Volume (V1)of first ball = 4/3 π 1.53
Volume (V2)of second ball = 4/3 π23
Let the radius of thethird ball = r cm
Volume of third ball(V3) = 4/3 πr3
Volume of thespherical ball is equal to the volume of the 3 small spherical balls.
Now,
Cancelling out thecommon part from both sides of the equation we get,
(3)3 =(2)3 + (1.5)3 + r3
r3 = 33–23– 1.53 cm3
r3 =15.6 cm3
r = (15.6)1/3 cm
r = 2.5 cm
As diameter = 2 xradius = 2 x 2.5 cm
= 5.0 cm.
Thus, the diameter ofthe third ball is 5 cm
Question - 28 : - A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic meters of gravel are required to grave the path to a depth of 20 cm?
Answer - 28 : -
Given,
Diameter of thecircular pond = 40 m
So, the radius of thepond = 40/2 = 20 m = r
Thickness (width ofthe path) = 2 m
As the whole view ofthe pond looks like a hollow cylinder.
And the height will be20 cm = 0.2 m
So,
Thickness (t) = R – r
2 = R – 20
R = 22 m
Volume of the hollowcylinder = π (R2– r2) × h
= π (222–202) × 0.2
= 52.8 m3
Therefore, the volumeof the hollow cylinder is the required amount of sand needed to spread acrossto a depth of 20 m.
Question - 29 : - A 16 m deep well with diameter 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7m. Find the height of the platform?
Answer - 29 : -
Let us assume the wellto be a solid right circular cylinder
Radius(r) of thecylinder = 3.5/2 m = 1.75 m
Depth of the well orheight of the cylinder (h) = 16 m
We know that,
Volume of the cylinder(V1) = πr2h
= π × 1.752 ×16
Given,
The length of theplatform (l) = 27.5 m
Breadth of theplatform (b) =7 m
Now, let the height ofthe platform be x m
We know that,
Volume of therectangle = l*b*h
V2 =27.5*7*x
As the earth dug up isspread evenly to form the platform
Volumes of both, thewell and the platform should be the same.
V1 = V2
π × 1.75 × 1.75 × 16 =27.5 × 7 × x
x = 0.8 m = 80 cm
Therefore, the heightof the platform is 80 cm.
Question - 30 : - A well of diameter 2 m is dug 14 m deep. The earth taken out of it is evenly spread all around it to form an embankment of height 40 cm. Find the width of the embankment?
Answer - 30 : -
Given,
Radius of the circularcylinder (r) = 2/2 m = 1 m
Height of the well (h)= 14 m
We know that,
Volume of the solidcircular cylinder = π r2h
= π × 12×14 …. (i)
And,
The height of theembankment (h) = 40 cm = 0.4 m
Let the width of theembankment be (x) m.
The embankment is ahollow cylinder with external radius = 1 + x and internal radius = 1
Volume of theembankment = π × r2 × h
= π × [(1 + x)2 –(1)2]× 0.4 ….. (ii)
As the well is spreadevenly to form embankment then the volumes will be same.
So, on equatingequations (i) and (ii), we get
π × 12 ×14 = π × [(1 + x)2 – (1)2] x 0.4
14/0.4 = 1 + x2 +2x – 1
35 = x2 +2x
x2 +2x – 35 = 0
Solving byfactorization method, we have
(x + 7) (x – 5) = 0
So, x = 5 m can onlybe the solution as it’s a positive value.
Therefore, the widthof the embankment is 5 m.