CBSE Class 10 Mathematics Surface Areas And Volume VBQs Set 09

Very Short Answer Type Questions

Question. Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes ? 
Answer: Let the heights of two cones be \( h \) and \( 3h \); and their base radii be \( 3r \) and \( r \).
As volume of cone, \( V_1 = \frac{1}{3} \pi r^2 h \)
Then, \( V_1 : V_2 = \frac{1}{3} \pi (3r)^2 (h) : \frac{1}{3} \pi (r)^2 (3h) \)
\( = 9 : 3 \) or \( 3 : 1 \)
Hence, the ratio of their volumes is 3 : 1.

Question. The volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere ? 
Answer: Let, the radius of the hemisphere be '\( r \)' cm.
Volume of the hemisphere \( = \frac{2}{3} \pi r^3 \) ...(i)
Surface area of the hemisphere \( = 2\pi r^2 + \pi r^2 \)
\( = 3\pi r^2 \) ...(ii)
According to the question:
\( \frac{2}{3} \pi r^3 = 3\pi r^2 \) [From (i) and (ii)]

\( \implies r = \frac{9}{2} \)
Hence, the diameter of the sphere = 9 cm

Question. Determine the volume of the largest possible cone that can be carved out from a hemisphere of radius '\( r \)' cm.
Answer: For the largest possible cone that can be carved out of a hemisphere of radius '\( r \)' cm has the following dimensions:
height = \( r \) cm and base radius = \( r \) cm
So, volume of the largest cone
\( = \frac{1}{3} \pi (r)^2 (r) = \frac{1}{3} \pi r^3 \)

Question. The volume of a right circular cylinder with the height equal to the radius is \( 25 \frac{1}{7} \text{ cm}^3 \). Find the height of the cylinder. (Use \( \pi = \frac{22}{7} \)) 
Answer: Let the height of the cylinder be '\( h \)' cm. Then, as per the question.
\( V = \pi (h)^2(h) = 25 \frac{1}{7} \quad [\because r = h] \)

\( \implies \frac{22}{7} \cdot h^3 = \frac{176}{7} \)

\( \implies h^3 = 8 \)
or \( h = 2 \text{ cm} \)
Thus, the height of the cylinder is 2 cm.

Question. A cone and a cylinder have the same radii but the height of the cone is 3 times that of the cylinder. Find the ratio of their volumes. 
Answer: Let '\( r \)' be the radii of the cone and the cylinder both.
Let '\( h \)' be the height of the cylinder. Then, the height of the cone will be '\( 3h \)'
Now, \( V_1 \) (Volume of cone) : \( V_2 \) (Volume of cylinder)
\( = \frac{1}{3} \pi r^2 (3h) : \pi r^2 h \)
\( = 1 : 1 \)

Question. How many cubes of side 2 cm can be made from a solid cube of side 10 cm ?
Answer: Volume of big cube of side, 10 cm = \( (10)^3 \) cu cm
\( = 1000 \) cu cm
Volume of a smaller cube of side 2 cm
\( = (2)^3 \) cu. cm
\( = 8 \text{ cm}^3 \)
Number of smaller cubes formed \( = \frac{1000}{8} = 125 \)

Question. The curved surface area of a cylinder is 264 m\(^2\) and its volume is 924 m\(^3\). Find the ratio of its height to its diameter. 
Answer: Curved surface area of cylinder \( = 2\pi r h \)
\( = 264 \text{ m}^2 \) ...(i)
Volume of cylinder \( = \pi r^2 h \)
\( = 924 \text{ m}^3 \)
According to question,
\( \frac{\pi r^2 h}{2\pi r h} = \frac{924}{264} \)
\( \frac{r}{2} = \frac{7}{2} \)
\( r = 7 \) m
Putting the value of \( r \) in equation (i), we have
\( 2 \times \frac{22}{7} \times 7 \times h = 264 \)
\( h = 6 \) m
\( \frac{h}{2r} = \frac{6}{14} = \frac{3}{7} \)
Hence, \( h : d = 3 : 7 \)

Question. If the area of three adjacent faces of a cuboid are X, Y, and Z respectively, then find the volume of cuboid. 
Answer: Let the length, breadth and height of the cuboid is \( l, b, \) and \( h \) respectively.
\( X = l \times b \)
\( Y = b \times h \)
\( Z = l \times h \)
\( XYZ = l^2 \times b^2 \times h^2 \)
Volume of cuboid = \( l \times b \times h \)
\( l^2 b^2 h^2 = XYZ \)
\( lbh = \sqrt{XYZ} \).

Question. A solid metallic cuboid of dimensions 9 m × 8 m × 2 m is melted and recast into solid cubes of edge 2 m. Find number of cubes so formed. 
Answer: Given: dimensions of the metallic cuboid,
length \( l = 9 \) m
breath \( b = 8 \) m
height \( h = 2 \) m
side of a cube formed, \( a = 2 \) m
Let, the number of cubes formed be '\( n \)'.
According to the question:
Volume of the metallic cuboid = \( n \times \) volume of the cube

\( \implies l \times b \times h = n \times a^3 \)

\( \implies 9 \times 8 \times 2 = n \times 2 \times 2 \times 2 \)

\( \implies n = 18 \)
So, the number of cubes formed is 18.

Question. How many solid spherical bullets can be made after melting a solid cube of iron whose edge measures 44 cm, each bullet being 4 cm in diameter ? [Take \( \pi = \frac{22}{7} \)] 
Answer: Given, a solid cube of iron of edge, \( a = 44 \) cm
Diameter of each bullet formed \( a = 4 \) cm
Then, radius of each bullet (\( r \)) = 2 cm
Let, the number of spherical bullets formed be '\( n \)'.
Then, volume of solid cube = \( n \times \) Volume of each spherical bullet
\( a^3 = n \times \frac{4}{3} \pi r^3 \)

\( \implies 44 \times 44 \times 44 = n \times \frac{4}{3} \times \frac{22}{7} \times (2)^3 \)

\( \implies n = \frac{44 \times 44 \times 44 \times 3 \times 7}{4 \times 22 \times 2 \times 2 \times 2} \)
\( = 2541 \)
Hence, the number of spherical bullets formed are 2,541.

Question. The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm\(^2\), find the volume of the cylinder. \( \left(\pi = \frac{22}{7}\right) \)
Answer: Let '\( r \)' be the radius and '\( h \)' be the height of the cylinder.
\( r + h = 37 \) ...(i)
\( 2\pi r(r + h) = 1628 \) ...(ii)
\( 2\pi r \times 37 = 1628 \)
\( 2\pi r = \frac{1628}{37} \)
\( r = 7 \) cm
putting the value of \( r \) in eqn (i), we have
\( h = 37 - 7 = 30 \) cm
volume of cylinder \( = \pi r^2 h \)
\( = \frac{22}{7} \times (7)^2 \times 30 \)
\( = 4620 \text{ cm}^3 \)
So, the volume of the cylinder is 4620 cm\(^3\).

Question. The ratio of the volumes of two spheres is 8 : 27. If \( r \) and \( R \) are the radii of sphere respectively, then find the \( (R - r) : r \).
Answer: Ratio of the volumes:
\( \frac{\text{Volume of } 1^{\text{st}} \text{ sphere}}{\text{Volume of } 2^{\text{nd}} \text{ sphere}} = \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{8}{27} \)

\( \implies \frac{r}{R} = \frac{2}{3} \)

\( \implies R = \frac{3}{2} r \)

\( \implies (R - r) : r = \left( \frac{3}{2} r - r \right) : r \)

\( \implies = \frac{r}{2} : r = 1 : 2 \).

Short Answer Type Questions

Question. A cone of height 24 cm and radius of base 6 cm is made up from modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere. Hence, find the surface area of this sphere.
Answer: Height of the cone, \( h = 24 \) cm
Radius of the cone, \( r = 6 \) cm
Let the radius of the sphere formed by reshaping cone be '\( R \)'.
As the cone is reshaped as sphere:
\(\therefore\) Volume of cone = Volume of sphere
\( \frac{1}{3} \pi r^2 h = \frac{4}{3} \pi R^3 \)

\( \implies 6 \times 6 \times 24 = 4 \times R^3 \)

\( \implies R^3 = 6 \times 6 \times 6 \)

\( \implies R = 6 \) cm
Radius (\( R \)) of the sphere formed = 6 cm
Surface area of the sphere \( = 4\pi R^2 \)
\( = 4 \times \pi (6)^2 \)
\( = 4 \times 36 \times \pi \)
\( = 144 \pi \text{ cm}^2 \)
Hence, the radius of the sphere is 6 cm and its surface area is \( 144 \pi \text{ cm}^2 \).

Question. Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed.
Answer: Given : side of \( 1^{\text{st}} \) cube, \( a_1 = 3 \) cm
side of \( 2^{\text{nd}} \) cube, \( a_2 = 4 \) cm
side of \( 3^{\text{rd}} \) cue, \( a_3 = 5 \) cm
We know that volume of cube \( = a^3 \)
[Where \( a \) is side of cube]
\(\therefore\) Volume of \( 1^{\text{st}} \) cube, \( V_1 = (3)^3 = 27 \text{ cm}^3 \)
Volume of \( 2^{\text{nd}} \) , \( V_2 = (4)^3 = 64 \text{ cm}^3 \)
Volume of \( 3^{\text{rd}} \) cube, \( V_3 = (5)^3 = 125 \text{ cm}^3 \)
Let edge of resulting cube be \( x \).
Volume of resulting cube = volume of (\( 1^{\text{st}} + 2^{\text{nd}} + 3^{\text{rd}} \)) cube
\( x^3 = (27 + 64 + 125) \text{ cm}^3 \)
\( x^3 = 216 \text{ cm}^3 \)

\( \implies x = 6 \) cm
Hence, the edge of cube so formed is 6 cm.

Question. The radius and height of a solid right circular cone are in the ratio of 5 : 12. If its volume is 314 cm\(^3\), find its total surface area. [Take \( \pi = 3.14 \)]
Answer: Let the radius (\( r \)) of the right circular cone be \( 5x \) and the height (\( h \)) of the right circular cone be \( 12x \).
Volume of the cone = 314 (given)
\( \frac{1}{3} \pi r^2 h = 314 \)

\( \implies \frac{1}{3} \times 3.14 \times (5x)^2 \times 12x = 314 \)

\( \implies x^3 = \frac{314 \times 3}{3.14 \times 25 \times 12} \)

\( \implies x^3 = 1 \)

\( \implies x = 1 \)
\(\therefore\) Radius of the cone (\( r \)) = 5 cm
Height of the cone (\( h \)) = 12 cm
Then, its total surface area \( = \pi r l + \pi r^2 \)
where, \( l = \sqrt{h^2 + r^2} = \sqrt{12^2 + 5^2} \)
\( = 13 \) cm
TSA \( = \pi r l + \pi r^2 \)
\(\therefore\) TSA \( = 3.14(5 \times 13 + 5^2) \)
\( = 3.14(65 + 25) \)
\( = 282.6 \text{ cm}^2 \)
Hence, the total surface area of the right circular cone is \( 282.6 \text{ cm}^2 \).

Question. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.
Answer: When \( n \) marbles are dropped into the beaker filled partially with water, the volume of water raised in the beaker, will be equal to the volume of \( n \) marbles, The shape of water raised in beaker is cylindrical.
Given: For marble
Diameter = 1.4 cm
Radius = 0.7 cm

\( \implies \) Volume of one marble \( = \frac{4}{3} \pi (0.7)^3 \)
\( = \frac{4}{3} \pi \times 0.343 \quad [\because \text{Volume of sphere} = \frac{4}{3} \pi r^3] \)
\( = \frac{1.372 \pi}{3} \text{ cm}^3 \)
For beaker
Diameter = 7 cm
Radius = 3.5 cm
Height of water level raised = 5.6 cm
\(\therefore\) Volume of raised water in beaker
\( = \pi r^2 h = \pi \times (3.5)^2 \times 5.6 \)
\( = 68.6 \pi \text{ cm}^3 \)
Volume of \( n \) spherical balls = Volume of water raised in cylinder
Required number of marbles, \( n \)
\( = \frac{\text{Volume of raised water in beaker}}{\text{Volume of one spherical marble}} \)
\( = \frac{68.6 \pi}{1.372 \pi} \times 3 = 150 \)
Hence, 150 marbles are required.

Question. The \( \frac{3}{4}^{\text{th}} \) part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
Answer: Given, a conical vessel of radius (\( r \)) = 5 cm
and height (\( h \)) of vessel = 24 cm
Radius (\( R \)) of the cylindrical vessel = 10 cm
Let the height of the water in cylindrical vessel be \( H \).
According to the question,

\( \implies \frac{3}{4} \times \) Volume of water in conical vessel
= Volume of water filled in the cylindrical vessel

\( \implies \frac{3}{4} \times \frac{1}{3} \pi r^2 h = \pi R^2 \times H \)

\( \implies \frac{(5)^2 \times 24}{4} = 10 \times 10 \times H \)

\( \implies H = \frac{5 \times 5 \times 6}{10 \times 10} = 1.5 \) cm
Hence, the height of water in the cylindrical vessel is 1.5 cm.

Question. Rampal decided to donate canvas for 10 tents, conical in shape with base diameter 14 m and height 24 m to a centre for handicapped persons' welfare. If the cost of 2m wide canvas is Rs. 40 per metre, find the amount by which Rampal helped the centre. 
Answer: Given: base diameter of the conical tent = 14 m
\(\therefore\) radius (\( r \)) of the conical tent \( = \frac{14}{2} = 7 \) cm
Height (\( h \)) of the tent = 24 m
Slant height of the conical tent, \( l = \sqrt{h^2 + r^2} \)
\( = \sqrt{h^2 + r^2} \)
\( = \sqrt{24^2 + 7^2} \)
\( = 25 \) m
C.S.A. of tent \( = \pi r l \)
\( = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^2 \)
C.S.A. of 10 tents = \( 550 \times 10 \text{ m}^2 = 5500 \text{ m}^2 \)
Cost of 2m wide canvas = Rs. 40 per meter
Cost of 5500 m\(^2\) canvas \( = \text{Rs. } 5500 \times \frac{40}{2} \)
= Rs. 1,10,000
Hence, the amount given by Rampal to help the centre is Rs. 1,10,000.

Question. A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel? 
Answer: Radius of sphere \( = \frac{\text{diameter}}{2} = \frac{6}{2} = 3 \) cm
Radius of cylinder vessel \( = \frac{12}{2} = 6 \) cm
Let, the level of water rise in cylinder be \( h \).
Volume of sphere, \( V = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \times \pi \times 3 \times 3 \times 3 = 36\pi \)
Volume of sphere = Increase volume in cylinder
\( 36 \pi = \pi \times 6 \times 6 \times h \)
\( h = 1 \) cm
Thus, level of water rise in vessel is 1 cm.

Question. How many silver coins 1.75 cm in diameter and thickness of 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm ? 
Answer: Let the number of coins taken to form a cuboid be '\( n \)'.
The radius (\( r \)) of a silver coin \( = \left(\frac{1.75}{2}\right) \) cm
Thickness or height (\( h \)) of a coin = 2 mm = 0.2 cm
Dimensions of a cuboid:
Length (\( l \)) = 5.5 cm,
Breadth (\( b \)) = 3.5 cm and height, (\( H \)) = 10 cm
Volume of one coin (cylindrical) \(\times\) number of coins = Volume of cuboid formed

\( \implies \pi r^2 h \times n = l \times b \times H \)

\( \implies \frac{22}{7} \times \left(\frac{1.75}{2}\right) \times \left(\frac{1.75}{2}\right) \times 0.2 \times n = 5.5 \times 10 \times 3.5 \)

\( \implies n = \frac{55 \times 3.5 \times 7 \times 2 \times 2}{22 \times 1.75 \times 1.75 \times 0.2} \)
\( = 400 \)
Hence, the number of coins needed are 400.

Question. A hemispherical tank of diameter 3 m is full of water. It is being empited by a pipe at the rate of \( 3 \frac{4}{7} \) litre per second. How much time will it take to make the tank half empty ? \(\left[ \text{Use } \pi = \frac{22}{7} \right]\) 
Answer: Given : diameter (\( d \)) of the hemispherical tank = 3 m
Then, the radius (\( r \)) of the hemispherical tank \( = \frac{3}{2} \) m
Half volume of a hemispherical tank :
\( V = \frac{1}{2} \times \frac{2}{3} \pi r^3 \)
\( = \frac{1}{3} \times \frac{22}{7} \times \left(\frac{3}{2}\right)^3 \)
\( = \frac{22 \times 9}{8 \times 7} = \frac{99}{28} \text{ m}^3 \)
Since, \( 1 \text{ m}^3 = 1000 \text{ l} \)
Then, \( V = \frac{99}{28} \times 1000 \text{ l} \)
Rate at which tank is emted \( = \frac{25}{7} \text{ l} / \text{sec} \)
Then, time taken \( = \frac{\frac{99}{28} \times 1000}{\frac{25}{7}} \text{ sec} \)
\( = \frac{99 \times 40}{4} \text{ sec} \)
\( = 990 \text{ sec} = 16.5 \text{ minutes.} \)

Question. A cylindrical tub whose diameter is 12 cm and height 15 cm is full of ice-cream. The whole ice-cream is to be divided into 10 children in equal ice-cream cones with conical base surmounted by a hemispherical top. If the height of the conical portion is twice the diameter of the base, find the diameter of conical part of ice-cream cone. 
Answer: Dimensions of cylindrical tub.
Diameter = 12 cm
Then, radius (\( R \)) = 6 cm
Height (\( H \)) = 15 cm
For ice-cream cones.
Let, the radius of cone and hemispherical top be '\( r \)'.
Then, height of conical portion,
\( h = 2(2r) = 4r \)
Volume of cylindrical tub = 10 \(\times\) volume of ice-cream cones

\( \implies \pi R^2 H = 10 \left( \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 \right) \)

\( \implies 6 \times 6 \times 15 = 10 \left( \frac{1}{3} \times r^2 \times 4r + \frac{2}{3} \times r^3 \right) \)

\( \implies 9 \times 6 \times 3 = (4r^3 + 2r^3) \)

\( \implies 6r^3 = 9 \times 6 \times 3 \)

\( \implies r^3 = 3 \times 3 \times 3 \)

\( \implies r = 3 \) cm
Diameter of conical part \( = 2(3) = 6 \) cm
Hence, the diameter of conical part is 6 cm.

Question. Sixteen glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions 20 cm × 10 cm × 10 cm and then the box is filled with water. Find the volume of water filled in the box.
Answer: Volume of the cubical box = \( (20 \times 10 \times 10) \) cu cm = 2000 cu cm
Volume of 16 glass spheres of radius 2 cm each
\( = \left[ 16 \times \frac{4}{3} \pi (2)^3 \right] \) cu cm
\( = 536.38 \) cu cm
\(\therefore\) Volume of water filled in the box
\( = (2000 - 536.38) \) cu cm
\( = 1463.62 \) cu cm

Question. Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs Rs. 100 per sq. m, find the amount, the associations will have to pay. What values are shown by these associations? \( \left[ \text{Use } \pi = \frac{22}{7} \right] \)
Answer: Slant height (\( l \)) \( = \sqrt{(2.8)^2 + (2.1)^2} = 3.5 \) cm
\(\therefore\) Area of canvas \( = 2 \times \frac{22}{7} \times (2.1) \times 4 + \frac{22}{7} \times 2.1 \times 3.5 \)
for one tent \( = 6.6(8 + 3.5) = 6.6 \times 11.5 \text{ m}^2 \)
\(\therefore\) Area for 100 tents \( = 66 \times 115 \text{ m}^2 \)
Cost of 100 tents = Rs. 66 \(\times\) 115 \(\times\) 100
50% Cost = 33 \(\times\) 11500 = Rs. 379500
Values: Helping the flood victims

Question. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer.
Answer: Volume of liquid in the bowl
\( = \frac{2}{3} \cdot \pi \cdot (18)^3 \text{ cm}^3 \)
Volume, after wastage \( = \frac{2\pi}{3} \cdot (18)^3 \cdot \frac{90}{100} \text{ cm}^3 \)
Volume of liquid in 72 bottles
\( = \pi (3)^2 \cdot h \cdot 72 \text{ cm}^3 \)
\( h = \frac{\frac{2}{3} \pi (18)^3 \cdot \frac{9}{10}}{\pi (3)^2 \cdot 72} = 5.4 \) cm.

Question. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. 5 per 100 sq. cm. [Use \( \pi = 3.14 \)]
Answer: Largest possible diameter of hemisphere = 10 cm.
\(\therefore\) radius = 5 cm.
Total surface area \( = 6(10)^2 + 3.14(5)^2 \)
Cost of painting \( = \frac{678.5 \times 5}{100} = \frac{\text{Rs. } 3392.50}{100} \)
= Rs. 33.9250 = Rs. 33.93

Question. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. \( \left[ \text{Use } \pi = \frac{22}{7} \right] \)
Answer: Volume of metal in 504 cones
\( = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{35}{20} \times \frac{35}{20} \times 3 \)
\( \frac{4}{3} \times \frac{22}{7} \times r^3 = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{35}{20} \times \frac{35}{20} \times 3 \)
\( r = 10.5 \) cm. diameter = 21 cm.
Surface area \( = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \)
\( = 1356 \text{ cm}^2 \)
[CBSE Marking Scheme 2015]

Question. In a cylindrical vessel of radius 10 cm, containing some water, 9000 small spherical balls are dropped which are completely immersed in water which raises the water level. If each spherical ball is of radius 0.5 cm, then the rise in the level of water in the vessel. 
Answer: Let '\( h \)' cm be the level of water in the cylindrical vessel
Then, volume of water in the vessel is,
\( V = \pi (10)^2 h \) cu cm. ...(i)
Volume of 9000 small spherical balls of radius 0.5 cm
\( V_1 = 9000 \times \frac{4}{3} \pi \left(\frac{5}{10}\right)^3 \) cu. cm.
\( = 1500 \pi \) cu. cm ...(ii)
Equating the equations, we get:
\( \pi (10)^2 h = 1500 \pi \)

\( \implies h = \frac{1500}{100} = 15 \)

Long Answer Type Questions

Question. A solid metallic hemisphere of radius 8 cm is melted and recast into a right circular cone of base radius 6 cm. Determine the height of the cone. 
Answer: For hemisphere, radius, \( r = 8 \) cm
Volume of hemisphere = \( \frac{2}{3} \pi r^3 \)
= \( \frac{2}{3} \times \pi \times (8)^3 = \frac{1024\pi}{3} \text{ cm}^3 \)
For cone, that is recasted from hemisphere
base radius, = 6 cm
Volume of cone = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 h \)
= \( 12\pi h \text{ cm}^3 \)
As volume remains the same, when a body is reformed to another body
\( \therefore \) Volume of hemisphere = Volume of cone
\( \frac{1024\pi}{3} = 12\pi h \)

\( \implies \) \( h = \frac{1024\pi}{3 \times 12\pi} = \frac{256}{9} = 28.44 \text{ cm} \)
Hence, the height of cone = 28.44 cm.

Question. A girl empties a cylindrical bucket full of sand, of base radius 18 cm and height 32 cm on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, find its slant height and correct it to one place of the decimal. 
Answer: Base radius of the cylindrical bucket, \( r = 18 \) cm
Height of the cylindrical bucket, \( h = 32 \) cm
Height of the conical heap, \( H = 24 \) cm
Let the base radius of the conical heap be \( R \).
Volume of the cylindrical bucket = Volume of the conical heap
[Since, the bucket of sand is emptied in the form of a conical heap]
\( \therefore \pi r^2 h = \frac{1}{3} \pi R^2 H \)

\( \implies \) \( 18 \times 18 \times 32 = \frac{1}{3} \times R^2 \times 24 \)

\( \implies \) \( R^2 = \frac{18 \times 18 \times 32}{8} = 18 \times 18 \times 4 \)

\( \implies \) \( R = \sqrt{18 \times 18 \times 2 \times 2} = 18 \times 2 = 36 \) cm
slant height, \( l = \sqrt{R^2 + H^2} \)
= \( \sqrt{(36)^2 + (24)^2} \)
= \( \sqrt{1296 + 576} \)
= \( \sqrt{1872} \)
= \( 43.27 = 43.3 \) cm
Hence, the slant height of the conical heap is 43.3 cm.

Question. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of a litre? 
Answer: Given: Shape of barrel of fountain pen is cylinder.
Length of barrel i.e., \( h = 7 \) cm
diameter of barrel i.e., \( d = 5 \) mm = \( \frac{5}{10} \text{ cm} = \frac{1}{2} \text{ cm} \)
\( \therefore \) radius, \( r = \frac{1}{2 \times 2} = \frac{1}{4} = 0.25 \text{ cm} \)

\( \implies \) Volume of barrel = \( \pi r^2 h \)
= \( \frac{22}{7} \times (0.25)^2 \times 7 \)
= \( 22 \times 0.0625 = 1.375 \text{ cm}^3 \)
According to question,
1.375 cm\(^3\) of ink can write 3300 words
\( \left(\frac{1}{5}\right)^{\text{th}} \) of a litre = \( \frac{1}{5} \times 1000 \text{ cm}^3 = 200 \text{ cm}^3 \).
\( \therefore 200 \text{ cm}^3 \) of ink can write = \( \frac{3300}{1.375} \times 200 \)
= 480000 words
Hence, \( \frac{1}{5}^{\text{th}} \) of a litre of ink can write 480000 words on an average.

Question. A solid metallic sphere, 3 cm in radius, is melted and recast into three spherical balls with radii 1.5 cm, 2 cm and \( x \) cm. Find the value of \( x \). 
Answer: Radius of a solid metallic sphere \( (R) = 3 \) cm
Radius of 3 spherical balls are:
First ball \( r_1 = 1.5 \) cm
Second ball \( r_2 = 2 \) cm
Third ball \( r_3 = x \) cm
According to the question volume of solid metallic sphere = Volume of 3 balls
\( \frac{4}{3} \pi R^3 = \frac{4}{3} \pi {r_1}^3 + \frac{4}{3} \pi {r_2}^3 + \frac{4}{3} \pi {r_3}^3 \)

\( \implies \) \( R^3 = {r_1}^3 + {r_2}^3 + {r_3}^3 \)

\( \implies \) \( (3)^3 = (1.5)^3 + (2)^3 + x^3 \)

\( \implies \) \( 27 = 3.375 + 8 + x^3 \)

\( \implies \) \( x^3 = 27 - 11.375 \)

\( \implies \) \( x^3 = 15.625 \)

\( \implies \) \( x = \sqrt[3]{\frac{15625}{1000}} \)
= \( \sqrt[3]{\frac{25 \times 25 \times 25}{10 \times 10 \times 10}} \)
= \( \frac{25}{10} = 2.5 \) cm
Hence, the value of \( x \) is 2.5 cm.

Question. Water flows at the rate of 10 m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm? 
Answer: When water flows through a pipe of a certain area of cross-section A with velocity v,
Then, volume of water coming from pipe in time t
= Area of cross-section \( \times \) Length
Given: speed of water flow = 10 m/min
= 1000 cm / min
For pipe
Diameter, \( d = 5 \) mm = \( \frac{5}{10} \) cm
Radius, \( r = \frac{5}{10 \times 2} \) cm = 0.25 cm
\( \therefore \) Amount of water that flows out of cylindrical pipe in 1 minute = \( \pi r^2 h \)
= \( \pi \times (0.25)^2 \times 10 \times 100 \)
= \( 62.5\pi \text{ cm}^3 \)
Amount of water required to fill conical vessel
= Volume of conical vessel
= \( \frac{1}{3} \pi r^2 h \)
Here, radius of conical vessel = \( \frac{40}{2} = 20 \) cm
Depth i.e., height of conical vessel = 24 cm

\( \implies \) Volume = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (20)^2 \times 24 \)
= \( 3200\pi \text{ cm}^3 \)
Time required to fill the vessel
= \( \frac{\text{Volume of conical vessel}}{\text{Volume of water that flows out in 1 minute}} \)
= \( \frac{3200\pi}{62.5\pi} = \frac{32000}{625} \)
= 51.2 minutes
= 51 minutes + \( \frac{2}{10} \times 60 \) seconds
= 51 minutes 12 seconds
Hence, the time required is 51 minutes 12 seconds.

Question. A factory manufactures 1,20,000 pencils daily. The pencils are cylindrical in shape, each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs. 0.05 per dm\(^2\). 
Answer: Given: Shape of the pencils is cylindrical.
Let radius of base be \( r \).
length of pencil, \( h = 25 \) cm
circumference of base = 1.5 cm

\( \implies \) \( 2\pi r = 1.5 \) cm

\( \implies \) \( r = \frac{1.5}{2\pi} \) cm
Curved surface area of 1 pencil
= \( 2\pi rh \)
= \( 2 \times \pi \times \frac{1.5}{2\pi} \times 25 = 37.5 \text{ cm}^2 \).
We know that
1 cm = 0.1 dm
1 cm\(^2\) = 0.01 dm\(^2\)
37.5 cm\(^2\) = \( 0.01 \times 37.5 \text{ dm}^2 \)
= 0.375 dm\(^2\)

\( \implies \) Curved surface area of 120000 pencils
= \( 0.375 \times 120000 \)
= 45000 dm\(^2\).
Given cost of colouring 1 dm\(^2\) curved surface area of pencil = Rs. 0.05
\( \therefore \) Cost of colouring 45000 dm\(^2\) CSA of pencil
= Rs. \( 0.05 \times 45000 \) = Rs. 2250

Question. From a rectangular block of wood, having dimensions 15 cm \( \times \) 10 cm \( \times \) 3.5 cm, a pen stand is made by making four conical depressions. The radius of each one of the depression is 0.5 cm and the depth is 2.1 cm. Find the volume of wood left in the pen-stand. 
Answer: Dimensions of the rectangular block of wood :
length \( (l) = 15 \) cm
breadth \( (b) = 10 \) cm
height \( (h) = 3.5 \) cm
Dimension of conical depression
radius of cone \( (r) = 0.5 \) m
height of cone \( (h) = 2.1 \) cm
Volume of wood left = Volume of cuboid - Total Volume of four conical depression
= \( l \times b \times h - 4 \times \frac{1}{3} \pi r^2 h \)
= \( 15 \times 10 \times 3.5 - 4 \times \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 2.1 \)
= \( 525 - 2.2 \)
= 522.8 cm\(^3\)
Hence, the volume of wood left in the pen stand is 522.8 cm\(^3\).

Question. In a hospital, used water is collected in a cylindrical tank diameter 2 m and height 5m. After recyling, this water is used to irrigate a park of a hospital whose length is 25 m and breadth is 20 m. If the tank is filled completely then what will be the height of standing water used for irrigating the park. 
Answer: Diameter of the cylindrical tank \( (d) = 2 \)m
\( \therefore \) Radius of the cylindrical tank \( (r) = 1 \) m
Height of the cylindrical tank \( (h) = 5 \) m
Then, volume of the cylindrical tank
= \( \pi r^2 h = \pi (1) \times 5 \)
= \( 5\pi \text{ m}^3 \) ...(i)
Length of the park \( (l) = 25 \) m
Breadth of the park \( (b) = 20 \) m
Let the height of standing water in the park be '\( h \)' m.
Then, the volume of water in the park
= \( lbh = 25 \times 20 \times h \) ...(ii)
Now, water from the tank is used to irrigate the park.
So, the volume of cylindrical tank = Volume of water in the park

\( \implies \) \( 5\pi = 25 \times 20 \times h \) [from (i) & (ii)]

\( \implies \) \( h = \frac{5\pi}{25 \times 20} \)
= \( \frac{\pi}{100} = \frac{3.14}{100} = 0.0314 \) m
Hence, the height of the standing water in park is 0.0314m.
By recycling water, better use of nature resource occurs without wastage. It helps in reducing and preventing pollution. It helps in saving and conserving water also.

Question. A solid metallic cylinder diameter 12 cm and height 15 cm is melted and recast into toys each in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed. 
Answer: Given: diameter of cylinder, \( d = 12 \) cm
\( \therefore \) radius of the cylinder, \( r = \frac{d}{2} = 6 \) cm
Height of the cylinder, \( h = 15 \) cm
radius of the cone, \( r' = 3 \) cm
radius of the cone, \( h' = 9 \) cm
According to the question:
Volume of cylinder = \( n \times \text{Volume of cone} \)
Where, \( n = \text{number of toys} \)
\( \therefore \pi r^2 h = n \times \frac{1}{3} \pi r'^2 h' \)
\( 6 \times 6 \times 15 = \frac{n}{3} \times 3 \times 3 \times 9 \)

\( \implies \) \( n = \frac{6 \times 6 \times 15}{9 \times 3} = 20 \)
Hence, the number of toys formed are 20.

Question. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread to form a platform of divisions 22 m \( \times \) 14 m. Find the height of the platform.
Answer: Let the height of the platform formed be '\( h \)' m.
For the cylindrical well:
height \( (h) = 20 \) m
Diameter \( (d) \) of the well = 7 m
Radius of the well \( (r) = \frac{7}{2} \) m
Dimensions of the platform formed,
length \( (l) = 22 \) m
Breadth \( (b) = 14 \) m
Now, according to the question :
Volume of the cylindrical well = volume of platform formed
\( \pi r^2 h = l \times b \times h \)

\( \implies \) \( \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 = 22 \times 14 \times h \)

\( \implies \) \( h = \frac{7 \times 20}{4 \times 14} = 2.5 \) m
Hence, the height of the platform formed is 2.5 m.

Question. Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. How much will the water level in the tank rise in half an hour?
Answer: Given: For cylindrical pipe Inner radius, \( r = 1 \) cm
Speed of water = 80 cm / sec
i.e., In 1 sec, water flows 80 cm
In 30 min i.e. \( 30 \times 60 \) seconds, water flows \( 80 \times 30 \times 60 = 144000 \) cm
Flowing water is filled in cylindrical tank. Hence, the volume of flowing water is equal to volume of water in cylindrical tank.
Volume of water that flows through cylindrical pipe in half an hour = \( \pi r^2 h \)
= \( \pi \times (1)^2 \times 144000 = 144000\pi \text{ cm}^3 \)
For cylindrical tank,
Base radius, \( r = 40 \) cm
Let height of water be raised by \( h \) cm
\( \therefore \) Volume of cylindrical tank = \( \pi (40)^2 \times h \)
= \( 1600\pi h \)
According to the question,
Volume of water in cylindrical tank = Volume of water flows through cylindrical pipe in half an hour

\( \implies \) \( 1600\pi h = 144000\pi \)

\( \implies \) \( h = \frac{144000\pi}{1600\pi} = 90 \) cm
Hence, the level of water rises to 90 cm in half an hour.

Question. The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having the base of diameter 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm. 
Answer: Let the rainfall be '\( a \)' cm = \( 0.01a \) m [because 1 m = 100 cm]
Given: For roof
Length, \( l = 22 \) m
Breadth, \( b = 20 \) m
\( \therefore \) Volume of water on roof = \( l \times b \times h \)
= \( 22 \times 20 \times \frac{a}{100} = \frac{22a}{5} \text{ m}^3 \)
For cylindrical vessel,
Diameter of base = 2 m
\( \therefore \) radius, \( r = 1 \) m
height, \( h = 3.5 \) m
\( \therefore \) Volume of water in cylindrical vessel = \( \pi r^2 h \).
= \( \frac{22}{7} \times (1)^2 \times 3.5 = 11 \text{ m}^3 \)
According to the question,
Volume of water on the roof = Volume of water in cylindrical vessel

\( \implies \) \( \frac{22a}{5} = 11 \)

\( \implies \) \( a = \frac{11 \times 5}{22} = 2.5 \) cm
Hence, the rainfall is 2.5 cm.

Question. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
Answer: Volume of the smaller sphere:
= \( \frac{4}{3} \pi (3)^3 \) cu cm
= \( 36\pi \) cu cm
Weight of the smaller sphere = 1 kg
\( \therefore \) density of metal = \( \frac{1}{36\pi} \)
Since, the weight of the greater sphere is 7 kg, its volume is \( \frac{7}{\frac{1}{36\pi}} \) i.e. \( 252\pi \)
Let R cm be the radius of the biggest sphere. Then,
\( \frac{4}{3} \pi R^3 = 36\pi + 252\pi \)

\( \implies \) \( \frac{4}{3} \pi R^3 = 288\pi \)

\( \implies \) \( R^3 = 216 \)

\( \implies \) \( R = 6 \)
Thus, the required diameter of the biggest sphere is 12 cm.

Question. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/hr, in how much time will the tank be filled ? 
Answer: Given: diameter \( (d) \) of the pipe = 20 cm
\( \therefore \) radius \( (r) \) of the pipe = 10 cm = 0.10 m
diameter \( (d) \) of the cylindrical tank = 10 m
\( \therefore \) radius \( (r) \) of the cylindrical tank = 5 m
height of the cylindrical tank = 2m
Rate of the flow of the water from the pipe = 3 km/hr
= \( \frac{3 \times 1000}{60 \times 60} \) m/sec
= \( \frac{5}{6} \) m / sec
Now, water required to fill the tank
= Volume of cylindrical tank = \( \pi r^2 h \)
= \( \pi (5)^2 \times 2 = 50\pi \text{m}^3 \)
Now, water that will flow from the pipe in 1 sec:
= \( \pi \times (0.1)^2 \times \frac{5}{6} \)
= \( \frac{5\pi}{600} \text{ m}^3 \)
Time taken to fill \( \frac{5\pi}{600} \text{ m}^3 = 1 \) sec
Then taken to fill \( 50\pi \text{ m}^3 = \frac{600}{5\pi} \times 50\pi \)
= 6000 sec = 100 min
Hence, the time taken to fill the tank is 100 min.

Question. Water is flowing at the rate of 5 km/hour through a pipe of diameter 14 cm into a rectangular tank of dimensions 50 m \( \times \) 44 m. Find the time in which the level of water in the tank will rise by 7 cm.
Answer: Given, diameter of the pipe = 14 cm
\( \therefore \) radius \( (r) \) of the pipe = 7 cm = \( \frac{7}{100} \) m
height of pipe, \( h = 5 \) km = 5000 m
[ \( \because \) distance covered in 1 hr = height of pipe]
Now, volume of water that flows out of the circular pipe in 1 hour
= \( \pi r^2 h \)
= \( \left( \frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times 5000 \right) \text{m}^3 \)
= 77 m\(^3\)
The dimensions of cuboidal tank,
\( l = 50 \) m, \( b = 44 \) m, \( h = \frac{7}{100} \) m
Volume of water to be filled in tank in '\( t \)' time
= \( lbh \)
= \( 50 \times 44 \times \frac{7}{100} \)
= 154 m\(^3\)
Total time in which 7 cm of water is filled in tank
= \( \frac{\text{Volume of cuboid}}{\text{Volume of water that flows through circular pipe in 1 hour}} \)
= \( \frac{154}{77} = 2 \) hr
Hence, the total time needed to rise the level of tank by 7 cm is 2 hours.

Question. A wall 24 m long, 0.4 m thick and 6 m high is constructed with bricks each of dimensions 25 cm \( \times \) 16 cm \( \times \) 10 cm. If the mortar occupies \( \left(\frac{1}{10}\right)^{\text{th}} \) of the volume of the wall, then find the number of bricks used in constructing the wall.
Answer: Given: Volume of cuboid = \( lbh \),
where, \( l = \text{length} \), \( b = \text{breadth} \) and \( h = \text{height} \)
For wall,
Length, \( l = 24 \) m
Breadth, \( b = 0.4 \) m
Height, \( h = 6 \) m
Volume of wall = \( l \times b \times h \)
= \( 24 \times 0.4 \times 6 \)
= \( \frac{24 \times 4 \times 6}{10} \text{ m}^3 \)
For 1 brick:
length \( l_b = 25 \) cm = 0.25 m
breadth, \( b_b = 16 \) cm = 0.16 m
height, \( h_b = 10 \) cm = 0.10 m
\( \therefore \) Volume of 1 brick = \( l \times b \times h \)
= \( 0.25 \times 0.16 \times 0.10 \text{ m}^3 \)
It is given that \( \left( \frac{1}{10} \right)^{\text{th}} \) of the volume of the wall is occupied by mortar, therefore \( \left( \frac{9}{10} \right)^{\text{th}} \) of the volume is covered by bricks.
\( \therefore \) Volume of wall covered by bricks = \( \frac{9}{10} \) of (Volume of wall)
= \( \frac{9}{10} \frac{(24 \times 4 \times 6)}{10} \)
No. of bricks
= \( \frac{\text{Volume of wall covered by bricks}}{\text{Volume of one brick}} \)
= \( \frac{\frac{9 \times 24 \times 4 \times 6}{10 \times 10}}{\frac{25 \times 16 \times 10}{1000000}} \)
= \( \frac{9 \times 24 \times 4 \times 6 \times 1000}{25 \times 16} \)
= \( 24 \times 6 \times 9 \times 10 = 12960 \)
Hence, the required no. of bricks in constructing the wall is 12960.

Question. Water in canal 6m wide and 1.5m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed for irrigation ?
Answer: Width of the canal = 6 m
Depth of the canal = 1.5 m
Length of the water column in 30 minutes i.e., \( \frac{1}{2} \) h
= \( \frac{10}{2} \text{ km} = \frac{10000}{2} \text{ m} = 5000 \text{ m} \)
Volume of water flows in \( \frac{1}{2} \) hour
= \( 6 \times 1.5 \times 5000 \text{ m}^3 \)
= \( 6 \times \frac{15}{10} \times 5000 = 45000 \text{ m}^3 \)
Since, the above amount (volume) of water is spread in the form of a cuboid of height 8 cm = \( \frac{8}{100} \) m
Let the area of the cuboid = \( a \text{ m}^2 \)
Volume of the cuboid
= Area \( \times \) Height = \( a \times \frac{8}{100} \text{ m}^3 \)
Thus, \( a \times \frac{8}{100} = 45000 \)
\( a = \frac{45000 \times 100}{8} = \frac{4500000}{8} = 562500 \text{ m}^2 \)
\( a = \frac{562500}{10000} \text{ hectares} \)
= 56.25 hectares
Thus, the required area = 56.25 hectares.