CBSE Class 10 Mathematics Surface Areas And Volume VBQs Set C

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 \( 9\text{ m} \times 8\text{ m} \times 2\text{ m} \) is melted and recast into solid cubes of edge \( 2\text{ m} \). Find number of cubes so formed. 
Answer: Given: dimensions of the metallic cuboid,
length \( l = 9\text{ m} \)
breath \( b = 8\text{ m} \)
height \( h = 2\text{ m} \)
side of a cube formed, \( a = 2\text{ m} \)
Let, the number of cubes formed be ‘\( n \)’.
According to the question:
Volume of the metallic cuboid \( = n \times \text{volume of the cube} \)
\( \Rightarrow l \times b \times h = n \times a^3 \)
\( \Rightarrow 9 \times 8 \times 2 = n \times 2 \times 2 \times 2 \)
\( \Rightarrow n = 18 \)
So, the number of cubes formed is 18.

Question. Find the ratio of the volume of a cube to that of the sphere which fits inside the cube.
Answer: Let the length of an edge of the cube is ‘\( a \)’ cm.
The radius of the sphere is \( \frac{a}{2} \)
Hence, Volume of cube : Volume of sphere
\( a^3 : \frac{4}{3}\pi \left(\frac{a}{2}\right)^3 \)
\( 6 : \pi \)

Question. How many solid spherical bullets can be made after melting a solid cube of iron whose edge measures \( 44\text{ cm} \), each bullet being \( 4\text{ cm} \) in diameter? [Take \( \pi = \frac{22}{7} \)] 
Answer: Given, a solid cube of iron of edge, \( a = 44\text{ cm} \)
Diameter of each bullet formed \( = 4\text{ cm} \)
Then, radius of each bullet \( (r) = 2\text{ cm} \)
Let, the number of spherical bullets formed be ‘\( n \)’.
Then, volume of solid cube \( = n \times \text{Volume of each spherical bullet} \)
\( a^3 = n \times \frac{4}{3}\pi r^3 \)
\( 44 \times 44 \times 44 = n \times \frac{4}{3} \times \frac{22}{7} \times (2)^3 \)
\( n = \frac{44 \times 44 \times 44 \times 3 \times 7}{4 \times 22 \times 2 \times 2 \times 2} \)
\( n = 2541 \)
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 \( 37\text{ cm} \). If the total surface area of the solid cylinder is \( 1628\text{ cm}^2 \), find the volume of the cylinder. (\( \pi = \frac{22}{7} \)) 
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)
\( 2\pi r \times 37 = 1628 \)
\( 2\pi r = \frac{1628}{37} \)
\( r = 7\text{ cm} \)
putting the value of \( r \) in eqn (i), we have
\( h = 37 - 7 = 30\text{ cm} \)
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 \( 4620\text{ cm}^3 \).

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:
\( \frac{\text{Volume of 1st sphere}}{\text{Volume of 2nd sphere}} = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{8}{27} \)
\( \Rightarrow \frac{r}{R} = \frac{2}{3} \Rightarrow R = \frac{3}{2}r \)
\( \Rightarrow (R - r) : r = \left(\frac{3}{2}r - r\right) : r \)
\( \Rightarrow \frac{r}{2} : r = 1 : 2 \).

SHORT ANSWER Type Questions

Question. A cone of height \( 24\text{ cm} \) and radius of base \( 6\text{ cm} \) 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, \( h = 24\text{ cm} \)
Radius of the cone, \( r = 6\text{ cm} \)
Let the radius of the sphere formed by reshaping cone be ‘\( R \)’.
As the cone is reshaped as sphere:
\( \therefore \text{Volume of cone} = \text{Volume of sphere} \)
\( \frac{1}{3} \pi r^2 h = \frac{4}{3} \pi R^3 \)
\( 6 \times 6 \times 24 = 4 \times R^3 \)
\( R^3 = 6 \times 6 \times 6 \)
\( R = 6\text{ cm} \)
Radius (\( R \)) of the sphere formed \( = 6\text{ cm} \)
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 \( 6\text{ cm} \) and its surface area is \( 144 \pi\text{ cm}^2 \).

Question. Three metallic solid cubes whose edges are \( 3\text{ cm}, 4\text{ cm} \) and \( 5\text{ cm} \) are melted and formed into a single cube. Find the edge of the cube so formed. 
Answer: Given : side of 1st cube, \( a_1 = 3\text{ cm} \)
side of 2nd cube, \( a_2 = 4\text{ cm} \)
side of 3rd cube, \( a_3 = 5\text{ cm} \)
We know that volume of cube \( = a^3 \) [Where \( a \) is side of cube]
\( \therefore \text{Volume of 1st cube, } V_1 = (3)^3 = 27\text{ cm}^3 \)
Volume of 2nd cube, \( V_2 = (4)^3 = 64\text{ cm}^3 \)
Volume of 3rd cube, \( V_3 = (5)^3 = 125\text{ cm}^3 \)
Let edge of resulting cube be \( x \).
Volume of resulting cube \( = \text{volume of (1st + 2nd + 3rd) cube} \)
\( x^3 = (27 + 64 + 125)\text{ cm}^3 \)
\( x^3 = 216\text{ cm}^3 \)
\( \Rightarrow x = 6\text{ cm} \)
Hence, the edge of cube so formed is \( 6\text{ cm} \).

Question. Two cones with the same base radius \( 8\text{ cm} \) and height \( 15\text{ cm} \) are joined together along their bases. Find the surface area of the shape so formed.
Answer: When two cones with the same base radius and height are joined, the shape so formed will be joined base to base, the total surface area of new solid becomes equal to the sum of curved surface areas of both the cones.
Given: radius of cone, \( r = 8\text{ cm} \), height of cone, \( h = 15\text{ cm} \).
slant height of cone, \( l = \sqrt{r^2 + h^2} = \sqrt{64 + 225} = \sqrt{289} = 17\text{ cm} \)
Surface area of shape formed, \( = \text{Curved surface area of 1st cone} + \text{curved surface area of 2nd cone} \)
\( = 2 \times \text{surface area of cone} \) [As both cones are identical]
\( = 2\pi rl = 2 \times \frac{22}{7} \times 8 \times 17 \)
\( = \frac{5984}{7} = 854.857\text{ cm}^2 \)
Hence, surface area of the shaped formed is \( 854.85\text{ cm}^2 \). \( = 855\text{ cm}^2 \) (approx.)

Question. From a solid cylinder, whose height is \( 2.4\text{ cm} \) and diameter \( 1.4\text{ cm} \), a conical cavity of the same height and same radius is hollowed out. Find the total surface area of the remaining solid. 
Answer: Here: the height of the cylinder \( (h) = \text{height of the cone } (h) = 2.4\text{ cm} \).
Radius of cylinder \( (r) = \text{Radius of cone } (r) = \frac{1.4}{2} = 0.7\text{ cm} \)
Then, slant height of the cone, \( l = \sqrt{h^2 + r^2} \)
\( = \sqrt{(2.4)^2 + (0.7)^2} \)
\( = \sqrt{5.76 + 0.49} \)
\( = \sqrt{6.25} = 2.5\text{ cm} \)
T.S.A. of the solid \( = \text{C.S.A. of cylinder} + \text{base area} + \text{C.S.A. of cone} \)
\( = 2\pi rh + \pi r^2 + \pi rl \)
\( = \pi r (2h + r + l) \)
\( = \frac{22}{7} \times 0.7 (2 \times 2.4 + 0.7 + 2.5) \)
\( = 2.2 \times 8 \)
\( = 17.6\text{ cm}^2 \)
Hence, the total surface area of the remaining solid is \( 17.6\text{ cm}^2 \).

Question. The radius and height of a solid right circular cone are in the ratio of 5 : 12. If its volume is \(314 \text{ cm}^3\), 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)
\(\frac{1}{3} \pi r^2 h = 314\)
\(\Rightarrow \frac{1}{3} \times 3.14 \times (5x)^2 \times 12x = 314\)
\(\Rightarrow x^3 = \frac{314 \times 3}{3.14 \times 25 \times 12}\)
\(\Rightarrow x^3 = 1 \Rightarrow x = 1\)
\(\therefore\) Radius of the cone \((r) = 5 \text{ cm}\)
Height of the cone \((h) = 12 \text{ cm}\)
Then, its total surface area = \(\pi rl + \pi r^2\)
where, \(l = \sqrt{h^2 + r^2} = \sqrt{12^2 + 5^2} = 13 \text{ cm}\)
\(\text{TSA} = \pi r(l + r)\)
\(\text{TSA} = 3.14(5 \times 13 + 5^2)\)
\(= 3.14(65 + 25)\)
\(= 282.6 \text{ cm}^2\)
Hence, the total surface area of the right circular cone is \(282.6 \text{ cm}^2\).

Question. A circus tent is in the shape of a cylinder surmounted by a conical top of the same diameter. If their common diameter is \(56 \text{ m}\), the height of the cylindrical part is \(6 \text{ m}\) and the total height of the tent above the ground is \(27 \text{ m}\), find the area of canvas used in making the tent. 
Answer: Diameter of conical and cylindrical part \((d) = 56 \text{ m}\)
\(\therefore\) Radius of conical & cylindrical part \((r) = 28 \text{ m}\)
Height of cylindrical part \((h) = 6 \text{ m}\)
Total height of tent = \(27 \text{ m}\)
\(\therefore\) Height of conical part \((h') = 27 - 6 = 21 \text{ m}\)
Slant height of conical part, \(l = \sqrt{h'^2 + r^2}\)
\(= \sqrt{21^2 + 28^2} = \sqrt{441 + 784} = \sqrt{1225} = 35 \text{ m}\)
Area of canvas used to make the tent:
= C.S.A. of cylindrical + C.S.A. of cone
\(= 2\pi rh + \pi rl\)
\(= \pi r (2h + l)\)
\(= \frac{22}{7} \times 28 (2 \times 6 + 35)\)
\(= 88 (12 + 35)\)
\(= 88 \times 47\)
\(= 4,136 \text{ m}^2\)

Question. Marbles of diameter \(1.4 \text{ cm}\) are dropped into a cylindrical beaker of diameter \(7 \text{ cm}\) containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by \(5.6 \text{ cm}\). 
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 = \(1.4 \text{ cm}\)
Radius = \(0.7 \text{ cm}\)
\(\Rightarrow \text{Volume of one marble} = \frac{4}{3} \pi(0.7)^3 = \frac{4}{3} \pi \times 0.343\) [\(\because\) Volume of sphere = \(\frac{4}{3} \pi r^3\)]
\(= \frac{1.372 \pi}{3} \text{ cm}^3\)
For beaker,
Diameter = \(7 \text{ cm}\)
Radius = \(3.5 \text{ cm}\)
Height of water level raised = \(5.6 \text{ cm}\)
\(\therefore\) 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, \(n = \frac{\text{Volume of raised water in beaker}}{\text{Volume of one spherical marble}}\)
\(= \frac{68.6 \pi}{1.372 \pi / 3} = \frac{68.6 \times 3}{1.372} = 150\)
Hence, 150 marbles are required.

Question. The \(\frac{3}{4}^{th}\) part of a conical vessel of internal radius \(5 \text{ cm}\) and height \(24 \text{ cm}\) is full of water. The water is emptied into a cylindrical vessel with internal radius \(10 \text{ cm}\). Find the height of water in cylindrical vessel. 
Answer: Given, a conical vessel of radius \((r) = 5 \text{ cm}\) and height \((h)\) of vessel = \(24 \text{ cm}\).
Radius \((R)\) of the cylindrical vessel = \(10 \text{ cm}\).
Let the height of the water in cylindrical vessel be \(H\).
According to the question,
\(\frac{3}{4} \times \text{Volume of water in conical vessel} = \text{Volume of water filled in the cylindrical vessel}\)
\(\frac{3}{4} \times \frac{1}{3} \pi r^2 h = \pi R^2 \times H\)
\(\frac{(5)^2 \times 24}{4} = 10 \times 10 \times H\)
\(H = \frac{5 \times 5 \times 6}{10 \times 10} = 1.5 \text{ cm}\)
Hence, the height of water in the cylindrical vessel is \(1.5 \text{ cm}\).

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

Question. A sphere of diameter \(6 \text{ cm}\) is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is \(12 \text{ cm}\). 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 = \(\frac{\text{diameter}}{2} = \frac{6}{2} = 3 \text{ cm}\)
Radius of cylinder vessel = \(\frac{12}{2} = 6 \text{ cm}\)
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
\(36 \pi = \pi \times 6 \times 6 \times h\)
\(h = 1 \text{ cm}\)
Thus, level of water rise in vessel is \(1 \text{ cm}\).

Question. How many silver coins \(1.75 \text{ cm}\) in diameter and thickness of \(2 \text{ mm}\), must be melted to form a cuboid of dimensions \(5.5 \text{ cm} \times 10 \text{ cm} \times 3.5 \text{ cm}\)? 
Answer: Let the number of coins taken to form a cuboid be ‘\(n\)’.
The radius \((r)\) of a silver coin = \(\frac{1.75}{2} \text{ cm}\)
Thickness or height \((h)\) of a coin = \(2 \text{ mm} = 0.2 \text{ cm}\)
Dimensions of a cuboid:
Length \((l) = 5.5 \text{ cm}\), Breadth \((b) = 3.5 \text{ cm}\) and height, \((H) = 10 \text{ cm}\)
Volume of one coin (cylindrical) \(\times\) number of coins = Volume of cuboid formed
\(\Rightarrow \pi r^2 h \times n = l \times b \times H\)
\(\Rightarrow \frac{22}{7} \times (\frac{1.75}{2}) \times (\frac{1.75}{2}) \times 0.2 \times n = 5.5 \times 10 \times 3.5\)
\(\Rightarrow n = \frac{5.5 \times 10 \times 3.5 \times 7 \times 2 \times 2}{22 \times 1.75 \times 1.75 \times 0.2}\)
\(= 400\)
Hence, the number of coins needed are 400.

Question. A hemispherical tank of diameter \(3 \text{ m}\) is full of water. It is being emptied by a pipe at the rate of \(3 \frac{4}{7} \text{ litre per second}\). How much time will it take to make the tank half empty? [Take \(\pi = \frac{22}{7}\)] 
Answer: Given : diameter \((d)\) of the hemispherical tank = \(3 \text{ m}\)
Then, the radius \((r)\) of the hemispherical tank = \(\frac{3}{2} \text{ m}\)
Half volume of a hemispherical tank:
\(V = \frac{1}{2} \times \frac{2}{3} \pi r^3 = \frac{1}{3} \times \frac{22}{7} \times (\frac{3}{2})^3 = \frac{22 \times 9}{8 \times 7} = \frac{99}{28} \text{ m}^3\)
Since, \(1 \text{ m}^3 = 1000 \text{ l}\), then \(V = \frac{99}{28} \times 1000 \text{ l}\)
Rate at which tank is emptied = \(\frac{25}{7} \text{ l/sec}\)
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 \(12 \text{ cm}\) and height \(15 \text{ cm}\) 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 = \(12 \text{ cm}\), Then, radius \((R) = 6 \text{ cm}\), Height \((H) = 15 \text{ cm}\)
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 = \(10 \times \text{volume of ice-cream cones}\)
\(\Rightarrow \pi R^2 H = 10 \left( \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 \right)\)
\(\Rightarrow 6 \times 6 \times 15 = 10 \left( \frac{1}{3} \times r^2 \times 4r + \frac{2}{3} \times r^3 \right)\)
\(\Rightarrow 9 \times 6 \times 3 = (4r^3 + 2r^3)\)
\(\Rightarrow 6r^3 = 9 \times 6 \times 3\)
\(\Rightarrow r^3 = 3 \times 3 \times 3 \Rightarrow r = 3 \text{ cm}\)
Diameter of conical part = \(2(3) = 6 \text{ cm}\).
Hence, the diameter of conical part is \(6 \text{ cm}\).

Question. A heap of rice is in the form of a cone of base diameter \(24 \text{ m}\) and height \(3.5 \text{ m}\). Find the volume of the rice. How much canvas cloth is required to just cover the heap? 
Answer: Conical heap of rice:
Dimensions: \(\text{diameter} = 24 \text{ m}\), \(\text{height} = 3.5 \text{ m}\), \(\Rightarrow \text{radius} = 12 \text{ m}\).
Volume of cone = \(\frac{1}{3} \pi r^2 h \text{ cu. units}\).
\(= \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 3.5 \text{ cu. m}\).
\(= 132 \times 4 = 528 \text{ cu. m}\).
The volume of the rice heap is \(528 \text{ cu. m}\).
Area of cloth required = Curved surface area.
CSA of cone = \(\pi rl \text{ sq. units}\) where \(l = \sqrt{h^2 + r^2} \text{ units}\).
Finding \(l\): \(l = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ m}\).
\(\Rightarrow \text{CSA} = \pi \times 12 \times 12.5 \text{ sq. units}\)
\(= \frac{22 \times 150}{7} = \frac{3300}{7} = 471.42857 \text{ m}^2\).
The area of canvas cloth required is \(471.42857 \text{ m}^2\).

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

Question. Water in a canal, \(5.4 \text{ m}\) wide and \(1.8 \text{ m}\) deep, is flowing with a speed of \(25 \text{ km/hour}\). How much area can it irrigate in \(40 \text{ minutes}\), if \(10 \text{ cm}\) of standing water is required for irrigation? 
Answer: speed of water in canal = \(25 \text{ km/hr}\).
in \(40 \text{ min} = \frac{40}{60} = \frac{2}{3} \text{ hr}\),
length of water = \(25 \times \frac{2}{3} = \frac{50}{3} \text{ km} = \frac{50000}{3} \text{ m}\).
volume of water in canal in \(40 \text{ minutes}\) = Volume of water for irrigation.
\(\frac{54}{10} \times \frac{18}{10} \times \frac{50000}{3} \text{ m}^3 = \frac{10}{100} \times l \times b \text{ m}^3\)
\(324 \times 5000 = l \times b \times \frac{1}{10}\)
\(1620000 = \text{area} \times \frac{1}{10}\)
\(\text{area irrigated in } 40 \text{ minutes is } 16200000 \text{ m}^2 = 1.62 \text{ km}^2 \text{ or } 162 \text{ hectares}\).

Question. The dimensions of a solid iron cuboid are \(4.4 \text{ m} \times 2.6 \text{ m} \times 1.0 \text{ m}\). It is melted and recast into a hollow cylindrical pipe of \(30 \text{ cm}\) inner radius and thickness \(5 \text{ cm}\). Find the length of the pipe.
Answer: For the hollow cylindrical pipe,
\(r = 30 \text{ cm}\) and \(R = 30 + 5 = 35 \text{ cm}\).
Let its length be \(h\).
Volume of the two is same.
\(\therefore 4.4 \times 2.6 \times 1.0 = \pi h (R^2 - r^2)\)
\(4.4 \times 100 \times 2.6 \times 100 \times 1.0 \times 100 = \frac{22}{7} \times h \times (35^2 - 30^2)\)
\(440 \times 260 \times 100 = \frac{22}{7} \times h \times (35 + 30)(35 - 30)\)
\(440 \times 260 \times 100 = \frac{22}{7} \times h \times 65 \times 5\)
\(h = \frac{440 \times 260 \times 100 \times 7}{22 \times 65 \times 5}\)
\(h = 11200 \text{ cm}\)
Pipe is \(11200 \text{ cm}\) or \(112 \text{ m}\) long. 

Question. A toy is in the form of a cone of radius \(3.5 \text{ cm}\) mounted on a hemisphere of same radius on its circular face. The total height of the toy is \(15.5 \text{ cm}\). Find the total surface area of the toy.
Answer: Height of hemisphere \( = r = 3.5 \text{ cm}\).
Height of cone \( = 15.5 \text{ cm} - 3.5 \text{ cm} = 12 \text{ cm} = h\).
Slant height of cone \( = \sqrt{r^2 + h^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm}\).
TSA of toy \( = \text{CSA of cone} + \text{CSA of hemisphere}\)
\(= \pi r l + 2\pi r^2\)
\(= \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times 3.5 \times 3.5\)
\(= 22 \times 12.5 \times 0.5 + 22 \times 3.5\)
\(= 22(6.25 + 3.5)\)
\(= 22(9.75) = 214.5 \text{ cm}^2\).
Total surface area of toy is \(214.5 \text{ cm}^2\). 

Question. A conical vessel, with base radius \(5 \text{ cm}\) and height \(24 \text{ cm}\), is full of water. This water is emptied into a cylindrical vessel of base radius \(10 \text{ cm}\). Find the height to which the water will rise in the cylindrical vessel. (Use \(\pi = \frac{22}{7}\))
Answer: Radius & height of conical vessel \( = 5 \text{ cm}\) & \(24 \text{ cm}\) resp.
Volume of cone \( = \frac{1}{3} \pi r^2 h\)
Volume of cone \( = \frac{1}{3} \pi \times 25 \times 24 \text{ cm}^3\)
Water is emptied of cylindrical vessel of \(r = 10 \text{ cm}\) & height \( = h\).
Volume of cone \( = \text{Volume of cylinder}\).
\(\Rightarrow \frac{1}{3} \pi \times 25 \times 24 = \pi \times 10 \times 10 \times h\)
\(\Rightarrow \frac{200}{100} = h\)
\(\Rightarrow 2 \text{ cm} = h\). 

Question. A sphere of diameter \(12 \text{ cm}\), is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by \(3 \frac{5}{9} \text{ cm}\). Find the diameter of the cylindrical vessel.
Answer: Diameter of sphere \( = 12 \text{ cm}\). Its radius \( = 6 \text{ cm}\).
Volume \( = \frac{4}{3} \pi \times 6^3 \text{ cm}^3\).
It is submerged into water in cylindrical vessel, thus water level rise by \(h = 3 \frac{5}{9} \text{ cm} = \frac{32}{9} \text{ cm}\).
Volume submerged \( = \text{Volume rise}\).
Let radius of cylinder be \(r \text{ cm}\).
\(\Rightarrow \frac{4}{3} \pi \times 6^3 = \pi \times r^2 \times \frac{32}{9}\)
\(\Rightarrow \frac{4 \times 216 \times 3}{3 \times 32} = r^2\)
\(\Rightarrow \frac{216 \times 3 \times 3}{32} = r^2\)
\(\Rightarrow r^2 = 81\)
\(\Rightarrow r = 9 \text{ cm}\).
Diameter \( = 2r = 2 \times 9 \text{ cm} = 18 \text{ cm}\). 

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 \(4.2 \text{ m}\) and height \(4 \text{ m}\) with the conical upper part of same diameter but of height \(2.8 \text{ m}\), and the canvas to be used costs \(₹ 100 \text{ per sq. m}\), find the amount, the associations will have to pay. What values are shown by these associations? Use \(\pi = \frac{22}{7}\).
Answer: Slant height \((l) = \sqrt{2.8^2 + 2.1^2} = 3.5 \text{ m}\).
\(\therefore \text{Area of canvas for one tent} = 2 \times \frac{22}{7} \times 2.1 \times 4 + \frac{22}{7} \times 2.1 \times 3.5\)
\(= 6.6(8 + 3.5) = 6.6 \times 11.5 \text{ m}^2 = 75.9 \text{ m}^2\).
\(\therefore \text{Area for 100 tents} = 75.9 \times 100 = 7590 \text{ m}^2\).
Cost of 100 tents \( = ₹ 7590 \times 100 = ₹ 7,59,000\).
\(50\% \text{ Cost} = \frac{759000}{2} = ₹ 3,79,500\).
Values: Helping the flood victims. 

Question. A hemispherical bowl of internal diameter \(36 \text{ cm}\) contains liquid. This liquid is filled into 72 cylindrical bottles of diameter \(6 \text{ cm}\). 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 \( = \frac{2}{3} \pi \cdot (18)^3 \cdot \frac{90}{100} \text{ cm}^3\).
Volume of liquid in 72 bottles \( = \pi(3)^2 \cdot h \cdot 72 \text{ cm}^3\).
\(h = \frac{\frac{2}{3} \pi (18)^3 \cdot \frac{9}{10}}{\pi (3)^2 \cdot 72} = 5.4 \text{ cm}\).

Question. A cubical block of side \(10 \text{ cm}\) 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 \(₹ 5 \text{ per 100 sq. cm}\). (Use \(\pi = 3.14\)).
Answer: Largest possible diameter of hemisphere \( = 10 \text{ cm}\).
\(\therefore \text{radius} = 5 \text{ cm}\).
Total surface area \( = 6(10)^2 + 3.14(5)^2 = 600 + 78.5 = 678.5 \text{ cm}^2\).
Cost of painting \( = \frac{678.5 \times 5}{100} = ₹ 33.9250 \approx ₹ 33.93\). 

Question. 504 cones, each of diameter \(3.5 \text{ cm}\) and height \(3 \text{ cm}\), are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. Use \(\pi = \frac{22}{7}\).
Answer: Volume of metal in 504 cones \( = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{35}{20} \times \frac{35}{20} \times 3\).
\(\frac{4}{3} \times \frac{22}{7} \times r^3 = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{35}{20} \times \frac{35}{20} \times 3\).
\(r = 10.5 \text{ cm}\). diameter \( = 21 \text{ cm}\).
Surface area \( = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 1386 \text{ cm}^2\). 

Question. In a cylindrical vessel of radius \(10 \text{ cm}\), 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 \(0.5 \text{ cm}\), then find 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,
\(V = \pi(10)^2h \text{ cu cm}\). ...(i)
Volume of 9000 small spherical balls of radius \(0.5 \text{ cm}\)
\(V_1 = 9000 \times \frac{4}{3} \pi \left(\frac{5}{10}\right)^3 \text{ cu. cm} = 1500\pi \text{ cu. cm}\). ...(ii)
Equating the equations, we get:
\(\pi(10)^2h = 1500\pi\)
\(h = \frac{1500}{100} = 15 \text{ cm}\).