CBSE Class 10 Maths HOTs Surface Area and Volumes Set B

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: Sol. Given, edges of three solid cubes are 3 cm, 4 cm and 5 cm, respectively.
\( \therefore \) Volume of first cube \( = (3)^3 = 27 \text{ cm}^3 \)
[\( \because \text{volume of cube} = (\text{side})^3 \)]
Volume of second cube \( = (4)^3 = 64 \text{ cm}^3 \)
and volume of third cube \( = (5)^3 = 125 \text{ cm}^3 \)
\( \therefore \) Sum of volume of three cubes \( = (27 + 64 + 125) = 216 \text{ cm}^3 \)
Let the edge of the resulting cube \( = R \text{ cm} \)
Then, volume of the resulting cube, \( R^3 = 216 \Rightarrow R = 6 \text{ cm} \)

Question. The volume of a right circular cylinder with its 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: Sol. Let \( h \) and \( r \) be the height and radius of right circular cylinder, respectively.
Given, height of cylinder \( = \) Radius of cylinder i.e. \( h = r \)
\( \because \text{Volume of cylinder} = \pi r^2 h \)
\( \therefore 25 \frac{1}{7} = \frac{22}{7} \times h^2 \times h \) [\( \because h = r \text{ and } V = 25 \frac{1}{7}, \text{ given} \)]
\( \Rightarrow \frac{176}{7} = \frac{22}{7} \times h^3 \)
\( \Rightarrow h^3 = 8 \)
\( \Rightarrow h^3 = 2^3 \Rightarrow h = 2 \) [taking cube root]
Hence, height of cylinder is 2 cm.

Question. An iron pole consists of a cylinder of height 240 cm and base diameter 26 cm, which is surmounted by another cylinder of height 66 cm and radius 10 cm. Find the mass of the pole given that \( 1 \text{ cm}^3 \) of iron has approximately 8 g mass. (take, \( \pi = 3.14 \))
Answer: Sol. Here, solid iron pole is a combination of two cylinders.
For first cylinder,
Height \( = 240 \text{ cm} \)
Base diameter \( = 26 \text{ cm} \)
\( \therefore \) Base radius \( = \frac{26}{2} \text{ cm} = 13 \text{ cm} \)
For second cylinder,
Height \( = 66 \text{ cm} \)
Radius \( = 10 \text{ cm} \)
We know that, Volume of cylinder \( = \pi r^2 h \)
\( \therefore \) Total volume of iron pole \( = \text{Volume of first cylinder} + \text{Volume of second cylinder} \)
\( = \pi (13)^2 \times 240 + \pi (10)^2 \times 66 \)
\( = \pi [169 \times 240 + 100 \times 66] \)
\( = 3.14 [40560 + 6600] \)
\( = 3.14 \times 47160 = 148082.4 \text{ cm}^3 \)
Hence, total mass of the iron pole \( = 148082.4 \times 8 \text{ g} = 1184659.2 \text{ g} \)
[\( \text{given, } 1 \text{ cm}^3 \approx 8 \text{ g} \)]
\( = \frac{1184659.2}{1000} \text{ kg} = 1184.66 \text{ kg} \) [\( \because 1 \text{ g} = \frac{1}{1000} \text{ kg} \)]

Question. A spherical metal ball of radius 8 cm is melted to make 8 smaller identical balls. The radius of each new ball is ......... cm.
Answer: Sol. Let radius of larger sphere be \( R = 8 \text{ cm} \) and radius of smaller sphere be \( r \text{ cm} \).
Let number of smaller sphere be \( n = 8 \).
According to the given condition,
Volume of larger sphere \( = n \times \text{volume of smaller sphere} \)
\( \therefore \frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3 \)
\( \therefore (8)^3 = 8 \times r^3 \)
\( \Rightarrow r^3 = \frac{8^3}{8} \Rightarrow r^3 = 8^2 \Rightarrow r^3 = 64 = (4)^3 \)
\( \Rightarrow r = 4 \text{ cm} \) [taking cube root]
Hence, radius of new ball is 4 cm.

Question. A solid is in the shape of a cone mounted on a hemisphere of same base radius. If the curved surface areas of the hemispherical part and the conical part are equal, then find the ratio of the radius and the height of the conical part.
Answer: Sol. Let radius, height and slant height of a cone are \( r, h \) and \( l \), respectively. Then, radius of hemisphere will be \( r \).
Now, curved surface area of cone \( C_1 = \pi r l \) and curved surface area of hemisphere, \( C_2 = 2 \pi r^2 \).
According to the question, \( C_1 = C_2 \)
\( \therefore \pi r l = 2 \pi r^2 \Rightarrow l = 2r \) ...(i)
Also, \( l = \sqrt{r^2 + h^2} \)
\( \Rightarrow 2r = \sqrt{r^2 + h^2} \) [\( \because \) from Eq. (i)]
On squaring both sides, we get
\( (2r)^2 = (\sqrt{r^2 + h^2})^2 \Rightarrow 4r^2 = r^2 + h^2 \)
\( \Rightarrow 3r^2 = h^2 \Rightarrow (\sqrt{3}r)^2 = h^2 \)
Taking square root both sides, we get
\( \sqrt{3}r = h \Rightarrow \frac{r}{h} = \frac{1}{\sqrt{3}} \)
Hence, the ratio of the radius and height of the conical part is \( 1 : \sqrt{3} \).

Question. A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of cone is 3.5 cm, find the volume of the solid. (take, \( \pi = \frac{22}{7} \))
Answer: Sol. Given, radius of hemisphere and cone is \( r = 7 \text{ cm} \).
And height of cone \( (h) = 3.5 \text{ cm} \).
Now, volume of cone \( V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 3.5 = 179.67 \text{ cm}^3 \)
and Volume of hemisphere, \( V_2 = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (7)^3 = 718.67 \text{ cm}^3 \)
\( \therefore \) The volume of solid figure \( = \text{Volume of cone} + \text{Volume of hemisphere} = V_1 + V_2 \)
\( = 179.67 + 718.67 = 898.34 \text{ cm}^3 \)
Hence, volume of solid shape is \( 898.34 \text{ cm}^3 \).

Question. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid.
Answer: Sol. Given radius and height of cylinder are \( r = 6 \text{ cm} \) and \( h = 14 \text{ cm} \).
Also, radius and height of cone will be \( r_1 = 6 \text{ cm} \) and \( h_1 = 14 \text{ cm} \).
Now, volume of cylinder, \( V_1 = \pi r^2 h = \frac{22}{7} \times (6)^2 \times 14 = 1584 \text{ cm}^3 \)
Volume of cone, \( V_2 = \frac{1}{3} \pi r_1^2 h_1 = \frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 14 = 528 \text{ cm}^3 \)
\( \therefore \) Volume of remaining solid \( = \text{Volume of cylinder} - \text{Volume of cone} = V_1 - V_2 \)
\( = 1584 - 528 = 1056 \text{ cm}^3 \)
Hence, volume of the remaining solid is \( 1056 \text{ cm}^3 \).

Question. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 0.5 cm in diameter. A full barrel of ink in the pen can be used for writing 275 words on an average. How many words would be written using a bottle of ink containing one-fourth of a litre? [CBSE 2015]
Answer: Sol. Given, height of cylindrical pen \( = 7 \text{ cm} \).
Radius \( = \frac{\text{Diameter}}{2} = \frac{0.5}{2} \text{ cm} \).
\( \therefore \) Volume of barrel of a fountain pen \( = \pi r^2 h = \frac{22}{7} \times \left( \frac{0.5}{2} \right)^2 \times 7 = \frac{22}{16} \text{ cm}^3 \).
It is given that, a pen can write 275 words by using the ink \( \frac{22}{16} \text{ cm}^3 \).
\( \therefore \) Volume of ink \( = 275 \text{ words} \)
\( \Rightarrow \frac{22}{16} \text{ cm}^3 = 275 \text{ words} \)
\( \Rightarrow \frac{1}{4} \times 1000 \text{ cm}^3 = \frac{275 \times 16}{22} \times \frac{1}{4} \times 1000 = 50000 \) [\( \because \) he will use \( \frac{1}{4} \text{ L} \) of ink to write words]
Hence, the pen can write 50000 words by \( \frac{1}{4} \text{ L} \) of ink.

Question. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is \( 0.04 \text{ m}^3 \)?
Answer: Sol. Let the rise of water level in the pond be \( h \text{ m} \) when 500 persons are taking a dip into a cuboidal pond.
Given that, Length of the cuboidal pond \( = 80 \text{ m} \), Breadth of the cuboidal pond \( = 50 \text{ m} \).
Now, volume for the rise of water level in the pond \( = \text{Length} \times \text{Breadth} \times \text{Height} = 80 \times 50 \times h = 4000h \text{ m}^3 \).
and the average displacement of the water by a person \( = 0.04 \text{ m}^3 \).
So, the average displacement of the water by 500 persons \( = 500 \times 0.04 \text{ m}^3 \).
Now, by given condition, Volume for the rise of water level in the pond \( = \) Average displacement of the water by 500 persons.
\( \Rightarrow 4000h = 500 \times 0.04 \)
\( \therefore h = \frac{500 \times 0.04}{4000} = \frac{20}{4000} = \frac{1}{200} \text{ m} = 0.005 \text{ m} = 0.005 \times 100 \text{ cm} = 0.5 \text{ cm} \)
Hence, the required rise of water level in the pond is 0.5 cm.

Question. A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions \( 25 \text{ cm} \times 16 \text{ cm} \times 10 \text{ cm} \). If the mortar occupies \( \frac{1}{10} \text{th} \) of the volume of the wall, then find the number of bricks used in constructing the wall.
Answer: Sol. Given that, a wall is constructed with the help of bricks and mortar.
\( \therefore \text{Number of bricks} = \frac{(\text{Volume of wall}) - (\frac{1}{10} \text{th volume of wall})}{\text{Volume of a brick}} \) ...(i)
Also, given that Length of a wall \( (l) = 24 \text{ m} \), Thickness of a wall \( (b) = 0.4 \text{ m} \), Height of a wall \( (h) = 6 \text{ m} \).
So, volume of a wall constructed with the bricks \( = l \times b \times h = 24 \times 0.4 \times 6 = \frac{24 \times 4 \times 6}{10} \text{ m}^3 \).
Now, \( \frac{1}{10} \text{th volume of a wall} = \frac{1}{10} \times \frac{24 \times 4 \times 6}{10} = \frac{24 \times 4 \times 6}{10^2} \text{ m}^3 \).
and Length of a brick \( (l_1) = 25 \text{ cm} = \frac{25}{100} \text{ m} \), Breadth of a brick \( (b_1) = 16 \text{ cm} = \frac{16}{100} \text{ m} \), Height of a brick \( (h_1) = 10 \text{ cm} = \frac{10}{100} \text{ m} \).
So, volume of a brick \( = l_1 \times b_1 \times h_1 = \frac{25}{100} \times \frac{16}{100} \times \frac{10}{100} = \frac{25 \times 16 \times 10}{10^6} = \frac{25 \times 16}{10^5} \text{ m}^3 \).
From Eq. (i),
Number of bricks \( = \frac{\frac{24 \times 4 \times 6}{10} - \frac{24 \times 4 \times 6}{100}}{\frac{25 \times 16}{10^5}} = \frac{\frac{24 \times 4 \times 6}{100} \times 9}{\frac{25 \times 16}{10^5}} \)
\( = \frac{24 \times 4 \times 6 \times 9 \times 1000}{25 \times 16} = 24 \times 6 \times 9 \times 10 = 12960 \).
Hence, the required number of bricks used in constructing the wall is 12960.

Question. Water is flowing at the rate of 5 km/h through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Determine the time in which the level of the water in the tank will rise by 7 cm.
Answer: Sol. Suppose, the level of the water in the tank will rise by 7 cm in \( x \text{ h} \).
Since, the water is flowing at the rate of 5 km/h. Therefore, length of the water flow in \( x \text{ h} = 5x \text{ km} = 5000x \text{ m} \) [\( \because 1 \text{ km} = 1000 \text{ m} \)].
We have, diameter of cylindrical pipe \( = 14 \text{ cm} \).
\( \therefore \) Radius of cylindrical pipe, \( r = \frac{14}{2} = 7 \text{ cm} = \frac{7}{100} \text{ m} \).
Volume of the water flowing through the cylindrical pipe in \( x \text{ h} = \pi r^2 h = \frac{22}{7} \times \left( \frac{7}{100} \right)^2 \times 5000x = 77x \text{ m}^3 \).
Also, volume of the water that falls into the tank in \( x \text{ h} = l \times b \times h = 50 \times 44 \times \frac{7}{100} = 154 \text{ m}^3 \)
[\( \because l = 50 \text{ m, } b = 44 \text{ m and } h = \text{radius} = \frac{7}{100} \text{ m} \)]
\( \because \text{Volume of the water flowing through the cylindrical pipe in } x \text{ h} = \text{Volume of water that falls in the tank in } x \text{ h} \)
\( \Rightarrow 77x = 154 \Rightarrow x = 2 \)
Hence, the level of water in the tank will rise by 7 cm in 2 h.

Question. Two identical cubes each of volume \( 64 \text{ cm}^3 \) are joined together end to end. What is the surface area of the resulting cuboid?
Answer: Let the length of a side of a cube = \( a \text{ cm} \).
Given, volume of the cube, \( a^3 = 64 \text{ cm}^3 \implies a = 4 \text{ cm} \).
On joining two cubes, we get a cuboid whose length, \( l = 2a = 8 \text{ cm} \), breadth, \( b = a = 4 \text{ cm} \), and height, \( h = a = 4 \text{ cm} \).
Now, surface area of the resulting cuboid \( = 2(lb + bh + hl) = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32 + 16 + 32) = 2(80) = 160 \text{ cm}^2 \).

Question. How many shots each having diameter \( 3 \text{ cm} \) can be made from a cuboidal lead solid of dimensions \( 9 \text{ cm} \times 11 \text{ cm} \times 12 \text{ cm} \)?
Answer: Given, dimensions of cuboidal solid = \( 9 \text{ cm} \times 11 \text{ cm} \times 12 \text{ cm} \).
Volume of cuboidal lead solid = \( 9 \times 11 \times 12 = 1188 \text{ cm}^3 \).
Diameter of shot = \( 3 \text{ cm} \). Radius of shot, \( r = 1.5 \text{ cm} \).
Volume of one shot = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.5)^3 = \frac{297}{21} \approx 14.143 \text{ cm}^3 \).
Required number of shots = \( \frac{1188}{14.143} \approx 84 \text{ shots} \).

Question. 16 glass spheres each of radius \( 2 \text{ cm} \) are packed into a cuboidal box of internal dimensions \( 16 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm} \) and then the box is filled with water. Find the volume of water filled in the box.
Answer: Given, dimensions of the cuboidal box = \( 16 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm} \).
Volume of the cuboidal box = \( 16 \times 8 \times 8 = 1024 \text{ cm}^3 \).
Radius of one glass sphere = \( 2 \text{ cm} \).
Volume of one glass sphere = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2)^3 = \frac{704}{21} \approx 33.523 \text{ cm}^3 \).
Now, volume of 16 glass spheres = \( 16 \times 33.523 \approx 536.37 \text{ cm}^3 \).
Required volume of water = Volume of cuboidal box \( - \) Volume of 16 glass spheres = \( 1024 - 536.37 = 487.6 \text{ cm}^3 \).

Question. If a solid piece of iron in the form of a cuboid of dimensions \( 49 \text{ cm} \times 33 \text{ cm} \times 24 \text{ cm} \), is moulded to form a solid sphere. Then, find radius of the sphere.
Answer: Given, dimensions of the cuboid = \( 49 \text{ cm} \times 33 \text{ cm} \times 24 \text{ cm} \).
Volume of the cuboid = \( 49 \times 33 \times 24 = 38808 \text{ cm}^3 \).
Let the radius of the sphere be \( r \), then volume of the sphere = \( \frac{4}{3} \pi r^3 \).
According to the question, Volume of the sphere = Volume of the cuboid.
\( \frac{4}{3} \pi r^3 = 38808 \implies \frac{4}{3} \times \frac{22}{7} \times r^3 = 38808 \implies r^3 = \frac{38808 \times 3 \times 7}{4 \times 22} = 441 \times 21 = 21 \times 21 \times 21 \).
\( r = 21 \text{ cm} \). Hence, the radius of the sphere is \( 21 \text{ cm} \).

Question. If volumes of two spheres are in the ratio \( 64 : 27 \), then find the ratio of their surface areas.
Answer: Let the radii of the two spheres be \( r_1 \) and \( r_2 \).
Volume of first sphere, \( V_1 = \frac{4}{3} \pi r_1^3 \); Volume of second sphere, \( V_2 = \frac{4}{3} \pi r_2^3 \).
Given, ratio of volumes \( \frac{V_1}{V_2} = \frac{64}{27} \implies \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{64}{27} \implies \left(\frac{r_1}{r_2}\right)^3 = \frac{64}{27} \implies \frac{r_1}{r_2} = \frac{4}{3} \).
Now, ratio of surface areas \( = \frac{4 \pi r_1^2}{4 \pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \).
Hence, the required ratio of their surface area is \( 16 : 9 \).

Question. From a solid cube of side \( 7 \text{ cm} \), a conical cavity of height \( 7 \text{ cm} \) and radius \( 3 \text{ cm} \) is hollowed out. Find the volume of the remaining solid.
Answer: Given that, side of solid cube, \( a = 7 \text{ cm} \).
Height of conical cavity, \( h = 7 \text{ cm} \); Radius of conical cavity, \( r = 3 \text{ cm} \).
Volume of cube = \( a^3 = (7)^3 = 343 \text{ cm}^3 \).
Volume of conical cavity = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3)^2 \times 7 = 66 \text{ cm}^3 \).
Volume of remaining solid = Volume of cube \( - \) Volume of conical cavity = \( 343 - 66 = 277 \text{ cm}^3 \).

Question. A hemispherical bowl of internal radius \( 9 \text{ cm} \) is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius \( 1.5 \text{ cm} \) and height \( 4 \text{ cm} \). How many bottles are needed to empty the bowl?
Answer: Given, radius of hemispherical bowl, \( r = 9 \text{ cm} \).
Radius of cylindrical bottles, \( R = 1.5 \text{ cm} \) and height, \( h = 4 \text{ cm} \).
Number of bottles = \( \frac{\text{Volume of hemispherical bowl}}{\text{Volume of one cylindrical bottle}} \)
\( = \frac{\frac{2}{3} \pi r^3}{\pi R^2 h} = \frac{\frac{2}{3} \times 9 \times 9 \times 9}{1.5 \times 1.5 \times 4} = \frac{486}{9} = 54 \text{ bottles} \).

Question. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter \( 2 \text{ cm} \) and height \( 16 \text{ cm} \). Find the diameter of each sphere.
Answer: Given, diameter of cylinder = \( 2 \text{ cm} \), so Radius = \( 1 \text{ cm} \) and height = \( 16 \text{ cm} \).
Volume of cylinder = \( \pi \times (1)^2 \times 16 = 16\pi \text{ cm}^3 \).
Let the radius of each sphere be \( r \). Volume of 12 solid spheres = \( 12 \times \frac{4}{3} \pi r^3 = 16 \pi r^3 \).
According to the question, Volume of 12 spheres = Volume of cylinder.
\( 16 \pi r^3 = 16 \pi \implies r^3 = 1 \implies r = 1 \text{ cm} \).
Diameter of each sphere, \( d = 2r = 2 \times 1 = 2 \text{ cm} \).

Question. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is \( 104 \text{ cm} \) and the radius of each hemispherical end is \( 7 \text{ cm} \), then find the cost of polishing its surface at the rate of \( \text{₹} 2 \text{ per dm}^2 \). [take, \( \pi = \frac{22}{7} \)]
Answer: Total length = \( 104 \text{ cm} \), Radius of hemisphere = \( 7 \text{ cm} \).
Length of cylindrical part, \( h = 104 - 2(7) = 90 \text{ cm} \).
Total surface area = \( 2 \times \text{CSA of hemisphere} + \text{CSA of cylinder} = 2(2 \pi r^2) + 2 \pi r h = 2 \pi r(2r + h) \).
\( \text{TSA} = 2 \times \frac{22}{7} \times 7 \times [2(7) + 90] = 44 \times 104 = 4576 \text{ cm}^2 \).
\( \text{TSA in } \text{dm}^2 = \frac{4576}{100} = 45.76 \text{ dm}^2 \).
Cost of polishing at \( \text{₹} 2 \text{ per dm}^2 = 45.76 \times 2 = \text{₹} 91.52 \).

Question. A solid metallic hemisphere of radius \( 8 \text{ cm} \) is melted and recasted into a right circular cone of base radius \( 6 \text{ cm} \). Determine the height of the cone.
Answer: Volume of hemisphere = \( \frac{2}{3} \pi (8)^3 \). Volume of cone = \( \frac{1}{3} \pi (6)^2 h \).
Equating volumes: \( \frac{1}{3} \pi \times 36 \times h = \frac{2}{3} \pi \times 512 \implies 12 h = \frac{1024}{3} \implies h = \frac{512}{18} \approx 28.44 \text{ cm} \).

Question. A rectangular water tank of base \( 11 \text{ m} \times 6 \text{ m} \) contains water upto a height of \( 5 \text{ m} \). If the water in the tank is transferred to a cylindrical tank of radius \( 3.5 \text{ m} \), find the height of the water level in the tank.
Answer: Volume of water in rectangular tank = \( 11 \times 6 \times 5 = 330 \text{ m}^3 \).
Let height in cylindrical tank be \( h \). Volume = \( \pi (3.5)^2 h \).
\( \frac{22}{7} \times 3.5 \times 3.5 \times h = 330 \implies 38.5 h = 330 \implies h = \frac{330}{38.5} \approx 8.57 \text{ m} \text{ or } 8.6 \text{ m} \).

Question. A copper rod of diameter \( 1 \text{ cm} \) and length \( 8 \text{ cm} \) is drawn into a wire of length \( 8 \text{ m} \) of uniform thickness. Find the thickness of the wire.
Answer: Volume of rod = \( \pi \times (0.5)^2 \times 8 = 2\pi \text{ cm}^3 \).
Wire length = \( 8 \text{ m} = 800 \text{ cm} \). Let radius of wire be \( r \).
Volume of wire = \( \pi r^2 \times 800 = 2\pi \implies r^2 = \frac{2}{800} = \frac{1}{400} \implies r = \frac{1}{20} \text{ cm} \).
Thickness (diameter) = \( 2 \times \frac{1}{20} = \frac{1}{10} \text{ cm} = 1 \text{ mm} \).

Question. The rain water from a roof of dimensions \( 22 \text{ m} \times 20 \text{ m} \) drains into a cylindrical vessel having diameter of base \( 2 \text{ m} \) and height \( 3.5 \text{ m} \). If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall (in cm).
Answer: Volume of cylindrical vessel = \( \pi \times (1)^2 \times 3.5 = \frac{22}{7} \times 1 \times 3.5 = 11 \text{ m}^3 \).
Let rainfall be \( a \text{ cm} = \frac{a}{100} \text{ m} \). Volume of water on roof = \( 22 \times 20 \times \frac{a}{100} = \frac{22a}{5} \text{ m}^3 \).
Equating volumes: \( \frac{22a}{5} = 11 \implies a = \frac{11 \times 5}{22} = 2.5 \text{ cm} \).

Question. A cylindrical bucket of height \( 32 \text{ cm} \) and base radius \( 18 \text{ cm} \) 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 \( 24 \text{ cm} \), find the radius and slant height of the heap.
Answer: Volume of sand in cylinder = \( \pi \times (18)^2 \times 32 = 10368\pi \text{ cm}^3 \).
Let radius of cone be \( r \). Volume of conical heap = \( \frac{1}{3} \pi r^2 \times 24 = 8\pi r^2 \).
Equating volumes: \( 8\pi r^2 = 10368\pi \implies r^2 = 1296 \implies r = 36 \text{ cm} \).
Slant height \( l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 36^2} = \sqrt{576 + 1296} = \sqrt{1872} \approx 43.267 \text{ cm} \).

Long Answer Type Questions

Question. If a hollow cube of internal edge \( 22 \text{ cm} \) is filled with spherical marbles of diameter \( 0.5 \text{ cm} \) and it is assumed that \( \frac{1}{8} \) space of the cube remains unfilled. Then, the number of marbles that the cube can accommodate is
Answer: Volume of cube = \( (22)^3 = 10648 \text{ cm}^3 \).
Filled space = \( 10648 \times \frac{7}{8} = 9317 \text{ cm}^3 \).
Radius of marble \( = 0.25 \text{ cm} \). Volume of one marble \( = \frac{4}{3} \times \frac{22}{7} \times (0.25)^3 \approx 0.0655 \text{ cm}^3 \).
Number of marbles \( = \frac{9317}{0.0655} \approx 142244 \).

Question. A solid iron cuboidal block of dimensions \( 4.4 \text{ m} \times 2.6 \text{ m} \times 1 \text{ m} \) is recast into a hollow cylindrical pipe of internal radius \( 30 \text{ cm} \) and thickness \( 5 \text{ cm} \). Find the length of the pipe.
Answer: Volume of cuboid = \( 4.4 \times 2.6 \times 1 = 11.44 \text{ m}^3 \).
Hollow pipe: \( r_i = 30 \text{ cm} = 0.3 \text{ m} \); \( \text{Thickness} = 0.05 \text{ m} \), so \( r_e = 0.35 \text{ m} \).
Volume of pipe = \( \pi(r_e^2 - r_i^2)h = \frac{22}{7}(0.35^2 - 0.3^2)h = \frac{22}{7}(0.0325)h = 0.1021h \text{ approx} \).
Using \( \frac{22}{7} \times \frac{0.455}{7} h = 11.44 \implies h = 112 \text{ m} \).

Question. A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains \( 41 \frac{19}{21} \text{ m}^3 \) of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?
Answer: Let radius be \( r \). Total height \( H = \text{diameter} = 2r \).
Height of cylinder \( h = H - r = 2r - r = r \).
Total volume \( = \text{cyl} + \text{hemi} = \pi r^2(r) + \frac{2}{3} \pi r^3 = \frac{5}{3} \pi r^3 \).
\( \frac{5}{3} \times \frac{22}{7} \times r^3 = \frac{880}{21} \implies r^3 = 8 \implies r = 2 \text{ m} \).
Total height \( H = 2r = 4 \text{ m} \).

Question. A medicine-capsule is in the shape of a cylinder of diameter \( 0.5 \text{ cm} \) with two hemispheres stuck to each of its ends. The length of entire capsule is \( 2 \text{ cm} \). The capacity of the capsule is
Answer: Radius \( r = 0.25 \text{ cm} \). Entire length = \( 2 \text{ cm} \).
Length of cylinder \( h = 2 - 2(0.25) = 1.5 \text{ cm} \).
Capacity = \( \text{Vol of cyl} + 2(\text{Vol of hemi}) = \pi r^2 h + \frac{4}{3} \pi r^3 \).
\( = \frac{22}{7} \times [ (0.25)^2 \times 1.5 + \frac{4}{3} \times (0.25)^3 ] = \frac{22}{7} [0.09375 + 0.0208] \approx 0.36 \text{ cm}^3 \).

Question. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are \( 6 \text{ cm} \) and \( 12 \text{ cm} \), respectively. If the slant height of the conical portion is \( 5 \text{ cm} \), then find the total surface area and volume of the rocket. [use \( \pi = 3.14 \)]
Answer: Radius \( r = 3 \text{ cm} \). Cylinder height \( H = 12 \text{ cm} \). Slant height cone \( l = 5 \text{ cm} \).
Cone height \( h = \sqrt{5^2 - 3^2} = 4 \text{ cm} \).
Volume = \( \text{Cyl} + \text{Cone} = \pi r^2 H + \frac{1}{3} \pi r^2 h = \pi(3^2)(12) + \frac{1}{3} \pi(3^2)(4) = 108\pi + 12\pi = 120\pi = 376.8 \text{ cm}^3 \).
TSA = \( \text{Base cyl} + \text{CSA cyl} + \text{CSA cone} = \pi r^2 + 2\pi r H + \pi r l \).
\( \text{TSA} = \pi(3^2) + 2\pi(3)(12) + \pi(3)(5) = 9\pi + 72\pi + 15\pi = 96\pi = 301.44 \text{ cm}^2 \).

Question. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is \( 3 \text{ cm} \) and the diameter of the base is \( 4 \text{ cm} \). Determine the volume of the solid toy. If a right circular cylinder circumscribes the toy, then find the difference of the volumes of the cylinder and the toy. [take \( \pi = 3.14 \)]
Answer: Radius \( r = 2 \text{ cm} \). Cone height \( h = 3 \text{ cm} \).
Volume toy = \( \text{Vol hemi} + \text{Vol cone} = \frac{2}{3}\pi(2^3) + \frac{1}{3}\pi(2^2)(3) = \frac{16\pi}{3} + 4\pi \approx 16.75 + 12.56 = 29.31 \text{ cm}^3 \).
Cylinder dimensions: \( r = 2 \text{ cm} \), height \( H = h + r = 3 + 2 = 5 \text{ cm} \).
Volume cylinder = \( \pi r^2 H = 3.14 \times 4 \times 5 = 62.8 \text{ cm}^3 \).
Difference = \( 62.8 - 29.31 = 33.49 \text{ cm}^3 \).

Case Base Questions

Mathematics teacher of a school took her 10th standard students to show Ram Mandir. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Ram Mandir to students. Ram mandir is a Hindu temple that is being built in Ayodhya, which is in Uttar Pradesh. The temple construction is being supervised by the Shri Ram Janmabhoomi Teerth Kshetra. Then the teacher said in this monuments one can find combination of solid figures. She pointed that there are cubical bases and in centre cylinder with the cone shape structure on the top is constructed.

Question. Ram Mandir is constructed in the form of the cubical base of \( 30 \text{ cm} \times 20 \text{ cm} \times 10 \text{ cm} \), then find the area covered.
Answer: Area covered by the cuboid base \( = L \times B + 2H(L + B) = 30 \times 20 + 2 \times 10(30 + 20) = 600 + 1000 = 1600 \text{ cm}^2 \).

Question. If the radius of the cylinder is \( 7 \text{ cm} \) and Height of the cylinder is \( 60 \text{ cm} \) and the radius of the cone is similar to that of cylinder and Height of the cone is \( 24 \text{ cm} \), then the ratio of curved surface area of cylinder to curved surface area of the cone.
Answer: \( \text{CSA Cylinder} = 2 \pi R H = 2 \times \pi \times 7 \times 60 = 840\pi \).
Cone slant height \( L = \sqrt{7^2 + 24^2} = 25 \text{ cm} \). \( \text{CSA Cone} = \pi R L = \pi \times 7 \times 25 = 175\pi \).
Ratio \( = \frac{840\pi}{175\pi} = \frac{24}{5} \text{ or } 24 : 5 \).

Question. Given structure is based on the concept of
(a) Area and perimeter
(b) Surface area and volume
(c) Both (a) and (b)
(d) None of the above
Answer: (b)