CBSE Class 10 Mathematics Surface Areas And Volume VBQs Set F

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 \(h_1\) and \(h_2\) be height and \(r_1, r_2\) be radii of two cones, then ratio of their volumes
\( = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} \)
Given, \(\frac{h_1}{h_2} = \frac{1}{3}\) and \(\frac{r_1}{r_2} = \frac{3}{1}\)
\( = \left(\frac{r_1}{r_2}\right)^2 \left(\frac{h_1}{h_2}\right) \)
\( = \left(\frac{3}{1}\right)^2 \left(\frac{1}{3}\right) = \frac{9}{1} \times \frac{1}{3} = \frac{3}{1}\)
Hence, ratio of their volumes is 3 : 1.

Question. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas.
Answer: Let the sides of two cubes be \(a\) and \(A\), then
\(\frac{a^3}{A^3} = \frac{1}{27}\) [Given]
\(\Rightarrow \frac{a}{A} = \frac{1}{3}\)
We know that the surface area of a cube is \(6(\text{side})^2\).
\(\therefore\) Ratio of surface areas \( = \frac{6a^2}{6A^2} = \left(\frac{a}{A}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)

Question. If the volume of a cube is \(8 \text{ cm}^3\), then what is the surface area of a cube.
Answer: Given, volume of a cube \( = 8 \text{ cm}^3\)
Let the side of a cube be \(a \text{ cm}\), then
\(a^3 = 8\)
\(\Rightarrow a = \sqrt[3]{8} = \sqrt[3]{2 \times 2 \times 2} = 2 \text{ cm}\)
Then, surface area of a cube \( = 6a^2 = 6 \times 2 \times 2 = 24 \text{ cm}^2\)

Question. Find the number of solid spheres of diameter 6 cm can be made by melting a solid metallic cylinder of height 45 cm and diameter 4 cm.
Answer: Let the number of spheres be \(n\).
Radius of sphere, \(r_1 = 3 \text{ cm}\)
Radius of cylinder, \(r_2 = 2 \text{ cm}\)
Volume of spheres = Volume of cylinder
\(n \times \frac{4}{3} \pi r_1^3 = \pi r_2^2 h\)
or, \(n \times \frac{4}{3} \times \frac{22}{7} \times (3)^3 = \frac{22}{7} \times (2)^2 \times 45\)
or, \(36n = 180\)
or, \(n = \frac{180}{36} = 5\)
Thus, the number of solid spheres = 5

Question. Three solid metallic spherical balls of radii 3 cm, 4 cm and 5 cm are melted into a single spherical ball, find its radius.
Answer: Let the radius of spherical ball be \(R\).
Volume of spherical ball = Volume of three balls
\(\frac{4}{3} \pi R^3 = \frac{4}{3} \pi [(3)^3 + (4)^3 + (5)^3]\)
or, \(R^3 = 27 + 64 + 125\)
or, \(R^3 = 216\)
or, \(R = 6 \text{ cm}\)

Question. 12 solid spheres of the same size are made by melting a solid metallic cone of base radius 1 cm and height 48 cm. Find the radius of each sphere.
Answer: No. of spheres = 12
Radius of cone, \(r = 1 \text{ cm}\)
Height of the cone = 48 cm
\(\therefore\) Volume of 12 spheres = Volume of cone
Let the radius of sphere be \(R \text{ cm}\),
\(12 \times \frac{4}{3} \pi R^3 = \frac{1}{3} \pi r^2 h\)
or, \(12 \times \frac{4}{3} \pi R^3 = \frac{1}{3} \pi \times (1)^2 \times 48\)
\(16R^3 = 16\)
or, \(R^3 = 1\)
or, \(R = 1 \text{ cm}\)

Question. The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm respectively. Find the curved surface area of the bucket.
Answer: Here, \(r_1 = 7 \text{ cm}, r_2 = 28 \text{ cm}, l = 45 \text{ cm}\)
Curved surface area of bucket \( = \pi l(r_1 + r_2)\)
\( = \frac{22}{7} \times 45(7 + 28)\)
\( = \frac{22}{7} \times 45 \times 35\)
\(\Rightarrow\) Curved surface area of bucket \( = 22 \times 45 \times 5 = 4,950 \text{ cm}^2\)

Short Answer Type Questions-I

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: Given, Volume of a right circular cylinder \( = 25\frac{1}{7} \text{ cm}^3\)
i.e., \(\pi r^2 h = \frac{176}{7}\)
where \(h\) is height and \(r\) is radius then \(r = h\),
\(\frac{22}{7} \times h^2 \times h = \frac{176}{7}\)
\(\Rightarrow h^3 = \frac{176}{22} = 8 = 2^3\)
Hence, height of the cylinder = 2 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: Let ABC be a cone, which is mounted on a hemisphere. Let radius be \(r \text{ cm}\) and height of cone be \(h\).
Curved surface area of the hemispherical part \( = \frac{1}{2}(4 \pi r^2) = 2 \pi r^2\)
Slant height of a cone, \(l = \sqrt{r^2 + h^2}\)
Curved surface area of cone \( = \pi r l = \pi r \sqrt{r^2 + h^2}\)
Given: \(2 \pi r^2 = \pi r \sqrt{r^2 + h^2}\)
\(\Rightarrow 2r = \sqrt{r^2 + h^2}\)
Squaring both the sides, we get
\(4r^2 = h^2 + r^2\)
\(\Rightarrow 3r^2 = h^2\)
\(\Rightarrow \frac{r^2}{h^2} = \frac{1}{3}\)
\(\Rightarrow \frac{r}{h} = \frac{1}{\sqrt{3}}\)
Hence, the ratio of the radius and the height is \(1 : \sqrt{3}\).

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 removed. Find the volume of the remaining solid.
Answer: Given, Height \((h) = 14 \text{ cm}\) and Base radius \((r) = 6 \text{ cm}\)
Volume of the remaining solid = Volume of a right circular cylinder – Volume of a right circular cone
\( = \pi r^2 h - \frac{1}{3} \pi r^2 h = \frac{2}{3} \pi r^2 h\)
\( = \frac{2}{3} \times \frac{22}{7} \times 6 \times 6 \times 14 = 1056 \text{ cm}^3\)

Aisha, a juice seller, was serving juice to her customers in two type of glasses. Both the glasses had inner radius 3 cm. The height of both the glasses was 10 cm. First type: A Glass with hemispherical raised bottom. Second type: A glass with conical raised bottom of height 1.5 cm. Isha insisted to have the juice in first type of glass and her father decided to have the juice in second type of glass.

Question. Out of the two, Isha or her father Suresh, who got more quantity of juice to drink and by how much ?
Answer: Capacity of first glass \( = \pi r^2 H - \frac{2}{3} \pi r^3\)
\( = \pi \times (3)^2 \times 10 - \frac{2}{3} \pi \times (3)^3\)
\( = 90\pi - 18\pi = 72\pi \text{ cm}^3\)
\( = 72 \times 3.14 = 226.08 \text{ cm}^3\)
Capacity of second glass \( = \pi r^2 H - \frac{1}{3} \pi r^2 h\)
\( = \pi(3)^2 \times 10 - \frac{1}{3} \pi (3)^2 \times 1.5\)
\( = 90\pi - 4.5\pi = 85.5\pi \text{ cm}^3\)
\( = 85.5 \times 3.14 = 268.47 \text{ cm}^3\)
Since, Suresh used second glass for drinking juice, so he got more quantity of juice.
Suresh got \(268.47 - 226.08 = 42.39 \text{ cm}^3\) more juice than Isha.

Question. A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the ratio between radius of their bases to their height is 3 : 4.
Answer: Let \(r\) be the radii of bases of cylinder and cone and \(h\) be the height.
Curved surface area of cylinder \( = 2 \pi r h\)
Curved surface area of cone \( = \pi r l = \pi r \sqrt{r^2 + h^2}\)
Given, \(\frac{2 \pi r h}{\pi r \sqrt{r^2 + h^2}} = \frac{8}{5}\)
\(\Rightarrow \frac{h}{\sqrt{r^2 + h^2}} = \frac{4}{5}\)
\(\Rightarrow \frac{h^2}{r^2 + h^2} = \frac{16}{25}\)
\(\Rightarrow 25h^2 = 16r^2 + 16h^2\)
\(\Rightarrow 9h^2 = 16r^2\)
\(\Rightarrow \frac{r^2}{h^2} = \frac{9}{16} \Rightarrow \frac{r}{h} = \frac{3}{4}\)
Hence, the ratio between radius of their bases to their heights is 3 : 4.

Question. A cylindrical glass tube with radius 10 cm has water upto a height of 9 cm. A metal cube of 8 cm edge is immersed completely. By how much the water level will rise in the glass tube ?
Answer: Let the height of water raised measured be \(h \text{ cm}\).
Volume of water displaced in cylinder \( = \pi(10)^2 h\)
Volume of cube \( = 8 \times 8 \times 8 \text{ cm}^3\)
\(\therefore \pi(10)^2 h = 8 \times 8 \times 8\)
\(\therefore h = \frac{8 \times 8 \times 8 \times 7}{22 \times 10 \times 10} = 1.629 \text{ cm}\)

Question. A solid metallic cuboid of dimensions 9 m \(\times\) 8 m \(\times\) 2 m is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.
Answer: Volume of cuboid \( = 9 \times 8 \times 2 \text{ m}^3\)
Volume of cube \( = 2 \times 2 \times 2 \text{ m}^3\)
Let number of recast cubes be \(n\).
\(n \times 2 \times 2 \times 2 = 9 \times 8 \times 2\)
\(n = \frac{9 \times 8 \times 2}{2 \times 2 \times 2} = 18\)
Hence, number of cubes recast = 18.

Question. A solid metallic cylinder of radius 3.5 cm and height 14 cm is melted and recast into a number of small solid metallic balls, each of radius \(\frac{7}{12} \text{ cm}\). Find the number of balls so formed.
Answer: Let the number of recast balls be \(n\).
Radius of cylinder, \(R = 3.5 \text{ cm}\), height, \(h = 14 \text{ cm}\).
Radius of balls, \(r = \frac{7}{12} \text{ cm}\).
\(n \times \frac{4}{3} \pi r^3 = \pi R^2 h\)
\(n \times \frac{4}{3} \times \left(\frac{7}{12}\right)^3 = (3.5)^2 \times 14\)
\(n = \frac{3.5 \times 3.5 \times 14 \times 3 \times 12 \times 12 \times 12}{4 \times 7 \times 7 \times 7} = 648\)
Hence, number of recast balls = 648.

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: Diameter of sphere = 6 cm \(\Rightarrow r = 3 \text{ cm}\).
Diameter of cylindrical vessel = 12 cm \(\Rightarrow R = 6 \text{ cm}\).
Volume of sphere \( = \frac{4}{3} \pi (3)^3 = 36 \pi \text{ cm}^3\).
Volume of sphere = Increase in volume of cylinder
\(36 \pi = \pi (6)^2 \times h\)
\(h = 1 \text{ cm}\).
\(\therefore\) Level of water rise in vessel = 1 cm.

Question. Find the number of coins of 1.5 cm diameter and 0.2 cm thickness to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer: Volume of coin \( = \pi (0.75)^2 \times 0.2 \text{ cm}^3\).
Volume of cylinder \( = \pi (2.25)^2 \times 10 \text{ cm}^3\).
No. of coins \( = \frac{\pi (2.25)^2 \times 10}{\pi (0.75)^2 \times 0.2} = 450\).

Question. A solid is in the form of a cylinder with hemispherical end. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. \((Use \pi = \frac{22}{7})\)
Answer: Radius of circular part \((r) = \frac{7}{2} \text{ cm}\).
Height of the cylinder \((h) = 20 - 7 = 13 \text{ cm}\).
Volume of solid = Volume of cylinder + 2 \(\times\) Volume of hemisphere
\(V = \pi r^2 h + 2 \times (\frac{2}{3} \pi r^3) = \pi r^2 (h + \frac{4}{3}r)\)
\(V = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times [13 + \frac{4}{3} \times \frac{7}{2}] = \frac{77}{2} \times (\frac{53}{3}) = 680.17 \text{ cm}^3\) (Approx).

Question. A cylindrical tank of radius 40 cm is filled upto height 3.15 m by another cylindrical pipe with the rate of 2.52 km/h in \(\frac{1}{2}\) hour. Calculate the diameter of cylindrical pipe ?
Answer: Radius of tank \((R) = 40 \text{ cm} = \frac{2}{5} \text{ m}\).
Height of tank filled \( = 3.15 \text{ m}\).
Volume of water in tank \( = \pi R^2 h = \pi (\frac{2}{5})^2 \times 3.15 \text{ m}^3\).
Let the radius of cylindrical pipe be \(r \text{ m}\).
Rate of flow \( = 2.52 \text{ km/h} = 0.7 \text{ m/s}\).
Time taken \( = \frac{1}{2} \text{ hour} = 1800 \text{ seconds}\).
Volume of water through pipe \( = \pi r^2 \times (\text{rate of flow}) \times \text{time}\)
\(\pi (\frac{2}{5})^2 \times 3.15 = \pi r^2 \times 0.7 \times 1800\)
\(r^2 = \frac{4 \times 315}{25 \times 100 \times 0.7 \times 1800} = \frac{1}{2500}\)
\(r = \frac{1}{50} \text{ m} = 2 \text{ cm}\).
Internal diameter of pipe \( = 2r = 4 \text{ cm}\) or 0.04 m.

Question. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap ?
Answer: Radius of conical heap \( = 12 \text{ m}\).
Volume of rice \( = \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 3.5 = 528 \text{ m}^3\).
Slant height \(l = \sqrt{12^2 + (3.5)^2} = 12.5 \text{ m}\).
Area of canvas required \( = \pi r l = \frac{22}{7} \times 12 \times 12.5 = 471.42 \text{ m}^2\).

Question. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Answer: Height of hemisphere \( = r = 3.5 \text{ cm}\).
Height of cone \( = 15.5 - 3.5 = 12 \text{ cm}\).
Slant height \(l = \sqrt{3.5^2 + 12^2} = 12.5 \text{ cm}\).
TSA of toy = CSA of cone + CSA of hemisphere
\( = \pi r l + 2 \pi r^2 = \pi r (l + 2r)\)
\( = \frac{22}{7} \times 3.5 \times (12.5 + 7) = 11 \times 19.5 = 214.5 \text{ cm}^2\).

Question. The \(\frac{3}{4}\)th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
Answer: Radius of conical vessel \( = 5 \text{ cm}\)
and its height \( = 24 \text{ cm}\)
Volume of this vessel \( = \frac{1}{3}\pi r^2 h\)
\( = \frac{1}{3} \times \pi \times 5 \times 5 \times 24\)
\( = 200\pi \text{ cm}^3\).
Internal radius of cylindrical vessel \( = 10 \text{ cm}\)
Let the height of emptied water be \(h\).
\(\therefore\) Volume of water in cylinder \( = \frac{3}{4} \times \text{Volume of cone}\)
\(\Rightarrow \pi r^2 h = \frac{3}{4} \times \text{Volume of cone}\)
\(\Rightarrow \pi \times 10 \times 10 \times h = 150\pi\)
\(\Rightarrow h = 1.5 \text{ cm}\)
Hence the height of water \( = 1.5 \text{ cm}\)

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 person's welfare. If the cost of 2 m wide canvas is ₹ 40 per metre, find the amount by which Rampal helped the centre.
Answer: Diameter of tent \( = 14 \text{ m}\) and height \( = 24 \text{ m}\)
\(\therefore\) radius of tent \( = 7 \text{ m}\)
Slant height \( = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2}\)
\( = \sqrt{576 + 49} = 25 \text{ m}\)
Surface area of the tent \( = \pi r l\)
\( = \frac{22}{7} \times 7 \times 25\)
\( = 550 \text{ m}^2\)
Surface area of 10 tents \( = 550 \times 10 = 5500 \text{ m}^2\)
Total cost \( = 5500 \times \frac{40}{2}\)
\( = Rs 110000\)
Hence, the amount by which Rampal helped the centre \( = Rs 110000\)

Question. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level into 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, then water level rise by \(3\frac{5}{9} \text{ cm} = \frac{32}{9} \text{ cm}\).
Volume submerged = Volume rise.
Let radius of cylinder be \(r \text{ cm}\).
\(\Rightarrow \frac{4}{3} \times \pi \times 6^3 = \pi \times r^2 \times \frac{32}{9}\)
\(\frac{4 \times 216 \times 3 \times 3}{3 \times 32} = r^2\)
\(\Rightarrow \frac{4 \times 27 \times 3}{4} = r^2 \Rightarrow 4 \times 81 \text{ cm}^2 = r^2\)
\(r = \frac{9}{2} \text{ cm}\) [Note: Correcting OCR transcription error based on math logic: \(\frac{4 \times 216 \times 3}{32 \times 3} = r^2 \Rightarrow r^2 = \frac{216 \times 3}{8} = 81 \Rightarrow r = 9\)]
Diameter \( = 2r = 2 \times 9 \text{ cm} = 18 \text{ cm}\).

Question. A well of diameter 4 m dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.
Answer: Diameter of earth dug out \( = 4 \text{ m}\)
Radius of earth dug out, \(r = 2 \text{ m}\)
Depth of the earth, \(h = 21 \text{ m}\),
Volume of earth dug out \( = \pi r^2 h = \frac{22}{7} \times 2 \times 2 \times 21 = 264 \text{ m}^3\)
Width of embankment \( = 3 \text{ m}\)
Outer radius of ring \( = 2 + 3 = 5 \text{ m}\)
Let the height of embankment be \(h\).
\(\therefore\) Volume of embankment = Volume of earth dug out
\(\pi(R^2 - r^2)h = 264\)
\(\frac{22}{7} \times (25 - 4) \times h = 264\)
\(h = \frac{264 \times 7}{22 \times 21} = 4\)
\(\therefore\) Height of embankment \( = 4 \text{ m}\)

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 sq. cm, find the volume of the cylinder. \(\left[ \text{Use } \pi = \frac{22}{7} \right]\)
Answer: Here \(r + h = 37\) and \(2\pi r(r + h) = 1628\)
or, \(2\pi r \times 37 = 1628\)
or, \(2\pi r = \frac{1628}{37}\)
or, \(r = 7 \text{ cm}\)
and \(h = 30 \text{ cm}\)
Hence, volume of cylinder \( = \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 30 = 4620 \text{ cm}^3\)

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 bowl \( = \frac{2}{3} \pi R^3\)
Volume of liquid in bowl \( = \frac{2}{3} \pi \times (18)^3 \text{ cm}^3\)
Volume of liquid after wastage \( = \frac{2}{3} \pi \times (18)^3 \times \frac{90}{100} \text{ cm}^3\)
Volume of one bottle \( = \pi r^2 h\)
Volume of liquid in 72 bottles \( = \pi \times (3)^2 \times h \times 72 \text{ cm}^3\)
Volume of bottles = Volume of liquid after wastage
\(\pi \times (3)^2 \times h \times 72 = \frac{2}{3} \pi \times (18)^3 \times \frac{90}{100}\)
or, \(h = \frac{\frac{2}{3} \pi \times (18)^3 \times \frac{90}{100}}{\pi \times (3)^2 \times 72}\)
Hence, the height of bottle \( = 5.4 \text{ cm}\)

Question. A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weights, a conical hole is drilled in the cylinder. The conical hole has a radius of \(\frac{3}{2} \text{ cm}\) and its depth \(\frac{8}{9} \text{ cm}\). Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.
Answer: Volume of cylinder \( = \pi r^2 h = \pi(3)^2 \times 5 = 45\pi \text{ cm}^3\)
Volume of conical hole \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{3}{2}\right)^2 \times \frac{8}{9} = \frac{2}{3} \pi \text{ cm}^3\)
Metal left in cylinder \( = 45\pi - \frac{2}{3}\pi = \frac{133\pi}{3} \text{ cm}^3\)
Again, the required ratio \( = \frac{\text{Volume of metal left}}{\text{Volume of metal taken out}} = \frac{\frac{133\pi}{3}}{\frac{2\pi}{3}} = 133 : 2\)
Hence, Volume of metal left : Volume of metal taken out \( = 133 : 2\).

Question. A solid right-circular cone of height 60 cm and radius 30 cm is dropped in a right-circular cylinder full of water of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder in cubic metre. \(\left[ \text{Use } \pi = \frac{22}{7} \right]\).
Answer: Volume of water in cylinder = Volume of cylinder \( = \pi r^2 h = \pi \times (60)^2 \times 180 = 648000\pi \text{ cm}^3\)
Water displaced on dropping cone = Volume of solid cone \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times (30)^2 \times 60 = 18000\pi \text{ cm}^3\)
Volume of water left in cylinder = Volume of cylinder – Volume of cone
\( = 648000\pi - 18000\pi = 630000\pi \text{ cm}^3\)
\( = \frac{630000 \times 22}{1000000 \times 7} \text{ m}^3 = 1.98 \text{ m}^3\)

Question. The rain water from 22 m \(\times\) 20 m roof drains into cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collected from the roof fills \(\frac{4}{5}\)th of cylindrical vessel then find the rainfall in cm.
Answer: Volume of water collected in cylindrical vessel \( = \frac{4}{5} \times \pi \times (1)^2 \times (3.5) \text{ m}^3\)
\( = \frac{4}{5} \times \frac{22}{7} \times 1 \times \frac{7}{2} = \frac{44}{5} \text{ m}^3\)
Let the rainfall is \(h \text{ m}\).
Volume of rain water from roof \( = 22 \times 20 \times h \text{ m}^3\)
or, \(22 \times 20 \times h = \frac{44}{5}\)
or, \(h = \frac{44}{5 \times 22 \times 20} = \frac{1}{50} \text{ m} = \frac{1}{50} \times 100 = 2 \text{ cm}\)

Question. A hollow cylindrical pipe is made up of copper. It is 21 dm long. The outer and inner diameters of the pipe are 10 cm and 6 cm respectively. Find the volume of copper used in making the pipe.
Answer: Height of cylindrical pipe, \(h = 21 \text{ dm} = 210 \text{ cm}\)
External radius, \(R = \frac{10}{2} = 5 \text{ cm}\)
Internal radius, \(r = \frac{6}{2} = 3 \text{ cm}\)
Volume of copper used in making the pipe = (Volume of external cylinder) – (Volume of internal cylinder)
\( = \pi R^2 h - \pi r^2 h = \pi h (R^2 - r^2)\)
\( = \frac{22}{7} \times 210 \times (5^2 - 3^2) = \frac{22}{7} \times 210 \times 16 = 10560 \text{ cm}^3\)

Question. A glass is in the shape of a cylinder of radius 7 cm and height 10 cm. Find the volume of juice in litre required to fill 6 such glasses. \(\left[ \text{Use } \pi = \frac{22}{7} \right]\)
Answer: Radius of the glass \( = 7 \text{ cm}\)
Height of the glass \( = 10 \text{ cm}\)
Volume of 1 glass \( = \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 10 = 1540 \text{ cm}^3\)
\(\therefore\) Volume of juice to fill 6 glasses \( = 6 \times 1540 = 9240 \text{ cm}^3\)
\(\therefore\) Volume in litre \( = \frac{9240}{1000} = 9.240 \text{ litre}\)

Question. A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is \(166\frac{5}{6} \text{ cm}^3\). Find the height of the toy. Also find the cost of painting the hemisphere part of the toy at the rate of ₹ 10 per \(cm^2\). \(\left[ \text{Use } \pi = \frac{22}{7} \right]\)
Answer: Given, radius of cone = radius of hemisphere = \(r = 3.5 \text{ cm}\).
Total volume, \(V = 166\frac{5}{6} \text{ cm}^3 = \frac{1001}{6} \text{ cm}^3\)
Let the height of cone be \(h\).
Total volume = Volume of cone + Volume of hemisphere
\(\frac{1001}{6} = \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3\)
or, \(\frac{1001}{6} = \frac{1}{3} \pi (3.5)^2 h + \frac{2}{3} \pi (3.5)^3\)
or, \(\frac{1001}{6} = \frac{1}{3} \pi [12.25h + 2 \times 42.875]\)
or, \(\frac{1001 \times 3 \times 7}{6 \times 22} = 12.25h + 85.75\)
or, \(\frac{21021}{132} = 12.25h + 85.75\)
or, \(12.25h = 159.25 - 85.75 \Rightarrow h = \frac{73.5}{12.25} = 6 \text{ cm}\)
Height of the toy \( = 6 + 3.5 = 9.5 \text{ cm}\).
Curved surface area of hemisphere \( = 2\pi r^2 = 2 \times \frac{22}{7} \times 3.5 \times 3.5 = 77 \text{ cm}^2\)
Cost of painting \( = Rs 10 \times 77 = Rs 770\)

Question. Two spheres of same metal weighs 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: Radius of first sphere \( = 3 \text{ cm}\). Mass \( = \frac{4}{3} \pi (3)^3 \cdot d = 1\) (where \(d = \text{density}\)).
Let radius of 2nd sphere be \(r \text{ cm}\). Mass \( = \frac{4}{3} \pi (r)^3 \cdot d = 7\).
\(\Rightarrow r^3 = 7(3)^3\).
\(\Rightarrow \frac{4}{3} \pi (3)^3 + \frac{4}{3} \pi (3)^3 \cdot 7 = \frac{4}{3} \pi R^3 \Rightarrow R^3 = (3)^3(1+7)\)
\(\Rightarrow R = 3(2) = 6\). \(\therefore\) Diameter \( = 12 \text{ cm}\).
Detailed Solution:
Weight of smaller sphere \( = 1 \text{ kg}\). Weight of larger sphere \( = 7 \text{ kg}\). Radius, \(r = 3 \text{ cm}\).
Volume of smaller sphere \( = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi(3)^3 = 36\pi \text{ cm}^3\).
Weight of recasted metal sphere \( = (1 + 7) \text{ kg} = 8 \text{ kg}\).
\(\therefore 8 \text{ kg}\) sphere occupies \(\frac{4}{3} \pi R^3\) where \(R\) is the radius of new sphere.
\(\Rightarrow 8 \times 36\pi = \frac{4}{3} \pi R^3 \Rightarrow R^3 = 216 \Rightarrow R = 6 \text{ cm}\).
Hence, diameter \( = 2R = 12 \text{ cm}\).

Question. A metallic solid sphere of radius 10.5 cm is melted and recasted into smaller solid cones each of radius 3.5 cm and height 3 cm. How many cones will be made ?
Answer: Radius of given sphere \( = 10.5 \text{ cm}\). Volume \( = \frac{4}{3} \pi \times 10.5^3\).
\( = 4\pi \times 3.5 \times 10.5 \times 10.5 \text{ cm}^3\).
Radius of one recasted cone \( = 3.5 \text{ cm}\) and height \( = 3 \text{ cm}\).
Volume \( = \frac{1}{3} \pi \times 3.5 \times 3.5 \times 3 = \pi \times 3.5 \times 3.5 \text{ cm}^3\).
Let the number of recasted cones be \(n\).
\(n \times \pi \times 3.5 \times 3.5 = 4 \times \pi \times 3.5 \times 10.5 \times 10.5\)
\(\Rightarrow n = 126\). Hence, number of recasted cones \( = 126\).

Question. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of same height and same base radius is hollowed out. Find the total surface area of the remaining solid. (Take \(\pi = 3.14\))
Answer: Height of cylinder = height of cone \( = 8 \text{ cm}\), radius \( = 6 \text{ cm}\).
Slant height of cone \( = \sqrt{8^2 + 6^2} = 10 \text{ cm}\).
Total surface area of remaining solid = Curved surface area of cylinder + Surface area of cone + Area of top cylinder.
\( = 2\pi r h + \pi r l + \pi r^2 = \pi r (2h + l + r)\)
\( = \frac{22}{7} \times 6 (16 + 10 + 6) = \frac{22}{7} \times 6 \times 32 = 603.43 \text{ cm}^2\).

Question. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recasted 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 cone \( = \frac{1}{3} \pi r^2 h\). Volume of metal in 504 cones \( = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \times 3\).
Volume of Sphere \( = \frac{4}{3} \pi r^3\). Since Volumes are equal:
\(\frac{4}{3} \times \frac{22}{7} \times r^3 = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{35}{20} \times \frac{35}{20} \times 3\)
\(r^3 = \left(\frac{21}{2}\right)^3 \Rightarrow r = 10.5 \text{ cm}\).
Diameter \( = 21 \text{ cm}\). Surface area \( = 4\pi r^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5 = 1386 \text{ cm}^2\).

Long Answer Type Questions 

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. \(\left[ \text{Take } \pi = \frac{22}{7} \right]\).
Answer: Radius \((r) = 7 \text{ cm}\), height of cone \((h) = 3.5 \text{ cm}\).
Volume of solid = Volume of hemisphere + volume of cone.
\( = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r + h)\)
\( = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times (14 + 3.5) = \frac{1}{3} \times 22 \times 7 \times 17.5\)
\( = \frac{1}{3} (2156 + 539) = 898.33 \text{ cm}^3\).

Question. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/hour. How much area will it irrigate in 30 minutes; if 8 cm standing water is needed ?
Answer: Length of canal covered in 30 min \( = 5000 \text{ m}\).
Volume of water flown \( = 6 \times 1.5 \times 5000 = 45000 \text{ m}^3\).
Area irrigated \( = \frac{\text{Volume}}{\text{Standing water height}} = \frac{45000}{0.08} = 562500 \text{ m}^2\).
Detailed Solution: Breadth \( = 6 \text{ m}\), Depth \( = 1.5 \text{ m}\), Speed \( = 10 \text{ km/hr}\).
Length in 30 min \( = 5 \text{ km} = 5000 \text{ m}\).
Volume \( = 5000 \times 6 \times 1.5 = 45000 \text{ m}^3\).
Volume = Area irrigated \(\times 8 \text{ cm} \Rightarrow 45000 = \text{Area} \times \frac{8}{100}\).
Area \( = 562500 \text{ m}^2 = 5.625 \times 10^5 \text{ m}^2\).

Question. Water is flowing at the rate of 15 km/h through a cylindrical pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time the level of water in pond rise by 21 cm ?
Answer: Quantity of water flowing through pipe in 1 hour \( = \pi \times \frac{7}{100} \times \frac{7}{100} \times 15000 \text{ m}^3\).
Required time \( = \left(50 \times 44 \times \frac{21}{100}\right) \div \left(\pi \times \frac{7}{100} \times \frac{7}{100} \times 15000\right) = 2 \text{ hours}\).
Detailed Solution: Speed \( = 15000 \text{ m/hr}\). Volume in 1 hr \( = \pi R^2 H = 231 \text{ m}^3\).
Volume in tank for 21 cm rise \( = 50 \times 44 \times 0.21 = 462 \text{ m}^3\).
Time taken \( = \frac{462}{231} = 2 \text{ hrs}\).

Question. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 \(cm^3\) of iron has approximately 8 gm mass. (Use \(\pi = 3.14\))
Answer: Volume of pole = Volume of small cylinder + Volume of large cylinder.
Small cylinder: \(r=8, h=60 \Rightarrow V_1 = 3.14 \times 8^2 \times 60 = 12057.6 \text{ cm}^3\).
Large cylinder: \(R=12, H=220 \Rightarrow V_2 = 3.14 \times 12^2 \times 220 = 99475.2 \text{ cm}^3\).
Total Volume \( = 111532.8 \text{ cm}^3\).
Mass \( = \frac{111532.8 \times 8}{1000} \text{ kg} = 892.262 \text{ kg}\).

Question. A vessel full of water is in the form of an inverted cone of height 8 cm and the radius of its top, which is open, is 5 cm. 100 spherical lead balls are dropped into vessel. One-fourth of the water flows out of the vessel. Find the radius of a spherical ball.
Answer: Volume of water in cone \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 5^2 \times 8 = \frac{200}{3} \pi \text{ cm}^3\).
Volume of water flown out \( = \frac{1}{4} \times \frac{200}{3} \pi = \frac{50}{3} \pi \text{ cm}^3\).
Let radius of ball be \(r\). \(100 \times \frac{4}{3} \pi r^3 = \frac{50}{3} \pi\).
\(r^3 = \frac{50}{400} = \frac{1}{8} \Rightarrow r = 0.5 \text{ cm}\).

Question. A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed.
Answer: Cylinder: \(r=6, h=15\). Cone: \(r=3, h=9\).
Let number of toys be \(n\).
\(n \times \frac{1}{3} \pi \times 3^2 \times 9 = \pi \times 6^2 \times 15\)
\(n = \frac{36 \times 15}{3 \times 9} = 20\). Hence, number of toys \( = 20\).

Question. From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. \(\left[ \text{Use } \pi = \frac{22}{7} \right]\)
Answer: Volume of cylinder \( = \pi r^2 h = \frac{22}{7} \times 4.2 \times 4.2 \times 10 = 554.40 \text{ cm}^3\).
Volume scooped out \( = 2 \times \text{Volume of hemisphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 4.2^3 = 310.46 \text{ cm}^2\).
Volume rest \( = 554.40 - 310.46 = 243.94 \text{ cm}^3\).
Wire: radius \( = 0.7 \text{ cm}\). \(\pi \times 0.7^2 \times l = 243.94\).
\(l = \frac{243.94 \times 10 \times 10}{22 \times 7} = 158.4 \text{ cm}\).