CBSE Class 10 Maths HOTs Surface Area and Volumes Set F

Very Short Answer Type Questions

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 any cylinder shape is \( \pi r^2 h \).
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 \( = \frac{\text{Volume of cylinder}}{\text{Volume of coin}} \)
\( = \frac{\pi (2.25)^2 \times 10}{\pi (0.75)^2 \times 0.2} = \frac{(3)^2 \times 10}{0.2} \)
\( = 450 \)

Question. A cone of height 24 cm and radius of base 6 cm is made up of clay. If we reshape it into a sphere, find the radius of sphere.
Answer: Volume of sphere = Volume of cone
\( \frac{4}{3} \pi r_1^3 = \frac{1}{3} \pi r^2 h \)
\( \frac{4}{3} \times r_1^3 = (6)^2 \times \frac{24}{3} \)
\( 4r_1^3 = 36 \times 24 \)
\( r_1^3 = 6^3 \)
\( r_1 = 6 \text{ cm} \)
Hence, radius of sphere is 6 cm.

Question. A metallic sphere of total volume \( \pi \) is melted and recast into the shape of a right circular cylinder of radius 0.5 cm. What is the height of cylinder?
Answer: Volume of cylinder = Volume of sphere,
\( \pi r^2 h = \pi \)
where \( r \) and \( h \) are radius of base and height of cylinder
\( (0.5)^2 h = 1 \)
\( (\frac{1}{2})^2 h = 1 \)
\( h = 4 \text{ cm} \).

Question. A metallic solid sphere of radius 4.2 cm is melted and recast into the shape of a solid cylinder of radius 6 cm. Find the height of the cylinder.
Answer: Volume of sphere = Volume of cylinder
\( \frac{4}{3} \pi R^3 = \pi r^2 h \)
\( \frac{4}{3} \times (4.2)^3 = 6^2 \times h \)
\( h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \)
Hence, height of cylinder \( h = 2.744 \text{ cm} \).

Question. A right circular cone of radius 3 cm, has a curved surface area of 47.1 cm2. Find the volume of the cone. (Use \( \pi = 3.14 \))
Answer: We have \( r = 3, \pi rl = 47.1 \)
Thus \( l = \frac{47.1}{3 \times 3.14} = 5 \)
\( h = \sqrt{5^2 - 3^2} = 4 \text{ cm} \)
Volume of cone \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times 3 \times 3 \times 4 = 37.68 \text{ cm}^3 \)

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. \( \pi = \frac{22}{7} \)
Answer: Here, \( r + h = 37 \) ...(1)
and \( 2\pi r(r + h) = 1628 \) ...(2)
Thus \( 2\pi r \times 37 = 1628 \)
\( 2\pi r = \frac{1628}{37} \)
\( r = 7 \text{ cm} \)
Substituting \( r = 7 \) in (1) we have \( h = 30 \text{ cm} \).
Here volume of cylinder \( = \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 30 = 4620 \text{ cm}^3 \)

Question. A glass cylinder with diameter 20 cm has water to a height of 9 cm. A metal cube of 8 cm edge is immersed in it completely. Calculate the height by which water will rise in the cylinder. Use \( \pi = \frac{22}{7} \)
OR
A cylinder glass tube with radius 10 cm has water upto a height of 9 cm. A metal cube of 8 cm edge is immersed in it completely. By how much the water will rise in the glass tube. Use \( \pi = \frac{22}{7} \)
Answer: Let \( h \) be the height of water raised measured.
Volume of water displaced in cylinder \( = \pi (10)^2 h \)
Volume of cube, \( \pi (10)^2 h = 8 \times 8 \times 8 \)
\( h = \frac{8 \times 8 \times 8 \times 7}{22 \times 10 \times 10} \)
\( = 1.629 \text{ cm.} \)

Question. Two cubes of 5 cm each are kept together joining edge to edge to form a cuboid. Find the surface area of the cuboid so formed.
Answer: Let \( l \) be the length of the cuboid so formed.
Thus \( l = 5 + 5 = 10 \text{ cm, } b = 5 \text{ cm; } h = 5 \text{ cm.} \)
Surface area \( = 2(l \times b + b \times h + h \times l) \)
\( = 2(10 \times 5 + 5 \times 5 + 5 \times 10) \)
\( = 2(50 + 25 + 50) \)
\( = 2 \times 125 \)
\( = 250 \text{ cm}^2 \)

Question. A sphere of maximum volume is cut out from a solid hemisphere of radius 6 cm. Find the volume of the cut out sphere.
Answer: Diameter of sphere = Radius of hemisphere = 6 cm
Radius of sphere = 3 cm
Volume \( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 3^3 \text{ cm}^3 \)
\( = 113.14 \text{ cm}^3 \)

Question. A solid metallic 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 = \( 9 \times 8 \times 2 \text{ cm}^3 \)
Volume of cube = \( 2 \times 2 \times 2 \text{ cm}^3 \)

Question. Let number of recast cubes be \( n \). Volume of \( n \) cubes = Volume of cuboid.
Answer: \( n \times 2 \times 2 \times 2 = 9 \times 8 \times 2 \)
\( n = \frac{9 \times 8 \times 2}{2 \times 2 \times 2} = 18 \)
Hence, number of cubes recast = 18

Question. The slant height of a bucket is 26 cm. The diameter of upper and lower circular ends are 36 cm and 16 cm. Find the height of the bucket.
Answer: Here, \( l = 26 \text{ cm} \), upper radius = 18 cm,
lower radius = 8 cm
\( d = \text{difference in radius} = 18 - 8 = 10 \text{ cm.} \)
Let \( h \) be the height of bucket
\( h = \sqrt{l^2 - d^2} = \sqrt{(26)^2 - (10)^2} \)
\( = \sqrt{676 - 100} = \sqrt{576} = 24 \text{ cm.} \)

Question. A cylinder and a cone have base radii 5 cm and 3 cm respectively and their respective heights are 4 cm and 8 cm. Find the ratio of their volumes.
Answer: Volume of cylinder = \( \pi (5)^2 \times 4 \text{ cm}^3 \)
\( = 100\pi \text{ cm}^3 \)
Volume of cone = \( \frac{1}{3} \pi \times 3^2 \times 8 \)
\( = 24\pi \)
Required ratio = \( 100\pi : 24\pi \)
\( = 25 : 6 \)

Question. A heap of wheat is in the form of cone of diameter 6 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? Use \(\pi = \frac{22}{7}\)
Answer: Volume of wheat in the form of cone
\(\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 3 \times 3 \times 3.5\)
\(= 11 \times 3 = 33 \text{ m}^3\)
\(l = \sqrt{3^2 + 3.5^2} = 4.609 \text{ m}\)
Canvas required to cover the heap
\(\pi rl = \frac{22}{7} \times 3 \times 4.609\)
\(= 43.45 \text{ m}^2\).

Question. The internal and external diameters of a hollow hemispherical vessel are 16 cm and 12 cm respectively. If the cost of painting 1 \( \text{cm}^2 \) of the surface area is Rs. 5.00, find the total cost of painting the vessel all over. (Use \( \pi = 3.14 \))
Answer: Here \( R = 8 \text{ cm}, r = 6 \text{ cm} \)
Surface area \( = 2\pi R^2 + 2\pi r^2 + \pi(R^2 - r^2) \)
\( = \pi [2 \times 8^2 + 2 \times 6^2 + (8^2 - 6^2)] \)
\( = \pi [64 \times 2 + 36 \times 2 + (64 - 36)] \)
\( = \pi [128 + 72 + 28] \)
\( = 228 \times 3.14 = 715.92 \text{ cm}^2 \)
Total cost \( = 715.92 \times 5 = \text{Rs. } 3579.60 \).

Question. A solid metallic cone of radius 2 cm and height 8 cm is melted into a sphere. Find the radius of sphere.
Answer: Let the radius of sphere be \( R \).
Volume of sphere = Volume of cone
\( \frac{4}{3} \pi R^3 = \frac{1}{3} \pi r^2 h \)
\( \frac{4}{3} \pi R^3 = \frac{1}{3} \pi \times 2 \times 2 \times 8 \)
\( R^3 = \frac{2 \times 2 \times 8}{4} \)
\( R^3 = 8 \)
\( R = 2 \) cm

Question. The radii of two right circular cylinders are in the ratio of 2 : 3 and their height are in the ratio of 5 : 4. Calculate the ratio of their curved surface area and radio of their volumes.
Answer: Let the radii of two cylinders be \(2x\) and \(3x\) and their heights be \(5y\) and \(4y\) respectively.
Ratio of their curved surface areas
\(= \frac{2\pi \times 2x \times 5y}{2\pi \times 3x \times 4y} = \frac{5}{6}\)
Since their curved surface areas are in the ratio of 5 : 6.
Ratio of their volumes
\(= \frac{\pi \times (2x)^2 \times 5y}{\pi \times (3x)^2 \times 4y} = \frac{4 \times 5}{9 \times 4} = \frac{5}{9}\)
Hence, their volumes are in the ratio of 5 : 9 and their C.S.A. are in the ratio of 5 : 6.

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

Question. Water is flowing at 7 m/s through a circular pipe of internal diameter of 4 cm into a cylindrical tank, the radius of whose base is 40 cm. Find the increase in water level in 30 minutes.
Answer: Volume of water in 30 minutes
\(= \pi \times (2)^2 \times 700 \times 60 \times 30 \, \text{cm}^3\)
Let height of water in tank = \(h \, \text{cm}\)
and radius = 40 cm
Volume of water in the tank
\(\pi \times 40^2 \times h = 700 \times 60 \times 30 \times 4 \times \pi\)
\(h = \frac{700 \times 60 \times 30 \times 4}{40 \times 40}\)
\(h = 3150 \, \text{cm}\)

Question. A decorative block, made up of two solids - a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the total surface area of the block. Use \(\pi = \frac{22}{7}\).
Answer: Surface area of block
\(= 216 - \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} + 2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\)
\(= 225.625 \text{ cm}^2\).

SHORT ANSWER TYPE QUESTIONS

Question. From a solid cylinder of height 24 cm and diameter 14 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
Answer: Height of the cylinder = height of the cone = 24 cm.
and radius of cylinder = radius of cone
\( = \frac{14}{2} = 7 \) cm
Slant height of cone \( = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = 25 \) cm
Total surface area of remaining part
= Surface area of cylinder + Surface area of cone + area of top
\( = 2 \pi rh + \pi rl + \pi r^2 \)
\( = \pi r(2h + l + r) \)
\( = \frac{22}{7} \times 7(2 \times 24 + 25 + 7) \)
\( = 22 \times 80 \)
\( = 1760 \text{ cm}^2 \)

Question. The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum be 8 cm, find the whole surface area of the frustum. (Use \( \pi = 3.14 \))
Answer: Let \( R \) and \( r \) be the radii of the circular ends of the frustum where \( R > r \)
Now \( 2 \pi R = 207.24 \)
\( R = \frac{207.24}{2 \times 3.14} = 33 \) cm
and \( 2 \pi r = 169.56 \) cm
\( r = \frac{169.56}{2 \times 3.14} = 27 = 27 \text{ cm} \)
Now \( l^2 = h^2 + (R - r)^2 = 8^2 + (33 - 27)^2 = 100 \)
\( l = 10 \text{ cm} \)
Whole surface area of the frustum
\( = \pi(R^2 + r^2 + (R + r)l) \)
\( = 3.14[(33)^2 + (27)^2 + (33 + 27)10] \)
\( = 3.14(1089 + 729 + 600) \)
\( = 3.14 \times 2418 \text{ cm}^2 \)
\( = 7592.52 \text{ cm}^2 \).

Question. A metal container, open from the top, is in the shape of a frustum of a cone of height 21 cm with radii of its lower and upper circular ends as 8 cm and 20 cm repetitively. Find the cost of milk which can completely fill the container at the rate of Rs. 35 per litre. Use \( \pi = \frac{22}{7} \).
Answer: As per question the figure is shown below.
If \( r_1 \) and \( r_2 \) be the radii of two circular ends and \( h \) be the height of frustum, then volume
\( = \frac{1}{3} \pi h [r_1^2 + r_2^2 + r_1 r_2] \)
We have \( r_1 = 8 \text{ cm} \)
\( r_2 = 20 \text{ cm} \)
and \( h = 21 \text{ cm} \)
Volume \( = \frac{1}{3} \times \frac{22}{7} \times 21 [(8)^2 + (20)^2 + 8 \times 20] \)
\( = 22[64 + 400 + 160] \)
\( = 22 \times 624 \)
\( = 13728 \text{ cm}^3 \)
\( = \frac{13728}{1000} \text{ lit (} \because 1000 \text{ cm}^3 = 1 \text{ lit.)} \)
\( V = 13.728 \text{ litres} \)
Total Cost \( = \text{Rs. } 13.728 \times 35 \)
\( = \text{Rs. } 480.48 \)

Question. A cone is cut by a plane parallel to the base and upper part is removed. If the curved surface area of upper cone is \( \frac{1}{9} \) times the curved surface of original cone. Find the ratio of line segment to which the con’s height is divided by the plane.
Answer: As per question the figure is shown below.
\( \frac{\text{Curved surface of upper cone}}{\text{Curved surface of original cone}} = \frac{1}{9} \)
\( \frac{\pi r l}{\pi R L} = \frac{1}{9} \)
\( \frac{rl}{RL} = \frac{1}{9} \) ...(1)
Since by AA similarity \( \Delta AOB \sim \Delta ACD \), thus
\( \frac{r}{R} = \frac{h}{H} = \frac{l}{L} \) ...(2)
Substituting (2) in (1) we have
\( \frac{h}{H} \times \frac{h}{H} = \frac{1}{9} \)
\( \frac{h^2}{H^2} = \frac{1}{9} \)
or, \( \frac{h}{H} = \frac{1}{3} \)
Hence \( \frac{\text{Height of upper cone}}{\text{Height of lower frustum}} = \frac{1}{3-1} = \frac{1}{2} \)
Ratio of the line segments \( OA : OC = 1 : 2 \)

Question. The slant height of a frustum of a cone is 4 cm and the perimeter (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. Use \( \pi = \frac{22}{7} \).
Answer: As per question the figure is shown below.
We have \( l = 4 \text{ cm} \)
\( 2\pi R = 18 \text{ cm} \)
\( R = \frac{18}{2\pi} = \frac{9}{\pi} \)
and \( 2\pi r = 6 \)
\( r = \frac{6}{2\pi} = \frac{3}{\pi} \text{ cm} \)
Curved surface area of frustum
\( \pi l(R + r) = \pi \times 4 \left( \frac{9}{\pi} + \frac{3}{\pi} \right) \)
\( = 4\pi \times \frac{12}{\pi} = 48 \text{ cm}^2 \).

Question. A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 m high embankment. Find the width of the embankment.
Answer: Depth of well \( = 14 \text{ m} \), radius \( = 2 \text{ m} \).
Volume of earth taken out
\( \pi r^2 h = \frac{22}{7} \times 2 \times 2 \times 14 = 176 \text{ m}^3 \)
Let \( r \) be the width of embankment. The radius of outer circle of embankment
\( = 2 + r \)
Area of upper surface of embankment \( = \pi[(2+r)^2 - (2)^2] \)
Volume of embankment = Volume of earth taken out
\( \pi[(2 + r)^2 - (2)^2] \times 0.4 = 176 \)
\( \pi[4 + r^2 + 4r - 4] \times 0.4 = 176 \)
\( r^2 + 4r = \frac{176 \times 7}{0.4 \times 22} \)
\( r^2 + 4r = 140 \)
\( r^2 + 4r - 140 = 0 \)
\( (r + 14)(r - 10) = 0 \)
\( r = 10 \)
Hence width of embankment \( = 10 \text{ m} \).

Question. A hemispherical depression is cut from one face of a cubical block, such that diameter 'l' of hemisphere is equal to the edge of cube. find the surface area of the remaining solid.
Answer: Let the radius of hemisphere be \( r \).
Therefore, \( r = \frac{l}{2} \)
Now, the required surface area
\( = \text{Surface area of cubical block} - \text{Area of base of hemisphere} + \text{Curved surface area of hemisphere} \)
\( = 6(l)^2 - \pi r^2 + 2\pi r^2 \)
\( = 6l^2 + \pi r^2 \)
\( = 6l^2 + \pi \left(\frac{l}{2}\right)^2 \)
\( = 6l^2 + \frac{\pi l^2}{4} \)
Surface area \( = \frac{1}{4}(24 + \pi)l^2 \text{ units} \).
\( = \frac{1}{4} \left(24 + \frac{22}{7}\right)l^2 \)
\( = \frac{1}{4} \times \frac{190}{7}l^2 \)
\( \approx 6.78 l^2 \text{ unit}^2 \)

Question. Water in a canal 6 m wide and 1.5 m deep is flowing with a speed of 10 km/h. How much area in hectare will it irrigate in 30 minutes if 8 cm of standing water is needed ?
Answer: As per question the figure is shown below.
Water flows in 1 hr \( = 10 \text{ km} \)
Water flows in \( \frac{1}{2} \text{ hr} = \frac{10}{2} = 5 \text{ km} = 5000 \text{ m} \)
Now volume of water flows in \( \frac{1}{2} \text{ hr} \)
\( lbh = 5000 \times 6 \times 1.5 \text{ m}^3 = 45000 \text{ m}^3 \).
According to the question,
Volume of water \( \frac{1}{2} \text{ hr} = \text{area of irrigated field} \times \frac{8}{100} \)
\( 45000 = \text{Area} \times \frac{8}{100} \)
Area \( = \frac{45000 \times 100}{8} = 562500 \text{ m}^2 = 56.25 \text{ hectare} \).

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: Diameter of pipe \( = 20 \text{ cm} \).
Radius of pipe \( = \frac{20}{2} = 10 \text{ cm} = 0.10 \text{ m} \)
Diameter of tank \( = 10 \text{ m} \)
radius of the tank \( = \frac{10}{2} = 5 \text{ m} \)
Depth of tank \( = 2 \text{ m} \)
Volume of tank \( = \pi r^2 h = \pi \times 5 \times 5 \times 2 = 50\pi \)
Speed of the water 3 km/hr \( = \frac{3000}{60} = 50 \text{ m/min} \)
Volume of water supplied in one minute
\( \pi r^2 h = \pi \times 0.10 \times 0.10 \times 50 \)
Time taken \( t = \frac{50\pi}{\pi \times 0.10 \times 0.10 \times 50} = 100 \)
Hence time taken to fill the tank is 100 minutes.

Question. Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
Answer: Volume of water flowing through pipe in 1 sec.
\( \pi R^2 H = \pi \times (1)^2 \times 0.4 \times 100 \text{ cm}^3 \)
Volume of water flowing in 30 min (\( 30 \times 60 \) sec)
\( = \pi \times (1)^2 \times 0.4 \times 100 \times 30 \times 60 \)
Volume of water in cylindrical tank in 30 min
\( \pi r^2 h = \pi \times (40)^2 \times h \)
Now
\( \pi \times (40)^2 \times h = \pi \times (1)^2 \times 0.4 \times 100 \times 30 \times 60 \)
Rise in water level
\( h = \frac{\pi \times (1)^2 \times 0.4 \times 100 \times 30 \times 60}{\pi \times 40 \times 40} \)
\( = 45 \text{ cm} \).

Question. A toy is in the form of a cylinder of diameter \(2\sqrt{2}\) m and height 3.5 m surmounted by a cone whose vertical angle is 90°. Find total surface area of the toy.
Answer: As per question the figure is shown below.
Here \( \angle C = 90^\circ \) and \( AC = BC = l \)
Thus \( AB^2 = AC^2 + BC^2 = l^2 + l^2 = 2l^2 \)
Now \( (2\sqrt{2})^2 = 2l^2 \)
\( 8 = 2l^2 \)
Thus \( l = 2 \) and \( r = \sqrt{2} \) m
Slant height of conical portion \( l = 2 \) m
Total surface area of toy
\( 2\pi rh + \pi r^2 + \pi rl = \pi r[7 + \sqrt{2} + 2] \) m²
\( = \pi\sqrt{2}[9 + \sqrt{2}] \) m²
\( = \pi[2 + 9\sqrt{2}] \) m²

Question. Find the volume of the largest solid right circular cone that can be cut out off a solid cube of side 14 cm.
Answer: The base of cone is the largest circle that can be inscribed in the face of the cube and the height will be equal to edge of the cube.
Radius of cone \( = \frac{14}{2} = 7 \) cm
Height of cone \( = 14 \) cm
Volume of cone \( = \frac{1}{3}\pi r^2h \)
\( = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 14 \)
\( = \frac{2156}{3} = 718.67 \) cm³.

Question. Water is flowing at the rate of 15 km/hr 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: Speed of water flowing through the pipe \( = 15 \text{ km/hr} = 15000 \text{ m/hr} \)
Volume of water flowing in 1 hr
\( \pi R^2 H = \frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times 15000 \text{ m}^3 \)
\( = 231 \text{ m}^3 \)
Volume of water in the tank when the depth is 21 cm
\( lbh = 50 \times 44 \times \frac{21}{100} \text{ m}^3 = 462 \text{ m}^3 \)
Time taken to fill 462 m³
\( = \frac{462}{231} = 2 \text{ hrs.} \)

Question. A medicine capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends, the length of the entire capsule is 15 mm and the diameter of the capsule is 5 mm. Find the Volume of the capsule.
Answer: As per question the figure is shown below.
Total height \( = 14 \) mm
Height of cylinder \( = 14 - 2 \times 2.5 \)
\( = 14 - 5 = 9 \) mm
Radius of cylinder \( = 2.5 \) mm
Radius of hemisphere \( = 2.5 \) mm
Volume of capsule = Volume of two hemispheres + Volume of cylinder
\( = 2 \times \frac{2}{3}\pi r^3 + \pi r^2h \)
\( = \frac{4}{3}\pi \left(\frac{5}{2}\right)^3 + \pi \left(\frac{5}{2}\right)^2 \times 9 \)
\( = \left(\frac{5}{2}\right)^2 \times \pi \left[\frac{4}{3} \times \frac{5}{2} + 9\right] \)
\( = \frac{25}{4}\pi \left[\frac{10}{3} + 9\right] = \frac{25}{4}\pi \left[\frac{10 + 27}{3}\right] \)
\( = \frac{25}{4}\pi \left[\frac{37}{3}\right] = \frac{25}{4} \times \frac{22}{7} \times \frac{37}{3} \)
\( = \frac{10175}{42} \text{ mm}^3 \)
\( = 242.26 \text{ mm}^3 \).

Question. A milk tanker cylindrical in shape having diameter 2 m and length 4.2 m supplies milk to the two booths in the ratio of 3 : 2. One of the milk booths has cuboidal vessel having base area 3.96 sq. m. and the other has a cylindrical vessel having radius 1 m. Find the level of milk in each of the vessels. Use \( \pi = \frac{22}{7} \)
Answer: Volume of milk \( = \frac{22}{7} \times 1 \times 1 \times 4.2 = 13.2 \text{ m}^3 \)
Supply of milk to booth I
\( = 13.2 \times \frac{3}{5} = 2.64 \times 3 = 7.92 \text{ m}^3 \)
Supply of milk to booth II
\( = 13.2 \times \frac{2}{5} = 2.64 \times 2 = 5.28 \text{ m}^3 \)
Height in 1st vessel \( = \frac{7.92}{3.96} = 2 \) m
Height in 2nd vessel \( = \frac{5.28}{\frac{22}{7} \times 1} = \frac{5.28 \times 7}{22} = 1.68 \) m

Question. In fig from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. Use \( \pi = \frac{22}{7} \) and \( \sqrt{5} = 2.236 \).
Answer: Let \( r \) be the radius of the top after cutting
\( h = 12 - 4 = 8 \) cm
Now \( \frac{4}{r} = \frac{12}{6} \Rightarrow r = 2 \) cm
Now slant length of frustum
\( l = \sqrt{h^2 + (R - r)^2} \)
\( = \sqrt{(8)^2 + (6 - 2)^2} \)
\( = \sqrt{64 + 16} = \sqrt{80} \)
\( = 4\sqrt{5} = 4 \times 2.236 \)
\( = 8.944 \) cm
Total surface area of frustum
\( = \pi[R^2 + r^2 + l(R + r)] \)
\( = \frac{22}{7}[ (6)^2 + (2)^2 + 8.944(6 + 2) ] \)
\( = \frac{22}{7}[ 36 + 4 + 71.552 ] \)
\( = \frac{22}{7} \times 111.552 \)
\( = 350.59 \text{ cm}^2 \).

Question. 150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.
Answer: Diameter of spherical marble = 1.4 cm
Radius \( r_1 = \frac{1.4}{2} = 0.7 = \frac{7}{10} \) cm
Diameter of cylindrical vessel = 7 cm
Radius \( R = \frac{7}{2} = 3.5 \) cm
Let \( h \) be the rise in water level then,
Volume of 150 spherical marbles = volume of water rise
\( 150 \times \frac{4}{3} \times \pi \times \frac{7}{10} \times \frac{7}{10} \times \frac{7}{10} = \pi \times \frac{7}{2} \times \frac{7}{2} \times h \)
\( h = \frac{4 \times 7}{5} \)
\( \frac{28}{5} = h \)
\( h = 5.6 \) cm
Thus 5.6 cm will be rise in the level of water.

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 formed so.
Answer: Height of cylinder = 15 cm
Diameter = 12 cm
Radius = 6 cm
Radius of cone = 3 cm
and height = 9 cm
Let the number of toys recast be \( n \).
Volume of \( n \) conical toys = Volume of cylinder
\( n \times \frac{1}{3}\pi \times 3 \times 3 \times 9 = \pi \times 6 \times 6 \times 15 \)
\( n = \frac{6 \times 6 \times 15}{3 \times 9} \)
\( = 20 \)
Hence the number of toys = 20.

Question. A well diameter 3 m is dug 14 m deep. The soil taken out of it is spread evenly around it to a width of 5 m. to form a embankment. Find the height of the embankment.
Answer: [The volume of soil taken out from the well
\( \pi r^2h = \left(\frac{3}{2}\right) \times \left(\frac{3}{2}\right) \times 14\pi \text{ m}^3 \)
The radius of embankment with well
\( = \frac{3}{2} + 5 = \frac{13}{2} \) m
Let the height of embankment be \( x \). Then the volume of soil used in embankment,
\( \pi(R^2 - r^2)x = \pi r^2h \)
\( \pi \left[ \left(\frac{13}{2}\right)^2 - \left(\frac{3}{2}\right)^2 \right]x = \left(\frac{3}{2}\right) \times \left(\frac{3}{2}\right) \times 14\pi \)
\( \frac{160}{4}x = \frac{3}{2} \times \frac{3}{2} \times 14 \)
\( x = \frac{3 \times 3 \times 14}{160} = 0.7875 \) m
Hence the height of embankment = 78.75cm

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 x 44 m. Find the time in which the level of water in the tank will rise by 7 cm.
Answer: Speed of water in pipe = \( 5 \text{ km/hour} \)
In an hour length of water = \( 5000 \text{ m} \)
Let time taken to fill the tank be \( t \).
Total length of water = \( t \times 5000 \text{ m} \)
Volume of water flown = Volume of water in tank
\( \pi r^2 h = l \times b \times h \)
\( \frac{22}{7} \times \left( \frac{7}{100} \right)^2 \times 5000t = 50 \times 44 \times \frac{7}{100} \)
\( \frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times 5000t = 50 \times 44 \times \frac{7}{100} \)
\( t = \frac{50 \times 44}{22 \times 50} = 2 \)
Hence, Time taken to fill the tank = 2 hours.

Question. A bucket open at the top is in form of a frustum of a cone with a capacity of 12308.8 cm\(^3\). The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use \( \pi = 3.14 \))
Answer: Here \( R = 20, r = 12, V = 12308.8 \)
\( V = \frac{1}{3} \pi (R^2 + r^2 + Rr)h \)
\( 12308.8 = \frac{1}{3} \times 3.14 (400 + 240 + 144)h \)
\( 12308.8 = \frac{1}{3} \times 3.14 \times 784 \times h \)
\( h = 15 \text{ cm} \)
Now \( l = \sqrt{(20 - 12)^2 + 15^2} = 17 \text{ cm} \)
Total area of metal sheet used,
= CSA + Base area
= \( \pi[(20 + 12) \times 17 + 12 \times 12] \)
= \( 2160.32 \text{ cm}^2 \)

Question. The radii of the circular ends of a frustum of cone of height 6 cm are 14 cm and 6 cm respectively. Find the lateral area and total surface area of the frustum.
Answer: We have \( r_1 = 14 \text{ cm}, r_2 = 6 \text{ cm}, h = 6 \text{ cm} \)
\( l = \sqrt{h^2 + (r_1 - r_2)^2} \)
\( l = \sqrt{6^2 + (14 - 6)^2} = \sqrt{6^2 + 8^2} \)
\( l = \sqrt{36 + 64} = 10 \text{ cm} \)
Lateral surface area,
\( \pi (r_1 + r_2)l = \frac{22}{7} \times (14 + 6) \times 10 \text{ cm}^2 \)
= \( 628.57 \text{ cm}^2 \)
Total surface area
\( \pi [r_1^2 + r_2^2 + l(r_1 + r_2)] = \frac{22}{7} \times [(196 + 36) + 20 \times 10] \)
= \( \frac{22}{7} \times 432 = 1357.71 \text{ cm}^2 \)

Question. A cone of radius 10 cm is divided into two parts by a plane parallel to its base through the mid-point of its height. Compare the Volume of the two parts.
Answer: As per question the figure is shown below.
Since \( \triangle ABC \sim \triangle APQ \) we have
\( \frac{h}{2h} = \frac{r_1}{10} \implies r_1 = 5 \text{ cm} \)
Volume of smaller cone
= \( \frac{1}{3} \pi (5)^2 \times h \)
Volume of frustum = \( \frac{1}{3} \pi \times h(5^2 + 10^2 + 5 \times 10) \)
= \( \frac{1}{3} \pi \times h \times 175 \)
Required ratio = \( \frac{\frac{1}{3} \pi \times 25 \times h}{\frac{1}{3} \pi \times 175 \times h} = \frac{1}{7} \)

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 2.1 cm. Find the volume of wood left in the pen stand.
Answer: Volume of cuboidal block
\( l \times b \times h = 15 \times 10 \times 3.5 = 525 \text{ cm}^3 \)
Volume of one cone
\( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 2.1 \text{ cm}^3 = 0.55 \text{ cm}^3 \)
Volume of 4 cones = \( 0.55 \times 4 = 2.2 \text{ cm}^3 \)
Volume of wood remaining in pen stand
= \( 525 - 2.2 = 522.80 \text{ cm}^3 \)

Question. The height of a cone is 30 cm. From its topside a small cone is cut by a plane parallel to its base. If volume of smaller cone is \( \frac{1}{27} \) of the cone then at what height it is cut from the base?
Answer: As per question the figure is shown below.
Let the radii of smaller cone and original cone be \( r_1 \) and \( r_2 \) respectively and the height of smaller cone be \( h \).
Since \( \triangle ABC \sim \triangle APQ \) we have
\( \frac{h}{30} = \frac{r_1}{r_2} \) ... (1)
Volume smaller cone = \( \frac{1}{27} \times \text{Volume of original cone} \)
\( \frac{1}{3} \pi r_1^2 h = \frac{1}{27} \times \frac{1}{3} \pi r_2^2 \times 30 \)
\( \left( \frac{r_1}{r_2} \right)^2 \times \frac{h}{30} = \frac{1}{27} \)
From (1) using \( \frac{r_1}{r_2} = \frac{h}{30} \) we have
\( \left( \frac{h}{30} \right)^2 \times \frac{h}{30} = \frac{1}{27} \)
\( \left( \frac{h}{30} \right)^3 = \frac{1}{27} \)
\( h^3 = \frac{30 \times 30 \times 30}{27} \)
\( h = 10 \text{ cm} \)
Hence, required height = \( 30 - 10 = 20 \text{ cm} \)

Question. In fig., from a cuboidal solid metallic block of dimensions 15 cm \(\times\) 10 cm \(\times\) 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. Use \(\pi = \frac{22}{7}\)
Answer: Total Surface area \(= 2(lb + bh + hl) + 2\pi rh\)
Here, \(l = 15 \text{ cm}, b = 10 \text{ cm}, h = 5 \text{ cm}, r = \frac{7}{2} \text{ cm}\)
TSA of Cuboidal block
\(= 2(15 \times 10 + 10 \times 5 + 5 \times 15)\)
\(= 550 \text{ cm}^2\).
Area of C.S. of Cylinder
\(2\pi rh = 2 \times \frac{22}{7} \times \frac{7}{2} \times 5\)
\(= 110 \text{ cm}^2\)
Area of two Circular bases \(= 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
\(= 77 \text{ cm}^2\)
Required area \(= 550 + 110 - 77 = 583 \text{ cm}^3\).

Question. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pipe, given that 1 \(\text{cm}^3\) of iron has approximately 8 g mass. (Use \(\pi = 3.14\))
Answer: As per question the figure is shown below.
Radius of lower cylinder \(R = 12 \text{ cm}\)
Radius of upper cylinder \(r = 8 \text{ cm}\)
Height of upper cylinder \(h = 60 \text{ cm}\)
Height of lower cylinder \(H = 220 \text{ cm}\)
Volume of solid iron pole,
\(\pi R^2 H + \pi r^2 h = 3.14 \times (12)^2 \times 220 + 3.14 \times (8)^2 \times 60\)
\(= 111532.8 \text{ cm}^3\)
Mass of pole \(= 111532.8 \times 8 \text{ g}\)
\(= 892.2624 \text{ kg}\).

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 flows out
\(= \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi \text{ cm}^3\)
Let the radius of one spherical ball be \(r \text{ cm}\)
\(\frac{4}{3}\pi r^3 \times 100 = \frac{50}{3}\pi\)
\(r^3 = \frac{50}{4 \times 100} = \frac{1}{8}\)
or, \(r = \frac{1}{2} = 0.5 \text{ cm}\)

Question. A cone is cut by a plane parallel to the base and upper part is removed. If the C.S.A. of the remainder is \(\frac{15}{16}\) of the C.S.A. of whole cone, find the ratio of the line segments to which cone’s height is divided by the plane.
Answer: As per question the figure is shown below.
Let the height of larger cone be \(H\) and height of smaller cone be \(h\). Let radius of larger and smaller cones be \(R\) and \(r\)
Since \(\Delta ONC \sim \Delta OMA\), we have
\(\frac{h}{H} = \frac{r}{R} = \frac{l}{L}\)
C.S.A. of the frustum \(= \frac{15}{16}(\text{C.S.A. of cone } OAB)\)
C.S.A. of cone \(OCD\)
\(= 1 - \frac{15}{16} = \frac{1}{16}(\text{C.S.A. of cone } OAB)\)
\(\frac{\text{C.S.A. of cone } OCD}{\text{C.S.A. of cone } OAB} = \frac{1}{16}\)
\(\frac{\pi rl}{\pi RL} = \frac{1}{16}\)
or, \(\left( \frac{r}{R} \right)\left( \frac{l}{L} \right) = \frac{1}{16}\)
\(\left( \frac{h}{H} \right)\left( \frac{h}{H} \right) = \frac{1}{16} \quad (\text{as } \frac{l}{L} = \frac{h}{H})\)
\(\frac{h}{H} = \frac{1}{4}\)
\(h = \frac{1}{4}H\)
\(ON = \frac{1}{4}H\)
\(MN = \frac{3}{4}H\)
\(ON : MN = 1 : 3\)

Question. A right angled triangle whose sides are 3 cm, 4 cm and 5 cm is revolved about the longest side. Find the surface area of figure obtained. Use \(\pi = \frac{22}{7}\)
Answer: As per question the figure is shown below.
By revolving right triangle about longest side double cone is generated. Let radius of double cone \(= x \text{ cm}\).
In \(\Delta ADE\) and \(\Delta ADC\),
\(\angle AED = \angle DAC = 90^\circ\)
\(\angle ADE = \angle ADC\) (common angle)
Thus due to AA symmetry we have
or, \(\Delta ADE \sim \Delta ADC\)
\(\frac{x}{AC} = \frac{AD}{DC} = \frac{DE}{AD}\)
\(\frac{x}{4} = \frac{3}{5} = \frac{DE}{3}\)
\(x = \frac{12}{5} = 2.4 \text{ cm}\)
\(DE = \frac{9}{5} = 1.8 \text{ cm}\)
Surface area of double cone
\(\pi rl_1 + \pi rl_2 = \pi r(l_1 + l_2)\)
\(= \frac{22}{7} \times 2.4 \times (3 + 4)\)
\(= 22 \times 2.4 = 52.8 \text{ cm}^2\).

Question. A circus tent is in the shape of a cylinder surmounted by a conical top of same diameter. If there common diameter is 56 m, the height of cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of canvas used in the tent.
Answer: Total height of tent \( = 27 \text{ m} \)
Height of cylindrical part \( = 6 \text{ m} \)
Height of conical part \( = 27 - 6 = 21 \text{ m} \)
Slant height of cone \( = \frac{56}{2} = 28 \text{ m} \)
Slant height of cone \( = \sqrt{r^2 + h^2} \)
\( = \sqrt{28^2 + 21^2} \)
\( = \sqrt{784 + 441} = \sqrt{1225} \)
\( = 35 \text{ m} \)
Area of canvas used \( = 2\pi rh + \pi rl \)
\( = \pi r(2h + l) \)
\( = \frac{22}{7} \times 28(2 \times 6 + 35) \)
\( = 22 \times 4 \times 47 \)
\( = 4136 \text{ m}^2 \)

Question. From a right circular cylinder of height 2.4 cm and radius 0.7 cm, a right circular cone of same radius is cut-out. Find the total surface area of the remaining solid.
Answer: Radius \( r = 0.7 \text{ cm} \)
and height \( h = 2.4 \text{ cm} \)
Slant height \( l = \sqrt{h^2 + r^2} = \sqrt{(2.4)^2 + (0.7)^2} \)
\( = 2.5 \text{ cm} \)
Total surface area of remaining solid
\( = \text{C.S.A. of cylinder} + \text{C.S.A. of cone} + \text{area of top.} \)
\( = 2\pi rh + \pi rl + \pi r^2 \)
\( = \frac{22}{7} \times 0.7(2 \times 2.4 + 2.5 + 0.7) \)
\( = \frac{22}{7} \times 0.7 \times 8 = \frac{176}{10} \)
Hence total surface area \( = 17.6 \text{ cm}^2 \)

Question. Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm, if the In crease in the level of the water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe.
Answer: Let the internal diameter of the pipe be \( r \) m.
Water flows in 1 hour \( = 2.52 \text{ km.} \)
Water flows in \( \frac{1}{2} \text{ hour} = \frac{2.52}{2} = 1.26 \text{ km} = 1260 \text{ m} \)
Volume of water flows in \( \frac{1}{2} \text{ hour} = \pi r^2h = \pi r^2 \times 1260 \)
Volume of the in cylindrical tank \( = \pi \times \left(\frac{40}{100}\right)^2 \times 3.15 \)
Volume of water flow \( = \) Volume of increase water
\( \pi r^2 \times 1260 = \pi \left(\frac{2}{5}\right)^2 \times 3.15 \)
or, \( 1260r^2 = \frac{2}{5} \times \frac{2}{5} \times 3.15 \)
or, \( r^2 = \frac{4}{25} \times \frac{315}{100} \times \frac{1}{1260} = \frac{1}{2500} \)
or, \( r = \frac{1}{50} \text{ m} = 2 \text{ cm} \)
Internal diameter of pipe \( = 4 \text{ cm.} \)

Long Answer Type Questions

Question. A hemispherical tank, of diameter 3 m, is full of water. It is being emptied 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 ? Use \( \pi = \frac{22}{7} \)
Answer: Diameter of tank \( = 3 \text{ m} \)
Radius \( r = \frac{3}{2} \text{ m} \)
Volume of hemispherical tank, \( V = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (\frac{3}{2})^3 \text{ m}^3 \)
\( = \frac{2}{3} \times \frac{22}{7} \times \frac{27}{8} = \frac{11}{7} \times \frac{9}{2} = \frac{99}{14} \text{ m}^3 \)
Since \( 1 \text{ m}^3 = 1000 \text{ litre} \), we have
Volume to be emptied \( = \frac{1}{2} \times \frac{99}{14} \times 1000 = \frac{99000}{28} \text{ litres} \).
Rate of emptying \( = \frac{25}{7} \text{ litres/sec} \).
Time taken \( = \frac{99000/28}{25/7} = \frac{99000}{28} \times \frac{7}{25} = 990 \text{ seconds} \)
\( = \frac{990}{60} = 16.5 \text{ minutes} \).
\( V = \frac{99}{14} \times 1000 \) litre
Volume of hemisphere
\( \frac{V}{2} = \frac{1}{2} \times \frac{99}{14} \times 1000 \) Litres
Let time taken for this volume to flow out be \( t \) sec.
Then according to question,
\( t \times 3 \frac{4}{7} = \frac{1}{2} \times \frac{99}{14} \times 1000 \)
\( t \times \frac{25}{7} = \frac{1}{2} \times \frac{99}{14} \times 1000 \)
\( t = \frac{7}{25} \times \frac{1}{2} \times \frac{99}{14} \times 1000 \)
\( = 990 \) sec
\( = 16 \) minutes 30 sec.

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. Use \( \pi = \frac{22}{7} \)
Answer: Volume of cone \( = \frac{1}{3} \pi r^2 h \)
Volume of metal in 504 cones
\( = 504 \times \frac{\pi}{3} \times \left(\frac{3.5}{2}\right)^2 \times 3 \)
Volume of Sphere \( = \frac{4}{3} \pi r^3 \)
\( \frac{4}{3} \pi r^3 = \text{Volume of 504 cones} \)
\( \frac{4}{3} \pi r^3 = 504 \times \frac{\pi}{3} \times \frac{35}{20} \times \frac{35}{20} \times 3 \)
\( r^3 = 126 \times \frac{7}{4} \times \frac{7}{4} \times 3 \)
\( = 7 \times 9 \times 2 \times \frac{7}{4} \times \frac{7}{4} \times 3 \)
\( = 3 \times 3 \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \times 3 \)
\( r = 3 \times \frac{7}{2} = 10.5 \) cm
Thus diameter is 21 cm.
Surface area \( 4\pi r^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5 \)
\( = 1386 \text{ cm}^2 \)

Question. The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the two parts.
Answer: As per question the figure is shown below.
Let the radius of cone be \( r_2 \) and cut off cone be \( r_1 \)
Height of the cone = 10 cm
And the height the cone cut off = 5 cm
Since \( \triangle AOC' \sim \triangle AOD \), we have
\( \frac{AO'}{AO} = \frac{r_1}{r_2} = \frac{5}{10} = \frac{1}{2} \)
\( r_2 = 2r_1 \)
Volume of cut off cone = \( \frac{1}{3} \pi r_1^2 \times 5 \)
= \( \frac{5}{3} \pi r_1^2 \text{ sq. units} \)
Volume of original cone = \( \frac{1}{3} \pi (2r_1)^2 \times 10 \)
= \( \frac{40}{3} \pi r_1^2 \text{ sq. units} \)
Volume of frustum
= Volume of original cone - Volume of cut off cone
= \( \frac{40}{3} \pi r_1^2 - \frac{5}{3} \pi r_1^2 = \frac{35}{3} \pi r_1^2 \text{ sq.units} \)
Ratio of two parts = \( \frac{\frac{35}{3} \pi r_1^2}{\frac{5}{3} \pi r_1^2} = \frac{7}{1} \)
Hence the ratio of two parts = 7 : 1

Question. A metallic right circular cone 20 cm high and whose vertical angle is 60\(^\circ\) is cut into two parts at the middle of its height by a plane parallel to its base if the frustum so obtained be drawn into a wire of uniform diameter \( \frac{1}{16} \) cm, find the length of the wire.
Answer: As per question the figure is shown below.
Total height of cone = 20 cm
and Vertex angle = 30\(^\circ\) (Semi-vertical angle)
Let the radius of cone be \( r_2 \). Then we have
\( \frac{r_2}{20} = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
\( r_2 = \frac{20}{\sqrt{3}} \text{ cm} \)
The height of the cone cut off = 10 cm
Let its radius be \( r_1 \). Then
\( \frac{r_1}{10} = \tan 30^\circ = \frac{1}{\sqrt{3}} \text{ cm} \)
\( r_1 = \frac{10}{\sqrt{3}} \text{ cm} \)
Let the length of wire be \( l \). Its radius is \( \frac{1}{32} \text{ cm} \).
Now Volume of frustum = Volume of wire
\( \frac{1}{3} \pi \times h [r_1^2 + r_2^2 + r_1 r_2] = \pi r^2 l \)
\( \frac{1}{3} \times 10 \times \pi \left[ \left( \frac{10}{\sqrt{3}} \right)^2 + \left( \frac{20}{\sqrt{3}} \right)^2 + \frac{10}{\sqrt{3}} \times \frac{20}{\sqrt{3}} \right] = \pi \left( \frac{1}{32} \right)^2 \times l \)
\( \frac{1}{3} \times 10 \times \left[ \frac{100}{3} + \frac{400}{3} + \frac{200}{3} \right] = \frac{1}{32 \times 32} \times l \)
\( \frac{1}{3} \times 10 \times \frac{700}{3} = \frac{1}{32 \times 32} \times l \)
\( l = \frac{32 \times 32 \times 700 \times 10}{3 \times 3} \)
\( l = 796444.4 \text{ cm} \)
Hence, the length of wire is 7964.44 m.

Question. A right circular cone is divided into three parts trisecting its height by two planes drawn parallel to the base. Show that volumes of the three portions starting from the top are in the ratio 1 : 7 : 19.
Answer: As per question the figure is shown below.
Let the radii of three cones from top be \( r_1, r_2 \), and \( r_3 \) respectively.
Let the height of given cone be \( 3h \). So, the height of cone ADE is \( 2h \) and height of cone ABC is \( h \).
Since \( \triangle ABC \sim \triangle ADE \),
\( \frac{r_1}{r_2} = \frac{h}{2h} \implies 2r_1 = r_2 \)
Since \( \triangle ADE \sim \triangle AFG \)
\( \frac{r_2}{r_3} = \frac{2h}{3h} \implies 3r_1 = r_3 \)
Volume of cone ABC = \( \frac{1}{3} \pi r_1^2 h \)
Volume of cone ADE = \( \frac{1}{3} \pi r_2^2 (2h) = \frac{1}{3} \pi (2r_1)^2 (2h) = \frac{1}{3} \pi (8r_1^2 h) \)
Volume of frustum BCED = \( \frac{1}{3} \pi \times 8 r_1^2 h - \frac{1}{3} \pi r_1^2 h = \frac{7}{3} \pi r_1^2 h \)
Volume of frustum DEGF = Volume of cone AFG - Volume of cone ADE
= \( \frac{1}{3} \pi r_3^2 (3h) - \frac{1}{3} \pi r_2^2 (2h) \)
= \( \frac{1}{3} \pi (3r_1)^2 (3h) - \frac{1}{3} \pi (2r_1)^2 (2h) \)
= \( \frac{1}{3} \pi r_1^2 h (27 - 8) = \frac{19}{3} \pi r_1^2 h \)
Ratio = \( \frac{1}{3} \pi r_1^2 h : \frac{7}{3} \pi r_1^2 h : \frac{19}{3} \pi r_1^2 h = 1 : 7 : 19 \)
Hence, required ratio = 1 : 7 : 19.

Question. A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. (Use \(\pi = 3.14\))
Answer: As per question the figure is shown below.
We have \(AC^2 = 20^2 + 15^2 = 625\)
\(AC = 25 \text{ cm}\)
\(ar(\Delta ABC) = ar(\Delta ABC)\)
\(\frac{1}{2} \times BC \times AB = \frac{1}{2} \times AC \times BD\)
\(15 \times 20 = 25 \times BD\)
\(BD = 12 \text{ cm}\)
Volume of double cone,
\(= \text{Volume of upper cone} + \text{Volume of lower cone}\)
\(= \frac{1}{3}\pi(BD)^2 \times AD + \frac{1}{3}\pi(BD)^2 \times CD\)
\(= \frac{1}{3}\pi(BD)^2 \{AD + CD\} = \frac{1}{3}\pi(BD)^2(AC)\)
\(= \frac{1}{3} \times 3.14 \times 144 \times 25 = 3768 \text{ cm}^2\)
Surface area = C.S.A. of upper cone + C.S.A. of lower cone
\(= \pi(12)(20) + \pi(12)(15)\)
\(= 12\pi \{20 + 15\}\)
\(= 12 \times 3.14 \times 35\)
\(= 1318.8 \text{ cm}^2\)