CBSE Class 10 Maths HOTs Surface Area and Volumes Set D

Very Short Answer Type Questions (VSA)

Question. Find the radius of a sphere (in cm) whose volume is \( 12\pi \text{ cm}^3 \).
Answer: Let radius of the sphere be \( r \) cm. According to question, \( \frac{4}{3} \pi r^3 = 12\pi \Rightarrow r^3 = \frac{3 \times 12}{4} = 9 = 3^2 \Rightarrow r = (3^2)^{1/3} = (3)^{2/3} \) cm.

Question. If the radius of the base of a right circular cylinder is halved, keeping the height same, then find the ratio of the volume of the cylinder thus obtained to the volume of original cylinder.
Answer: Let the radius and height of the original cylinder be \( r \) and \( h \) respectively. \( \therefore \) Volume of the original cylinder = \( \pi r^2 h \). According to question, Radius of the new cylinder = \( r/2 \), Height of the new cylinder = \( h \). \( \therefore \) Volume of the new cylinder = \( \pi \left(\frac{r}{2}\right)^2 h = \frac{\pi r^2 h}{4} \). Hence, required ratio = \( \frac{\text{Volume of the new cylinder}}{\text{Volume of original cylinder}} = \frac{\pi r^2 h / 4}{\pi r^2 h} = 1:4 \).

Question. A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of the two parts are equal, then what will be the ratio of its radius and the slant height of the conical part?
Answer: Let \( r \) be the radius of hemisphere and conical part. Also, let \( l \) be the slant height of conical part. Given that, surface area of hemisphere = surface area of conical part \( \Rightarrow 2\pi r^2 = \pi r l \Rightarrow 2r = l \Rightarrow \frac{r}{l} = \frac{1}{2} \) i.e., \( r:l = 1:2 \).

Question. Find the number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm.
Answer: Let \( n \) be the number of solid spheres formed by melting the solid metallic cylinder. \( n \times \text{Volume of one solid sphere} = \text{Volume of the solid cylinder} \Rightarrow n \times \frac{4}{3}\pi R^3 = \pi r^2 h \) (where \( R, r \) be the radius of sphere and cylinder respectively and \( h \) is height of cylinder) \( \Rightarrow n \times \frac{4}{3}(3)^3 = (2)^2 \times 45 \Rightarrow n \times \frac{4}{3} \times 27 = 180 \Rightarrow n = \frac{180 \times 3}{4 \times 27} = 5 \). Thus, the number of solid spheres that can be formed is 5.

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 is half the conical part, then find the ratio of the radius and the height of the conical part.
Answer: Let \( r \) be the radius of cone and hemispherical part and \( h \) be the height of cone. According to question, \( 2\pi r^2 = \frac{1}{2} \pi r l \), where \( l \) is slant height of cone \( \Rightarrow 4r = l \Rightarrow 16r^2 = l^2 \Rightarrow 16r^2 = r^2 + h^2 \Rightarrow 15r^2 = h^2 \Rightarrow \frac{r^2}{h^2} = \frac{1}{15} \Rightarrow \frac{r}{h} = \frac{1}{\sqrt{15}} \).

Question. A cuboidal solid block of metal 49 cm \(\times\) 44 cm \(\times\) 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.
Answer: Let \( r \) be radius of sphere. Now, volume of block = volume of sphere \( \Rightarrow 49 \times 44 \times 18 = \frac{4}{3} \times \frac{22}{7} \times r^3 \Rightarrow \frac{49 \times 44 \times 18 \times 3 \times 7}{4 \times 22} = r^3 \Rightarrow (7 \times 3)^3 = r^3 \) or \( r = 21 \) cm. \( \therefore \) Radius of sphere = 21 cm.

Question. The total surface area of a cube is \( 32 \frac{2}{3} \text{ m}^2 \). Find the volume of cube.
Answer: Let the side of the cube be \( x \) m. \( \therefore \) Total surface area = \( 6x^2 \text{ m}^2 \Rightarrow 6x^2 = 32 \frac{2}{3} \Rightarrow 6x^2 = \frac{98}{3} \Rightarrow x^2 = \frac{98}{3 \times 6} = \frac{49}{9} \Rightarrow x = \frac{7}{3} \) m. \( \therefore \) Volume of the cube = \( x^3 \text{ m}^3 = (\frac{7}{3})^3 \text{ m}^3 = \frac{343}{27} \text{ m}^3 = 12 \frac{19}{27} \text{ m}^3 \).

Question. If curved surface area of cylinder is equal to its volume. What is the radius of cylinder?
Answer: Let \( h \) and \( r \) be the height and radius of the base of the right circular cylinder respectively, then \( 2\pi r h = \pi r^2 h \) i.e., \( r = 2 \).

Question. How many cubes of side 2 cm can be made from a solid cube of side 10 cm?
Answer: Let \( n \) be the number of solid cubes of side 2 cm made from a solid cube of side 10 cm. \( \therefore n \times \text{Volume of one small cube} = \text{Volume of big cube} \Rightarrow n \times (2)^3 = (10)^3 \Rightarrow 8n = 1000 \Rightarrow n = \frac{1000}{8} = 125 \). Thus, the number of solid cubes formed of side 2 cm each is 125.

Question. A cone and a cylinder have the same radii but the height of the cone is 3 times that of the cylinder. Find the ratio of their volumes.
Answer: Let the radius and the height of the cylinder are \( r \) and \( h \) respectively. So, radius of the cone is \( r \) and height of the cone is \( 3h \). \( \therefore \) Volume of the cylinder = \( \pi r^2 h \). Volume of the cone = \( \frac{1}{3} \pi r^2 (3h) = \pi r^2 h \). So, required ratio = \( \frac{\pi r^2 h}{\pi r^2 h} = 1:1 \).

Question. If the total surface area of a solid hemisphere is \( 462 \text{ cm}^2 \), find its volume. \( [ \text{Take } \pi = \frac{22}{7} ] \)
Answer: Let radius of hemisphere = \( r \) cm. Total surface area of hemisphere = \( 462 \text{ cm}^2 \Rightarrow 3\pi r^2 = 462 \Rightarrow r^2 = \frac{462 \times 7}{3 \times 22} = 49 \Rightarrow r = 7 \). \( \therefore \) Volume of hemisphere = \( \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = 718.67 \text{ cm}^3 \).

Question. The volume of a hemisphere is \( 2425 \frac{1}{2} \text{ cm}^3 \). Find its curved surface area. \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Given, volume of hemisphere = \( \frac{4851}{2} \text{ cm}^3 \). Let the radius of the hemisphere be \( r \) cm. \( \therefore \) Volume of hemisphere = \( \frac{2}{3} \pi r^3 \Rightarrow \frac{2}{3} \frac{22}{7} r^3 = \frac{4851}{2} \Rightarrow r^3 = \frac{4851 \times 3 \times 7}{2 \times 2 \times 22} \Rightarrow r = \frac{21}{2} \) cm. Now, curved surface area of hemisphere = \( 2\pi r^2 = 2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 693 \text{ cm}^2 \).

Question. A cube of side 6 cm is cut into a number of cubes, each of side 2 cm. Find the number of cubes formed.
Answer: Number of cubes formed = \( \frac{\text{Volume of given cube}}{\text{Volume of each small cube}} = \frac{6 \times 6 \times 6}{2 \times 2 \times 2 = 27} \).

Question. A cubical ice-cream brick of edge 22 cm is to be distributed among some children by filling ice-cream cones of radius 2 cm and height 7 cm upto its brim. How many children will get the ice-cream cones?
Answer: Volume of cubical ice cream brick = \( (22)^3 \text{ cm}^3 \). Radius of ice cream cone \( (r) = 2 \) cm. Height of ice cream cone \( (h) = 7 \) cm. Volume of ice cream cone = \( \frac{1}{3}\pi r^2 h = \left( \frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7 \right) \text{ cm}^3 \). \( \therefore \) Number of ice cream cones = \( \frac{\text{Volume of cubical ice cream brick}}{\text{Volume of each ice cream cone}} = \frac{22 \times 22 \times 22}{\frac{1}{3} \times \frac{22}{7} \times (2)^2 \times 7} = 363 \). \( \therefore \) 363 children will get the ice cream cones.

Question. Two cubes each of volume \( 27 \text{ cm}^3 \) are joined end to end to form a solid. Find the surface area of the resulting cuboid.
Answer: Let the edge of each cube be \( a \) cm. \( \therefore \) Volume of each cube = \( a^3 \text{ cm}^3 \) i.e., \( a^3 = 27 = (3)^3 \Rightarrow a = 3 \). \( \therefore \) Surface area of the cuboid = \( 2(lb + bh + hl) = 2(6 \times 3 + 3 \times 3 + 3 \times 6) = 2 \times 45 = 90 \text{ cm}^2 \).

Question. A cone of height 21 cm and radius of base 7 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.
Answer: Let the radius of the sphere be \( r \) and \( R, h \) are the radius, height of cone respectively. Radius of the cone = 7 cm. Height of the cone = 28 cm. Now, Volume of sphere = Volume of cone \( \Rightarrow \frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 h \Rightarrow \frac{4}{3}\pi r^3 = \frac{1}{3}\pi (7)^2 \times (28) \Rightarrow r^3 = 7^3 \Rightarrow r = 7 \) cm. \( \therefore \) Diameter of the sphere = \( 2r = 14 \) cm.

Question. The dimensions of a metallic cuboid are 100 cm \(\times\) 80 cm \(\times\) 64 cm. It is melted and recast into a cube. Find the surface area of the cube.
Answer: Volume of given cuboid = \( 100 \times 80 \times 64 = 512000 \text{ cm}^3 \). Now, cuboid is melted and recast into a cube. Let side of the cube = \( a \) cm. Also, volume of the cube = volume of the cuboid \( \Rightarrow a^3 = 512000 \Rightarrow a = 80 \). Surface area of cube = \( 6a^2 = 6 \times (80)^2 = 38400 \text{ cm}^2 \).

Short Answer Type Questions

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 cylinder is 1628 sq. cm, find the volume of the cylinder. \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Let height of cylinder = \( h \) cm and radius of cylinder = \( r \) cm. Given, \( r + h = 37 \) cm. Total surface area of cylinder = \( 2\pi r h + 2\pi r^2 \Rightarrow 2\pi r(r + h) = 1628 \Rightarrow 2\pi r(37) = 1628 \Rightarrow r = \frac{1628 \times 7}{2 \times 22 \times 37} = 7 \). \( \therefore r + h = 37 \Rightarrow 7 + h = 37 \Rightarrow h = 30 \). Hence, volume of cylinder = \( \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 30 = 4620 \text{ cm}^3 \).

Question. A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of rs 25 per metre. \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Radius \( (r) \) of conical tent = 7 m. Height \( (h) \) of conical tent = 24 m. \( \therefore \) Slant height \( (l) = \sqrt{(24)^2 + (7)^2} = \sqrt{625} = 25 \) m. Curved surface area of tent = \( \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^2 \). Width of cloth used = 5 m. Let length of cloth used is \( x \) m. \( \therefore \) Area of cloth = curved surface area of tent \( \Rightarrow 5 \times x = 550 \Rightarrow x = 110 \). Hence, cost of cloth used at rate of rs 25 per metre = rs \( (110 \times 25) = rs 2750 \). 

Question. A toy is in the form of a cone of base radius 3.5 cm mounted on a hemisphere of base diameter 7 cm. If the total height of the toy is 15.5 cm, find the total surface area of the toy. \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Radius \( (r) \) of cone = Radius \( (r) \) of hemisphere = 3.5 cm. Height \( (H) \) of toy = 15.5 cm. \( \therefore \) Height \( (h) \) of cone = \( 15.5 - 3.5 = 12 \) cm. Hence, slant height \( (l) \) of cone = \( \sqrt{h^2 + r^2} = \sqrt{(12)^2 + (3.5)^2} = \sqrt{144 + 12.25} = 12.5 \) cm. \( \therefore \) Total surface area of toy = \( \pi r l + 2\pi r^2 = \frac{22}{7} \times 3.5(12.5) + 2 \times \frac{22}{7} \times (3.5)^2 = \frac{22 \times 3.5}{7}(12.5 + 7) = \frac{1501.5}{7} = 214.5 \text{ cm}^2 \).

Question. From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.
Answer: We have, radius of cylinder \( (r_1) = \frac{7}{2} \) cm. Height of cylinder \( (h_1) = 14 \) cm. Radius of both cones, \( (r_2) = 2.1 \) cm. Height of both cones, \( (h_2) = 4 \) cm. Volume of the remaining solid = Volume of cylinder – 2 \(\times\) Volume of cone = \( \pi r_1^2 h_1 - 2 \times \frac{1}{3}\pi r_2^2 h_2 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14 - 2 \times \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4 = 539 - 36.96 = 502.04 \text{ cm}^3 \).

Question. A girl empties a cylindrical bucket full of sand, of base radius 18 cm and height 32 cm on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct to one place of decimal.
Answer: Radius \( (r) \) of bucket = 18 cm. Height \( (h) \) of bucket = 32 cm. Height \( (H) \) of conical heap = 24 cm. Let radius of conical heap be \( R \). Volume of cylindrical bucket = Volume of conical heap \( \Rightarrow \pi r^2 h = \frac{1}{3}\pi R^2 H \Rightarrow 18 \times 18 \times 32 = \frac{1}{3} \times R^2 \times 24 \Rightarrow R^2 = \frac{18 \times 18 \times 32}{8} = 1296 \Rightarrow R = 36 \) cm. Slant height \( (l) \) of conical heap = \( \sqrt{R^2 + H^2} = \sqrt{(36)^2 + (24)^2} = \sqrt{1872} \approx 43.3 \) cm.

Question. The \(\left(\frac{3}{4}\right)^{th}\) part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
Answer: We have, radius of cone \( (r) = 5 \) cm. Height of cone \( (h) = 24 \) cm. Radius of cylinder \( (R) = 10 \) cm. Let height of water in cylinder be \( h_1 \). \( \therefore \) According to question, \( \frac{3}{4} \) of volume of cone = Volume of water in cylinder \( \Rightarrow \frac{3}{4} \times \frac{1}{3}\pi r^2 h = \pi (R)^2 h_1 \Rightarrow \frac{1}{4}(5)^2 \times 24 = (10)^2 \times h_1 \Rightarrow h_1 = \frac{150}{100} = 1.5 \) cm.

Question. A hemispherical tank, of diameter 3 m, is full of water. It is being emptied by a pipe at the rate of \( 3 \frac{4}{7} \) litres per second. How much time will it take to make the tank half empty? \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Radius of hemispherical tank = \( \frac{3}{2} \) m. Volume of tank = \( \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times \left(\frac{3}{2}\right)^3 = \frac{99}{14} \text{ m}^3 = \frac{99000}{14} \) litres. Volume of half tank = \( \frac{99000}{2 \times 14} = \frac{24750}{7} \) litres. Time taken to remove \( 3 \frac{4}{7} \) litres of water = 1 sec. \( \therefore \) Time taken to remove \( \frac{24750}{7} \) litres of water = \( \frac{7}{25} \times \frac{24750}{7} = 990 \) sec = 16.5 minutes.

Question. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
Answer: Given, radius of smaller sphere \( (r) = 3 \) cm. \( \therefore \) Volume of smaller sphere = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(3)^3 = 36\pi \text{ cm}^3 \). Volume of smaller sphere \(\times\) Density = Mass \( \Rightarrow 36\pi \times \text{Density of metal} = 1 \Rightarrow \text{Density of metal} = 1/36\pi \). Let radius of bigger sphere be \( R \) cm. \( \therefore \) Volume of bigger sphere \(\times\) density = Mass \( \Rightarrow \frac{4}{3}\pi(R)^3 \times \frac{1}{36\pi} = 7 \Rightarrow R^3 = \frac{7 \times 36 \times 3}{4} = 189 \). Volume of new sphere = Volume of smaller sphere + Volume of bigger sphere \( \Rightarrow \frac{4}{3}\pi (R')^3 = \frac{4}{3}\pi r^3 + \frac{4}{3}\pi R^3 \Rightarrow \frac{4}{3}\pi(R')^3 = \frac{4}{3}\pi(3)^3 + \frac{4}{3}\pi(189) \Rightarrow (R')^3 = 27 + 189 = 216 \Rightarrow R' = 6 \) cm. \( \therefore \) Diameter of new sphere = \( 2 \times 6 = 12 \) cm.

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 increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe.
Answer: Rate of flow of water = 2.52 km/h = \( \frac{2.52 \times 1000 \times 100}{60 \times 60} \) cm/s = 70 cm/s. \( \therefore \) Length of standing water in the pipe in 30 min = \( 70 \times 30 \times 60 \) cm = 126000 cm = \( h_1 \) (say). Let internal radius of pipe be \( r_1 \). Height of water level in the tank in half an hour, \( (h_2) = 3.15 \) m = 315 cm. Radius of tank \( (r_2) = 40 \) cm. Volume of water that flows from the cylindrical pipe in half an hour = Increased volume of water in cylindrical tank in half an hour \( \Rightarrow \pi r_1^2 h_1 = \pi r_2^2 h_2 \Rightarrow r_1^2 \times 126000 = 40 \times 40 \times 315 \Rightarrow r_1^2 = \frac{40 \times 40 \times 315}{126000} = 4 \Rightarrow r_1 = 2 \) cm. \( \therefore \) Internal diameter of pipe = \( 2r_1 = 2 \times 2 = 4 \) cm.

Long Answer Type Questions

Question. Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs rs 120 per sq.m, find the amount shared by each school to set up the tents. \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Radius \( (r) \) of cylindrical part = radius \( (r) \) of conical part = 2.8 m. Height \( (h) \) of cylindrical part = 3.5 m. Height \( (h_1) \) of conical part = 2.1 m. \( \therefore l = \sqrt{r^2 + (h_1)^2} = \sqrt{(2.8)^2 + (2.1)^2} = 3.5 \) m. Area of canvas required, per tent = CSA of cone + CSA of cylinder = \( \pi r l + 2\pi r h = \pi r [3.5 + 2 \times 3.5] = \frac{22}{7} \times \frac{28}{10} \times \frac{105}{10} = \frac{462}{5} \text{ m}^2 \). Cost of canvas per tent = rs \( \left(\frac{462}{5} \times 120\right) = rs 11088 \). Total cost of 1500 tents = rs \( (11088 \times 1500) \). Amount shared by each school = rs \( \left(\frac{11088 \times 1500}{50}\right) = rs 332640 \).

Question. A 21 m deep well with diameter 6 m is dug and the earth from digging is evenly spread to form a platform 27 m \(\times\) 11 m. Find the height of the platform. \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Height of the well, \( (H) = 21 \) m. Radius of the well, \( (r) = \frac{6}{2} = 3 \) m. Length of platform, \( l = 27 \) m. Breadth of platform, \( b = 11 \) m. Let height of platform = \( h \) m. According to question, Volume of platform = Volume of the cylindrical well \( \Rightarrow l \times b \times h = \pi r^2 H \Rightarrow 27 \times 11 \times h = \frac{22}{7} \times 3 \times 3 \times 21 \Rightarrow h \left( 297 \right) = 594 \Rightarrow h = \frac{594}{297} = 2 \) m.

Question. Water is flowing at the rate of 6 km/h through a pipe of diameter 14 cm into a rectangular tank which is 60 m long and 22 m wide. Determine the time in which the level of the water in the tank will rise by 7 cm. \( [ \text{Use } \pi = \frac{22}{7} ] \)
Answer: Radius of pipe, \( (r) = \frac{14}{2} = 7 \) cm = \( \frac{7}{100} \) m. Rate of flow of water, \( (v) = 6 \) km/h = 6000 m/h. Length of tank, \( (l) = 60 \) m. Breadth of tank, \( (b) = 22 \) m. Height of tank, \( (h) = 7 \) cm = \( \frac{7}{100} \) m. Let \( t \) hours be the time taken to fill the tank up to the height of 7 cm. \( \therefore \) Water flowing through pipe in \( t \) hours = Volume of water in the tank upto the height of 7 cm \( \Rightarrow \pi r^2 \times v \times t = l \times b \times h \Rightarrow \frac{22}{7} \times \left( \frac{7}{100} \right)^2 \times 6000 \times t = 60 \times 22 \times \frac{7}{100} \Rightarrow t = 1 \). Thus, the level of the water in the tank will rise by 7 cm in 1 hour.