CBSE Class 10 Mathematics Surface Areas And Volumes Worksheet Set 05

Multiple Choice Questions

 

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 the ratio of its radius and the slant height of the conical part is
(a) 1:2
(b) 2:3
(c) 2:5
(d) 3:2
Answer: (a) 1:2
Sol. C.S.A of conical part = C.S.A. of spherical part
\( \pi rl = 2\pi r^2 \)
\( l = 2r \)
\( \implies \frac{r}{l} = \frac{1}{2} \)
\( \implies r : l = 1 : 2 \)

 

Question. Two cubes each with 6 cm edge are joined end to end. The surface area of the resulting cuboid is
(a) 180 cm\(^2\)
(b) 360 cm\(^2\)
(c) 300 cm\(^2\)
(d) 260 cm\(^2\)
Answer: (b) 360 cm\(^2\)
Hint: Length of resulting cuboid = 12 cm.
Width = 6 cm, height = 6 cm.
\( \therefore \) Surface area = \( 2(12 \times 6 + 12 \times 6 + 6 \times 6) \)
\( = 2(72 + 72 + 36) = 2(180) = 360 \text{ cm}^2 \).

 

Question. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas. 
(a) 1:2
(b) 1:3
(c) 1:4
(d) 1:9
Answer: (d) 1:9
Sol. Let \( r_1 \) and \( r_2 \) be the edges of the two cubes respectively.
Ratio of their volumes = \( \frac{r_1^3}{r_2^3} = \frac{1}{27} \)
\( \implies \frac{r_1}{r_2} = \frac{1}{3} \)
Ratio of their surface areas = \( \frac{6r_1^2}{6r_2^2} = \left(\frac{r_1}{r_2}\right)^2 \)
\( = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \)

 

Question. A conical military tent having diameter of the base 24 m and slant height of the tent is 13 m, find the curved surface area of the cone. [\( \pi = \frac{22}{7} \)]
(a) \( \frac{3442}{7} \text{ m}^2 \)
(b) \( \frac{3423}{7} \text{ m}^2 \)
(c) \( \frac{3432}{7} \text{ m}^2 \)
(d) \( \frac{3348}{7} \text{ m}^2 \)
Answer: (c) \( \frac{3432}{7} \text{ m}^2 \)
Sol. Diameter of tent = 24 m \( \therefore \) radius = 12 m
Slant height = 13 m
Curved surface area = \( \pi rl = \frac{22}{7} \times 12 \times 13 = \frac{3432}{7} \text{ m}^2 \)

 

Question. The diameter of the hemisphere whose curved surface area is 308 cm\(^2\) is
(a) 7 cm
(b) 0.7 cm
(c) 14 cm
(d) 70 cm
Answer: (c) 14 cm
Sol. Let, radius = \( r \)
\( \therefore \text{ CSA } = 308 \text{ cm}^2 \)
\( \implies 2\pi r^2 = 308 \)
\( \implies 2 \times \frac{22}{7} \times r^2 = 308 \)
\( \implies r = 7 \text{ cm} \)
\( \therefore \text{ diameter } = 2r = 2 \times 7 = 14 \text{ cm} \)

 

Question. A joker’s cap is in the form of a right circular cone of base radius 7 cm and the slant height is 25 cm. The area of the cap is [\( \pi = \frac{22}{7} \)]
(a) 225 cm\(^2\)
(b) 350 cm\(^2\)
(c) 450 cm\(^2\)
(d) 550 cm\(^2\)
Answer: (d) 550 cm\(^2\)
Sol. Base radius of Joker’s cap = 7 cm; slant height = 25 cm
Curved surface area = \( \pi rl = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2 \)

 

Very Short Answer Type Questions

Question. Two cubes each of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.
Answer: Sol. Length of resulting cuboid, \( l = 4 \text{ cm} + 4 \text{ cm} = 8 \text{ cm} \), breadth, \( b = 4 \text{ cm} \), height, \( h = 4 \text{ cm} \)
Surface area of cuboid = \( 2(l \times b + b \times h + h \times l) \)
\( = 2(8 \times 4 + 4 \times 4 + 8 \times 4) = 160 \text{ cm}^2 \)

 

Question. A wallpaper, 312 m long and 25 cm wide is required to cover the walls of a room. Length of the room is 7 m and its breadth is twice its height. Determine the height of the room.
Answer: Sol. Area of wallpaper = \( 312 \times 0.25 = 78.00 \text{ m}^2 \)
Length of room = 7 m
Let height be \( x \) m, then breadth be \( 2x \) m
Surface area of four walls = \( 2[l \times h + b \times h] \)
\( = 2(7x + 2x^2) = 14x + 4x^2 \)
According to the question,
\( 4x^2 + 14x = 78 \)
\( \implies 2x^2 + 7x – 39 = 0 \)
\( \implies 2x^2 + 13x – 6x – 39 = 0 \)
\( \implies x(2x + 13) – 3(2x + 13) = 0 \)
\( \implies (x – 3)(2x + 13) = 0 \)
\( \implies x – 3 = 0 \) or \( 2x + 13 = 0 \)
\( \implies x = 3 \) or \( x = -\frac{13}{2} \) (rejected)
\( \implies x = 3 \text{ m} \)
Hence, height of the room is 3 m

 

Short Answer Type Questions

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: Sol. Conical heap of rice:
Dimensions: diameter = 24 m, height 3.5 m. \( \implies \text{radius} = 12 \text{m} \).
Volume of cone = \( \frac{1}{3} \pi r^2 h \text{ cu. units} \).
\( = \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 3.5 \text{ cu. m.} \)
\( = 11 \times 4 \times 12 \)
\( = 132 \times 4 \)
\( = 528 \text{ cu-m.} \)
The volume of the rice heap is 528 cu-m.
Area of cloth required = Curved surface area.
CSA of cone = \( \pi rl \text{ sq. unit} \) where \( l = \sqrt{h^2 + r^2} \text{ units} \).
Finding \( l \): \( l = \sqrt{(3.5)^2 + (12)^2} \text{ units} \)
\( = \sqrt{12.25 + 144} \)
\( = \sqrt{156.25} \)
\( = 12.5 \text{ m} \).
\( \implies \text{CSA} = \pi rl \text{ sq. units} \)
\( = \frac{22}{7} \times 12 \times 12.5 \)
\( = \frac{22 \times 150}{7} = \frac{3300}{7} \approx 471.428571 \text{ m}^2 \).
The area of canvas cloth required is 471.428571 m\(^2\).

 

Question. The total surface area of a solid cylinder is 231 cm\(^2\). If the curved surface area of this solid cylinder is \( \frac{2}{3} \) of its total surface area, find its radius and height. [Use \( \pi = \frac{22}{7} \)] 
Answer: Sol. Let radius of the base of cylinder be \( r \) cm and height be \( h \) cm
Total surface area = \( 2\pi r (r + h) = 231 \text{ cm}^2 \).
Curved surface area = \( 2\pi rh \)
According to the question,
Curved surface area = \( \frac{2}{3} \text{ total surface area} \)
\( \implies 2\pi rh = \frac{2}{3} \times 2\pi r (r + h) \)
\( \implies 2\pi rh = \frac{2}{3} \times 231 \)
\( \implies 2\pi rh = 154 \text{ cm}^2 \)
Now \( 2\pi r (r + h) = 231 \implies 2\pi r^2 + 2\pi rh = 231 \)
\( \implies 2\pi r^2 + 154 = 231 \implies 2\pi r^2 = 231 – 154 = 77 \)
\( \implies 2 \times \frac{22}{7} \times r^2 = 77 \implies r^2 = \frac{77 \times 7}{22 \times 2} \implies r = \frac{7}{2} \text{ cm} \)
\( \because 2\pi rh = 154 \implies 2 \times \frac{22}{7} \times \frac{7}{2} \times h = 154 \)
\( h = \frac{154}{22} = 7 \text{ cm} \)

 

Question. The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling it at the rate of 30 paise per square metre.
Answer: Sol. Diameter of roller = 84 cm \( \implies \) radius of roller = 42 cm
Area levelled in one revolution = \( 2 \times \frac{22}{7} \times 120 \times 42 \text{ cm}^2 = \frac{221760}{7} \text{ cm}^2 \)
Area levelled in 500 revolutions
\( = 500 \times \frac{221760}{7} \text{ cm}^2 = 500 \times \frac{221760}{7 \times 10000} \text{ m}^2 \)
Cost of levelling 1 m\(^2\) = 30 paise
Total cost of levelling = Rs. \( \frac{30}{100} \times 500 \times \frac{221760}{7 \times 10000} \)
= Rs. 475.20

 

Question. An iron pipe 20 cm long has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm, find the whole surface area of the pipe.
Answer: Sol. We have, \( R \) = external radius = 12.5 cm
\( r \) = internal radius = (external radius – thickness) = (12.5 – 1) cm = 11.5 cm
\( h \) = height of the pipe = 20 cm
\( \therefore \) Total surface area of the pipe = external curved surface + internal curved surface + 2(area of the base of the ring)
\( = 2\pi Rh + 2\pi rh + 2 (\pi R^2 - \pi r^2) \)
\( = 2\pi (R + r) h + 2\pi (R^2 - r^2) \)
\( = 2\pi (R + r) h + 2\pi (R + r) (R - r) = 2\pi (R + r) (h + R - r) \)
\( = 2 \times \frac{22}{7} \times (12.5 + 11.5) \times (20 + 12.5 – 11.5) \text{ cm}^2 \)
\( = 2 \times \frac{22}{7} \times 24 \times 21 \text{ cm}^2 = 3168 \text{ cm}^2 \)

 

Question. The circumference of the base of 10 m high conical tent is 44 m. Calculate the length of canvas used in making the tent if width of canvas is 2 m.
Answer: Sol. Circumference of the base = 44 m
\( \implies 2\pi r = 44 \text{ m} \implies r = 7 \text{ m} \)
\( h = 10 \text{ m}, l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 10^2} = \sqrt{49 + 100} = \sqrt{149} \text{ m} \)
Area of canvas required = \( \pi rl = \frac{22}{7} \times 7 \times \sqrt{149} \text{ m}^2 = 22\sqrt{149} \text{ m}^2 \)
Length of canvas required = \( \frac{\text{area of canvas}}{\text{width of canvas}} \)
\( = \frac{22\sqrt{149}}{2} \text{ m} = 11\sqrt{149} \text{ m} \)
\( = 11 \times 12.206 \text{ m} = 134.27 \text{ m} \)

 

Long Answer Type Questions

 

Question. The dimensions of a room are 8 m × 6 m × h. It has two doors each of size 2 m × 1 m and one almirah of size 3 m × 2 m. The cost of covering the walls by wallpaper which is 40 cm wide at Rs. 1.25 per m is Rs. 362.50. Find height.
Answer: Sol. Total cost of covering the walls with wallpaper = Rs. 362.50
Cost of paper per m = Rs. 1.25
\( \implies \text{Length of paper} = \frac{\text{Rs. 362.50}}{\text{Rs. 1.25}} = 290 \text{ m} \)
Area of paper required = area of 4 walls – area of 2 doors – area of 1 almirah
\( \implies 290 \times \frac{40}{100} = [2h(8 + 6) – 2 \times 2 \times 1 – 3 \times 2] \text{ m}^2 \)
\( \implies 29 \times 4 = (28h – 4 – 6) \text{ m}^2 \)
\( \implies 116 = 28h – 10 \)
\( \implies 28h = 126 \)
\( \implies h = \frac{126}{28} = 4.5 \text{ m} \)

 

PRACTICE QUESTIONS

 

Question. A cube whose edge is 20 cm long, has circles on each of its faces painted black. What is the total area of the unpainted surface of the cube if the circles are of the largest possible areas?
(a) 90.72 \( cm^2 \)
(b) 256.72 \( cm^2 \)
(c) 330.3 \( cm^2 \)
(d) 514.28 \( cm^2 \)
Answer: (d) 514.28 \( cm^2 \)

 

Question. A cube of side 4 cm is cut into cubes of side 1 cm, then the total surface area of all the small cubes is
(a) 384 \( cm^2 \)
(b) 382 \( cm^2 \)
(c) 64 \( cm^2 \)
(d) 32 \( cm^2 \)
Answer: (a) 384 \( cm^2 \)

 

Question. How many metres of cloth 1 m 10 cm wide, will be required to make a conical circus tent whose height is 12 m and the radius of whose base is 10 m? Also determine the cost of the cloth at Rs. 7 per m.
Answer: [Length of cloth = 285.45 m (approx), Cost = Rs. 1998.15 (approx)]

 

Question. If a right circular cylinder just encloses a sphere of radius r. Find curved surface area of the cylinder.
Answer: \( 4\pi r^2 \)

 

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: \( 1 : \sqrt{3} \)

 

Question. A solid cube is cut into two cuboids of equal volumes. Find the ratio of the total surface area of the given cube and that of one of the cuboids.
Answer: \( 3 : 2 \)

 

Question. The cost of preparing the walls of a room 12 m long at the rate of Rs. 1.35 per \( m^2 \) is Rs. 340.20 and the cost of matting the floor at 85 paise per \( m^2 \) is Rs. 91.80. Find the height of the room.
Answer: 6 m

 

Question. The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. The cost to paint 1 \( cm^2 \) of the surface is Rs. 0.05. Find the total cost of painting the vessel all over.
Answer: Rs. 96.28 (approx)

 

Question. There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Answer: \( 4 : 1 \)

 

Question. A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per \( m^2 \).
Answer: Rs. 5365.80

 

Question. A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is 8/9 of the curved surface of the whole cone, find the ratio of the line segments into which the cone’s altitude is divided by the plane.
Answer: \( 1 : 2 \)

 

Question. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm.
Answer: 770 \( cm^2 \)

 

Question. A class room is 7 m long, 6.5 m wide and 4 m high. It has one door 3 m \( \times \) 1.4 m and three windows each measuring 2 m \( \times \) 1 m. The interior walls are to be colour washed. The contractor charges Rs. 5.25 per \( m^2 \). Find the cost of colour washing.
Answer: Rs. 513.45

SECTION A

Choose and write the correct option in the following questions.

 

Question. The ratio of volume of a cone and a cylinder of equal diameter and equal height is
(a) 3 : 1
(b) 1 : 3
(c) 1 : 2
(d) 2 : 1
Answer: (b) 1 : 3

 

Question. A cylinder, a cone and a hemisphere are on the same base and having same height. The ratio of their volumes is
(a) 1 : 2 : 3
(b) 2 : 1 : 3
(c) 3 : 1 : 2
(d) 3 : 2 : 1
Answer: (c) 3 : 1 : 2

 

Question. The radii of bases of cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3, then ratio between the volumes of cylinder and that of cone is
(a) 8 : 9
(b) 9 : 8
(c) 5 : 7
(d) 7 : 5
Answer: (b) 9 : 8

 

Solve the following questions.

 

Question. What is the curved surface area of a right circular cone of height 24 cm and base diameter 14 cm?
Answer: \( 550 \text{ cm}^2 \)

SECTION B

Solve the following questions.

 

Question. What is the ratio of the volume of a cube to that of a sphere which will fit inside it?
Answer: \( 6 : \pi \)

 

Question. The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. If the cost of painting \( 1 \text{ cm}^2 \) of the surface area is Rs. 0.05, find the total cost of painting the vessel all over. \( \left( \text{Take } \pi = \frac{22}{7} \right) \)
Answer: \( 1925.79 \text{ cm}^2 \)

 

Question. The rain water from a roof of dimensions 22 m \( \times \) 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall (in cm).
Answer: 2.5 cm

 

Question. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 100 cm and the diameter of the hemispherical ends is 28 cm. Find the cost of polishing the surface of the solid at the rate of 5 paise per sq. cm.
Answer: Rs. 440

 

Solve the following questions.

 

Question. A cylindrical tub, whose diameter is 12 cm and height 15 cm is full of ice cream. The whole ice-cream is to be divided among 10 children in equal ice-cream cones, with conical base surmounted by hemispherical top. If the height of conical portion is twice the diameter of base, find the diameter of conical part of ice-cream cone.
Answer: 6 cm

 

Question. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth of a litre? 
Answer: 480000 words

 

Question. A cylinder, whose height is two-thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
Answer: 4 cm

 

Question. Five hundred persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is \( 0.04 \text{ m}^3 \)?
Answer: 0.5 cm

 

Solve the following questions.

 

Question. Water is flowing at the rate of 15km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm? 
Answer: 2 hours

 

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 the vessel. One-fourth of the water flows out of the vessel. Find the radius of a spherical ball. 
Answer: \( \frac{1}{2} \text{ cm} \)
 

 

 

Please click on below link to download CBSE Class 10 Mathematics Surface Areas And Volumes Worksheet Set E