CBSE Class 10 Mathematics Surface Areas And Volumes Worksheet Set 07

Question. The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 3 : 4
(b) 5 : 3
(c) 27 : 20
(d) 20 : 27
Answer: (d) 20 : 27

Question. Two metallic right circular cones having their heights 4.1 cm and 4.3 cm and radii of their bases 2.1 cm each, have been melted together and recast into a sphere. The diameter of the sphere is
(a) 3.5 cm
(b) 4.2 cm
(c) 4.9 cm
(d) 5.6 cm
Answer: (b) 4.2 cm

Question. A right circular cylindrical vessel is full of water. How many right cones having the same radius and height as those of the right cylinder will be required to store that water?
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3

Question. A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Find the ratio of their volumes.
Answer: Let the radius and height of each be \( r \). (Note: For a hemisphere, height = radius).
\( V_{cone} : V_{hemi} : V_{cyl} = \frac{1}{3}\pi r^3 : \frac{2}{3}\pi r^3 : \pi r^3 \)
\( \implies \) Ratio \( = \frac{1}{3} : \frac{2}{3} : 1 = 1 : 2 : 3 \)

Question. A factory manufactures 1,20,000 pencils daily. The pencils are cylindrical in shape, each of length 25 cm and circumference 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs. 0.05 per \( \text{dm}^2 \).
Answer: Circumference \( 2\pi r = 1.5 \text{ cm} \), Length \( h = 25 \text{ cm} \).
Curved Surface Area of one pencil \( = 2\pi rh = 1.5 \times 25 = 37.5 \text{ cm}^2 = 0.375 \text{ dm}^2 \).
Total CSA for 1,20,000 pencils \( = 1,20,000 \times 0.375 = 45,000 \text{ dm}^2 \).
Cost \( = 45,000 \times 0.05 = \text{Rs. } 2250 \).

Question. A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of the canvas use in making the tent, if the breadth of the canvas is 1.5 m. 
Answer: Radius \( r = 15 \text{ m} \), Height of cylinder \( h = 5.5 \text{ m} \), Total height \( = 8.25 \text{ m} \).
Height of cone \( H = 8.25 - 5.5 = 2.75 \text{ m} \).
Slant height \( l = \sqrt{15^2 + 2.75^2} \approx 15.25 \text{ m} \).
Total Area of canvas \( = 2\pi rh + \pi rl = \frac{22}{7} \times 15 \times (2 \times 5.5 + 15.25) = \frac{22 \times 15 \times 26.25}{7} = 1237.5 \text{ m}^2 \).
Length of canvas \( = \frac{\text{Area}}{\text{Breadth}} = \frac{1237.5}{1.5} = 825 \text{ m} \).

Question. A solid is hemispherical at the bottom and conical above. If the surface areas of two parts are equal, then find the ratio of the radius and the height of the conical part. 
Answer: CSA of hemisphere = CSA of cone
\( \implies 2\pi r^2 = \pi rl \implies 2r = l \)
\( \implies 2r = \sqrt{r^2 + h^2} \)
\( \implies 4r^2 = r^2 + h^2 \implies 3r^2 = h^2 \)
\( \implies \frac{r^2}{h^2} = \frac{1}{3} \implies \frac{r}{h} = \frac{1}{\sqrt{3}} \)
Ratio is \( 1 : \sqrt{3} \).

Question. Two tanks are of the same capacity. The dimension of the first tank are 12 cm × 8 cm × 4 cm. The second tank has a square base with depth 6 cm. Find the side of the square.
Answer: Volume of first tank \( = 12 \times 8 \times 4 = 384 \text{ cm}^3 \).
Volume of second tank \( = \text{side}^2 \times \text{depth} = a^2 \times 6 = 384 \)
\( \implies a^2 = \frac{384}{6} = 64 \implies a = 8 \text{ cm} \).
The side of the square base is 8 cm.

Question. Areas of three adjacent faces of a cuboid are \( 24 \text{ cm}^2 \), \( 8 \text{ cm}^2 \) and \( 12 \text{ cm}^2 \) respectively. Find the volume of the cuboid.
Answer: Let dimensions be \( l, b, h \). Given \( lb = 24, bh = 8, hl = 12 \).
Volume \( V = lbh \).
\( V^2 = (lb)(bh)(hl) = 24 \times 8 \times 12 = 2304 \)
\( \implies V = \sqrt{2304} = 48 \text{ cm}^3 \).

Question. The area of the base of a cone is \( 770 \text{ cm}^2 \) and the curved surface area is \( 814 \text{ cm}^2 \). Find the volume of the cone. 
Answer: Base area \( \pi r^2 = 770 \implies \frac{22}{7}r^2 = 770 \implies r^2 = 245 \).
CSA \( \pi rl = 814 \implies \sqrt{\frac{22}{7}} \times \sqrt{245} \times l = 814 \). From \( r^2=245 \), \( r = 7\sqrt{5} \).
\( \implies \frac{22}{7} \times 7\sqrt{5} \times l = 814 \implies 22\sqrt{5}l = 814 \implies l = \frac{37}{\sqrt{5}} \).
\( h^2 = l^2 - r^2 = \frac{1369}{5} - 245 = \frac{1369 - 1225}{5} = \frac{144}{5} \implies h = \frac{12}{\sqrt{5}} \text{ cm} \).
Volume \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 770 \times \frac{12}{\sqrt{5}} = \frac{3080}{\sqrt{5}} \text{ cm}^3 \approx 1377.4 \text{ cm}^3 \).

Question. The curved surface area of a cylinder is \( 264 \text{ m}^2 \) and its volume is \( 924 \text{ m}^3 \). Find the ratio of its height to its diameter.
Answer: \( 2\pi rh = 264 \) and \( \pi r^2 h = 924 \).
Dividing: \( \frac{\pi r^2 h}{2\pi rh} = \frac{924}{264} \implies \frac{r}{2} = 3.5 \implies r = 7 \text{ m} \).
\( 2 \times \frac{22}{7} \times 7 \times h = 264 \implies 44h = 264 \implies h = 6 \text{ m} \).
Diameter \( d = 14 \text{ m} \).
Ratio height to diameter \( = 6 : 14 = 3 : 7 \).

Question. 2.2 cubic cm of brass is to be drawn into a cylindrical wire 0.50 cm in diameter. Find the length of the wire. (Use \( \pi = \frac{22}{7} \))
Answer: Volume \( V = 2.2 \text{ cm}^3 \), diameter \( d = 0.5 \text{ cm} \), radius \( r = 0.25 \text{ cm} \).
\( \pi r^2 h = 2.2 \implies \frac{22}{7} \times (0.25)^2 \times h = 2.2 \)
\( \implies \frac{22}{7} \times \frac{1}{16} \times h = \frac{22}{10} \)
\( \implies h = \frac{7 \times 16}{10} = 11.2 \text{ cm} \).

Question. The radii of the circular bases of a right circular cylinder and a cone are in the ratio of 3:4 and their heights are in the ratio of 2:3. What is the ratio of their volumes?
Answer: Let radii be \( 3x, 4x \) and heights be \( 2y, 3y \).
\( V_{cyl} : V_{cone} = \pi(3x)^2(2y) : \frac{1}{3}\pi(4x)^2(3y) \)
\( \implies 18\pi x^2 y : 16\pi x^2 y = 18 : 16 = 9 : 8 \).

Question. The radius and slant height of a right circular cone are in the ratio of 7 :13 and its curved surface area is \( 286 \text{ cm}^2 \). Find the radius of the cone. (use \( \pi = \frac{22}{7} \))
Answer: Let \( r = 7x, l = 13x \).
CSA \( = \pi rl = 286 \implies \frac{22}{7} \times 7x \times 13x = 286 \)
\( \implies 22 \times 13 \times x^2 = 286 \implies 286x^2 = 286 \implies x^2 = 1 \implies x = 1 \).
Radius \( r = 7(1) = 7 \text{ cm} \).

Question. A solid rectangular piece of iron measures 5 m × 6 m × 2 m. Find the weight of this piece in kg if \( 1 \text{ m}^3 \) of iron weighs 50 grams. Also, find the cost of the piece of iron at the rate of Rs. 750/ kg.
Answer: Volume \( = 5 \times 6 \times 2 = 60 \text{ m}^3 \).
Weight \( = 60 \times 50 = 3000 \text{ grams} = 3 \text{ kg} \).
Cost \( = 3 \times 750 = \text{Rs. } 2250 \).

Question. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 108 cm and the diameter of the hemispherical ends is 36 cm, find the cost of polishing its surface at the rate of 70 paise per square cm.
Answer: Radius \( r = 18 \text{ cm} \). Total length \( = 108 \text{ cm} \).
Height of cylinder \( h = 108 - 2(18) = 72 \text{ cm} \).
Total Surface Area \( = 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r) \)
\( \implies TSA = 2 \times \frac{22}{7} \times 18 \times (72 + 36) = \frac{85536}{7} \text{ cm}^2 \).
Cost \( = \frac{85536}{7} \times \frac{70}{100} = \text{Rs. } 8553.6 \).

Question. A solid wooden toy is in the form of a cone mounted on a hemisphere. If the radii of hemisphere and base of cone are 4.2 cm each and the total height of toy is 10.2 cm, find the volume of wood used in the toy. Also, find the total surface area of the toy.
Answer: \( r = 4.2 \text{ cm} \). Total height \( = 10.2 \implies \) Cone height \( H = 10.2 - 4.2 = 6 \text{ cm} \).
Volume \( = V_{hemi} + V_{cone} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 H = \frac{1}{3}\pi r^2(2r + H) \)
\( \implies V = \frac{1}{3} \times \frac{22}{7} \times 4.2^2 \times (8.4 + 6) = 266.11 \text{ cm}^3 \).
Slant height \( l = \sqrt{4.2^2 + 6^2} = \sqrt{17.64 + 36} = \sqrt{53.64} \approx 7.32 \text{ cm} \).
TSA \( = 2\pi r^2 + \pi rl = \pi r(2r + l) = \frac{22}{7} \times 4.2 \times (8.4 + 7.32) = 207.50 \text{ cm}^2 \).

Question. Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio of heights of x and y.
Answer: Let \( r_y = R, r_x = 3R \). Also \( V_y = \frac{1}{2} V_x \implies V_x = 2V_y \).
\( \implies \frac{1}{3} \pi (3R)^2 h_x = 2 \times \frac{1}{3} \pi R^2 h_y \)
\( \implies 9R^2 h_x = 2 R^2 h_y \implies 9h_x = 2h_y \)
\( \implies \frac{h_x}{h_y} = \frac{2}{9} \). Ratio is 2:9.

Question. If per cent increase in radius of a cylinder 3500%, then by how much per cent will the volume of cylinder change (keeping height of cylinder constant)?
Answer: New radius \( R = r + 35r = 36r \).
New volume \( V' = \pi R^2 h = \pi (36r)^2 h = 1296 (\pi r^2 h) = 1296 V \).
Percentage change \( = \frac{1296V - V}{V} \times 100 = 1295 \times 100 = 129500\% \).

Question. Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs Rs. 100 per sq m, find the amount, the associations will have to pay. [Use \( \pi = \frac{22}{7} \)]
Answer: Radius \( r = 2.1 \text{ m} \). Cylinder height \( h = 4 \text{ m} \). Cone height \( H = 2.8 \text{ m} \).
Slant height \( l = \sqrt{2.1^2 + 2.8^2} = 3.5 \text{ m} \).
Area of 1 tent \( = 2\pi rh + \pi rl = \frac{22}{7} \times 2.1 \times (2 \times 4 + 3.5) = 0.66 \times 11.5 \times 100 / 10 = 75.9 \text{ m}^2 \). (Wait, \( \frac{22}{7} \times 2.1 \times 11.5 = 6.6 \times 11.5 = 75.9 \)).
Total area for 100 tents \( = 7590 \text{ m}^2 \).
Total cost \( = 7590 \times 100 = \text{Rs. } 7,59,000 \).
Amount associations pay (50%) \( = \text{Rs. } 3,79,500 \).

Question. A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of \( \frac{3}{2} \text{ cm} \) and its depth is \( \frac{8}{9} \text{ cm} \). 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 \( = \pi (3)^2 (5) = 45\pi \text{ cm}^3 \).
Volume taken out (cone) \( = \frac{1}{3} \pi (\frac{3}{2})^2 (\frac{8}{9}) = \frac{1}{3} \pi (\frac{9}{4}) (\frac{8}{9}) = \frac{2}{3} \pi \text{ cm}^3 \).
Volume left \( = 45\pi - \frac{2}{3} \pi = \frac{133}{3} \pi \text{ cm}^3 \).
Ratio \( = \frac{133}{3} \pi : \frac{2}{3} \pi = 133 : 2 \).

Assertion and reason questions

Question. Assertion (A): The sum of the length, breadth and height of a cuboid is 19 cm and its diagonal is \( 5\sqrt{5} \text{ cm} \). Its surface area is \( 236 \text{ cm}^2 \).
Reason (R): The lateral surface area of a cuboid is 2(l + b).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (c) A is true but R is false.

Question. Assertion (A): The measurements of the cuboidal box are 10 m × 7 m × 3 m. Then, the volume of the box is \( 206 \text{ m}^3 \).
Reason (R): Volume of the cuboid = length × breadth × height.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.