CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set 08

SURFACE AREA & VOLUME

FORMULAE

1. If \( \ell \), \( b \) and \( h \) denote respectively the length, breadth and height of a cuboid, then -

(i) total surface area of the cuboid = \( 2 (\ell b+bh+\ell h) \) square units.

(ii) Volume of the cuboid = Area of the base \( \times \) height = \( \ell bh \) cubic units.

(iii) Diagonal of the cuboid or longest rod = \( \sqrt{\ell^2 + b^2 + h^2} \) units.

(iv) Area of four walls of a room = \( 2 (\ell + b) h \) sq. units.

2. If the length of each edge of a cube is ‘\( a \)’ units, then-

(i) Total surface area of the cube = \( 6a^2 \) sq. units.

(ii) Volume of the cube = \( a^3 \) cubic units

(iii) Diagonal of the cube = \( \sqrt{3} a \) units.

3. If \( r \) and \( h \) denote respectively the radius of the base and height of a right circular cylinder, then -

(i) Area of each end = \( \pi r^2 \)

(ii) Curved surface area = \( 2\pi rh = (\text{circumference}) \text{ height} \)

(iii) Total surface area = \( 2\pi r (h + r) \) sq. units.

(iv) Volume = \( \pi r^2h = \text{Area of the base} \times \text{height} \)

4. If \( R \) and \( r \) (\( R > r \)) denote respectively the external and internal radii of a hollow right circular cylinder, then -

(i) Area of each end = \( \pi(R^2 – r^2) \)

(ii) Curved surface area of hollow cylinder = \( 2\pi (R + r) h \)

(iii) Total surface area = \( 2\pi (R + r) (R + h – r) \)

(iv) Volume of material = \( \pi h (R^2 – r^2) \)

5. If \( r \), \( h \) and \( \ell \) denote respectively the radius of base, height and slant height of a right circular cone, then-

(i) \( \ell^2 = r^2 + h^2 \)

(ii) Curved surface area = \( \pi r\ell \)

(iii) Total surface area = \( \pi r^2 + \pi r\ell \)

(iv) Volume = \( \frac{1}{3} \pi r^2h \)

6. For a sphere of radius \( r \), we have

(i) Surface area = \( 4\pi r^2 \)

(ii) Volume = \( \frac{4}{3} \pi r^3 \)

7. If \( h \) is the height, \( \ell \) the slant height and \( r_1 \) and \( r_2 \) the radii of the circular bases of a frustum of a cone then -

(i) Volume of the frustum = \( \frac{\pi}{3} (r_1^2 + r_1r_2 + r_2^2) h \)

(ii) Lateral surface area = \( \pi (r_1 + r_2) \ell \)

(iii) Total surface area = \( \pi\{(r_1 + r_2) \ell + r_1^2 + r_2^2\} \)

(iv) Slant height of the frustum = \( \sqrt{h^2 + (r_1 - r_2)^2} \)

(v) Height of the cone of which the frustum is a part = \( \frac{hr_1}{r_1 - r_2} \)

(vi) Slant height of the cone of which the frustum is a part = \( \frac{\ell r_1}{r_1 - r_2} \)

(vii) Volume of the frustum = \( \frac{h}{3} \{A_1 + A_2 + \sqrt{A_1A_2}\} \), where \( A_1 \) and \( A_2 \) denote the areas of circular bases of the frustum.

Question. A cylinder and a cone have same base area. But the volume of cylinder is twice the volume of cone. Find the ratio between their heights.
Answer: Since, the base areas of the cylinder and the cone are the same.
\( \Rightarrow \) their radius are equal (same).
Let the radius of their base be \( r \) and their heights be \( h_1 \) and \( h_2 \) respectively.
Clearly, volume of the cylinder = \( \pi r^2h_1 \)
and, volume of the cone = \( \frac{1}{3} \pi r^2h_2 \)
Given :
Volume of cylinder = 2 \( \times \) volume of cone
\( \Rightarrow \pi r^2h_1 = 2 \times \frac{1}{3} \pi r^2h_2 \)
\( \Rightarrow h_1 = \frac{2}{3} h_2 \Rightarrow \frac{h_1}{h_2} = \frac{2}{3} \)
i.e., \( h_1 : h_2 = 2 : 3 \)

Question. The radius of a sphere increases by 25%. Find the percentage increase in its surface area.
Answer: Let the original radius be \( r \).
\( \Rightarrow \) Original surface area of the sphere = \( 4\pi r^2 \)
Increase radius = \( r + 25\% \text{ of } r \)
\( = r + \frac{25}{100}r = \frac{5r}{4} \)
\( \Rightarrow \) Increased surface area
\( = 4\pi \left(\frac{5r}{4}\right)^2 = \frac{25\pi r^2}{4} \)
Increased in surface area
\( = \frac{25\pi r^2}{4} – 4\pi r^2 = \frac{25\pi r^2 – 16\pi r^2}{4} = \frac{9\pi r^2}{4} \)
and, percentage increase in surface area
\( = \frac{\text{Increase in area}}{\text{Original aera}} \times 100\% \)
\( = \frac{\frac{9\pi r^2}{4}}{4\pi r^2} \times 100\% = \frac{9}{16} \times 100\% \)
= 56.25%
Alternative Method :
Let original radius = 100
\( \Rightarrow \) Original C.S.A. = \( \pi(100)^2 = 10000\pi \)
Increased radius = \( 100 + 25\% \text{ of } 100 = 125 \)
\( \Rightarrow \) Increased C.S.A. = \( \pi(125)^2 = 15625\pi \)
Increase in C.S.A. = \( 15625\pi – 10000\pi = 5625\pi \)
\( \therefore \) Percentage increase in C.S.A.
\( = \frac{\text{Increase in C.S.A.}}{\text{Original C.S.A.}} \times 100\% \)
\( = \frac{5625\pi}{10000\pi} \times 100\% = 56.25\% \)
Conversely, if diameter decreases by 20%, the radius also decreases by 20%.

Question. Three solid spheres of radii 1 cm, 6 cm and 8 cm are melted and recasted into a single sphere. Find the radius of the sphere obtained.
Answer: Let radius of the sphere obtained = \( R \text{ cm} \).
\( \therefore \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (1)^3 + \frac{4}{3} \pi (6)^3 + \frac{4}{3} \pi (8)^3 \).
\( R^3 = 1 + 216 + 512 = 729 \)
\( \therefore R = (9^3)^{1/3} = 9 \text{ cm} \)

Question. The decorative block shown in figure is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take \( \pi = \frac{22}{7} \))
Answer: The total surface area of the cube = \( 6 \times (\text{edge})^2 = 6 \times 5 \times 5 \text{ cm}^2 = 150 \text{ cm}^2 \).
Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So, the surface area of the block
= TSA of cube – base area of hemisphere + CSA of hemisphere
\( = 150 – \pi r^2 + 2\pi r^2 = (150 + \pi r^2) \text{ cm}^2 \)
\( = 150 \text{ cm}^2 + \left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \text{ cm}^2 \)
\( = (150 + 13.86) \text{ cm}^2 = 163.86 \text{ cm}^2 \)

Question. A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in figure. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the best diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take \( \pi = 3.14 \))
Answer: Denote radius of cone by \( r \), slant height of cone by \( \ell \), height of cone by \( h \), radius of cylinder by \( r' \) and height of cylinder by \( h' \). Then \( r = 2.5 \text{ cm}, h = 6 \text{ cm}, r' = 1.5 \text{ cm}, h' = 26 – 6 = 20 \text{ cm} \) and
\( \ell = \sqrt{r^2 + h^2} = \sqrt{2.5^2 + 6^2} \text{ cm} = 6.5 \text{ cm} \)
Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So a part of the base of the cone (a ring) is to be painted.
So, the area to be painted orange
= CSA of the cone + base area of the cone – base area of the cylinder
\( = \pi r\ell + \pi r^2 – \pi(r')^2 \)
\( = \pi[(2.5 \times 6.5) + (2.5)^2 – (1.5)^2] \text{ cm}^2 \)
\( = \pi[20.25] \text{ cm}^2 = 3.14 \times 20.25 \text{ cm}^2 \)
\( = 63.585 \text{ cm}^2 \)
Now, the area to be painted yellow
= CSA of the cylinder + area of one base of the cylinder
\( = 2\pi r'h' + \pi(r')^2 \)
\( = \pi r' (2h' + r') \)
\( = (3.14 \times 1.5) (2 \times 20 + 1.5) \text{ cm}^2 = 4.71 \times 41.5 \text{ cm}^2 \)
\( = 195.465 \text{ cm}^2 \)

Question. A cone of height 24 cm and radius of base 6 cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Answer: Volume of cone = \( \frac{1}{3} \times \pi \times 6 \times 6 \times 24 \text{ cm}^3 \)
If \( r \) is the radius of the sphere, then its volume is \( \frac{4}{3} \pi r^3 \).
Since the volume of clay in the form of the cone and the sphere remains the same, we have.
\( \frac{4}{3} \times \pi \times r^3 = \frac{1}{3} \times \pi \times 6 \times 6 \times 24 \)
\( r^3 = 3 \times 3 \times 24 = 3^3 \times 2^3 \)
\( r = 3 \times 2 = 6 \)
Therefore, the radius of the sphere is 6 cm.

Question. Selvi's house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m \( \times \) 1.44 m \( \times \) 95 cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use \( \pi = 3.14 \))
Answer: The volume of water in the overhead tank equals the volume of the water removed from the shump.
Now the volume of water in the overhead tank (cylinder) = \( \pi r^2h \)
\( = 3.14 \times 0.6 \times 0.6 \times 0.95 \text{ m}^3 \)
The volume of water in the sump when full
= \( \ell \times b \times h = 1.57 \times 1.44 \times 0.95 \text{ m}^3 \)
The volume of water left in the sump after filling the tank
\( = (1.57 \times 1.44 \times 0.95) – (3.14 \times 0.6 \times 0.6 \times 0.95)] \text{ m}^3 \)
\( = (1.57 \times 0.6 \times 0.6 \times 0.95 \times 2) \text{ m}^3 \)
So, the height of the water left in the sump
\( = \frac{\text{volume of water left in the sump}}{\ell \times b} \)
\( = \frac{1.57 \times 0.6 \times 0.6 \times 0.95 \times 2}{1.57 \times 1.44} \text{ m} \)
= 0.475 m = 47.5 cm
Also, \( \frac{\text{Capacity of tank}}{\text{Capacity of sump}} \)
\( = \frac{3.14 \times 0.6 \times 0.6 \times 0.95}{1.57 \times 1.44 \times 0.95} = \frac{1}{2} \)
Therefore, the capacity of the tank is half the capacity of the sump.

Question. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.
Answer: The volume of the rod = \( \pi \times \left(\frac{1}{2}\right)^2 \times 8 \text{ cm}^3 = 2\pi \text{ cm}^3 \).
The length of the new wire of the same volume = 18 m = 1800 cm
If \( r \) is the radius (in cm) of cross section of the wire, its volume = \( \pi \times r^2 \times 1800 \text{ cm}^3 \)
Therefore, \( \pi \times r^2 \times 1800 = 2\pi \)
i.e. \( r^2 = \frac{1}{900} \)
i.e. \( r = \frac{1}{30} \)
So, The diameter of the cross section i.e. the thickness of the wire is \( \frac{1}{15} \) cm, i.e. 0.67 mm (approx.)

EXERCISE # 1

Question. A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm. Find the rise in water level in the vessel.
Answer: 4.05 cm

Question. Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively to form a single cube. Find its edge. Also, find the surface area of the new cube.
Answer: 6 m, \( 216 \text{ cm}^2 \)

Question. Water flows in a tank 150 m \( \times \) 100 m at the base, through a pipe whose crosssection is 2 dm by 1.5 dm at the speed of 15 km per hour. In what time, will the water be 3 metres deep.
Answer: 100 Hours

Question. A right circular cone is 3.6 cm high and radius of its base is 1.6 cm. It is melted and recast into a right circular cone with radius of its base as 1.2 cm. Find its height.
Answer: 6.4 cm

Question. A solid cube of side 7 cm is melted to make a cone of height 5 cm, find the radius of the base of the cone.
Answer: 8.09 cm

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 height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if height of the conical part is 12 cm.
Answer: \( 770 \text{ cm}^2 \)

Question. A solid sphere of radius 3 cm is melted and then cast into small spherical balls each of diameter 0.6 cm. Find the number of balls thus obtained.
Answer: 1000

Question. Three solid spheres of iron whose diameters are 2 cm, 12 cm and 16 cm, respectively, are melted into a single solid sphere. Find the radius of the solid sphere.
Answer: 9 cm

Question. How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter.
Answer: 9.2541

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: 1 cm

Question. A spherical canon ball, 28 cm in diameter is melted and cast into a right circular conical mould, the base of which is 35 cm in diameter. Find the height of the cone, correct to one placed of decimal.
Answer: 35.84 cm

Question. Length of a class-room is two times its height and its breadth is \( 1\frac{1}{2} \) times its height. The cost of white-washing the walls at the rate of -j 1.60 per \( \text{m}^2 \) is -j 179.20. Find the cost of tiling the floor at the rate of -j 6.75 per \( \text{m}^2 \).
Answer: -j . 324

Question. A room is half as long again as it is broad. The cost of carpeting the room at -j 3.25 per \( \text{m}^2 \) is -j 175.50 and the cost of papering the walls at -j 1.40 per \( \text{m}^2 \) is -j 240.80. If 1 door and 2 windows occupy \( 8 \text{ m}^2 \), find the dimensions of the room.
Answer: \( \lambda = 9 \text{ m}, b = 6 \text{cm}, h = 6 \text{ cm} \)

Question. A rectangular tank is 225 m by 162 m at the base. With what speed must water flow into it through an aperture 60 cm by 45 cm that the level may be raised 20 cm in 5 hours?
Answer: 5400 m/h

Question. An agricultural field is in the form of a rectangle of length 20 m and width 14 m. A pit 6 m long, 3 m wide and 2.5 m deep is dug in a corner of the field and the earth taken out of the pit is spread uniformly over the remaining area of the field. Find the extent to which the level of the field has been raised.
Answer: \( \frac{1}{21} \text{kg}, \frac{1}{9} \text{kg} \)

Question. A rectangular sheet of paper 44 cm \( \times \) 18 cm is rolled along its length and a cylinder is formed. Find the radius of the cylinder.
Answer: 7 cm

Question. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the cost of painting the toy at the rate of -j 5 pr 1000 \( \text{cm}^2 \).
Answer: 51 paise

Question. A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is four times the radius of its base, find the radius of the ice-cream cone.
Answer: 3 cm

Question. A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the wooden toy.
Answer: \( 266.11 \text{ cm}^3 \)

Question. A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the sphere is 14 cm and the total height of the vessel is 13 cm. Find its capacity. (Take \( \pi = 22/7 \)).
Answer: \( 1642.66 \text{ cm}^3 \)

Question. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid. (use \( \pi = 22/7 \)).
Answer: \( 641.66 \text{ cm}^3, 418 \text{ cm}^2 \)

Question. Find what length of canvas 2m in width is required to make a conical tent 20 m in diameter and 42m in slant height allowing 10% for folds and stitching. Also find the cost of canvas at the rate of -j 60 per metre.
Answer: 726 m, -j .43560

Question. From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the cone and of the remaining material.
Answer: \( 718 \frac{2}{3} \text{ cm}^3, 2025 \frac{1}{3} \text{ cm}^3 \)

Question. A cone of maximum volume is carved out of a block of wood of size 20 cm \( \times \) 10 cm \( \times \) 10 cm. Find the volume of the cone carved out correct to one decimal place. Take \( \pi = 3.1416 \).
Answer: \( 523.6 \text{ cm}^3 \)

Question. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of the wood in the entire stand.
Answer: \( 523.53 \text{ cm}^3 \)

Question. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest \( \text{cm}^2 \).
Answer: \( 18 \text{ cm}^2 \)

Question. From a solid cylinder whose height is 8 cm and radius is 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to 4 significant figures. (\( \pi = 3.1416 \)). Also find the total surface area of the remaining solid.
Answer: \( 603.2 \text{ cm}^3, 603.2 \text{ cm}^2 \)

Question. A metallic cylinder has radius 3 cm and height 5 cm. It is made of a metal A. To reduce its weight, a conical hole is drilled in the cylinder as shown and it is completely filled with a lighter metal B. The conical hole has a radius of \( \frac{3}{2} \) cm and its depth is \( \frac{8}{9} \) cm. Calculate the ratio of the volume of the metal A to the volume of the metal B in the solid.
Answer: 133 : 2

Question. An open cylinder vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is \( \frac{7}{2} \) cm and height 8 cm. Find the volume of water required to fill the vessel.
Answer: \( 282 \frac{1}{3} \text{ cm}^3 \)

Question. A hemispherical tank full of water is emptied by a pipe at the rate of \( 3 \frac{4}{7} \) litres per second. How much time will it take to empty half the tank, if it is 3 m in diameter ?
Answer: 16.5 minutes

Question. The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.
Answer: 10 cm

Question. The entire surface of solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of (i) their curved surface; (ii) their volumes.
Answer: (i) 15 : 16 (ii) 3 : 4

Question. A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is 24 cm, find (i) The radius and (ii) The slant height of the heap. Leave your answer in square root form.
Answer: (i) 36 cm (ii) \( 4\sqrt{117} \text{ cm} \)

Question. A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is \( \frac{1}{2} \) cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone correct to one decimal place.
Answer: 5.4 cm

Question. A right circular cone of height 20 cm and base diameter 30 cm is cast into smaller cones of equal sizes with base radius 10 cm and height 9 cm. Find how many cones are made.
Answer: 5.