CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 17

Excluded from NCERT: Exercise 1

Question 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Answer:
Volume of the sphere \( = \) volume of the cylinder: \( \frac{4}{3}\pi(4.2)^3 = \pi(6)^2 h \).
\( h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 36} = \frac{296.352}{108} = 2.744 \) cm.
Hence, the height of the cylinder is 2.744 cm (about 2.74 cm).
In simple words: Equate the volume of the sphere and the cylinder.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast, so equate the volumes.
2. Convert all measurements to the same unit before calculating.

Question 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Answer:
Volume of the new sphere \( = \) sum of the volumes of the three spheres: \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(6^3 + 8^3 + 10^3) \).
\( r^3 = 216 + 512 + 1000 = 1728 \Rightarrow r = 12 \) cm.
Hence, the radius of the resulting sphere is 12 cm.
In simple words: Add the cubes of the radii and take the cube root.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast, so equate the volumes.
2. Convert all measurements to the same unit before calculating.

Question 3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer:
Volume of earth dug out \( = \pi r^2 h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 = 770 \) m³.
Volume of the platform \( = 22 \times 14 \times H = 308H \) m³. Equating: \( 308H = 770 \Rightarrow H = 2.5 \) m.
Hence, the height of the platform is 2.5 m.
In simple words: The well is a cylinder and the platform is a cuboid with the same volume.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast, so equate the volumes.
2. Convert all measurements to the same unit before calculating.

Question 4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Answer:
Radius of the well \( r_1 = \frac{3}{2} \) m, depth 14 m. Outer radius of the embankment \( r_2 = 4 + \frac{3}{2} = \frac{11}{2} \) m.
Volume of earth \( = \pi r_1^2 \times 14 = \pi(r_2^2 - r_1^2)H \Rightarrow \frac{9}{4} \times 14 = \left(\frac{121}{4} - \frac{9}{4}\right)H = 28H\).
\( H = \frac{31.5}{28} = 1.125 \) m.
Hence, the height of the embankment is 1.125 m.
In simple words: The embankment is a hollow cylinder (ring) whose inner radius equals the well's radius.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast, so equate the volumes.
2. Convert all measurements to the same unit before calculating.

Question 5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Answer:
Volume of ice cream in the cylinder \( = \pi \times 6^2 \times 15 = 540\pi \) cm³.
Volume of one cone with the hemispherical top \( = \frac{1}{3}\pi(3)^2(12) + \frac{2}{3}\pi(3)^3 = 36\pi + 18\pi = 54\pi \) cm³.
Number of cones \( = \frac{540\pi}{54\pi} = 10 \).
Hence, 10 ice cream cones can be filled.
In simple words: Divide the volume of the cylinder by the volume of one cone with its hemisphere.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast, so equate the volumes.
2. Convert all measurements to the same unit before calculating.

Question 6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm \( \times \) 10 cm \( \times \) 3.5 cm?
Answer:
Coin: radius \( = 0.875 \) cm, thickness \( = 0.2 \) cm. Volume of one coin \( = \pi(0.875)^2(0.2) \) cm³. Volume of the cuboid \( = 5.5 \times 10 \times 3.5 = 192.5 \) cm³.
Number of coins \( = \frac{192.5}{\frac{22}{7} \times 0.765625 \times 0.2} = 400 \).
Hence, 400 coins must be melted.
In simple words: Divide the cuboid's volume by the volume of one coin.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast, so equate the volumes.
2. Convert all measurements to the same unit before calculating.

Question 7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Answer:
Volume of sand in the bucket \( = \pi \times 18^2 \times 32 \). Volume of the conical heap \( = \frac{1}{3}\pi r^2 \times 24 = 8\pi r^2 \).
\( 8r^2 = 18^2 \times 32 \Rightarrow r^2 = 1296 \Rightarrow r = 36 \) cm.
Slant height \( l = \sqrt{36^2 + 24^2} = \sqrt{1872} = 12\sqrt{13} \) cm.
Hence, the radius is 36 cm and the slant height is \( 12\sqrt{13} \) cm (about 43.27 cm).
In simple words: Equate the volume of the cylinder and the cone, then use the Pythagoras theorem.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast, so equate the volumes.
2. Convert all measurements to the same unit before calculating.

Question 8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Answer:
In 30 minutes the water covers a length of \( 10 \times \frac{1}{2} = 5 \) km \( = 5000 \) m. Volume of water \( = 5000 \times 6 \times 1.5 = 45000 \) m³.
Area irrigated \( = \frac{45000}{0.08} = 562500 \) m² \( = \frac{562500}{10000} \) hectares \( = 56.25 \) hectares.
Hence, the area irrigated is 56.25 hectares.
In simple words: Volume of water divided by the required depth gives the area.

Teacher's Note:
1. Use 1 hectare = 10000 m².
2. 8 cm = 0.08 m.

Question 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Answer:
Radius of the pipe \( = 0.1 \) m, so its cross-section \( = \pi(0.1)^2 = 0.01\pi \) m². Speed \( = 3 \) km/h \( = 50 \) m/min.
Water flowing in 1 minute \( = 50 \times 0.01\pi = 0.5\pi \) m³.
Volume of the tank \( = \pi(5)^2(2) = 50\pi \) m³. Time \( = \frac{50\pi}{0.5\pi} = 100 \) minutes.
Hence, the tank will be filled in 100 minutes.
In simple words: Volume of the tank divided by the volume flowing in per minute.

Teacher's Note:
1. Convert km/h to m/min.
2. Pi cancels.

Excluded from NCERT: Exercise 3

Question 10. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³. (Use \( \pi = 3.14 \))
Answer:
Diameter of the wire \( = 3 \) mm \( = 0.3 \) cm, so the number of turns needed to cover 12 cm \( = \frac{12}{0.3} = 40 \).
Length of wire in one turn \( = 2\pi r = 2 \times 3.14 \times 5 = 31.4 \) cm. Length of the wire \( = 40 \times 31.4 = 1256 \) cm \( = 12.56 \) m.
Volume of the wire \( = \pi(0.15)^2 \times 1256 = 3.14 \times 0.0225 \times 1256 \approx 88.74 \) cm³.
Mass \( = 88.74 \times 8.88 \approx 788 \) g.
Hence, the length of the wire is 1256 cm and its mass is about 788 g.
In simple words: Count the turns first, then use the volume and the density to get the mass.

Teacher's Note:
1. Convert mm to cm before dividing.
2. Mass = volume times density.

Question 11. A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Use \( \pi = 3.14 \))
Answer:
In right triangle PQR, \( PQ = 3 \) cm, \( PR = 4 \) cm, so \( QR = 5 \) cm. Let PD be perpendicular to QR; PD is the common radius of the two cones. From the similar triangles, \( PD = \frac{PQ \times PR}{QR} = \frac{12}{5} \) cm, \( DQ = \frac{PQ^2}{QR} = \frac{9}{5} \) cm and \( DR = \frac{16}{5} \) cm.
Volume \( = \frac{1}{3}\pi (PD)^2 (DQ + DR) = \frac{1}{3} \times 3.14 \times \frac{144}{25} \times 5 \approx 30.14 \) cm³.
Surface area \( = \pi (PD)(PQ + PR) = 3.14 \times \frac{12}{5} \times 7 = 52.75 \) cm².
Hence, the volume is 30.14 cm³ and the surface area is 52.75 cm².
In simple words: The radius of both cones is the altitude from the right angle to the hypotenuse.

Teacher's Note:
1. PD times QR equals PQ times PR (twice the area).
2. The surface of the double cone is the two curved surfaces.

Question 12. A cistern, internally measuring 150 cm \( \times \) 120 cm \( \times \) 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each being 22.5 cm \( \times \) 7.5 cm \( \times \) 6.5 cm?
Answer:
Volume of the cistern \( = 150 \times 120 \times 110 = 1980000 \) cm³. Empty space \( = 1980000 - 129600 = 1850400 \) cm³.
Let n bricks be put in. Volume of one brick \( = 22.5 \times 7.5 \times 6.5 = 1096.875 \) cm³. Each brick absorbs \( \frac{1}{17} \) of its volume of water, so the water-filled space needed is \( n \times \frac{16}{17} \) of the bricks' volume.
Condition: \( \frac{n}{17} \times 1096.875 + 1850400 = n \times 1096.875 \), that is \( \frac{16n}{17} \times 1096.875 = 1850400 \), so \( n = 1792.41 \).
Since the number of bricks must be a whole number and the water must not overflow, \( n = 1792 \).
Hence, 1792 bricks can be put in.
In simple words: The bricks take up space but also absorb some water, so the free volume is shared.

Teacher's Note:
1. Round the answer down, because a fraction of a brick cannot be placed.
2. Absorbed water frees up space in the cistern.

Question 13. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km², show that the total rainfall was approximately equivalent to the addition of the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Answer:
Volume of rain water \( = 7280 \times \frac{10}{100000} = 0.728 \) km³ (10 cm \( = \frac{10}{100000} \) km).
Volume of water in 3 rivers \( = 3 \times \left(1072 \times \frac{75}{1000} \times \frac{3}{1000}\right) = 0.7236 \) km³.
The two volumes, 0.728 km³ and 0.7236 km³, are nearly equal.
Hence, the total rainfall in the valley was approximately equal to the normal water of the three rivers.
In simple words: Compare the two volumes in the same unit.

Teacher's Note:
1. Express everything in km before multiplying.
2. 10 cm is 0.0001 km.