CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 13

Question. If the length of each edge of a cube is doubled, how many times it becomes:
(i) in volume
(ii) in S.A.
Answer: (i) Let the edge of the cube be \( a \). Its volume is \( V_1 = a^3 \).
If the length of each edge is doubled, the new edge is \( 2a \).
New volume \( V_2 = (2a)^3 = 8a^3 = 8V_1 \).
Thus, the volume becomes 8 times.

(ii) The original surface area is \( S_1 = 6a^2 \).
New surface area \( S_2 = 6(2a)^2 = 6(4a^2) = 24a^2 = 4S_1 \).
Thus, the surface area becomes 4 times.

Question. Two cubes each of side \( 8\text{ cm} \) are joined end to end. Find the S.A. of the resulting Cuboid.
Answer: When two cubes each of side \( 8\text{ cm} \) are joined end to end, the dimensions of the resulting cuboid are:
Length \( (l) = 8 + 8 = 16\text{ cm} \)
Breadth \( (b) = 8\text{ cm} \)
Height \( (h) = 8\text{ cm} \)

Surface Area (S.A.) of the resulting cuboid:
\( \text{S.A.} = 2(lb + bh + hl) \)
\( \text{S.A.} = 2(16 \times 8 + 8 \times 8 + 8 \times 16) \)
\( \text{S.A.} = 2(128 + 64 + 128) \)
\( \text{S.A.} = 2(320) = 640\text{ cm}^2 \)

Therefore, the surface area of the resulting cuboid is \( 640\text{ cm}^2 \).

Question. A river \( 2\text{ m} \) deep and \( 45\text{ m} \) wide is flowing at the rate of \( 3\text{ km/hr} \). Find the amount of water that runs into the sea per minute.
Answer: Depth of the river \( (h) = 2\text{ m} \)
Width of the river \( (b) = 45\text{ m} \)
Rate of water flow \( = 3\text{ km/hr} = \frac{3 \times 1000\text{ m}}{60\text{ min}} = 50\text{ m/min} \)

Amount of water running into the sea per minute (Volume of water flowing per minute):
\( V = \text{Length} \times \text{Breadth} \times \text{Height} \)
\( V = 50\text{ m} \times 45\text{ m} \times 2\text{ m} = 4500\text{ m}^3 \)

Therefore, the amount of water that runs into the sea per minute is \( 4500\text{ m}^3 \).

Question. In a shower, \( 5\text{ cm} \) of rain falls. Find the volume of water that falls on \( 2\text{ hectares} \) of the ground.
Answer: Area of the ground \( (A) = 2\text{ hectares} = 2 \times 10000\text{ m}^2 = 20000\text{ m}^2 \)
Depth of rainfall \( (h) = 5\text{ cm} = \frac{5}{100}\text{ m} = 0.05\text{ m} \)

Volume of water that falls on the ground:
\( \text{Volume} = \text{Area} \times \text{Height} \)
\( \text{Volume} = 20000\text{ m}^2 \times 0.05\text{ m} = 1000\text{ m}^3 \)

Therefore, the volume of water is \( 1000\text{ m}^3 \).

Question. A wall \( 15\text{ m} \) long, \( 3\text{ dm} \) wide and \( 4\text{ m} \) high is made up of bricks, each measuring \( 22\text{ cm} \times 12.5\text{ cm} \times 7.5\text{ cm} \). If \( \frac{1}{12} \) of the total volume of the wall consists of mortar, how many bricks are there in the wall?
Answer: Dimensions of the wall:
Length \( (L) = 15\text{ m} = 1500\text{ cm} \)
Breadth \( (B) = 3\text{ dm} = 30\text{ cm} \) (since \( 1\text{ dm} = 10\text{ cm} \))
Height \( (H) = 4\text{ m} = 400\text{ cm} \)

Total Volume of the wall \( (V) = 1500 \times 30 \times 400 = 18,000,000\text{ cm}^3 \)
Since \( \frac{1}{12} \) of the wall consists of mortar, the volume of bricks is:
\( \text{Volume of bricks} = \left(1 - \frac{1}{12}\right) \times V = \frac{11}{12} \times 18,000,000 = 16,500,000\text{ cm}^3 \)

Dimensions of each brick:
\( 22\text{ cm} \times 12.5\text{ cm} \times 7.5\text{ cm} \)
Volume of one brick \( = 22 \times 12.5 \times 7.5 = 2062.5\text{ cm}^3 \)

Number of bricks in the wall:
\( \text{Number of bricks} = \frac{\text{Volume of bricks}}{\text{Volume of one brick}} = \frac{16,500,000}{2062.5} = 8000 \)

Therefore, there are 8000 bricks in the wall.

Question. The T.S.A of a solid right circular cylinder is \( 231\text{ cm}^2 \). Its Curved Surface is \( \frac{2}{3}\text{rd} \) of T.S.A. Find the radius of its base and height.
Answer: Total Surface Area (T.S.A.) of the cylinder \( = 231\text{ cm}^2 \)
Curved Surface Area (C.S.A.) \( = \frac{2}{3} \times \text{T.S.A.} = \frac{2}{3} \times 231 = 154\text{ cm}^2 \)

We know that:
\( \text{T.S.A.} = \text{C.S.A.} + 2\pi r^2 \)
\( 231 = 154 + 2\pi r^2 \)
\( 2\pi r^2 = 231 - 154 = 77 \)
\( 2 \times \frac{22}{7} \times r^2 = 77 \)
\( r^2 = \frac{77 \times 7}{44} = \frac{49}{4} \)
\( r = \frac{7}{2}\text{ cm} \)

Since \( \text{C.S.A.} = 2\pi r h = 154\text{ cm}^2 \):
\( 2 \times \frac{22}{7} \times \frac{7}{2} \times h = 154 \)
\( 22h = 154 \)
\( h = 7\text{ cm} \)

Therefore, the radius of its base is \( r = \frac{7}{2}\text{ cm} \) and height is \( h = 7\text{ cm} \).

Question. A well, \( 14\text{ m} \) deep, is \( 2\text{ m} \) in radius. Find the cost of cementing the inner Curved surface at the rate of Rs 2 per square metre.
Answer: Depth of the well \( (h) = 14\text{ m} \)
Radius of the well \( (r) = 2\text{ m} \)

Inner Curved Surface Area (C.S.A.) of the well:
\( \text{C.S.A.} = 2\pi r h = 2 \times \frac{22}{7} \times 2 \times 14 = 176\text{ m}^2 \)

Rate of cementing \( = \text{Rs } 2\text{ per m}^2 \)
Cost of cementing \( = 176 \times 2 = \text{Rs } 352 \)

Therefore, the cost of cementing is \( \text{Rs } 352 \).

Question. A solid cylinder of radius \( 7\text{ cm} \) and \( h = 20\text{ cm} \) is made of material whose density is \( 6\text{ g} \) per \( \text{cm}^3 \). Find its cost at the rate of Rs 100 per kilogram.
Answer: Radius of the cylinder \( (r) = 7\text{ cm} \)
Height of the cylinder \( (h) = 20\text{ cm} \)

Volume of the cylinder:
\( V = \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 20 = 3080\text{ cm}^3 \)

Density of the material \( = 6\text{ g/cm}^3 \)
Mass of the cylinder \( = \text{Volume} \times \text{Density} = 3080 \times 6 = 18480\text{ g} = 18.48\text{ kg} \)

Rate of material \( = \text{Rs } 100\text{ per kg} \)
Cost of the cylinder \( = 18.48 \times 100 = \text{Rs } 1848 \)

Therefore, the cost is \( \text{Rs } 1848 \).

Question. The Cost of polishing the T.S.A of a closed cylindrical oil tank at \( 25\text{ P} \) per \( \text{dm}^2 \) is Rs 77. Its height is 3 times that of the radius of its base. Find radius of the base and height.
Answer: Let the radius of the base be \( r\text{ dm} \) and height be \( h\text{ dm} \).
Given: \( h = 3r \)

Total cost of polishing \( = \text{Rs } 77 = 7700\text{ Paise} \)
Rate of polishing \( = 25\text{ Paise/dm}^2 \)
Total Surface Area (T.S.A.) \( = \frac{7700}{25} = 308\text{ dm}^2 \)

We know that:
\( \text{T.S.A.} = 2\pi r(r + h) = 308 \)
Substituting \( h = 3r \):
\( 2\pi r(r + 3r) = 308 \)
\( 2\pi r(4r) = 308 \)
\( 8\pi r^2 = 308 \)
\( 8 \times \frac{22}{7} \times r^2 = 308 \)
\( r^2 = \frac{308 \times 7}{176} = \frac{49}{4} \)
\( r = \frac{7}{2}\text{ dm} \)

Height \( h = 3r = 3 \times \frac{7}{2} = \frac{21}{2}\text{ dm} \)

Therefore, the radius of the base is \( r = \frac{7}{2}\text{ dm} \) and the height is \( h = \frac{21}{2}\text{ dm} \).

Question. An iron pipe \( 20\text{ cm} \) long has exterior diameter equal to \( 25\text{ cm} \). If the thickness of the pipe is \( 1\text{ cm} \), find the whole surface of the pipe.
Answer: Length of the pipe \( (h) = 20\text{ cm} \)
Exterior diameter \( = 25\text{ cm} \implies \) Exterior radius \( (R) = 12.5\text{ cm} \)
Thickness \( = 1\text{ cm} \implies \) Interior radius \( (r) = 12.5 - 1 = 11.5\text{ cm} \)

The whole surface area (Total Surface Area) of the hollow pipe is given by:
\( \text{T.S.A.} = \text{External C.S.A.} + \text{Internal C.S.A.} + 2 \times \text{Area of base rings} \)
\( \text{T.S.A.} = 2\pi R h + 2\pi r h + 2\pi (R^2 - r^2) \)
\( \text{T.S.A.} = 2\pi [h(R + r) + (R - r)(R + r)] \)
\( \text{T.S.A.} = 2\pi (R + r)(h + R - r) \)
\( \text{T.S.A.} = 2 \times \frac{22}{7} \times (12.5 + 11.5) \times (20 + 12.5 - 11.5) \)
\( \text{T.S.A.} = 2 \times \frac{22}{7} \times 24 \times 21 \)
\( \text{T.S.A.} = 2 \times 22 \times 24 \times 3 = 3168\text{ cm}^2 \)

Therefore, the whole surface area of the pipe is \( 3168\text{ cm}^2 \).