Mensuration – Volume and Area(level-1)
Q1–10 of 55If V be the volume and S be the surface area of a cuboid of dimensions a, b, c, then 1/V is equal to
- $$\text{Volume } V = abc$$$$\text{Surface Area } S = 2(ab + bc + ca)$$
Evaluation:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ca + ab}{abc} = \frac{S/2}{V} = \frac{S}{2V}$$$$\implies \frac{1}{V} = \frac{2}{S} \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$
Correct Option: (b) 2/S (1/a + 1/b + 1/c)
It is required to construct a big rectangular hall to accommodate 500 persons, allowing 22.5 m³ space per person. The height of the hall is to be kept at 7.5 m, while the total inner surface area of the walls must be 1200 sq. m. Then the length and breadth of the hall respectively are
Volume required: $V = 500 \times 22.5 = 11250 \text{ m}^3$
Area of base: $\text{Area} = \frac{V}{h} = \frac{11250}{7.5} = 1500 \text{ m}^2 \implies l \times b = 1500$
Inner wall area (2h(l + b)):
$$2(7.5)(l + b) = 1200 \implies 15(l + b) = 1200 \implies l + b = 80$$Solving for $l$ and $b$:
$$l = 50\text{ m}, \quad b = 30\text{ m}$$
Correct Option: (c) 50 m and 30 m
A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m by 2.5 m. The earth dug out is evenly spread over the remaining portion of the field. How much is the level of the field raised?
Volume of earth dug out: $V = 5 \times 4.5 \times 2.1 = 47.25 \text{ m}^3$
Area of the field: $13.5 \times 2.5 = 33.75 \text{ m}^2$
Remaining area: $33.75 - (5 \times 4.5) = 33.75 - 22.5 = 11.25 \text{ m}^2$
Height raised:
$$h = \frac{47.25}{11.25} = 4.2 \text{ m}$$
Correct Option: (c) 4.2 m
The length, breadth and height of a cuboid are in the ratio 1 : 2 : 3. The length, breadth and height of the cuboid are increased by 100%, 200% and 200% respectively. Then the increase in the volume of the cuboid is
Initial dimensions: $l, 2l, 3l \implies V_1 = l \times 2l \times 3l = 6l^3$
New dimensions:
$l' = l + 100\% \text{ of } l = 2l$
$b' = 2l + 200\% \text{ of } 2l = 6l$
$h' = 3l + 200\% \text{ of } 3l = 9l$
New volume: $V_2 = 2l \times 6l \times 9l = 108l^3$
Increase in volume: $108l^3 - 6l^3 = 102l^3$
Ratio of increase: $\frac{102l^3}{6l^3} = 17 \text{ times}$
If a metallic cuboid weighs 16 kg, how much would a miniature cuboid of metal weigh, if all dimensions are reduced to one-fourth of the original?
Scale factor for dimensions: $k = \frac{1}{4}$
Scale factor for volume/weight: $k^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$
Weight of miniature: $16 \times \frac{1}{64} = 0.25 \text{ kg}$
Correct Option: (a) 0.25 kg
A rectangular water tank is open at the top. Its capacity is 24 m³. Its length and breadth are 4 m and 3 m respectively. Ignoring the thickness of material used for building the tank, the total cost of painting the inner and outer surfaces of the tank at the rate of ₹10 per m² is
Volume: $l \times b \times h = 24 \implies 4 \times 3 \times h = 24 \implies h = 2 \text{ m}$
Surface area of open tank:
$$\text{Area} = lb + 2h(l + b) = (4 \times 3) + 2(2)(4 + 3) = 12 + 28 = 40 \text{ m}^2$$Total area to paint (inner + outer): $2 \times 40 = 80 \text{ m}^2$
Total cost: $80 \times 10 = \text{₹}800$
Correct Option: (d) ₹800
If the areas of three adjacent faces of a cuboid are x, y, z respectively, then the volume of the cuboid is
Given $x = lb$, $y = bh$, $z = hl$.
- $$x \cdot y \cdot z = (lb)(bh)(hl) = (lbh)^2 = V^2$$
- $$V = \sqrt{xyz}$$
Correct Option: (c) $\sqrt{xyz}$
The length of an edge of a hollow cube open at one face is √3 metres. What is the length of the largest pole that it can accommodate?
The largest pole in a hollow open cube corresponds to the space diagonal of the cube.
Length of space diagonal:
$$d = a\sqrt{3} = \sqrt{3} \times \sqrt{3} = 3 \text{ m}$$
Correct Option: (b) 3 m
If the total length of diagonals of a cube is 12 cm, then what is the total length of the edges of the cube?
A cube has 4 space diagonals.
Length of 1 diagonal: $\frac{12}{4} = 3 \text{ cm}$
Edge of the cube ($a$): $a\sqrt{3} = 3 \implies a = \sqrt{3} \text{ cm}$
Total length of 12 edges: $12 \times \sqrt{3} = 12\sqrt{3} \text{ cm}$
Correct Option: (d) $12\sqrt{3}$ cm
A larger cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. The ratio of the total surface areas of the smaller cubes and the larger cube is
Sum of volumes of 3 smaller cubes:
$$V = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216 \text{ cm}^3$$Side of larger cube ($A$): $A^3 = 216 \implies A
= 6 \text{ cm}$
Sum of surface areas of smaller cubes:
$$S_1 = 6(3^2 + 4^2 + 5^2) = 6(9 + 16 + 25) = 6(50)= 300 \text{ cm}^2$$Surface area of larger cube:
$$S_2 = 6(6^2) = 216 \text{ cm}^2$$Ratio: $\frac{300}{216} = \frac{25}{18}$
Correct Option: (c) 25 : 18