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31. The sum of perimeters of the six faces of a cuboid is 72 cm and the total surface area of the cuboid is 16 cm². Find the longest possible length that can be kept inside the cuboid :

A. 5.2 cm

B. 7.8 cm

C. 8.05 cm

D. 8.36 cm

Discuss

Solution:

Sum of perimeters of the six faces :

\begin{aligned} = 2 [2 (l + b) + 2 (b + h) + 2 (l + h)] \\ = 4 (2 l + 2 b + 2 h) \\ = 8 (l + b + h) \end{aligned}

Total surface area

= 2 (l b + b h + l h)

\begin{aligned} ∴ 8 (l + b + h) = 72 \\ \Rightarrow l + b + h = 9 \\ 2 (l b + b h + l h) = 16 \\ \Rightarrow l b + b h + l h = 8 \end{aligned}

Now,

{(l + b + h)}^{2} = l^{2} + b^{2} + h^{2} + 2

(l b + b h + l h)

\begin{aligned} \Rightarrow {(9)}^{2} = l^{2} + b^{2} + h^{2} + 16 \\ \Rightarrow l^{2} + b^{2} + h^{2} = 81 - 16 \\ \Rightarrow l^{2} + b^{2} + h^{2} = 65 \end{aligned}

Required length :

\begin{aligned} = \sqrt{l^{2} + b^{2} + h^{2}} \\ = \sqrt{65} \\ = 8.05 c m \end{aligned}

32. The surface area of a cube is 150 cm². Its volume is :

A. 64 $c m^{3}$

B. 125 $c m^{3}$

C. 150 $c m^{3}$

D. 216 $c m^{3}$

Discuss

Solution:

\begin{aligned} 6 a^{2} = 150 \\ \Rightarrow a^{2} = 25 \\ \Rightarrow a = 5 \end{aligned}

∴ Volume :

a^{3} = 5^{3} = 125 c m^{3}

33. If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes will be ?

A. 1 : 3

B. 2 : 3

C. 5 : 9

D. 7 : 9

Discuss

Solution:

Let the length of each edge of each cube be a
Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a
Surface area of the cuboid :

\begin{aligned} = 2 [3 a \times a + a \times a + 3 a \times a] \\ = 2 [3 a^{2} + a^{2} + 3 a^{2}] \\ = 14 a^{2} \end{aligned}

Sum of surface area of 3 cubes :

\begin{aligned} = (3 \times 6 a^{2}) \\ = 18 a^{2} \end{aligned}

∴ Required ratio :

\begin{aligned} = 14 a^{2} : 18 a^{2} \\ = 7 : 9 \end{aligned}

34. The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. Find the ratio of its diameter to its height.

A. 3 : 7

B. 7 : 3

C. 6 : 7

D. 7 : 6

Discuss

Solution:

\begin{aligned} \frac{π r^{2} h}{2 π r h} = \frac{924}{264} \\ \Rightarrow r = (\frac{924}{264} \times 2) \\ \Rightarrow r = 7 m \end{aligned}

And,

\begin{aligned} ∴ 2 π r h = 264 \\ \Rightarrow h = (264 \times \frac{7}{22} \times \frac{1}{2} \times \frac{1}{7}) \\ \Rightarrow h = 6 m \end{aligned}

∴ Required ratio :

= \frac{2 r}{h} = \frac{14}{6} = 7 : 3

35. It is required to fix a pipe such that water flowing through it at a speed of 7 metres per minute fills a tank of capacity 440 cubic metres in 10 minutes. The inner radius of the pipe should be :

A. $\sqrt{2} m$

B. $2 m$

C. $\frac{1}{2} m$

D. $\frac{1}{\sqrt{2}} m$

Discuss

Solution:

Let the inner radius of the pipe be r metres
Then,
Volume of water flowing through the pipe in 10 minutes :

\begin{aligned} = [(\frac{22}{7} \times r^{2} \times 7) \times 10] m^{3} \\ = (220 r^{2}) m^{3} \\ ∴ 220 r^{2} = 440 \\ \Rightarrow r^{2} = 2 \\ \Rightarrow r = \sqrt{2} m \end{aligned}

36. Which one of the following figures will generate a cone when rotated about one of its straight edges ?

A. An equilateral triangle

B. A sector of a circle

C. A segment of a circle

D. A right-angled triangle

Discuss

Solution:

Answer :- A right-angled triangle.

37. If the heights of two cones are in the ratio 7 : 3 and their diameters are in the ratio 6 : 7, what is the ratio of their volumes ?

A. 3 : 7

B. 4 : 7

C. 5 : 7

D. 12 : 7

Discuss

Solution:

Let the heights of two cones be 7x and 3x and their radii be 6y and 7y respectively
Then,
Ratio of volume :

\begin{aligned} = \frac{\frac{1}{3} π \times {(6 y)}^{2} \times 7 x}{\frac{1}{3} π \times {(7 y)}^{2} \times 3 x} \\ = \frac{36 \times 7}{49 \times 3} \\ = \frac{12}{7} O r 12 : 7 \end{aligned}

38. Consider the volumes of the following
1. A parallelepiped of length 5 cm, breadth 3 cm and height 4 cm
2. A cube of each side 4 cm
3. A cylinder of radius 3 cm and length 3 cm
4. A sphere of radius 3 cm
The volumes of these in the decreasing order is :

A. 1, 2, 3, 4

B. 1, 3, 2, 4

C. 4, 2, 3, 1

D. 4, 3, 2, 1

Discuss

Solution:

Volume of parallelepiped :

\begin{aligned} = (5 \times 3 \times 4) c m^{3} \\ = 60 c m^{3} \end{aligned}

Volume of cube :

\begin{aligned} = {(4)}^{3} c m^{3} \\ = 64 c m^{3} \end{aligned}

Volume of cylinder :

\begin{aligned} = (\frac{22}{7} \times 3 \times 3 \times 3) c m^{3} \\ = 84.86 c m^{3} \end{aligned}

Volume of spare :

\begin{aligned} = (\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3) \\ = 113.14 c m^{3} \end{aligned}

So, Option D is correct decreasing order

39. If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to from a single sphere, the diameter of the new sphere will be :

A. 12 cm

B. 24 cm

C. 30 cm

D. 36 cm

Discuss

Solution:

Volume of new sphere :

\begin{aligned} = [\frac{4}{3} π \times {(6)}^{3} + \frac{4}{3} π \times {(8)}^{3} + \frac{4}{3} π \times {(10)}^{3}] c m^{3} \\ = [\frac{4}{3} π {{(6)}^{3} + {(8)}^{3} + {(10)}^{3}}] c m^{3} \\ = (\frac{4}{3} π \times 1728) c m^{3} \\ = [\frac{4}{3} π \times {(12)}^{3}] c m^{3} \end{aligned}

Let the radius of the new sphere be R
Then,

\begin{aligned} \frac{4}{3} π R^{3} = \frac{4}{3} π \times {(12)}^{3} \\ \Rightarrow R = 12 c m \end{aligned}

∴ Diameter :

\begin{aligned} = 2 R \\ = 2 \times 12 \\ = 24 c m \end{aligned}

40. The diameter of a spare is 8 cm. It is melted and drawn into a wire of diameter 3 mm. The length of the wire is :

A. 36.9 m

B. 37.9 m

C. 38.9 m

D. 39.9 m

Discuss

Solution:

Let the length of the wire be h
Then,

\begin{aligned} π \times \frac{3}{20} \times \frac{3}{20} \times h = \frac{4}{3} π \times 4 \times 4 \times 4 \\ \Rightarrow h = (\frac{4 \times 4 \times 4 \times 4 \times 20 \times 20}{3 \times 3 \times 3}) c m \\ \Rightarrow h = (\frac{102400}{27}) c m \\ \Rightarrow h = 3792.5 c m \\ \Rightarrow h = 37.9 m \end{aligned}

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