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51. What is the weight of water contained in a conical vessel 21 cm deep and 16 cm in diameter ?

A. 1.256 kg

B. 1.408 kg

C. 2.480 kg

D. 3.875 kg

Discuss

Solution:

Volume of water :

\begin{aligned} = (\frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 21) c m^{3} \\ = 1408 c m^{3} \\ = (\frac{1408}{1000}) k g \\ = 1.408 k g \end{aligned}

52. If the ratio of volumes of two cones is 2 : 3 and the ratio of the radii of their bases is 1 : 2, then the ratio of their height will be :

A. 3 : 4

B. 4 : 3

C. 3 : 8

D. 8 : 3

Discuss

Solution:

let their radii be x and 2x, and their heights be h and H respectively
Then,

\begin{aligned} \frac{\frac{1}{3} \times π \times x^{2} \times h}{\frac{1}{3} \times π \times {(2 x)}^{2} \times H} = \frac{2}{3} \\ \Leftrightarrow \frac{h}{H} = \frac{2}{3} \times 4 \\ \Leftrightarrow \frac{h}{H} = \frac{8}{3} O r 8 : 3 \end{aligned}

53. The volume of a sphere is $2145 \frac{11}{21} c m^{3} .$ Its radius is equal to :

A. 7 cm

B. 8 cm

C. 9 cm

D. None of these

Discuss

Solution:

\begin{aligned} \frac{4}{3} \times \frac{22}{7} \times r^{3} = \frac{45056}{21} \\ \Rightarrow r^{3} = (\frac{45056}{21} \times \frac{3}{4} \times \frac{7}{22}) \\ \Rightarrow r^{3} = 512 \\ \Rightarrow r = \sqrt[3]{512} \\ \Rightarrow r = 8 c m \end{aligned}

54. A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical ball. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is :

A. 2.5 cm

B. 2.66 cm

C. 3 cm

D. 3.5 cm

Discuss

Solution:

Let the radius of the third ball be R cm
Then,

\frac{4}{3} π \times {(\frac{3}{4})}^{3} + \frac{4}{3} π \times {(1)}^{3}

+ \frac{4}{3} π \times R^{3}

= \frac{4}{3} π \times {(\frac{3}{2})}^{3}

\begin{aligned} \Rightarrow \frac{27}{64} + 1 + R^{3} = \frac{27}{8} \\ \Rightarrow R^{3} = \frac{125}{64} = \frac{{(5)}^{3}}{{(4)}^{3}} \\ \Rightarrow R = \frac{5}{4} \end{aligned}

∴ Diameter of the third ball :

= 2 R = \frac{5}{2} c m = 2.5 c m

55. The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is :

A. 1 : 1

B. 1 : 2

C. 2 : 1

D. 2 : 3

Discuss

Solution:

Let the radius of the sphere be r
Then, radius of the cylinder = r
Height of the cylinder = 2r
Surface area of sphere =

4 π r^{2}

Surface area of the cylinder =

2 π r (2 r) = 4 π r^{2}

∴ Required ratio :
=

4 π r^{2}

4 π r^{2}

= 1 : 1

56. A hemispherical bowl of internal radius 12 cm contains liquid. This liquid is to be filled into cylindrical container of diameter 4 cm and height 3 cm. The number of containers that is necessary to empty the bowl is :

A. 80

B. 96

C. 100

D. 112

Discuss

Solution:

Volume of hemispherical bowl :

= (\frac{2}{3} \times π \times 12 \times 12 \times 12) c m^{3}

Volume of 1 cylindrical container :

= (π \times 2 \times 2 \times 3) c m^{3}

∴ Number of containers required :

\begin{aligned} = \frac{2}{3} \times \frac{12 \times 12 \times 12}{2 \times 2 \times 3} \\ = 96 \end{aligned}

57. Length of each edge of a regular tetrahedron is 1 cm. It volume is :

A. $\frac{\sqrt{3}}{12} c m^{3}$

B. $\frac{1}{4} \sqrt{3} c m^{3}$

C. $\frac{\sqrt{2}}{6} c m^{3}$

D. $\frac{1}{12} \sqrt{2} c m^{3}$

Discuss

Solution:

Length of each edge of a regular tetrahedron = 1 cm
Volume of regular tetrahedron :

\begin{aligned} = \frac{a^{3}}{6 \sqrt{2}} c m^{3} \\ = \frac{1}{6 \sqrt{2}} \\ = \frac{\sqrt{2}}{6 \sqrt{2} \times \sqrt{2}} c m^{3} \\ = \frac{\sqrt{2}}{12} Or \frac{1}{12} \sqrt{2} c m^{3} \end{aligned}

58. The base of a right prism is a trapezium whose lengths of two parallels sides are 10 cm and 6 cm and distance between them is 5 cm. If the heights of the prism is 8 cm, its volume is :

A. 320 cm³

B. 300 cm³

C. 310 cm³

D. 300.5 cm³

Discuss

Solution:

Length of parallel sides of prism = 10 cm and 6 cm
Height of prism = 8 cm
∴ Volume of prism :

\begin{aligned} = \frac{1}{2} (10 + 6) \times 5 \times 8 \\ = \frac{1}{2} \times 16 \times 5 \times 8 \\ = 320 c m^{3} \end{aligned}

59. A rectangular water reservoir contains 42000 litres of water. If the length of reservoir is 6 m and breadth of the reservoir is 3.5 m, then the depth of the reservoir will be :

A. 2 m

B. 5 m

C. 6 m

D. 8 m

Discuss

Solution:

Volume of the reservoir = 42000 litres = 42 m³
Let the depth of the reservoir be h metres
then,

\begin{aligned} 6 \times 3.5 \times h = 42 \\ O r, h = \frac{42}{6 \times 3.5} = 2 m \end{aligned}

60. A cistern, open at the top, is to be lined with sheet of lead which weights 27 kg/m². The cistern is 4.5 m long and 3 m wide and holds 50 m³. The weight of lead required is :

A. 1660.5 kg

B. 1764.5 kg

C. 1860.5 kg

D. 1864.5 kg

Discuss

Solution:

Let the depth of the cistern be h metresThen,

\begin{aligned} 4.5 \times 3 \times h = 50 \\ \Rightarrow h = \frac{50}{13.5} \\ \Rightarrow h = \frac{100}{27} \end{aligned}

Area of sheet required :

\begin{aligned} = l b + 2 (b h + l h) \\ = l b + 2 h (l + b) \\ = [4.5 \times 3 + 2 \times \frac{100}{27} \times (4.5 + 3)] m^{2} \\ = (13.5 + \frac{200}{27} \times 7.5) m^{2} \\ = (\frac{27}{2} + \frac{500}{9}) m^{2} \\ = \frac{1243}{18} m^{2} \end{aligned}

∴ Weight of lead :

\begin{aligned} = (27 \times \frac{1243}{18}) k g \\ = (\frac{3729}{2}) k g \\ = 1864.5 k g \end{aligned}

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