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Aptitude

71). The ratio of the surface area of two spheres is 4 : 9, then the ratio of their radius is

A). 3 : 2

B). 2 : 3

C). 1 : 1

D). 4 : 3

View Answer

Correct Answer: 2 : 3

Let the surface are of the first sphere = \( \Large 4 \pi r^{2} \) and

The surface area of the second sphere = \( \Large 4 \pi R^{2} \)

Their ratio = \( \Large 4 \pi r^{2} : 4 \pi R^{2}= r^{2}:R^{2}=4:9 \) (given)

Ratio of their radius = 2 : 3

72). If the volume and surface area of a sphere are numerically the same, then its radius is

A). 1 unit

B). 2 units

C). 3 units

D). 4 units

View Answer

Correct Answer: 3 units

Volume of the sphere = \( \Large \frac{4}{3} \pi r^{3}\ cm^{3} \)

Surface area of the sphere \( \Large 4 \pi r^{2}\ cm^{2} \)

Volume of the sphere and the surface area of the sphere are

numerically equal (given)

\( \Large \frac{4}{3} \pi r^{3} = 4 \pi r^{2} \)

r = 3

73). The internal diameter of a hollow sphere f metal is 18 cm. and its thickness is 2 cm. Find the weight of the sphere if a cubic cm. of metal weighs 50 grams.

A). 126.50 kg.

B). 126.13 kg.

C). 136.50 kg.

D). 136.13 kg.

View Answer

Correct Answer: 126.13 kg.

Internal diameter of the sphere 18 cm

radius = 9 cm

Internal volume of the sphere = \( \Large \frac{4}{3} \times \pi \times 9 \times 9 \times 9\ cm^{3} \)

Thickness of the sphere 2 cm

External radius = 9 + 2 = 11 cm

External volume = \( \Large \frac{4}{3} \times \pi \times 11 \times 11 \times 11\ cm^{3} \)

Volume of the metal used = External volume.

= \( \Large \frac{4}{3} \pi \times 11^{3} - \frac{4}{3} \pi \times 9^{3} \)

=\( \Large \frac{4}{3} \times \frac{22}{7} \left(11^{3} - 9^{3}\right) \)

=\( \Large \frac{4}{3} \times \frac{22}{7} \left(1331 - 729\right) \)

=\( \Large \frac{4}{3} \times 22 \times 86 = \frac{7568}{3}\ cm^{3} \)

Weight of the metal =\( \Large 1\ cm^{3}=50 gram \)

Weight of \( \Large \frac{7568}{3}\ cm^{3}=\frac{7568}{3} \times \frac{50}{1000}kg \) = 126.13 kg.

74). The volume of a sphere 1s \( \Large 38808\ m^{3} \) what is its radius.

A). 41 mts.

B). 42 mts.

C). 27 mts.

D). 21 mts.

View Answer

Correct Answer: 21 mts.

Volume of the sphere = \( \Large \frac{4}{3} \pi r^{3} = 38808\ m^{3} \) (given)

\( \Large \frac{4}{3} \pi r^{3}=38808 \)

\( \Large \pi r^{3}=38808 \times \frac{3}{4} \)

\( \Large r^{3} = 38808 \times \frac{3}{4} \times \frac{7}{22} = 441 \times 21 \)

\( \Large r = \left(441 \times 21\right)^{\frac{1}{3}} \)

= \( \Large \left(21^{3}\right)^{\frac{1}{3}} \) = 21 mts.

75). A cylindrical boiler has a hemispherical top of radius 3.5 mts. If the height of the cylindrical portion is 4 mts., then what is the volume of the boiler.

A). \( \Large 243.83\ m^{3} \)

B). \( \Large 263.83\ m^{3} \)

C). \( \Large 283.83\ m^{3} \)

D). \( \Large 163.83\ m^{3} \)

View Answer

Correct Answer: \( \Large 243.83\ m^{3} \)

Radius of the hemisphere = 3.5 mts.

Volume of the hemisphere = \( \Large \frac{2}{3} \times \pi r^{3} \)

=\( \Large \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \)

=\( \Large \frac{11 \times 49}{6} = 89.83\ m^{3} \)

Radius of the cylinder = 3.5 mts.

Height of the cylinder = 4 mts.

Volume of the cylinder = \( \Large \pi r^{2}h \)

=\( \Large \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4 \)

=\( \Large 77 \times 2 = 154\ m^{3} \)

Volume = Volume of the hemisphere + Volume of the cylinder.

= 89.43 + 154 = 243.83 \( \Large m^{3} \)

76). A metallic sphere with radius 3 cm. and a metallic one with the same radius and height 15 cms. are melted and formed a cylinder with the same radius. Find the height of the metallic cylinder.

A). 9 cm.

B). 19 cm.

C). 29 cm.

D). 7 cm.

View Answer

Correct Answer: 9 cm.

Volume of the metallic sphere = \( \Large \frac{4}{3} \times \pi \times 3 \times 3 \times 3 = 36 \pi \ cm^{3} \)

Volume of the metallic cone = \( \Large \frac{1}{3} \times \pi \times r^{2}h \)

=\( \Large \frac{1}{3} \times \pi \times 3 \times 3 \times 15 = 45 \pi \ cm^{3} \)

Volume of the metallic cylinder = Volume of sphere + Volume of cone

=\( \Large 36 \pi +45 \pi = 81 \pi \ cm^{3} \)

Radius of the metallic cylinder = 3 cm (given)

=\( \Large \pi \times 3^{2} \times h = 81 \pi \)

\( \Large h = \frac{81}{9} \) 9 cm

Height of the metallic cylinder = 9 cm.

77). 12 metallic spheres each has a radius of 7 cms. are melted and formed a metallic cylinder rod with radius 3.5 cm. What is the height of the cylindrical rod?

A). 448 cm.

B). 600 cm.

C). 56 cm.

D). 112 cm.

View Answer

Correct Answer: 448 cm.

Volume of a metallic sphere = \( \Large \frac{4}{3} \times \pi \times 7 \times 7 \times 7\ cm^{3} \)

Volume of 12 metallic sphere = \( \Large 12 \times \frac{4}{3} \times \pi \times 7 \times 7 \times 7 \)

= \( \Large 16 \times \pi \times 7 \times 7 \times 7\ cm^{3} \)

Volume of metallic cylinder = \( \Large \pi \times \frac{7}{2} \times \frac{7}{2} \times h\ cm^{3} \)

Volume of metallic cylinder rod = Volume of 12 metallic sphere (given)

\( \Large \pi \times \frac{7}{2} \times \frac{7}{2} \times h=16 \times \pi \times 7 \times 7 \times 7 \)

\( \Large h = \frac{16 \times \pi \times 7 \times 7 \times 7}{ \pi \times \frac{7}{2} \times \frac{7}{2}} \)

=\( \Large \frac{16 \times 7 \times 7 \times 7 \times 2 \times 2}{7 \times 7} \)

=\( \Large 112 \times 4 \) = 448 cm.

78). The diameter of a circular cone is 14 metres and its height is 12 metres. Find its volume.

A). \( \Large 564\ m^{3} \)

B). \( \Large 500\ m^{3} \)

C). \( \Large 616\ m^{3} \)

D). \( \Large 418\ m^{3} \)

View Answer

Correct Answer: \( \Large 616\ m^{3} \)

Diameter of the base of the cone = 14 mts.

radius = 7 mts.

height = 12 mts.

Volume of the cone = \( \Large \frac{1}{3} \pi r^{2}h \)

=\( \Large \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \)

= \( \Large 22 \times 28 = 616\ cm^{3} \)

79). If the length of diagonal of cube is \( \Large 8\sqrt{3} \) then its volume is.

A). \( \Large 512\ cm^{3} \)

B). \( \Large 384\ cm^{3} \)

C). \( \Large 64\ cm^{3} \)

D). \( \Large 192\ cm^{3} \)

View Answer

80). For a sphere of radius 10 cm., the numerical value of surface area is what percent of the numerical value of its volume?

A). 24%

B). 26.5%

C). 30%

D). 45%

View Answer

Correct Answer: 30%

Radius of the sphere = 10 cm.

Surface area of the sphere = \( \Large 4 \pi r^{2} \)

Volume of the sphere = \( \Large 4 \pi r^{3} \)

Percentage of the surface area of the sphere to its volume.

= \( \Large \frac{4 \pi r^{2}}{\frac{4}{3} \pi r^{3}} \times 100 = \frac{100 \times 3}{10} \) = 30%

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