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Surface Areas and Volumes

Multiple Choice Questions

301. The volumes of two spheres are in the ratio 64 : 27. The ratio of their diameters is equal to:

4 : 3
3 : 4
16 : 9
16 : 9

92 Views

302. The curved surface area of a cone of slant height

straight x over 2

is πrx. Then area of its base is:

2m²
4πx²
πx²
πx²

96 Views

303. If the radius and height of a cone are both increased by 10%, then the volume of the cone is increased by:

136 Views

304. If the area of the base of a right circular cylinder is 38.5 cm² and its height is 6 cm, then its volume (in cm³) is:

231
115.5
462
462

109 Views

305. Area of base of a solid hemisphere is 36π cm². Then its volume is:

108 $straight pi$ cm³
72 $straight pi$ cm³
288 $straight pi$ cm³
288 $straight pi$ cm³

148 Views

306. Height of a cone is 12 cm and its base diameter is 10 cm. Its slant height is:

12.5 cm
313 cm
15 cm
15 cm

107 Views

307. If the sum of all the edges of a cube is 36 cm, then the volume of the cube is

9 cm³
27 cm³
216 cm³
216 cm³

95 Views

Short Answer Type

308. The radius of the base and the height of a right circular cone are 7 cm and 24 cm respectively. Find the volume and total surface area of the cone.

211 Views

309.
If the radius of the base of a right circular cone is halved keeping the height same, what is the ratio of the volume of the reduced cone to that of the original one?

Let the radius of the base and the height of the original cone be r and h respectively.
∴ Volume of the original cone (v₁)

$equals 1 third πr squared straight h space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
For the reduced cone

Radius = $straight r over 2$
Height = h

$therefore$ Volume of the reduced cone (v₂)

$equals 1 third straight pi open parentheses straight r over 2 close parentheses squared straight h equals space 1 fourth open parentheses 1 third πr squared straight h close parentheses equals 1 fourth straight v subscript 1 space space space space space space space space space space space space vertical line space From space left parenthesis 1 right parenthesis therefore space space space straight v subscript 2 over straight v subscript 1 equals 1 fourth equals 1 space colon space 4$

Hence, the ratio of the volume of the reduced cone to that of the original one is 1 : 4.