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Surface Areas and Volumes

Short Answer Type

191. Find the capacity in litres of a conical vessel with
(ii) height 12 cm, slant height 13 cm.

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192. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use $straight pi$ = 3.14)

Let the radius of the base of the cone be r cm.
h = 15 cm Volume = 1570 cm³ $rightwards double arrow space space space space space space 1 third πr squared straight h space equals space 1570 rightwards double arrow space space space space space space 1 third cross times 3.14 cross times straight r squared cross times 15 equals 1570 rightwards double arrow space space space space space space space straight r squared equals fraction numerator 1570 cross times 3 over denominator 3.14 cross times 15 end fraction rightwards double arrow space space space space space space space straight r squared equals 100 rightwards double arrow space space space space space space space straight r equals square root of 100 rightwards double arrow space space space space space space space straight r equals 10 space cm$

Hence, the radius of the base of the cone is 10 cm.

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193. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

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194. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

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Long Answer Type

195.

The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find

(i) height of the cone,
(ii) slant height of the cone,
(iii) curved surface area of the cone.

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Short Answer Type

196. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

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197. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Question 7 and 8.

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198. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

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