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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)

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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.

For heap of wheat
Diameter = 10.5 m

$therefore$ Radius (r) = $fraction numerator 10.5 over denominator 2 end fraction space cm space equals space 5.25 space straight m$

Height (h) = 3 m

$therefore$ Volume = $1 third πr squared straight h$

$space space space space space space space space space space space space space space space space space space space equals 1 third cross times 22 over 7 cross times left parenthesis 5.25 right parenthesis squared cross times 3 space space space space space space space space space space space space space space space space space space space space equals space 86.625 space straight m cubed Slant space height space left parenthesis straight l right parenthesis space equals space square root of straight r squared plus straight h squared end root space space space space space space space space space space space space space space space space equals square root of left parenthesis 5.25 right parenthesis squared plus left parenthesis 3 right parenthesis squared end root space space space space space space space space space space space space space space space space equals space square root of 27.5625 plus 9 end root space space space space space space space space space space space space space space space space space equals square root of 36.5625 end root equals 6.05 space straight m left parenthesis approx. right parenthesis$

$therefore$ Curved surface area = $πrl$

$equals 22 over 7 cross times 5.25 cross times 6.05 equals 99.825 space straight m squared$
Hence, the area of the canvas required is 99.825 m². .