A tent is in the form of a cylinder surmounted by a conical top.

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 Multiple Choice QuestionsShort Answer Type

541. An iron pole consisting of a cylindrical portion 110 cm high and base diameter 12cm is surmounted by a cone 9cm high. Find the mass of the pole it is given that 1cm3of iron has 8 gm mass.

open parentheses bold Use bold space bold pi bold space bold equals bold space bold 355 over bold 113 close parentheses
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 Multiple Choice QuestionsLong Answer Type

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542. A tent is in the form of a cylinder surmounted by a conical top. If the height and diameter of the cylinderical part are 2.1 m and 4m respectively, and the height of the top is 2.8m, find the area of the canvas used for making the tent. Find the cost of the canvas used for making the tent. Find the cost of the canvas of the tent at the rate of Rs. 500 per m2. Also, find the volume of air enclosed in the tent.


Let ‘r’ met. be the radius, l be the slant height and ‘h’ m be the height of the conical part. Then, r 2m, l = 2.8 m.

h equals space square root of l to the power of italic 2 minus straight r squared end root
equals square root of left parenthesis 2.8 right parenthesis squared minus left parenthesis 2 right parenthesis squared end root equals square root of left parenthesis 2.8 plus 2 right parenthesis left parenthesis 2.8 minus 2 right parenthesis end root
equals space square root of 4.8 space straight x space 0.8 end root equals square root of 48 over 10 straight x 8 over 10 end root equals square root of 384 over 100 end root
equals space square root of 3.84 space end root equals space 1.96 space straight m

Let ‘R’ met. be the radius and ‘H’ met. be the height of the cylindrical part.
Thus,
R = 2m and H = 2.1m

Let ‘r’ met. be the radius, l be the slant height and ‘h’ m b

Now,
Area of the canvas used
= curved surface area of conical part + curved surface area of cylindrical part = straight pirl + 2straight piRH = straight pirl + 2straight pirH
[∵ r = R] = straight pir (l + 2H)
equals 22 over 7 straight x space 2 space left parenthesis 2.8 space plus space 1 space straight x space 2.1 right parenthesis space equals space 22 over 7 straight x space left parenthesis 2.8 space plus 4.2 right parenthesis
equals space 22 over 7 straight x space 2 space straight x space 7 space equals space 44 straight m squared
Cost of the canvas of the tent = Rs. 500 x 44 = Rs. 22,000
And.
Volume of air enclosed in the tent = Volume of the conical part + Volume of cylindrical part

equals 1 third πr squared straight h space plus space πr squared straight h space equals space 1 third πr squared straight h space plus space space space space space space left square bracket because space straight r space equals space straight R right square bracket
equals space πr squared space open square brackets 1 third straight h space plus space straight H close square brackets space equals 22 over 7 space straight x space 2 space straight x space 2 space open square brackets 1 third straight x space 1.96 space plus space 2.1 close square brackets
equals space 22 over 7 space straight x space 4 space straight x space open square brackets fraction numerator 1.96 plus 6.3 over denominator 3 end fraction close square brackets equals 22 over 7 straight x space 14 space straight x space fraction numerator 8.26 over denominator 3 end fraction
equals space 34.61 space straight m cubed space




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543. A toy is in the form of a cone mounted on a hemisphere with same radius. The diameter of the base of the conical portion is 7.0 cm and the total height of the toy is 14.5 cm. Find

the volume of the toy open square brackets Use space straight pi space equals space 22 over 7 close square brackets.
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544. An iron pillar has lower part in the form of a right circular cylinder and the upper part in the form of a right circular cone. The radius of the base of each of the cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if 1 cm3 of iron weighs 7.5 grams.

open parentheses Take space straight pi space equals space 22 over 7 close parentheses
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 Multiple Choice QuestionsShort Answer Type

545. A bucket is made up of a metal sheet is in the form of a frustum of a cone of height 16cm with radii of its lower and upper ends at 8cm and 20cm respectively. Find the cost of the bucket, if the cost of metal sheet used is Rs. 15 per 100 cm2.


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 Multiple Choice QuestionsLong Answer Type

546. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 1 over 27 of the volume of the given cone, at which height above the base is the section made?
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547. A hollow cone is cut by a plane parallel to the base and the upper portul is removed. If the curved surface of the remainder is  8 over 9 of the curved surface of the whole cone, find the ratio of the line-segment into which the cones altitude is divided by the plane.




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548. If the radii of the circular ends of a conical bucket, which is 16 cm high, are 20 cm and 8 cm, find the capacity and total surface area of the bucket.


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549. A bucket is in the form of a frustum of a cone and holds 28.49 litres of milk. The radii of the top and bottom are 28 cm and 21 cm. Find the height of the bucket.
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 Multiple Choice QuestionsShort Answer Type

550.

A bucket of height 8 cm made up of copper sheets is in the form of frustum of a right circular cone with radii of its lower ends as 3 cm and 9 cm respectively. Calculate
(i) the height of the cone of which the bucket is a part.
(ii) the volume of water which can be filled in the bucket.
(iii) the area of copper sheet required to make the bucket.

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