A metallic right circular cone 20 cm high and whose vertical ang

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 Multiple Choice QuestionsShort Answer Type

431.

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

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432. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
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433. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
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434.

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

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435.

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

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436.

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. 

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437.

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it. 

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438.

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ` 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ` 8 per 100 cm2 . (Take π = 3.14)

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 Multiple Choice QuestionsLong Answer Type

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439.

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1 over 16 cm, find the length of the wire.


In fig. cone ABC is cut out by a plane parallel to the base FG. DEFG is the frustum so obtained. Let O be the centre of the base of the cone and O’ the centre of the base of the frustum.


In fig. cone ABC is cut out by a plane parallel to the base FG. DEFG

It is given that ∠BAC = 60° ∠OAC = 30°

In right triangle AOC, tan 30 degree space equals space OC over OA
rightwards double arrow space space Oc space space equals space OA space straight x space tan space 30 degree
space space space space space space space space space space space space equals space 10 space straight x space fraction numerator 1 over denominator square root of 3 end fraction equals fraction numerator 10 over denominator square root of 3 end fraction space cm

And, C =  equals fraction numerator 10 over denominator square root of 3 end fraction space cm. space space space space... left parenthesis straight i right parenthesis
Since space space space increment AOC space minus space increment AO apostrophe straight F
space space space space space space space space space space space space space space left parenthesis Using space AA space similar space condition right parenthesis
therefore space space space fraction numerator AO over denominator AO apostrophe end fraction space equals space fraction numerator OC over denominator straight O apostrophe straight F end fraction
rightwards double arrow space space space 10 over 20 equals fraction numerator begin display style fraction numerator 10 over denominator square root of 3 end fraction end style over denominator straight O apostrophe straight F end fraction
rightwards double arrow space space space space straight O apostrophe straight F space equals space fraction numerator 20 over denominator square root of 3 end fraction space space space space space space space space space... left parenthesis ii right parenthesis
Height of the frustum = P'O = 1 half space AO apostrophe space equals space 10 space cm

therefore volume of the frustum = πh over 3 left square bracket straight R squared plus Rr plus straight r squared right square bracket

                                         equals fraction numerator straight pi space straight x space 10 over denominator 3 end fraction
left square bracket OF squared plus straight O apostrophe straight F space straight x space PE space plus space PE squared right square bracket
equals space fraction numerator straight pi space straight x space 10 over denominator 3 end fraction
open square brackets open parentheses fraction numerator 20 over denominator square root of 3 end fraction close parentheses plus fraction numerator 20 over denominator square root of 3 end fraction straight x fraction numerator 10 over denominator square root of 3 end fraction plus open parentheses fraction numerator 10 over denominator square root of 3 end fraction close parentheses squared close square brackets
space space space space space space space space space space space space space space space space space space left square bracket space Using space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis right square bracket
equals space fraction numerator πx 10 over denominator 3 end fraction
open square brackets fraction numerator 400 plus 200 plus 100 over denominator 3 end fraction close square brackets
equals space fraction numerator straight pi space straight x space 10 space straight x space 700 over denominator 9 end fraction space cm cubed
equals space fraction numerator straight pi space 7000 over denominator 9 end fraction space cm cubed space space space space space space space space space space space space space space space space space space space space space... left parenthesis iii right parenthesis

space space space space space space space space space space space space space space space space space space space space
Radius of the wire = 1 over 32 space cm
Let h be length of the wire
therefore         Volume of wire = straight p space straight x space open parentheses 1 over 32 close parentheses squared space straight x space straight h space space space space space space space space space space space space space space space space.... left parenthesis iv right parenthesis
From (iii) and (iv), we get

straight pi space straight x space open parentheses 1 over 32 close parentheses squared space straight x space straight h space equals space fraction numerator straight pi space straight x space 7000 over denominator 9 end fraction
therefore space space space space space space space straight h space equals space fraction numerator straight pi space straight x space 7000 over denominator 9 end fraction space straight x space fraction numerator 32 space straight x space 32 space over denominator straight pi end fraction
cm
space space space space space space space space space space space space space space space space space space space space equals space 7168000 over 9 equals 796444.4
cm space left parenthesis approx right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space equals space 7964.44 space straight m
space space space space space space space space space space space space space space space space space space space space space space space space equals space 7964 space straight m. space left parenthesis approx right parenthesis space



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 Multiple Choice QuestionsShort Answer Type

440.

A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

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