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Question

A bucket of height 3 cm and made up of metal sheet is in the form of frustum of a right circular cone with radii of its lower and upper ends as 6 cm and 10 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 slant height of the bucket.
(iv) The area of the metal sheet required to make the bucket.

A
7.5 cm, 196π cm3, 5 cm, 116π cm2
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B
5 cm, 192π cm3, 7 cm, 118π cm2
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C
5.2 cm, 192π cm3, 5.6 cm, 108π cm2
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D
6.3 cm, 156π cm3, 25 cm, 16π cm2
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Solution

The correct option is A 7.5 cm, 196π cm3, 5 cm, 116π cm2
Let ABCD be the bucket which is frustum of a cone with vertex 0 .
Let ON=xcm
OAB=OMC
x3+x=610 [since ONOM=NBMC]
=>x3+x=35
=>5x=3(3+x)
=>5x=9+3x
=>x=92
Therefore,
ON=92cm and OM=92+3=152cm
=7.5
Therefore
The height of the cone = 7.5cm
Volume of the bucket=13π×102×7.513π×62×92[Volume of the large cone Volume of the small cone]
=13π(750162)cm3
=196πcm3
Slant height of cone of radius 10cm
102+7.52cm
=156.25cm
=12.5cm
Slant height of cone of radius 6cm
922+(6)2cm
=814+36cm
=152cm
Therefore,
Slant height of bucket=(12.5152)cm
=102cm
i.el=5cm
The area of the metal sheet=π×l(R+r)+πr2
=π×5(10+6)+π62cm2
=80π+36πcm2
=116πcm2
299946_242400_ans.png

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