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, $196\pi$ cm${}^3$, $5$ cm, $116\pi$ cm${}^2$
• B. $5$ cm, $192\pi$ cm${}^3$, $7$ cm, $118\pi$ cm${}^2$
• C. $5.2$ cm, $192\pi$ cm${}^3$, $5.6$ cm, $108\pi$ cm${}^2$
• D. $6.3$ cm, $156\pi$ cm${}^3$, $25$ cm, $16\pi$ cm${}^2$

Answer 1

The correct option is A $7.5$ cm, $196\pi$ cm${}^3$, $5$ cm, $116\pi$ cm${}^2$

Let $ABCD$ be the bucket which is frustum of a cone with vertex 0 .
Let $ON=xcm$
$△OAB=△OMC$
$\frac{x}{3+x}=\frac{6}{10}$ [since $\frac{ON}{OM}=\frac{NB}{MC}$]
$=>\frac{x}{3+x}=\frac{3}{5}$
$=>5x=3(3+x)$
$=>5x=9+3x$
$=>x=\frac{9}{2}$
Therefore,
$ON=\frac{9}{2}cm$ and $OM=\frac{9}{2}+3=\frac{15}{2}cm$
$=7.5$
Therefore
The height of the cone = $7.5cm$
Volume of the bucket=$\frac{1}{3}\pi \times {10}^2\times 7.5-\frac{1}{3}\pi \times 6^2\times \frac{9}{2}$[Volume of the large cone Volume of the small cone]
$=\frac{1}{3}\pi (750-162)c{m}^3$
$=196\pi c{m}^3$
Slant height of cone of radius $10cm$
$\sqrt{{10}^2+{7.5}^2}cm$
$=\sqrt{156.25}cm$
$=12.5cm$
Slant height of cone of radius $6cm$
$\sqrt{{\frac{9}{2}}^2+(6{)}^2}cm$
$=\sqrt{\frac{81}{4}+36}cm$
$=\frac{15}{2}cm$
Therefore,
Slant height of bucket=$(12.5-\frac{15}{2})cm$
=$\frac{10}{2}cm$
i.e$l=5cm$
The area of the metal sheet=$\pi \times l(R+r)+\pi r^2$
=$\pi \times 5(10+6)+\pi 6^{2}c{m}^2$
=$80\pi +36\pi c{m}^2$
=$116\pi c{m}^2$