Question

A cylindrical container of radius 6 cm and height 15 cm is filled with chocolate. The chocolate is to be filled into cones of height 12 cm and radius 3 cm having a hemispherical shape on the top. Find the number of cones that can be filled with the chocolate.

• A. 20
• B. 15
• C. 25
• D. 10

Answer 1

The correct option is D

10

Volume of the cylinder
= $\pi r^{2}h=\frac{22}{7}\times (6{)}^2\times 15=1695.6c{m}^3$

Volume of the cone= $\frac{1}{3}\pi r^{\prime 2}h=\frac{1}{3}\pi \times (3{)}^2\times 12=113.04c{m}^3$

Volume of a hemisphere = $\frac{2}{3}\pi r^{\prime \prime 3}=\frac{2}{3}\pi (3{)}^3=56.52c{m}^3$

Volume of new shape = volume of cone + volume of hemisphere = 169.56 $c{m}^3$

Let the number of cones that can be filled be $x$.
$x$ = $\frac{\text{volume of the cylinder}}{\text{Total volume of the solid(cone+ hemisphere)}}$

= $\frac{1695.6}{169.56}$

= 10

Therefore, 10 cones can be filled.