Question

(a) A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top which is open is 5 cm. It is filled with water up to the brim. When spherical lead shots, each of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

(b) A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones. [6 MARKS]

Answer 1

Each Part: 3 Marks

(a) Given: Radius of the cone r = 5 cm and height of the cone, h = 8 cm.
$\therefore$ Volume of water in cone = Volume of cone
$=\frac{1}{3}\pi r^{2}h=(\frac{1}{3}\times \frac{22}{7}\times 5^2\times 8)c{m}^3$
$=\frac{4400}{21}c{m}^3$
$\therefore$ Volume of water flows $=\frac{1}{4}\times \frac{4400}{21}c{m}^3$
$=\frac{1100}{21}c{m}^3$
Now, volume of one spherical lead shot
$=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (0.5{)}^{3}c{m}^3$
$=(\frac{4}{3}\times \frac{22}{7}\times 0.125)c{m}^3=\frac{11}{21}c{m}^3$
$\therefore$ Number of lead shots
$=\frac{Volumeofwaterflows}{Volumeofonesphericalleadshot}$
$=\frac{\frac{1100}{21}}{\frac{11}{21}}=100$


(b) Volume of cone $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \pi \times 4\times 3=4\pi$
Volume of sphere $=\frac{4}{3}\pi r^3=\frac{4}{3}\times \pi \times 216=288\pi$
No. of cones $=\frac{Volumeofsphere}{Volumeofcone}=\frac{288\pi }{4\pi }=72$