Question

A vessel is in the form of an inverted cone. It's height is 11cm and the radius of its top which is open is 2.5cm. It is filled with water upto the rim. When lead shots each of which is a sphere of radius 0.25cm are dropped into the vessel. 2/5 of the water flows out. Find the number of lead shots dropped into the vessel​

Answer 1

Given that -

$\; \; \; \; \;{\bullet{\leadsto}}$ A vessel is in the form of an inverted cone.

$\; \; \; \; \;{\bullet{\leadsto}}$ Vessel (cone) height = 11 cm

$\; \; \; \; \;{\bullet{\leadsto}}$The radius of it's (vessel) (cone) top which is open = 2.5 cm.

$\; \; \; \; \;{\bullet{\leadsto}}$ Lead shots each of which is a sphere of radius 0.25cm are dropped into the vessel.

(Radius of lead shots = 0.25 cm)

$\; \; \; \; \;{\bullet{\leadsto}}$ 2/5 of the water flows out.

To find -

$\; \; \; \; \;{\bullet{\leadsto}}$ The number of lead shots dropped into the vessel.

Solution -

$\; \; \; \; \;{\bullet{\leadsto}}$ 440 are the number of lead shots dropped into the vessel.

Using formula -

${\bullet} \: \: \:{\boxed{\boxed{\sf{\longrightarrow Volume_{(cone)} \: = \: \dfrac{1}{3} \pi r^{2} h}}}}$

${\bullet} \: \: \:{\boxed{\boxed{\sf{\longrightarrow Volume_{(sphere)} \: = \dfrac{4}{3} \pi r^{3}}}}}$

Where,

● Value of π is ${\sf{\dfrac{22}{7}}}$ or 3.14

● π pronounced as pi

● r denotes radius

● h denotes height

Full solution -

~ Let us find the volume of vessel firstly,

${\sf{\longmapsto Volume_{(cone)} \: = \: \dfrac{1}{3} \pi r^{2} h}}$

Henceforth,

${\sf{\longmapsto Volume_{(vessel)} \: = \: \dfrac{1}{3} \pi r^{2} h}}$

$\: \: \: \:{\sf{\longmapsto Volume_{(vessel)} \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \times 2.5^{2} \times 11}}$

$\: \: \:{\sf{\longmapsto Volume_{(vessel)} \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \times 2.5 \times 2.5 \times 11}}$

$\:{\sf{\longmapsto Volume_{(vessel)} \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \times 6.25 \times 11}}$

$\: \: \: \: \:{\sf{\longmapsto Volume_{(vessel)} \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \times 68.75}}$

${\sf{\longmapsto Volume_{(vessel)} \: = \: \dfrac{22}{21} \times 68.75}}$

$\; \; \; \; \; \; \; \; \;{\bf{\longmapsto Volume_{(vessel)} \: = 71.95 \: cm^{3}}}$

Now,

As the question says that 2/5 of the water flows out or this is the displaced volume.

So,

Now as we already know that volume added = volume displaced

Henceforth, the formed equation is

${\longmapsto} x × 0.0654 = {\sf{\dfrac{2}{5}}}$

Here,

x is assumption at the place of the number of lead shots dropped into the vessel

0.0654 is volume of lead shots

${\sf{\dfrac{2}{5}}}$ is flow out volume

71.95 is volume of vessel

${\longmapsto}$ x × 0.0654 = 28.78

${\longmapsto} x = {\sf{\dfrac{28.78}{0.0654}}}$

${\longmapsto}$ x = 440

${\longmapsto}$ The lead shots dropped into the vessel = 440

${\red{\frak{Henceforth, \: 440 \: is \: no. \: of \: lead \: shot \: drop \: in \: vessel}}}$