Math, asked by pratikdhakate, 4 months ago

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​

Answers

Answered by Steph0303
6

Answer:

  • Number of lead shots dropped = 440

Given:

  • Height of the conical vessel = 11 cm
  • Radius of the top of the vessel = 2.5 cm
  • Radius of the lead shots = 0.25 cm

When 'n' number of balls are dropped inside the vessel, 2/5 th of the volume flows outside the vessel.

To Find:

  • n = ?

Steps:

Hence let's first calculate the volume of the conical vessel.

\implies \text{Volume of a cone} = \dfrac{1}{3} \pi r^2h\\\\\\\implies \text{Volume of the vessel} = \dfrac{1}{3} \times 3.14 \times 2.5 \times 2.5 \times 11\\\\\\\implies \text{Volume of the vessel} = 71.95\:\:cm^3

Volume of the sphere is calculated as:

\implies \text{Volume of Sphere} = \dfrac{4}{3} \pi r^3\\\\\\\implies \text{Volume of the lead shots} = \dfrac{4}{3} \times 3.14 \times ( 0.25 )^3\\\\\\\implies \text{Volume of the lead shots} = 0.0654\:\:cm^3

Amount of volume displaced = 2/5th of Vessel Volume.

We know that,

Volume Added = Volume Displaced

Hence writing in terms of equation we get:

→ n × Volume of 1 lead shot = 2/5th Volume of Cone

\implies n \times 0.0654 = \dfrac{2}{5} \times 71.95\\\\\\\implies n \times 0.0654 = 28.78\\\\\\\implies n = \dfrac{28.78}{0.0654}\\\\\\\implies \boxed{ n \approx 440 }

Hence the number of lead shots dropped into the vessel is 440.

Answered by Anonymous
24

{\large{\rm{\underline{Required \; Solution}}}}

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 concepts -

\; \; \; \; \;{\bullet{\leadsto}} Formula to find volume of cone.

\; \; \; \; \;{\bullet{\leadsto}} Formula to find volume of sphere.

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,

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

Henceforth,

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

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

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

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

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

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

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

{\pink{\frak{Henceforth, \: v \: of \: vessel \: is \: 71.95 \: cm^{3}}}}

~ Let us find the volume of lead shots

{\bf{\longmapsto Volume_{(sphere)} \: = \: \dfrac{4}{3} \pi r^{3}}}

Henceforth,

{\bf{\longmapsto Volume_{(lead \: shots)} \: = \: \dfrac{4}{3} \pi r^{3}}}

\: \: \: \:{\bf{\longmapsto Volume_{(sphere)} \: = \: \dfrac{4}{3} \times \dfrac{22}{7} \times \:0.25^{3}}}

\: \: \:{\bf{\longmapsto Volume_{(sphere)} \: = \: \dfrac{4}{3} \times \dfrac{22}{7} \times 0.25 \times 0.25 \times 0.25}}

\: \: \:{\bf{\longmapsto Volume_{(sphere)} \: = \: \dfrac{4}{3} \times \dfrac{22}{7} \times 0.0625 \times 0.25}}

\: \: \:{\bf{\longmapsto Volume_{(sphere)} \: = \: \dfrac{4}{3} \times \dfrac{22}{7} \times 0.015725}}

\: \: \: \: \: \:{\bf{\longmapsto Volume_{(sphere)} \: = \: \dfrac{88}{21} \times 0.015725}}

\:{\bf{\longmapsto Volume_{(sphere)} \: = \: 0.0654 cm^{3}}}

{\pink{\frak{Henceforth, \: v \: of \: sphere \: is \: 0.0654 \: 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}}} × 71.95

⚔ 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

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

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