Question

$\huge\color{cyan}\boxed{\colorbox{black}{Question :-}}$



ᴀ ɢʀᴀɪɴ ꜱɪʟᴏ ɪꜱ ʙᴜɪʟᴛ ꜰʀᴏᴍ ᴛᴡᴏ ʀɪɢʜᴛ ᴄɪʀᴄᴜʟᴀʀ ᴄᴏɴᴇꜱ ᴀɴᴅ ᴀ ʀɪɢʜᴛ ᴄɪʀᴄᴜʟᴀʀ ᴄʏʟɪɴᴅᴇʀ ᴡɪᴛʜ ɪɴᴛᴇʀɴᴀʟ ᴍᴇᴀꜱᴜʀᴇᴍᴇɴᴛꜱ ʀᴇᴘʀᴇꜱᴇɴᴛᴇᴅ ʙʏ ᴛʜᴇ ꜰɪɢᴜʀᴇ ᴀʙᴏᴠᴇ. ᴏꜰ ᴛʜᴇ ꜰᴏʟʟᴏᴡɪɴɢ, ᴡʜɪᴄʜ ɪꜱ ᴄʟᴏꜱᴇꜱᴛ ᴛᴏ ᴛʜᴇ ᴠᴏʟᴜᴍᴇ ᴏꜰ ᴛʜᴇ ɢʀᴀɪɴ ꜱɪʟᴏ, ɪɴ ᴄᴜʙɪᴄ ꜰᴇᴇᴛ?
(ʀᴇꜰᴇʀ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ ꜰᴏʀ ᴅɪᴀɢʀᴀᴍ)

ᴏᴘᴛɪᴏɴꜱ :

ᴀ) 261.8
ʙ) 785.4
ᴄ) 916.3
ᴅ) 1047.2




➳ ɴᴏ ꜱᴘᴀᴍꜱ !
➳ ɴᴏ ᴄᴏᴘɪᴇᴅ ᴀɴꜱᴡᴇʀꜱ !
➳ ᴄʟᴇᴀʀ ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ ɴᴇᴇᴅᴇᴅ !
➳ ᴛʜᴀɴᴋꜱ ɪɴ ᴀᴅᴠᴀɴᴄᴇ ꜰᴏʀ ᴀʟʟ ʙʀᴀɪɴʟʏ ᴀɴꜱᴡᴇʀꜱ !




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Answer 1

Given : Radius of the grain silo is 5 ft. , Height of the two cones of the grain silo is 5 ft. & Height of the Cylinder of the grain silo is 10 ft. . [ From the given attachment ( image ) ]

Exigency To Find : Volume of Grain Silo in cubic feet .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀As , We know that Volume of Grain Silo will be :

$\qquad \dag\:\:\bigg\lgroup \sf{ Volume _{(Grain\:Silo)} \:: Volume \:_{(Cylinder)} + 2 \times ( Volume_{(Cone)})}\bigg\rgroup$

⠀⠀⠀⠀⠀Here ,

$\qquad \leadsto \sf Volume_{(Cone )} \: =\:\bf \dfrac{1}{3} \: \pi \:r^2 h \:$

$\qquad \leadsto \sf Volume_{(Cylinder )} \: =\:\bf \: \pi \:r^2 h \:$

⠀⠀⠀⠀⠀Here , r is the Radius , h is the Height & $\sf \pi \:= \:\dfrac{22}{7} \:$ or 3.14 .

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= Volume \:_{(Cylinder)} + 2 \times ( Volume_{(Cone)} ) \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= Volume \:_{(Cylinder)} + 2 \times ( Volume_{(Cone)} ) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( \pi r^2 h \bigg) + \bigg( 2 \times ( \dfrac{1}{3} \pi \:r^2 \:h ) \bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( \pi \times (5)^2 \times 10 \bigg) + \bigg( 2 \times ( \dfrac{1}{3}\times \pi\times \:(5)^2 \:\times 5 ) \bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( \pi \times 25 \times 10 \bigg) + \bigg( 2 \times ( \dfrac{1}{3}\times \pi\times \:25 \:\times 5 )\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( 3.14 \times 25 \times 10 \bigg) + \bigg( 2 \times ( \dfrac{1}{3} \times 3.14 \times \:25 \:\times 5 )\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( 3.14 \times 250 \bigg) + \bigg( 2 \times ( \dfrac{1}{3} \times 3.14 \times \:125 \: )\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( 785 \bigg) + \bigg( 2 \times ( \dfrac{1}{3} \times 392.5 \: )\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( 3.14 \times 250 \bigg) + \bigg( 2 \times ( \dfrac{1}{3} \times 392.5 \: )\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( 3.14 \times 250 \bigg) + \bigg( 2 \times ( \dfrac{1}{\cancel {3}} \times \cancel {392.5} \: )\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( 3.14 \times 250 \bigg) + \bigg( 2 \times ( 130.8\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= \bigg( 785 \bigg) + \bigg( 2 \times ( 130.8\bigg) \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= 785 + 262.6 \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:= 1047.6 \:$

$\qquad \dashrightarrow \sf \:\: Volume _{(Grain\:Silo)} \:\approx 1047.6 \:$

$\qquad \dashrightarrow \pmb{\underline{\purple{\:Volume _{(Grain\:Silo)} \:\approx 1047.6 \: ft.^3 }} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Hence,\:Volume \:\:of\:Grain\:Silo \:is\:\bf{1047.6\:ft^3}}\:(\:Approx \:)\:.}}$