Question

${\huge{\underline{\small{\mathbb{\pink{ QUESTION \: -: \: }}}}}}$


ᴛʜᴇ ʜᴇɪɢʜᴛ ʜ ᴏꜰ ᴡᴀᴛᴇʀ ɪɴ ᴀ ᴄʏʟɪɴᴅʀɪᴄᴀʟ ᴄᴏɴᴛᴀɪɴᴇʀ ᴡɪᴛʜ ʀᴀᴅɪᴜꜱ ʀ = 5 ᴄᴍ ɪꜱ ᴇQᴜᴀʟ ᴛᴏ 10 ᴄᴍ. ᴘᴇᴛᴇʀ ɴᴇᴇᴅꜱ ᴛᴏ ᴍᴇᴀꜱᴜʀᴇ ᴛʜᴇ ᴠᴏʟᴜᴍᴇ ᴏꜰ ᴀ ꜱᴛᴏɴᴇ ᴡɪᴛʜ ᴀ ᴄᴏᴍᴘʟɪᴄᴀᴛᴇᴅ ꜱʜᴀᴘᴇ ᴀɴᴅ ꜱᴏ ʜᴇ ᴘᴜᴛꜱ ᴛʜᴇ ꜱᴛᴏɴᴇ ɪɴꜱɪᴅᴇ ᴛʜᴇ ᴄᴏɴᴛᴀɪɴᴇʀ ᴡɪᴛʜ ᴡᴀᴛᴇʀ. ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏꜰ ᴛʜᴇ ᴡᴀᴛᴇʀ ɪɴꜱɪᴅᴇ ᴛʜᴇ ᴄᴏɴᴛᴀɪɴᴇʀ ʀɪꜱᴇꜱ ᴛᴏ 13.2 ᴄᴍ. ᴡʜᴀᴛ ɪꜱ ᴛʜᴇ ᴠᴏʟᴜᴍᴇ ᴏꜰ ᴛʜᴇ ꜱᴛᴏɴᴇ ɪɴ ᴄᴜʙɪᴄ ᴄᴍ?

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Answer 1

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☛ Given :-

☛ Given :-Cylinder with R = 5 , H = 10

☛ Given :-Cylinder with R = 5 , H = 10∴ Volume of water in Cylinder Container

☛ Given :-Cylinder with R = 5 , H = 10∴ Volume of water in Cylinder Container➣ \: π \times 5 \times 5 \times 10➣π×5×5×10

☛ Given :-Cylinder with R = 5 , H = 10∴ Volume of water in Cylinder Container➣ \: π \times 5 \times 5 \times 10➣π×5×5×10➣ \: 250πcm {}^{3}➣250πcm3

☛ Given :-Cylinder with R = 5 , H = 10∴ Volume of water in Cylinder Container➣ \: π \times 5 \times 5 \times 10➣π×5×5×10➣ \: 250πcm {}^{3}➣250πcm3Height of Water after stone is immersed :

☛ Given :-Cylinder with R = 5 , H = 10∴ Volume of water in Cylinder Container➣ \: π \times 5 \times 5 \times 10➣π×5×5×10➣ \: 250πcm {}^{3}➣250πcm3Height of Water after stone is immersed :∴ New \: volume = πr {}^{2}h

∴Newvolume=πr2h

∴Newvolume=πr2h➟ \: π \times 5 \times 5 \times 13.2➟π×5×5×13.2

∴Newvolume=πr2h➟ \: π \times 5 \times 5 \times 13.2➟π×5×5×13.2➟ \: 330π \: cm {}^{3}

∴ Volume of the one = New Volume of Cylinder -

∴ Volume of the one = New Volume of Cylinder -Volume of water in Cylinder Container

∴ Volume of the one = New Volume of Cylinder -Volume of water in Cylinder Container➱ \: 330π \: cm {}^{3} - 250π \: cm {}^{3}➱330πcm3−250πcm3

∴ Volume of the one = New Volume of Cylinder -Volume of water in Cylinder Container➱ \: 330π \: cm {}^{3} - 250π \: cm {}^{3}➱330πcm3−250πcm3\large\pink{\underline{{\boxed{\textbf{ = 80π \: cm}}}}}{ {}^{3} \: \: } = 80π cm3

∴ Volume of the one = New Volume of Cylinder -Volume of water in Cylinder Container➱ \: 330π \: cm {}^{3} - 250π \: cm {}^{3}➱330πcm3−250πcm3\large\pink{\underline{{\boxed{\textbf{ = 80π \: cm}}}}}{ {}^{3} \: \: } = 80π cm3▃▃▃▃

Answer 2

Answer:

80π cm³

$\rule{70mm}{2pt}$

Step-by-step explanation:

Given that the The height 'h' of water in a cylindrical container with radius 'r' = 5 cm is equal to 10 cm. Peter needs to measure the volume of a stone with a complicated shape and so he puts the stone inside the container with water. The height of the water inside the container rises to 13.2 cm.

$\underline{\underline{\orange{\sf{As\:per\:given\: statement:}}}}$

“The height 'h' of water in a cylindrical container with radius 'r' = 5 cm is equal to 10 cm.” Here, the radius of cylinder is 5 cm. But it's said that cylindrical container with radius 'r' = 5 cm is equal to 10 cm. It doesn't mean that radius of cylinder is 5 cm at first and later 10 cm. If you read carefully, you'll notice that at first the questioner is talking about the height of the cylinder. So, obviously it's height is 10 cm.

We know-

$\huge{\green{\underline{ \boxed{\red{\sf{Vol.\:of\: cylinder\:=\:\pi r^{2}h}}}}}}$

Therefore, the volume of water in cylindrical container is πr²h (where r = 5 cm and h = 10 cm). Substitute the values in this formula.

$\implies\:\sf{\pi\:\times\:(5)^{2}\:\times\:10}$

$\implies\:\sf{\pi\:\times\:25(10)}$

$\implies\:\sf{250\pi}$

Later he puts the stone inside the container with water, due to this the height of the water inside the container rises to 13.2 cm.

$\underline{\underline{\orange{\sf{As\:per\:given\: statement:}}}}$

New volume of the cylindrical container is πr²h (there is no change in formula.) But this time, height of the container is 13.2 cm.

Substitute the values,

$\implies\:\sf{\pi\:\times\:(5)^{2}\:\times\:13.2}$

$\implies\:\sf{\pi\:\times\:25(13.2)}$

$\implies\:\sf{330\pi}$

We need to find out the volume of the stone in cubic cm. So, volume of the stone can by find by subtracting the new volume of the cylindrical container from old value of cylindrical container.

Therefore,

$\green{\underline{ \boxed{\red{\sf{Vol.\:of\: stone\:= \: vol. \: of \:cylindrical \: container(new - old)}}}}}$

Substitute the values,

$\implies\:\sf{330\pi\:-\:250\pi}$

$\implies\:\sf{80\pi}$

Therefore, the volume of the stone is 80π or 251.2 cm³.