Question

if the base of the cylinder is 4 radius and height 10cm. how many coin of radius 3cm can be filled in cylinder

Answer 1

$\huge\tt{Question \: :}$

If the base radius of the cylinder is 4 cm and height is 10 cm, how many coins of base radius 3 cm and height 1 cm can be filled in the cylinder?

$\huge\tt{Theory \: :}$

For solving this question, we will have to deal with the volumes of the two shapes. But, here the data is incomplete. We know that the coin is in the shape of a cylinder and its height is not given. For solving and letting you know the method, we take the height of the coin as 1 cm.

$\sf \: r _{1} = 4 \: cm$

$\sf \: r_{2} = 3 \: cm$

$\sf \: h_{1} = 10 \: cm$

$\sf \: h_{2} = 1 \: cm$

Now, let's suppose that the number of coins that fit into the cylinder is 'n'.

Then, the volume of the cylinder and the volume of n number of coins fitting into the cylinder will be equal,
i.e. Volume of the cylinder = n × Volume of 1 coin

So, let's start with the calculations.

$\sf \: { \cancel\pi} {r _{1}}^{2} h_{1} = n \times { \cancel\pi} {r_{2}}^{2} h_{2}$

$\sf \: {4}^{2} \times 10 = n \times {3}^{2} \times 1 \\ \\ \sf \implies \: 16 \times 10 = 9n \\ \\ \sf \implies \: n = 17.78$

Rounding off, the value of n is 18.

$\huge \tt{Answer \: :}$

Thus, the number of coins that can fit in the cylinder are 18 coins.

Answer 2

ǫ) ɪғ ᴛʜᴇ ʙᴀsᴇ ʀᴀᴅɪᴜs ᴏғ ᴛʜᴇ ᴄʏʟɪɴᴅᴇʀ ɪs 4 ᴄᴍ ᴀɴᴅ ʜᴇɪɢʜᴛ ɪs 10 ᴄᴍ, ʜᴏᴡ ᴍᴀɴʏ ᴄᴏɪɴs ᴏғ ʙᴀsᴇ ʀᴀᴅɪᴜs 3 ᴄᴍ ᴀɴᴅ ʜᴇɪɢʜᴛ 1 ᴄᴍ ᴄᴀɴ ʙᴇ ғɪʟʟᴇᴅ ɪɴ ᴛʜᴇ ᴄʏʟɪɴᴅᴇʀ?


ᴀɴs-✰ɪɴ ᴛʜɪs ᴛʏᴘᴇ ᴏғ ǫᴜᴇsᴛɪᴏɴ, ᴛʜᴇ ᴅᴀᴛᴀ ɪs ᴜsᴜᴀʟʟʏ ɪɴᴄᴏᴍᴘʟᴇᴛᴇ, ᴀɴᴅ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ᴄᴏɪɴ ɪs ɴᴏᴛ ɢɪᴠᴇɴ.sᴏ, ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ᴛᴀᴋᴇ ɪᴛ ᴀs 1 ᴄᴍ (ᴛʜɪs ɪs ᴛʜᴇ ᴛʀɪᴄᴋ) ᴀɴᴅ ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ғɪɴᴅ ᴛʜᴇ ᴠᴏʟᴜᴍᴇs ᴏғ ᴄʏʟɪɴᴅᴇʀ ᴀɴᴅ ᴄᴏɪɴ sᴇᴘᴀʀᴀᴛᴇʟʏ ᴀɴᴅ ᴛʜᴇɴ ᴅɪᴠɪᴅᴇ ᴛʜᴇ ᴄʏʟɪɴᴅᴇʀ's ᴠᴏʟᴜᴍᴇ ʙʏ ᴄᴏɪɴ's ᴠᴏʟᴜᴍᴇ.

✰ʀᴇᴍᴇᴍʙᴇʀ:-ᴄᴏɪɴ ɪs ᴀʟsᴏ ᴀ ᴛʏᴘᴇ ᴏғ ᴄʏʟɪɴᴅᴇʀ.


✰ɪɴ ᴄʏʟɪɴᴅᴇʀ-

ʙᴀsᴇ ʀᴀᴅɪᴜs= 4 ᴄᴍ (ʟᴇᴛ ɪᴛ ʙᴇ ʀ1)
ʜᴇɪɢʜᴛ=10 ᴄᴍ (ʟᴇᴛ ɪᴛ ʙᴇ ʜ1)
sᴏ, ɪᴛs ᴠᴏʟᴜᴍᴇ=[π×(ʀ1)²×ʜ1] ᴄᴜʙɪᴄ ᴜɴɪᴛs
=(22/7×4²×10) ᴄᴍ³
=(22/7×16×10) ᴄᴍ 3
=3520/7 ᴄᴍ³

✰ɪɴ ᴄᴏɪɴ-

ʀᴀᴅɪᴜs=3 ᴄᴍ (ʟᴇᴛ ɪᴛ ʙᴇ ʀ2)
ʜᴇɪɢʜᴛ=1 ᴄᴍ (ʟᴇᴛ ɪᴛ ʙᴇ ʜ2)
sᴏ, ɪᴛs ᴠᴏʟᴜᴍᴇ=[π×(ʀ2)²×ʜ2] ᴄᴜʙɪᴄ ᴜɴɪᴛs
=(22/7×3²×1) ᴄᴍ³
=(22/7×9×1) ᴄᴍ³
=198/7 ᴄᴍ³

ɴᴏᴡ, ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏғ ᴄᴏɪɴs ᴛʜᴀᴛ ᴄᴀɴ ʙᴇ ғɪʟʟᴇᴅ ᴛʜᴇ ᴄʏʟɪɴᴅᴇʀ=ᴠᴏʟᴜᴍᴇ ᴏғ ᴛʜᴇ ᴄʏʟɪɴᴅᴇʀ/ᴠᴏʟᴜᴍᴇ ᴏғ ᴛʜᴇ ᴄᴏɪɴ
=3520/7 ᴄᴍ³ ÷ 198/7 ᴄᴍ³
=3520/7 ᴄᴍ³ × 7/198 ᴄᴍ³
=160/9 ( ᴄᴍ³ ɢᴇᴛs ᴄᴜᴛ)
=17.7
=18 ( ᴀғᴛᴇʀ ʀᴏᴜɴᴅɪɴɢ ᴏғғ)

sᴏ, 18 ᴄᴏɪɴs ᴄᴀɴ ʙᴇ ғɪʟʟᴇᴅ ᴛʜᴇ ᴄʏʟɪɴᴅᴇʀ.