Question

2)An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet as shown in the figure. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.
(Take π = 22/7)



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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet.

• The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm.

Now, we have Dimensions of frustum as

• Height of frustum, H = 34 cm

• Radius of lower end, r = 12.5 cm

• Radius of upper end, R = 22.5 cm

Dimensions of cylinder as

• Radius of cylinder, r = 12.5 cm

• Height of cylinder, h = 6 cm

Now, Slant height (l) of frustum is given by

$\rm :\longmapsto\:l = \sqrt{ {(R - r)}^{2} + {H}^{2} }$

$\rm :\longmapsto\:l = \sqrt{ {(22.5 - 12.5)}^{2} + {34}^{2} }$

$\rm :\longmapsto\:l = \sqrt{ 100 + 1156 }$

$\rm :\longmapsto\:l = \sqrt{1256 }$

$\bf\implies \:l \: = \: 35.44 \: cm$

Now, The area of metal required to make the bucket is

$\rm \:  =  \: CSA_{Frustum} + CSA_{Cylinder} + Area_{Base}$

$\rm \:  =  \: \pi(R + r)l + 2\pi \: rh + \pi {r}^{2}$

$\rm \:  =  \: \pi\bigg[(R + r)l + 2 \: rh + {r}^{2} \bigg]$

$\rm \:  =  \: \dfrac{22}{7}\bigg[(22.5 + 12.5)35.44 + 2(12.5)(6) + {(12.5)}^{2} \bigg]$

$\rm \:  =  \: \dfrac{22}{7}\bigg[(35)35.44 + 150 + 156.25 \bigg]$

$\rm \:  =  \: \dfrac{22}{7}\bigg[1240.4 + 306.25 \bigg]$

$\rm \:  =  \: \dfrac{22}{7} \times 1546.65$

$\rm \:  =  \: 4860.9 \: {cm}^{2}$

Now, Volume of water hold in bucket =

$\rm \:  =  \: Volume_{Frustum}$

$\rm \:  =  \: \dfrac{\pi \: H}{3}( {R}^{2} + Rr + {r}^{2})$

$\rm \:  =  \: \dfrac{22 \times 36}{7 \times 3}( {22.5}^{2} + (22.5)(12.5) + {12.5}^{2})$

$\rm \:  =  \: \dfrac{22 \times 12}{7}( 506.25 + 281.25 + 156.25)$

$\rm \:  =  \: \dfrac{264}{7}( 943.75)$

$\rm \:  =  \: 35592.86 \: {cm}^{3}$

$\rm \:  =  \: 35.59286 \: litres \: \approx \: 35.6 \: litres$

Answer 2

$\huge\color{pink}\boxed{\colorbox{Black}{☆Answer☆}}$

• ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴄᴏɴᴇ = 45 ᴄᴍ

• ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴄʏʟɪɴᴅʀɪᴄᴀʟ ʙᴀsᴇ = 25 ᴄᴍ

• ᴠᴇʀᴛɪᴄᴀʟ ʜᴇɪɢʜᴛ ᴏғ ᴄᴏɴᴇ = 40 ᴄᴍ

• ʜᴇɪɢʜᴛ ᴏғ ᴄʏʟɪɴᴅʀɪᴄᴀʟ ʙᴀsᴇ = 16 ᴄᴍ

• ɴᴏᴡ , ʀᴀᴅɪᴜs ᴏғ ᴄᴏɴᴇ = ᴅ/2 = 45/2 ᴄᴍ = 22.5 ᴄᴍ

• ʀᴀᴅɪᴜs ᴏғ ʙᴀsᴇ = ᴅ/2 = 25/2 = 12.5 ᴄᴍ

ᴛʜᴇʀᴇғᴏʀᴇ , ʟᴀᴛᴇʀᴀʟ sᴜʀғᴀᴄᴇ ᴏғ ᴄᴏɴᴇ :-

$= \: \pi \: r(l + r) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{22}{7} \times22.5 \: (40 + 22.5) \: \\ \: \ \\ \: \: \: \: = \: 4419.64 {cm}^{2}$

• ʟᴀᴛᴇʀᴀʟ sᴜʀғᴀᴄᴇ ᴏғ ᴄʏʟɪɴᴅʀɪᴄᴀʟ ʙᴀsᴇ = 2π(ʀ+ʜ)

= 2×22/7(16+12.5)

= 179.14 ᴄᴍ ^2

• ᴛʜᴜs , ᴀʀᴇᴀ ᴏғ ᴍᴇᴛᴀʟʟɪᴄ sʜᴇᴇᴛ ᴜsᴇᴅ ᴛᴏ ᴍᴀᴋᴇ ʙᴜᴄᴋᴇᴛ :-

$4419.64 \: + \: 179.14 \: = \: 4598.78 \: {cm}^{2}$

•

$ᴠᴏʟᴜᴍᴇ \: \: ᴏғ \: \: ᴄᴏɴᴇ \: = \: \frac{1}{3} \: \pi \: {r}^{2} \: h \\ \\ = \frac{1}{3} \: \times \: \frac{22}{7} \: \times \: {(22.5)}^{2} \: \times \: 40 \\ \\ = \: 21214.2 {cm}^{3}$

ᴠᴏʟᴜᴍᴇ ᴏғ ᴡᴀᴛᴇʀ ᴛʜᴇ ʙᴜᴄʟᴇᴛ ᴄᴀɴ ʜᴏʟᴅ = 21214.2 ᴄᴍ^2

ʜᴇʏ ᴍᴀᴛᴇ , ᴀʙᴏᴠᴇ ɪs ʏᴏᴜʀ ʀᴇǫᴜɪʀᴇᴅ ᴀɴsᴡᴇʀ ... ɪғ ʏᴏᴜ ʟɪᴋᴇᴅ ᴛʜᴇ ɢᴜᴠᴇɴ ᴀɴsᴡᴇʀ sᴏ ᴍᴀʀᴋ ɪᴛ ᴀs ᴀ ʙʀᴀɪɴʟɪᴇsᴛ ᴀɴsᴡᴇʀ...!!!