Question

It cost Rs 4400 to paint the inner curved surface of cylindrical
can 10 m deep. If the cost of painting is at the rate of Rs 20 per
square metre. Find
5 marks
(1) inner curved surface area of the vessel,
(ii) radius of the vessel
(iii) capacity of the vessel​

Answer 1

Given,

It cost Rs 4400 to paint the inner curved surface of cylindrical  can 10 m deep. The cost of painting is at the rate of Rs 20 per  square meter.

To find :

(i) inner curved surface area of the vessel,

(ii) radius of the vessel

(iii) capacity of the vessel

Solution :

Cost of painting = Rs. 4400

Rate of painting​  = Rs. 20/m²

Now,

⇒ Inner CSA of vessel = Cost of painting/Rate of painting

⇒ Inner CSA of vessel = 4400/20

⇒ Inner CSA of vessel = 220 m²

∴ Inner curved surface area of vessel = 220 m²

Now, CSA of cylinder = 2πrh

So atq,

⇒ 2πrh = 220

⇒ 2 * 22/7 * r * 10 = 220

⇒ 440r/7 = 220

⇒ 440r = 220 * 7

⇒ 440r = 1540

⇒ r = 1540/440

⇒ r = 3.5 m

∴ Radius of vessel = 3.5 m.

Now, we have, r = 3.5m , h = 10 m.

⇒ Capacity of cylinder = πr²h

⇒ Capacity of vessel = 22/7 * (3.5)² * 10

⇒ Capacity of vessel = 22/7 * 12.25 * 10

⇒ Capacity of vessel = 385 m³

∴ Capacity of vessel = 385 m³

Answer 2

Answer :-

$\:\large{\boxed{\sf{Firstly,\;let's\;understand\;the\;concept\;used\;:-}}}$

Here the concept of CSA and Volume of Cylinder has been used. We see we are given the Total cost of painting and its rate of painting. So we can divide the total cost of painting by rate of painting to find out the Inner Curved Surface area. After finding CSA we can apply values and find the Radius. And after we got radius we can find the volume.

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★ Formula Used :-

$\:\\\large{\boxed{\sf{CSA\;of\;Can\;=\;\bf{\dfrac{Total\;cost\;of\;Painting}{Rate\;of\;Painting}}}}}$

$\:\\\large{\boxed{\sf{2 \pi rh \;= \; \bf{CSA\;of\;Cylinder}}}}$

$\:\\\large{\boxed{\sf{Volume\;of\;Cylinder\;=\;\bf{\pi r^{2} h}}}}$

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★ Question :-

It cost Rs 4400 to paint the inner curved surface of cylindrical

can 10 m deep. If the cost of painting is at the rate of Rs 20 per

square metre. Find :-

(i) inner curved surface area of the vessel,

(ii) radius of the vessel

(iii) capacity of the vessel

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★ Solution :-

Given,

» Total cost of painting the can = Rs. 4400

» Rate of painting the can at per m² = Rs. 20

» Depth of the can = h = 10 m

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i.) ~ For the Inner CSA of Vessel :-

$\:\\\qquad\large{\sf{:\rightarrow\;\;\: CSA\;of\;Can\;=\;\bf{\dfrac{Total\;cost\;of\;Painting}{Rate\;of\;Painting}}}}$

$\:\qquad\large{\sf{:\rightarrow\;\;\: CSA\;of\;Can\;=\;\bf{\dfrac{4400}{20}\:\;=\;\:\underline{\underline{220\;\;m^{2}}}}}}$

$\:\\\large{\underline{\underline{\rm{Thus,\;CSA\;of\;the\;Vessel\;is\;\;\boxed{\bf{220\;\;m^{2}}}}}}}$

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ii.) ~ For the Radius of Vessel :-

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\: 2 \pi rh \;= \; \bf{CSA\;of\;Cylinder}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\: 2\:\times\:\dfrac{22}{7}\:\times\:r\:\times\:10 \;= \; \bf{220\;\;m^{2}}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:r\:\;= \; \bf{\dfrac{220\;\times\;7}{10\;\times\;22}\;\;=\;\:\underline{\underline{3.5\;\;m}}}}}$

$\:\\\large{\underline{\underline{\rm{Thus,\;radius\;of\;the\;Vessel\;is\;\;\boxed{\bf{3.5\;\;m}}}}}}$

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iii.) ~ For the Volume of Vessel :-

$\:\\\qquad\large{\sf{:\Longrightarrow\;\;\:Volume\;of\;Cylinder\;=\;\bf{\pi r^{2} h}}}$

$\:\\\qquad\large{\sf{:\Longrightarrow\;\;\:Volume\;of\;Cylinder\;=\;\bf{\dfrac{22}{7}\:\times\: (7)^{2}\:\times\:10\;\:=\;\:\underline{\underline{385\;\;m^{3}}}}}}$

$\:\\\large{\underline{\underline{\rm{Thus,\;capacity\;of\;the\;Vessel\;is\;\;\boxed{\bf{385\;\;m^{3}}}}}}}$

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★ More to know :-

$\:\\\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\:\times\:\pi r^{2}h}$

$\:\\\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;Length\:\times\:Breadth\:\times\:Height}$

$\:\\\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

$\:\\\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\:\times\:\pi r^{3}}$

$\:\\\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Sphere\;=\;\dfrac{4}{3}\:\times\:\pi r^{3}}$

$\:\\\;\;\;\;\sf{\leadsto\;\;\;TSA\;of\;Cylinder\;=\;2 \pi rh \:+\: 2 \pi r^{2}}$

$\:\\\;\;\;\;\sf{\leadsto\;\;\;CSA\;of\;Cylinder\;=\;2 \pi rh}$

$\:\\\;\;\;\;\sf{\leadsto\;\;\;TSA\;of\;Cone\;=\;\pi rL \:+\: \pi r^{2}}$