State lost more than 2100 sq. km. of forest area in
past 2 decades
Nagpur, Dec. 26 : Maharashtra has lost a staggering 2116 sq. km. of forest area or an equivalent of three Tadoba forest reserves in the span of two decades, reveals a report tracking climate change in India released recently by the Ministry of Statistics and Programme Implementation.
Informal Letter :
You are Pritesh/Preeti Desai,
staying at 1/17, Gharkul, P.K.
Road, Meher Colony, Nagpur.
Write a letter to your uncle
giving your views about the
conservation of forest.
Answers
i) The curved surface area of cylinderical petrol tank is 59.4 m².
ii) The steel actually used to make the tank is 95.04 m².
Step-by-step-explanation:
We have given that,
For a cylinderical petrol tank,
Diameter ( d ) = 4.2 m
∴ Radius ( r ) = d ÷ 2 = 4.2 ÷ 2 = 2.1 m
Height ( h ) = 4.5 m
We have to find,
i) Curved surface area of tank
ii) Total steel used to make tank
i)
Now, we know that,
\displaystyle{\pink{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:h}}Curvedsurfaceareaofcylinder=2πrh
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:\times\:4.5}⟹CSAcylinder=2×722×2.1×4.5
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:4.5}⟹CSAcylinder=2×22×0.3×4.5
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:9}⟹CSAcylinder=22×0.3×9
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:198\:\times\:0.3}⟹CSAcylinder=198×0.3
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:59.4\:m^2}⟹CSAcylinder=59.4m2
\therefore\:\underline{\boxed{\red{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:59.4\:m^2\:}}}∴Curvedsurfaceareaofcylinder=59.4m2
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ii)
Now,
The tank is closed with steel.
We have to find how much steel was used to make the tank.
Now,
\displaystyle{\pink{\sf\:Total\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:(\:r\:+\:h\:)}}Totalsurfaceareaofcylinder=2πr(r+h)
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:(\:2.1\:+\:4.5\:)}⟹TSAcylinder=2×722×2.1(2.1+4.5)
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:6.6}⟹TSAcylinder=2×22×0.3×6.6
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:13.2}⟹TSAcylinder=22×0.3×13.2
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:3.96}⟹TSAcylinder=22×3.96
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:87.12\:m^2}⟹TSAcylinder=87.12m2
\therefore\:\underline{\boxed{\green{\sf\:Total\:surface\:area\:of\:cylinder\:=\:87.12\:m^2\:}}}∴Totalsurfaceareaofcylinder=87.12m2
Now, from the given condition,
\displaystyle{\pink{\sf\:Steel\:used\:to\:make\:tank\:-\:Steel\:wasted\:=\:Total\:surface\:area\:of\:cylinder}}Steelusedtomaketank−Steelwasted=Totalsurfaceareaofcylinder
\displaystyle{\implies\sf\:Steel\:used\:-\:\dfrac{1}{12}\:Steel\:used\:=\:TSA_{cylinder}}⟹Steelused−121Steelused=TSAcylinder
\displaystyle{\implies\sf\:Steel\:used\:\left(\:1\:-\:\dfrac{1}{12}\:\right)\:=\:87.12}⟹Steelused(1−121)=87.12
\displaystyle{\implies\sf\:Steel\:used\:\left(\:\dfrac{12\:-\:1}{12}\:\right)\:=\:87.12}⟹Steelused(1212−1)=87.12
\displaystyle{\implies\sf\:Steel\:used\:\times\:\dfrac{11}{12}\:=\:87.12}⟹Steelused×1211=87.12
\displaystyle{\implies\sf\:Steel\:used\:=\:\dfrac{\cancel{87.12}\:\times\:12}{\cancel{11}}}⟹Steelused=