Question

$\sf \: An \: electric \: geyser \: is \: cylindrical \: in \: shape, \: having \: a \: diameter \: of \: 35 \: cm \: and \: height \: 1.2 \: \\ \sf m. \: Neglecting \: the \: thickness \: of \: its \: walls, \: calculate \: \\ \\ \sf i) \: Its \: outer \: lateral \: surface \: area, \\ \\ \sf ii) \: Its \: capacity \: in \: litres.$

Chapter Name :- Mensuration​

Answer 1

Question-

An electric geyser is cylindrical in shape, having a diameter of 35cm and height 1.2m

Neglecting the thickness of its walls, calculate

(i) Its outer lateral surface area,

(ii) Its capacity in litres

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

Required Answer-

• i)Its outer Lateral Surface Area is $\sf{13200cm^2}$
• ii)Its Capacity in litres is 1155 litres

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

Given-

• diameter=35m,radius$\sf{\frac{35}{2}}$=17.5
• height=1.2m$\sf{1.2 \times 100}$=120cm

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

Formula Using-

${\large{\underline{\underline{\pmb{\frak{L \sf{S}A}}}}}}$

${\underline{\boxed{\red{\sf{LSA_{(cylinder)}=2 \pi rh}}}}}$

${\large{\underline{\underline{\pmb{\frak{V\sf{olume}}}}}}}$

${\underline{\boxed{\red{\sf{Volume_{(cylinder)}=\pi r^2 h}}}}}$

Here-

• LSA=Lateral Surface Area
• $\sf{\pi = \dfrac{22}{7}}$
• r=radius
• h=height

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

$\qquad\qquad\quad{\large{\underline{\underline{\blue{\sf{Solution}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\large{\underline{\pmb{\frak{Calculating\:LSA\:of\:geyser}}}}}$

${\large{\pink{\dashrightarrow}{\qquad{\sf{LSA_{(cylinder)}=2 \pi rh}}}}}$

${\large{\pink{\dashrightarrow}{\qquad{\sf{LSA_{(cylinder)}=2 \times \dfrac{22}{7} \times 1750 \times 120}}}}}$

${\large{\pink{\dashrightarrow}{\qquad{\sf{LSA_{(cylinder)}=2 \times 22 \times 2.5\times 120}}}}}$

${\large{\pink{\dashrightarrow}{\qquad{\sf{LSA_{(cylinder)}=2 \times 22 \times 2.5\times 120}}}}}$

${\large{\blue{\rightsquigarrow}{\qquad{\underline{\sf{LSA_{(cylinder)}=\purple{13200cm^2}}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\large{\underline{\pmb{\frak{Calculating\:Capacity}}}}}$

${\large{\purple{\dashrightarrow}{\qquad{\sf{Volume_{(cylinder)}=\pi r^2 h}}}}}$

${\large{\purple{\dashrightarrow}{\qquad{\sf{Volume_{(cylinder)}=\dfrac{22}{7} 17.5\times 17.5\times 120}}}}}$

${\large{\purple{\dashrightarrow}{\qquad{\sf{Volume_{(cylinder)}=22 \times 2.5\times 17.5\times 120}}}}}$

${\large{\blue{\rightsquigarrow}{\qquad{\underline{\sf{Volume_{(cylinder)}\purple{115500cm^2}}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\large{\underline{\sf{Calculating\:Capacity}}}}$

• We know that $\sf{1 litres=100cm^2}$

${\large{\green{\dashrightarrow}{\therefore{\qquad{\sf {\dfrac{115500}{100}}}}}}}$

${\large{\green{\rightsquigarrow}{\therefore{\qquad{\sf {\cancel\dfrac{1155}{100}=\red{1155 litres}}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

~Therefore

• i)Its outer Lateral Surface Area is $\sf{13200cm^2}$
• ii)Its Capacity in litres is 1155 litres

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$