Question

The radius and slant height of Cone is 2 cm and 6cm respectively its curved surface

area is ……….........cm².

a) 12 π b) 4 π c) 8 π d) 6 π​

Answer 1

Given{ Radius of cone=2cm

⠀⠀⠀⠀ Slant height of cone = 6cm

To find: Curved Surface Area of cone?

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

⠀⠀⠀

$\begin{gathered}\dag\;{\underline{\frak{As\;we\;know\;that,}}}\\ \\\end{gathered}$

$\begin{gathered}\star\;{\boxed{\sf{\pink{Curved\:surface\:area_{\;(cone)} = \pi rl}}}}\\ \\\end{gathered}$

where,

r = radius = 2 cm

l = Slant height = 6 cm

⠀⠀⠀

$\begin{gathered}\dag\;{\underline{\frak{Putting\:values\:in\:formula,}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf CSA_{\;(cone)} = \pi \times 2 \times 6\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf CSA_{\;(cone)} = \pi \times 12\\ \\\end{gathered}$

$\begin{gathered}:\implies{\underline{\boxed{\frak{\purple{CSA_{\;(cone)} = 12 \pi cm^2 }}}}}\;\bigstar\\ \\\end{gathered}$

$\therefore\:{\underline{\sf{Curved\:surface\: area\:of\:cone\:is\: \bf{12 \pi\:or\: 37.714\:cm^2}.}}}$

Answer 2

$\huge\sf \pmb{\orange {\underline \pink{\underline{\:Ꭺ ꪀ \mathfrak ꕶ᭙ꫀя \: }}}}$

$\large{\underline{\underline{\sf{ \: {★Given :-}}}}}$

• Weight = 100 N

• Height = 10 m

$\large{\underline{\underline{\sf{ \: {★To \: find:-}}}}}$

• Potential energy = ?

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

➊ We know that :-

$\qquad \bull \bf \: {Weight = Mass \times Acceleration }$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: {100 = mass \times 10 } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: { \dfrac{100}{10} = mass } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: { mass = \cancel\dfrac{100}{10} } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered}\end{gathered}$

➋ We know that :-

$\qquad \bull \bf \: { Potential \: energy=mass \times acceleration \times height }$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: {potential \: energy = 10 \times 10 \times 10 } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: {potential \: energy = 10 \times 10 \times 10 } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow \underline{ \boxed{\sf \: {potential \: energy = 1000 \: J}}} \\\end{gathered}\end{gathered} \end{gathered} \end{gathered} \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\large \quad \therefore\sf \: { The \: correct \: option \: is \: \underline{\underline{1000 \: J}}} \\\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

• The potential energy is 1000 J.

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