Physics, asked by tiktok3519, 5 months ago

If the ratio between the curved surface area and the total surface area of a right-circular cylinder is 3:4, find the ratio of height to the radius of the cylinder.

Don't spam plz !!​

Answers

Answered by Anonymous
1

Given: The ratio between the CSA and the TSA of a right circular cylinder is Given:

To Find: Ratio of height to the radius of the cylinder?

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━

☯ Let radius and height of a right circular cylinder be 'r' and 'h' respectively.

⠀⠀⠀

\begin{gathered}\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}\\ \\\end{gathered}

Ratio between the Curved surface area and the Total surface area of a right circular cylinder is 3:4.

⠀⠀⠀

\begin{gathered}:\implies\sf \dfrac{CSA_{\;(cylinder)}}{TSA_{\;(cylinder)}} = \dfrac{3}{4}\\ \\\end{gathered}

\begin{gathered}:\implies\sf \dfrac{ \cancel{2 \pi r} h}{ \cancel{2 \pi r} (r + h)} = \dfrac{3}{4}\\ \\\end{gathered}

\begin{gathered}:\implies\sf \dfrac{h}{r + h} = \dfrac{3}{4}\\ \\\end{gathered}

\begin{gathered}:\implies\sf 4h = 3(r + h)\\ \\\end{gathered}

\begin{gathered}:\implies\sf 4h = 3r + 3h\\ \\\end{gathered}

\begin{gathered}:\implies\sf 3r = 4h - 3h\\ \\\end{gathered}

\begin{gathered}:\implies\sf 3r = h\\ \\\end{gathered}

\begin{gathered}:\implies\sf{\boxed{\sf{\pink{ \dfrac{h}{r} = 3}}}}\;\bigstar\\ \\\end{gathered}

∴ The ratio between height and radius of right circular cylinder is 3/1.

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━

\begin{gathered}\qquad\boxed{\underline{\underline{\purple{\bigstar \: \bf\:Formulas\:related\:to\:SA\:and\:Volume\:\bigstar}}}} \\ \\\end{gathered}

\begin{gathered}\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area\ formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}\end{gathered}

Similar questions