Math, asked by anureetkaurbrar2006, 4 days ago

if cylinders radius is increased by 50%,then by how much %height should decrease so that volume remains same​

Answers

Answered by arpitmondal96
6

Let the radius of the original cylinder =r cm and

Height=h cm

Original volume=πr2h

New Radius after increase = 150/100r=3/2r

New height after decrease=80/100h=4/5h

Now, new volume of the cylinder=π×(3r/2)2×4/5h=9πr2/5h

% increase=Changeinvolume__ =dfrac9πr2 5h-pir2h__ ×100=4πr2__ h

                    Original volume              πr2h                                            5__

                                                                                                                πr2h ×100=4/5×100=80%  

 

 

                   

 

Answered by mathdude500
28

\large\underline{\sf{Solution-}}

Let assume that

Radius of cylinder be r units

Height of cylinder be h units

We know, Volume (V) of cylinder of radius r and height h is given by

\boxed{ \rm{ \:V = \pi \:  {r}^{2} \: h \: }} -  -  -  - (1) \\

Now, given that

Radius is increased by 50 % of the original radius.

So,

\rm \: r' \:  =  \: r \:  +  \: 50\% \: of \: r \\

\rm \: r' \:  =  \: r \:  +  \: \dfrac{r}{2}  \\

\rm \: r' \:  =  \: \dfrac{3r}{2}  \\

Let assume that height of cylinder be decreased by x %.

So,

\rm \: h' \:  =  \: h \:  -   \: x\% \: of \: h \\

\rm \: h' \:  =  \: h \: - \:  \frac{xh}{100}  \\

\rm \: h' \:  =  \: h \: \bigg(1- \:  \frac{x}{100} \bigg) \\

So, Volume V' of radius r' and height h' is given by

\boxed{ \rm{ \:V = \pi \:  {r'}^{2} \: h'\: }} -  -  -  - (2) \\

According to statement,

\rm \: V = V' \\

\rm \: \pi \:  {r}^{2}h \:  =  \: \pi \:  {r'}^{2} h' \\

\rm \:{r}^{2}h\:=\:{r'}^{2} h' \\

On substituting the value of r' and h', we get

\rm \:{r}^{2}h\:=\:\dfrac{3r}{2} \times \dfrac{3r}{2}   \times h\bigg(1 - \dfrac{x}{100} \bigg)  \\

\rm \:1\:=\:\dfrac{3}{2} \times \dfrac{3}{2}   \times \bigg(1 - \dfrac{x}{100} \bigg)  \\

\rm \: \frac{4}{9} \:=\:1 - \dfrac{x}{100}   \\

\rm \: \frac{x}{100} \:=\:1 - \dfrac{4}{9}   \\

\rm \: \frac{x}{100} \:=\:\dfrac{9 - 4}{9}   \\

\rm \: \frac{x}{100} \:=\:\dfrac{5}{9}   \\

\rm\implies \:x = \dfrac{500}{9} \:  \% \\

\rm\implies \:Height \: of \: cylinder \: is \: decreased \: by \: \dfrac{500}{9} \:  \% \\

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r  \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} =  \dfrac{4}{3}\pi {r}^{3}  }\\ \\ \bigstar \: \bf{Volume_{(cube)} =  {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

Similar questions