Question

A solid metal cylinder of 10 centimetre height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3:4 keeping the height 10 cm.What would be the percentage change in the flat surface area before and after?​

Answer 1

Answer:

Step-by-step explanation:

Given Conditions :-

A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3:4 keeping the height 10 cm.

To Find :-

Percentage change in the flat surface area.

Solution :-

Let radius and volume of cylinder be R and V.

And cone and volume 1 be r1 and v1.

And cone and volume 2 be r2 and v2.

Since, Cylinder is melted and recast into cone 1 and cone 2.

Then, V = v1 + v2.

Given Ratio = 3 : 4

So v1 = 3x and v2 = 4x

Volume of cylinder V = 7x.

Percentage = 1/2 × 100 = 50 %

Hence, The percentage change in the flat surface area is 50 %.

Answer 2

AnswEr :

$\:\bullet$ Height of cylinder = 10cm

$\:\bullet$ Radius of cylinder = $\bf\frac{D}{2} = \frac{14}{2} = 7cm$

$\:\bullet$ Volume of two cones = 3:4

$\underline{\dag\:\textsf{According \: to \: given \: in question:}}$

$\:\bullet$ Let the Volume of cone 1 be 3x

$\:\bullet$ Let the Volume of cone 2 be 4x

$\therefore$Hence, total volume = 7x [i.e. 3x + 4x = 7x = Volume of cylinder]

$\rule{100}2$

$\underline{\dag\:\textsf{Let's \: head \: to \: the \: question \: now:}}$

$\normalsize\star{\boxed{\sf{Volume \: of \: cylinder = \pi r^2h }}}$

$\normalsize\ : \implies\sf\ V = \pi \times\ 7 \times\ 7 \times\ 10 \\ \\ \normalsize\ : \implies\sf\ V = 490 \pi$

$\normalsize\sf\ Now, \: [Volume_{cyl} = \: V_{cone \: 1} + V_{cone \: 2} ] \: So,$

$\normalsize\ : \implies\sf\ V_{cyl} = 3x + 4x \\ \\ \normalsize\ : \implies\sf\ 490 \pi = 7x \\ \\ \normalsize\ : \implies\sf\ x = \frac{490 \pi}{7} \\ \\ \normalsize\ : \implies\sf\ x = 70 \pi$

$\rule{100}{1}$

$\:\star\:\sf\ Volume \: of \: cone \: 1 -$

$\normalsize\ : \implies\sf\ V_{c1} = 3x = 3 \times\ 70 \pi \\ \\ \normalsize\ : \implies\sf\ V_{c1} = 210 \pi$

$\:\star\:\sf\ Volume \: of \: cone \: 2-$

$\normalsize\ : \implies\sf\ V_{c2} = 4x = 4 \times\ 70 \pi \\ \\ \normalsize\ : \implies\sf\ V_{c2} = 280 \pi$

$\rule{100}2$

$\normalsize\sf\ Now; \: find \: radius \: of \: cones \: by \: using \: formulae$

$\normalsize\star{\boxed{\sf{Volume \: of \: cone = \frac{1}{3} \pi r^2 h }}}$

$\normalsize\ : \implies\sf\ V_{c1} = 210 \pi \\ \\ \normalsize\ : \implies\sf\ r_{1}^{2} = \frac{210 \pi \times\ 3}{\pi \times\ 10} \\ \\ \normalsize\ : \implies\sf\ r_{1}^{2} = 63$

$\normalsize\ : \implies\sf\ V_{c2} = 280 \pi \\ \\ \normalsize\ : \implies\sf\ r_{2}^{2} = \frac{280 \pi \times\ 3}{\pi \times\ 10} \\ \\ \normalsize\ : \implies\sf\ r_{2}^{2} = 84$

$\rule{100}2$

$\normalsize\sf\ Now; \: Area \: of \: flat \: surfaces \: before \: melting \: of \: cylinder;$

$\normalsize\star{\boxed{\sf{Flat \: surface \: area_{B} = 2 \pi r^2 }}}$

$\normalsize\ : \implies\sf\ F.S.A_{B} = 2 \pi \times\ 49 \\ \\ \normalsize\ : \implies\sf\ F.S.A_{B} = 98 \pi$

$\normalsize\ : \implies{\boxed{\sf \green{F.S.A_{c} = 98 \pi }}}$

$\normalsize\sf\ Area \: of \: flat \: surface \: area \: after \: melting;$

$\normalsize\star{\boxed{\sf{F.S.A_{A} = \pi [ V_{c1} + V_{c2} ] }}} \\ \\ \normalsize\ : \implies\sf\ F.S.A_{A} = \pi [63 + 84] = 147 \pi$

$\normalsize\ : \implies{\boxed{\sf \blue{F.S.A_{A} = 147 \pi }}}$

$\rule{100}{1}$

$\normalsize\sf\ Percentage \: of \: change \: in \: flat_{surface \: area} \: before \: and \: after$

$\normalsize\ : \implies\sf\ Percentage = \frac{F.S.A_{A} - F.S.A_{B}}{98 \pi} \times\ 100 \\ \\ \normalsize\ : \implies\sf\ Percentage = \frac{1}{\cancel{2}} \times\ \cancel{100} = 50 \%$

$\therefore\underline{\textsf{Hence, \: the \: percentage \: of \: change \: in \: surface \: areas \: is \: 50 \%}}$