Question

A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30 cm and 10 cm respectively. Find the cost of milk which can completely fill the bucket at the rate of Rs. 20 per litre and the cost of metal sheet used if it costs Rs. 10 per 100cm2. [Use π=3.14]

Answer 1

$\\ \huge\bold{\purple{ \mathtt{ Given:}}} \\ Depth \: of \: the \: bucket = 24cm \\ Diamter \: of \: the \: top = 30cm \\ Radius( R = 15) \\ Diamter \: of \: the \: bottom = 10cm \\ Radius( r= 5cm)$

$\huge{ \bold{ To \: find- }}$

$= > Cost \: of \: milk \: at \: the \: rate \: of \\ Rs \:.20 \: per \: litre$

$\huge{ \mathtt{ \pink{SOLUTION:-}}}$

$Volume \: of \: frustum \: of \: cone \\ = > \frac{1}{3} \pi \: h \times 24{{r}^{2} + {r}^{2} + (r)(r)} \\ = > \frac{1}{3} \pi \times 24cm( {15}^{2} + {5}^{2} + 15 \: \times 5) \\ = > 8\pi(225 + 25 + 75) \\ = > 8\pi(325) \\ = > 8 \times 3.14 \times 325 \\ = > 8164 {cm}^{3}$

$\large{ \boxed{ \blue{1 \: l = 1000 \ {cm}^{3} }}}$

$Therefore \\ = > 8164 {cm}^{3} = 8164 \times \frac{1}{1000} l \\ = > 8.164 \: l$

$\large{\boxed{ \green{ Cost \: of \: milk \: per \: l = Rs20}}}$

$Cost \: of \:8.164 \: l \: milk = 8.164 \times 20 \\ = > Rs = 163.28$

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$TSA\: of \: frustum \: of \: cone \\ = > \pi \: l(R+ r) + \pi \: {R}^{2} + \pi \: {r}^{2} \\ = > l = \sqrt{ {h}^{2} } + {(R - r)}^{2} \\ TSA = \pi \: l(R + r) + \pi \: {r}^{2} \\ h = 24cm \\ R = 15cm \\ r = 5cm \\ l = \sqrt{ {24}^{2} } + {(15 - 5)}^{2} \\ = > \sqrt{576 + 100 } \\ = > \sqrt{676} \\ l = 26cm$

$Putting \: values... \\ = 3.14 \times 26(5 + 15) + 3.14 \times {5}^{2} \\ = 3.14(26 \times 20 + 25) \\ = 3.14(520 + 25) \\ = 3.14 \times 545 \\ TSA = 1711.3 \\ Total \: cost \: of \: metal \: sheet \: used \\ = > \frac{10 \times 1711.3}{100} \\ = > Rs = 1711.3$

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$\large{ \bold{ \red{Thanks...♡}}}$