Question

An open metal bucket is in the shape of a frustum of a cone of height 21cm with radii of it's lower and upper ends 10cm and 20cm respictevely. Find the cost of milk which can completely fill the bucket at rs 30 per liter

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Cost\:of\:milk=115.5\:rupees}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: {\implies Height \: of \: glass = 21\: cm} \\ \\ \tt: {\implies Diameter \: of \: glass ( d_{1})=20\: cm} \\ \\ \tt: {\implies Diameter \: of \: glass ( d_{2})=10\: cm} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Volume \: of \: Bucket = ?$

• According to given question :

$\tt \circ \: r_{1} = 10 \: cm \\ \\ \tt \circ \: r_{2} = 5 \: cm \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Volume \: of \: glass = \frac{1}{3} \pi h(( r_{1})^{2} + {( r_{2} )}^{2} + ( r_{1} \times r_{2})) \\ \\ \tt: \implies Volume \: of \: Bucket = \frac{1}{3} \times \frac{22}{7} \times 21\times ( {10}^{2} + {5}^{2} + (10 \times 5)) \\ \\ \tt: \implies Volume \: of \:Bucket = \frac{1}{3} \times 22 \times 3 \times (100 + 25+ 50) \\ \\ \tt: \implies Volume \: of \: Bucket= \frac{1}{3} \times 22\times 3\times 175 \\ \\ \green{\tt: \implies Volume \: of \: Bucket = 3850 \: {cm}^{3} }$

$\bold{As \: we \: know \: that} \\ \tt: \implies {1 \: cm}^{2} = \frac{1}{1000} \: litre \\ \\ \tt: \implies 3850 \: {cm}^{2} = \frac{1}{1000} \times 3850 \\ \\ \green{ \tt\therefore Bucket \: can \: hold \: 3.85 \: litre \: water}$

$\bold{for \: cost \: of \: milk} \\ \tt: \implies cost \: of \: 1 \: litre \: milk = 30 \: rupees \\ \\ \tt: \implies cost \: of \:3.85 \: litre \: milk = 30 \: rupees 3.85 \\ \\ \green{\tt: \implies cost \: of \:3.85 \: litre \: milk =115.5 \: rupees}$

Answer 2

$</p><p>\tt{\huge{\purple{Hello!!! }}}$

$</p><p>\tt{\red{Given \: - }}$

• Height of frustum = 21 cm.

• Radius of lower end = 10 cm.

• Radius of upper end = 21 cm.

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We Know That :

$\tt{ \orange{{Volume_{Frustum} \: = \: \frac{1}{3} \pi \: h \:( {R}^{2} + {r}^{2} + Rr) }}}$

Substituting the values, We Get The Value Of V as :

$</p><p>\tt{\blue{\implies{Volume_{Frustum} = 1.54\: litres. }}}$

• The cost of milk is Rs. 30 per litre.

Hence,

$\tt{ \purple{\therefore{ \implies{Cost \: of \: milk \: =RS. \:462 }}}}$