Question

A container is in the shape of a frustum of a cone having diameters of its two circular faces are 35cm and 30 cm and vertical height 14cm is completely filled with oil . if each cm³ of oil has mass 1.2g then find the cost of oil in container when it cost RS.40 per kg​

Answer 1

QuestioN:

• A container in the shape of of a frustum of a cone having diameters of its two circular faces as 35 cm and 30 cm and vertical height 14 cm, is completely filled with oil. if each of oil has mass 1.2 g, then find the cost of oil in the container if it costs Rs. 40 per kg.

${\large{\frak{\pmb{\underline{ Clearly,\:we\: have}}}}}$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\succ\:r = \bf\dfrac{30}{2}\:cm=15\:cm$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\succ\:R = \bf\dfrac{35}{2}\:cm=17.5\:cm$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\succ\:h = 14\:cm$

$\:━━━━━━━━━━━━$

$\:$

$\Large{\underline{\mathcal{To\: Find }}}$

• Cost of oil in the container @ ₹40 /kg.

$\:\:\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━$

$\:$

${\bold { \underline{\large{Now,}}}}$

$\:\:\large{\ddagger}\underline {\footnotesize{\boxed{\frak{Volume_{(container)}={\frak{\orange{\bigg\{\bf\dfrac{1}{3}\pi (R^2+r^2+Rr)h\bigg\}cu}}}}}}}$

$\:\:$

$\:$

$\sf {\scriptsize\Bigg[\bf\dfrac{1}{3}\times \bf\dfrac{22}{7}\times\Bigg\{\bigg(\dfrac{35}{2}\bigg)^2+(15)^2+\dfrac{35}{2}\times 15\Bigg\}\times 14 \Bigg]cm^3}$

$\:\:$

$\:$

$\sf \:\:\:\:\:\rightarrowtail\: \dfrac{44}{3}\times\bigg(\dfrac{1225}{4}+225+\dfrac{525}{2}\bigg)cm^3$

$\:\:$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrowtail\: \bigg( \dfrac{44}{3}\times \dfrac{3175}{4}\bigg)cm^3$

$\:\:$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrowtail\: \bigg( \dfrac{3175 \times 11}{3}\bigg)cm^3$

$\:\:$

$\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\small{\bf{\underline{\mathcal{\red{Mass\:of\:1\:cm^3\:of\:oil\:=\:1.2\:g }}}}}$

$\:\:$

$\:$

$\:\:\:\:\:\:\:\sf \footnotesize{Mass_{(oil\:in\:the\: container)}=\bigg( \dfrac{3175 \times 11}{3}\times1.2 \bigg)\:g}$

$\:\:$

$\:$

$\sf\:\:\:\:\:\:\:\:\rightarrowtail\:\bigg( \dfrac{3175 \times 11}{3}\times\dfrac{12}{10}\times\dfrac{1}{1000}\bigg)\:kg$

$\:\:$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrowtail\:\dfrac{1397}{100}\:kg\:=13.97\:kg$

$\:\:$

$\:$

• $\bf\twoheadrightarrow$Cost of oil at ₹ 40 per kg = ₹(13.97×40) $\:\:=\:\:\large{\sf{\bf {\underline{\purple{ ₹558.80.}}}}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\dag\small{\mathcal{\fcolorbox{#D4E157}{#D4E157}{Ras}\frak{cle}}}$

Answer 2

Required Solution :

Here we're provided with the diameters of two circular faces of a frustum of a cone and its height. So our first aim should be calculating the volume of frustum of cone by its formula.

★ Formula of frustum of cone :-

• $\red{\boxed{ \bf{V \: = \: \dfrac{1}{3} \pi \: H \bigg(R {}^{2} +Rr + r {}^{2} \bigg)}}} \: \bigstar$

Here,

• R is radius of upper part
• r is radius of lower part (base)
• H is the height between both the radius

$\pink\bigstar$ As we are given with the diameters so we would get to know the radius !

• Diameter of upper base = 35 cm
• Diameter of lower base = 30 cm

As we know, (Radius = Diameter / 2)

Therefore,

• R = 17.5 cm
• r = 15 cm

$\pink\bigstar$ Substituting all the values in the above given formula inorder to get the volume of frustum of cone :

$\longmapsto \: \sf{{V \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: \times 14 \bigg(17.5 {}^{2} +17.5 \: . \: 15+ 15 {}^{2} \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: \times 14 \bigg(17.5 \times 17.5 +17.5 \times 15+ 15 {}^{2} \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: \times 14 \: \bigg(306.25 + 17.5 \times 15+ 15 {}^{2} \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: \times 14 \: \bigg(306.25 + 17.5 \times 15+ 15 \times 15 \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: \times 14 \: \bigg(306.25 + 17.5 \times 15+ 225 \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: \times 14 \: \bigg(306.25 +262.5+ 225 \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{1}{3} \times \dfrac{22}{ \cancel{7}} \: \times \cancel{14} \: \bigg(306.25 +262.5+ 225 \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{22 \times 2}{3} \: \bigg(306.25 +262.5+ 225 \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{44}{3} \: \bigg(568.75 \: + \: 225 \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{44}{3} \: \bigg(793.75 \bigg)}} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{44}{3} \times \: 793.75 }} \\ \\ \longmapsto \: \sf{{V \: = \: \dfrac{34925}{3} }} \: cm {}^{3}$

$\pink\bigstar$As we know that 1cm³ of oil has mass of 1.2g and the frustum of cone is fully filled by the oil so we would be finding out the amount of oil which is filled in container.

• $\sf{Amount \: of \: oil} = \: \dfrac{34925}{3} \times 1.2$

• $\sf{Amount \: of \: oil} = \: 13970 \: g$

$\pink\bigstar$ Converting the unit :

(As we know, 1kg = 1000g)

• $\sf{Amount \: of \: oil} = \: \dfrac{13970}{1000}$

• $\sf{Amount \: of \: oil} = \: 13.97 \: kg$

$\pink\bigstar$ Finally calculating the cost of oil in container when it costs ₹40 per kg.

• $\sf{Cost \: of \: oil \: = \: 13.97 \: \times \: 40 }$

• $\sf{Cost \: of \: oil \: = \: ₹558.80 }$

$\underline{\bf{ \therefore \:Cost \: of \: oil \: in \: container \: when \: it \: cost \: ₹40 \: per \: kg \: is \: ₹558.80</p><p>}}$