Question

A drinking glass vessel is in the shape of furstom of a cone of height 14cm the diameter of its two circular ends are 4cm and 2cm. find the capacity of the vessel​

Answer 1

$\bf \underline{Given} :$ ↴

$\sf \implies Height \: of \: frustum = 14 \: cm$

$\sf \implies Diameter \: of \: bigger \: circular \: end = 4 \: cm$

$\sf \implies Diameter \: of \: smaller \: circular \: end = 2 \: cm$

$\bf \underline{To \: find} :$ ↴

$\sf \implies Volume \: or \: Capacity \: of \: glass.$

$\bf \: \underline{ Formula \: to \: be \: used},$

$\sf \implies \boxed{ \pink{\sf \: \dfrac{1}{3} \pi h \: \bigg((r_1)^{2} + (r_2)^{2} + (r_1 \times r_1) \bigg)}}$

$\sf \: Where,$

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

$\sf \implies r_1 = \dfrac{Diameter \: of \: bigger \: end }{2} = \dfrac{4}{2} = 2 \:cm$

$\sf \implies r_2 = \dfrac{Diameter \: of \: smaller \: end }{2} = \dfrac{2}{2} = 1 \:cm$

$\underline{ \sf \: Now, we \: will \: find \: the \: volume \: of \: glass. }$

$\sf \implies \sf \: \dfrac{1}{3} \pi h \: \bigg((r_1)^{2} + (r_2)^{2} + (r_1 \times r_1) \bigg)$

$\sf \: \underline{Putting \: the \: values},$

$\sf \implies \sf \: \dfrac{1}{3} \: \times \dfrac{22}{7} \times 14 \: \bigg(2^{2} + 1^{2} + 2 \times 1 \bigg)$

$\sf \implies \sf \: \dfrac{1}{3} \: \times \dfrac{22}{7} \times 14 \: \bigg(4 + 1 + 2\bigg)$

$\sf \implies \sf \: \dfrac{1}{3} \: \times \dfrac{22}{7} \times \dfrac{14}{1} \times \dfrac{7}{1}$

$\sf \implies \dfrac{22 \times 14}{3}$

$\sf \implies \dfrac{308}{3}$

$\red{ \sf \implies 102.66 \: cm^{3} }$

$\underline{\boxed{ \pink{ \sf \:Hence, capacity \: of \: the \: glass = 102.66 \: cm^{3}}}}$