Question

a sphere of diameter 15.6cm is melted and cast into a right circular cone of height 31.2cm.Find the diameter of the base of the cone?

Answer 1

hey mate

Here is your answer

Diameter of a sphere = 15.6 cm

Radius of the sphere = 15.6 / 2 = 7.8 cm

Volume of a sphere = 4/3 πr3

Volume of the sphere = 4/3 x 22/7 x 7.8 x 7.8 x 7.8

Height of the cone = 31.2 cm

Let the radius of the cone be 'r' cm.

Volume of a cone = 1/3 x πr2h

Volume of the cone = 1/3 x 22/7 x r2 x 31.2

Volume of the cone = Volume of the sphere

1/3 x 22/7 x r2 x 31.2 = 4/3 x 22/7 x 7.8 x 7.8 x 7.8

⇒ r2 = (7.8)2

⇒ r = 7.8 cm

Radius of the cone = 7.8 cm

∴ Diameter of the cone = 2 x 7.8 = 15.6 cm





Glad to help you !

Answer 2

$\: \: \: \: \: \: \: \green {\underline{ \huge \frak{Hola! \: Mate}}} \\ \\ \underline{ \tiny\bf{As \: we \: know \: the \: volume \: of \: sphere \: is \: equal \:t o \rightarrow \: \frac{4}{3}\pi {r}^{3} }} \\ \\ \bf{the \: value \: of \:( \tiny{ \pi \: = \frac{22}{7} }}) \tiny{and \: r = \frac{d}{2} = \frac{15.6}{2} = 7.8 } \\ \\ \rightarrow \frac{4}{3} \times \frac{22}{7} \times ( {7.8})^{3 } =1,988.6 \\ \\ \tiny \bf{then \: the \: sphere \: changed \: into \: a \: cone \: and \: the \: hight \: is \: given : 31.2 \: cm} \\ \\ \tiny\underline{ \bf{AS \: we \: know \: the \: volume \: of \: right \: circular \: cone}} : \frac{1}{3} \pi {r}^{2} h \\ \\ \bf{ we \:got \: the \: volume \: is \: 1,988.6} \\ \\ > > \small\bf{1,988.6 = \frac{1}{ \cancel3} \times \frac{22}{7} \times {(r)}^{2} } \times \cancel{ 31.2} ^{10.4} \\ \\ \bf{ > >1,988.6 = 32.7 \times {r}^{2}} \\ \\ \bf{> > {r}^{2} = \frac{1,988.6}{32.7} } \\ \\ > > \: {r}^{2} = 60.81 \\ \\ > > \: r = \sqrt{60.81} = 7.8 \\ d=7.8×2=15.6$