Question

A metallic sphere of radius 4.2 cm is melted and recasted as a cylinder of radius 6 cm,find its height

Answer 1

${\huge{\underline{\rm{GIVEN}}}}$

• Radius of metallic Sphere = 4.2cm

• Radius of Cylinder = 6cm.

${\huge{\underline{\rm{TO\:FIND}}}}$

• The Height of the Cylinder

${\huge{\underline{\rm{FORMULAE\:USED}}}}$

• ${\huge{\boxed{\sf{\blue{\pi r^2h}}}}}$

• ${\huge{\boxed{\sf{\red{\dfrac{4}{3}\pi r^3}}}}}$

Now,

• If it is recasted then the volume of both the figure will be same.

So,

$\implies\rm{Volume\:of\:sphere=Volume\:of\: Cylinder}$

$\implies\rm{\dfrac{4}{3}\times{\dfrac{22}{7}}\times{4.2}\times{4.2}\times{4.2} = \dfrac{22}{7}\times{6}\times{6}\times{h}}$

$\implies\rm{\dfrac{\cancel{4}}{3}\times{\dfrac{\cancel{22}}{\cancel{7}}}\times{4.2}\times{4.2}\times{4.2} = \dfrac{\cancel{22}}{\cancel{7}}\times{\cancel{36}}\times{h}}$

$\implies\rm{\dfrac{74.08}{3}= 9h}$

$\implies\rm{24.696 = 9h}$

$\implies\rm{h =\dfrac{24.696}{9}}$

$\implies\rm{ h = 2.744m}$

MORE FORMULAES:-

$\boxed{\begin{minipage}{5cm}\\ \\ \sf\bf{Area\:of\:circle= \tt \pi r^2 }\\ \\ \tt\bf{T.S.A\:of\:Cylinder= \tt 2\pi r(r+h) }\\ \\ \rm\bf{LSA of Cylinder= \tt 2\pi rh }\end{minipage}}$

Answer 2

$\sf{\underline{\underline{\red{Question:-}}}}$

A metallic sphere of radius 4.2 cm is melted and recasted as a cylinder of radius 6 cm,find its height.

$\sf{\underline{\underline{\red{Note:-}}}}$

• If a solid is recasted in any other solid then volume will remains same.

So,

→ Volume of sphere = Volume of cylinder

$\sf{\underline{\underline{\red{Given:-}}}}$

• sphere's radius = 4.2cm
• cylinder's radius = 6cm

$\large\sf → \frac{4}{3}πr^3 = 2πr^{2}h$

$\large\sf→ \frac{4}{3}×\frac{22}{7}×4.3^3= 2×\frac{22}{7}×6×h$

$\large\sf→ \frac{98.784}{36}=h$

$\sf→ h = 2.7444$

$\sf{\underline{\underline{\red{Hence:-}}}}$

• hieght = 2.7444cm