Question

A solid is in the form of a cone of vertical height h mounted on the top base of a right circular cylinder of height 1/3 h. The circumference of the base of the cone and that of the cylinder are both equal to C. If V be the volume of the solid, prove that
C = 4√(ЗπV/7h)​

Answer 1

Correct Statement :-

A solid is in the form of a cone of vertical height h mounted on the top base of a right circular cylinder of height 1/3 h. The circumference of the base of the cone and that of the cylinder are both equal to C. If V be the volume of the solid, prove that C = 4√(ЗπV/8h)

Solution :-

Let radius of the base of cylinder and cone be 'r' units.

Since,

It is given that

• Circumference of base is C,

Therefore,

$\rm :\longmapsto\:C = 2\pi \: r$

$\bf\implies \:r = \dfrac{C}{2\pi} - - - (1)$

Dimensions of cone,

• Radius of cone = r

• Height of cone = h

Thus,

$\rm :\longmapsto\:Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2}h - - - (2)$

Dimensions of cylinder,

• Radius of cylinder = r

• Height of cylinder = h/3

Thus,

$\rm :\longmapsto\:Volume_{(cylinder)} = \pi \: {r}^{2} \dfrac{h}{3}$

$\rm :\longmapsto\:Volume_{(cylinder)} =\dfrac{1}{3} \pi \: {r}^{2} h - - - - (3)$

Now,

• Total volume of solid,

$\rm :\longmapsto\:Volume_{(solid)} = Volume_{(cone)} + Volume_{(cylinder)}$

$\rm :\longmapsto\:V =\dfrac{1}{3}\pi \: {r}^{2}h + \dfrac{1}{3}\pi \: {r}^{2}h$

$\rm :\longmapsto\:V =\dfrac{2}{3}\pi \: {r}^{2}h$

$\rm :\longmapsto\:V =\dfrac{2}{3}\pi \: {\bigg(\dfrac{C}{2\pi}\bigg) }^{2}h$

$\rm :\longmapsto\:3V =2 \pi \: \dfrac{ {C}^{2} }{ {4\pi}^{2} } h$

$\rm :\longmapsto\:3V = \: \dfrac{ {C}^{2} }{ {2\pi}} h$

$\rm :\longmapsto\: {C}^{2} = \dfrac{6\pi \: V}{h}$

$\rm :\longmapsto\:C = \sqrt{\dfrac{6\pi \: V}{h} }$

can re written as,

$\rm :\longmapsto\:C = \sqrt{\dfrac{6\pi \: V \times 8}{h \times 8} }$

$\rm :\longmapsto\:C = \sqrt{\dfrac{48\pi \: V}{8h} }$

$\rm :\longmapsto\:C = 4\sqrt{\dfrac{3\pi \: V}{8h} }$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(cylinder)} = \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(sphere)} = \dfrac{4}{3} \pi \: {r}^{3} }}$

$\boxed{ \sf{ \: Volume_{(hemi - sphere)} = \dfrac{2}{3} \pi \: {r}^{3} }}$

$\boxed{ \sf{ \: Volume_{(frustum)} = \dfrac{\pi \: h}{3}({r}^{2} + {R}^{2} + rR)}}$