Question

The base radius and height of a cylindrical block of wood are 15cm and 40cm. what is the volume of the largest cone that can be carved out of this?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• Radius of cylindrical block, r = 15 cm

and

• Height of cylindrical block, h = 40 cm

According to statement,

• The cone of maximum volume is carved out from this cylindrical block.

Therefore,

$\rm :\longmapsto\:Radius_{(cone)} = Radius_{(cylinder)} = 15 \: cm$

$\rm :\longmapsto\:Height_{(cone)} = Height_{(cylinder)} = 40 \: cm$

Therefore,

Dimensions of cone are

• Radius of cone, r = 15 cm

• Height of cone, h = 40 cm

We know that

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

$\rm :\longmapsto\:Volume_{(cone)} \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: \times \: {15}^{2} \: \times 40$

$\bf\implies \:Volume_{(cone)} = \dfrac{66000}{7} \: {cm}^{3}$

Additional Information :-

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length²+breadth²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²