Question

From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed, then the volume of remaining solid is (a) 280pie cm3

(c) 240 pie cm3
(b) 330 pie cm3

(d) 140 pie cm3​

Answer 1

Here,

☆ Height of solid cylinder, h = 10 cm

☆ Radius of solid cylinder, r = 6 cm

☆ So, Volume of cylinder is

$\tt \longrightarrow \: V_{(Cylinder)} \: = \pi \: {r}^{2} h$

$\tt \longrightarrow \: V_{(Cylinder)} = \pi \: {(6)}^{2} \times 10$

$\tt\implies \: \boxed{ \purple{ \tt \: V_{(Cylinder)} = 360 \: \pi \: {cm}^{3}}} - - (i)$

Now, cone of same base and height is removed from solid cylinder.

So,

☆ Height of cone, h = 10 cm

☆ Radius of cone, r = 6 cm

☆ So, Volume of cone is

$\tt \longrightarrow \: V_{(Cone) } = \dfrac{1}{3} \pi \: {r}^{2} h$

$\tt \longrightarrow \: V_{(Cone) } = \dfrac{1}{3} \times \pi \: \times {(6)}^{2} \times 10$

$\tt \longrightarrow \: V_{(Cone) } = \dfrac{1}{3} \times \: \pi \times 6 \times 6 \times 10$

$\tt\implies \: \boxed{ \purple{ \tt \: V_{(Cone)} = 120 \: \pi \: {cm}^{3} }} - - (ii)$

$\red{ \bf \: Now, }$

$\boxed{ \purple{ \tt \: V_{(remaning \: solid)} \: = V_{(Cylinder)} - V_{(Cone)} }}$

$\tt \longrightarrow \: V_{(remaning \: solid)} = 360\pi \: - \: 120\pi$

$\tt\implies \: \boxed{ \green{ \bf \: V_{(remaning \: solid)} = 240 \: \pi \: {cm}^{3} }}$

$\large{\boxed{\boxed{\bf{Hence, option \: (c) \: is \: correct}}}}$