Question

a solid toy is in the form of a hemisphere surmounted by a cone of height and radius of base are 7cm This toy is completely in a cylindrical vessel containing full of water if the radius of base of the vessel is 7 cm and its height is 20 cm find the volume of water left in the vessel after immersion of the toy( take pie =22 / 7 )​

Answer 1

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

Dimensions of Cone :-

• Radius of cone, r = 7 cm

• Height of cone, h = 7 cm

We know,

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

$\rm \: \: = \: \: \dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 7$

$\rm \: \: = \: \: \dfrac{22}{3} \times 49$

$\rm \: \: = \: \: \dfrac{1078}{3} \: {cm}^{3}$

Hence,

$\: \: \: \: \: \: \red{\bf\implies \:Volume_{(cone)} = \dfrac{1078}{3} \: {cm}^{3}}$

Dimensions of hemisphere :-

• Radius of hemisphere, r = 7 cm

We know,

$\blue{\bf :\longmapsto\:Volume_{(hemisphere)} = \dfrac{2}{3}\pi {r}^{3}}$

$\rm \: \: = \: \: \dfrac{2}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 7$

$\rm \: \: = \: \: \dfrac{44}{3} \times 49$

$\rm \: \: = \: \: \dfrac{2156}{3} \: {cm}^{3}$

Hence,

$\: \: \: \: \: \: \blue{\bf\implies \:Volume_{(hemisphere)} = \dfrac{2156}{3} \: {cm}^{3}}$

Dimensions of Cylinderical Vessel :-

• Radius of cylinderical vessel, r = 7 cm

• Height of cylinderical vessel, H = 20 cm

We know,

$\green{\bf :\longmapsto\:Volume_{(cylinder)} = \pi {r}^{2}H}$

$\rm \: \: = \: \: \dfrac{22}{7} \times 7 \times 7 \times 20$

$\rm \: \: = \: \: 22 \times 7 \times 20$

$\rm \: \: = \: \: 3080 \: {cm}^{3}$

Hence,

$\: \: \: \: \: \: \: \: \: \green{\bf :\longmapsto\:Volume_{(cylinder)} =3080 \: {cm}^{3} }$

☆ Now, this toy having base hemispherical surmounted by a cone is completely submerged in a cylindrical vessel containing full of water, then the water left in the vessel after immersion of the toy is

$\bf :\longmapsto\:Volume_{(water \: left \: in \: cylinderical \: vessel)}$

$\rm \: \: = \: \: Volume_{(cylinder)} - Volume_{(cone)} - Volume_{(hemisphere)}$

$\rm \: \: = \: \: 3080 - \dfrac{2156}{3} - \dfrac{1078}{3}$

$\rm \: \: = \: \: \dfrac{9240 - 1078 - 2156}{3}$

$\rm \: \: = \: \: \dfrac{9240 - 3234}{3}$

$\rm \: \: = \: \: \dfrac{6006}{3}$

$\rm \: \: = \: \: 2002 \: {cm}^{3}$

$\bf :\implies\:Volume_{(water \: left \: in \: cylinderical \: vessel)} = 2002 \: {cm}^{3}$

More information :-

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

Answer 2

Step-by-step explanation:

hope this will help you ☺️