Question

A sphere of diameter 6cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water by how much will the level of water rise in the cylindrical vessel?

Answer 1

HELLO DEAR,

the diameter of sphere = 6 cm

=> radius of sphere = R = 6/2 = 3 cm

Similarly,

diameter of cylindrical vessel = 12 cm

=> radius of cylindrical vessel = r = 12/2 = 6 cm

Now,

volume of sphere=and volume of cylinder

.

$\frac{4}{3} \pi {r}^{3} = \pi \times {r}^{2} h \\ = > h = \frac{4 \times {3}^{3} }{3 \times 6 \times 6} = 1cm$
THE WATER LEVEL RISESE TO 1CM

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

1 ) diameter of the sphere ( d ) = 6cm

    radius of the sphere ( r ) = d/2 = 6/2 = 3cm

2 ) diameter of the cylindrical vessel ( D ) = 12 cm

radius of the cylindrical vessel ( R ) = 12/2 = 6 cm

level of water raised = h cm

according to the problem given ,

volume of the water raised in the vessel = volume of the sphere

πR²h = 4/3 × πr³

h = ( 4 × r³ ) / ( 3 × R² )

h = ( 4 × 3 × 3 × 3 ) / ( 3 × 6 × 6 )

h = 1 cm