Question

In a cylindrical vessel of radius 10 cm, containing some water, 9000 small
spherical balls are dropped which are completely immersed in Water
which raises the water level. If each spherical ball is of radius 0.5 cm,
then find the rise in the level of water in the vessel.​

Answer 1

Answer:

HOPE IT'S HELPFUL......

Answer 2

The rise in water level in vessel is 14.9 cm

Step-by-step explanation:

Given as :

The radius of cylindrical vessel = r = 10 cm

The height of vessel = h cm

The vessel  containing water , 9000 spherical balls dropped which raises the water level .

The radius of spherical ball = R = 0.5 cm

The height of water rise = height of vessel = h cm

According to question

The volume of spherical ball = $\dfrac{4}{3}$ × π × radius³

Or, V =  $\dfrac{4}{3}$ × 3.14 × (0.5 cm)³

∴    V = 0.523  cm³

Again

volume of cylindrical vessel = π × radius² × h

Or,  v = 3.14 × (10 cm)² × h

Or, v = 314 h

Now,

number of spherical ball dropped = $\dfrac{volume of cylinder}{volume of sphere}$

i.e  9000 = $\dfrac{314 h}{0.523}$

Or, 314 h = 0.523 × 9000

or, 314 h = 4707

∴       h = $\dfrac{4707}{314}$

i.e     h = 14.9

So, The rise in water level = h = 14.9 cm

Hence, The rise in water level in vessel is 14.9 cm Answer