Question

150 spherical marbles, each of radius 1.4 cm, are dropped in a cylinderical vessel fo radius 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vassel.

Answer 1

Given:-

• 150 spherical marbles, each of radius 1.4 cm, are dropped in a cylinderical vessel fo radius 7 cm containing some water, which are completely immersed in water.

To find:-

• The rise in the level of water in the vessel.

Solution:-

Here,

• Diameter of spherical ball = 1.4 cm
• Radius of spherical ball = 0.7 cm
• Diameter of cylinder = 7 cm
• Radius of cylinder = 3.5 cm
• No. of spherical balls = 150

Let,

• the rise in water be h.

According to the question,

→ 150 × [4/3 πr³] = πr²h

→ 150 × 4/3 × π × 0.7 × 0.7 × 0.7 = π × 3.5 × 3.5 × h

→ h = 150 × 4/3 × [0.7 × 0.7 × 0.7/3.5 × 3.5]

→ h = 28/5

→ h = 5.6 cm

Hence,

• Rise of water level is 5.6 cm.

Answer 2

Given,

• 150 spherical marbles, each of radius 1.4 cm.
• They are dropped in a cylinderical vessel fo radius 7 cm containing some water.

To Find,

• The rise in the level of water in the vassel.

Solution,

The Diameter of Spherical Balls

= 1.4cm •••(Given)

So,

The Radius Of Spherical Balls (r) = 1.4cm/2

= 0.7cm

The Radius Of Cylinderical vessel (R)

= 7cm •••(Given)

Let's,

The Rise Of Water = X

Volume of All Spheres

Volume of All Spheres= Volume Of Cylinderical vessel

$:\implies 150 \times \frac{4}{3} \pi {r}^{3} = \pi {R}^{2} h \\ \\ :\implies 150 \times \frac{4}{3} \pi {(0.7cm)}^{3} = \pi {(7cm)}^{2} x \\ \\ :\implies 150 \times \frac{4}{3} \times \pi \times {( \frac{7}{10} cm)}^{3} = \pi \times 49 {cm}^{2} \times x \\ \\:\implies 200 \times \frac{343}{1000} {cm}^{3} = 49 {cm}^{2} \times x \\ \\ :\implies \frac{343 {cm}^{3} }{5} \times \frac{1}{49 {cm}^{2} } = x \\ \\ :\implies \frac{7}{5} cm = x$

Required Answer,

The Rise Of Water level 7/5cm.