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Volume And Surface Area Of Solids | Volume And Surface Area Of Solids Exercise 15D

Question 21

A cylindrical bucket with base radius 15cm is filled with water up to a height of 20cm. A heavy iron spherical ball of radius 9cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.

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ANSWER:

It is given that

Radius of the cylindrical bucket = 15cm

Height of the cylindrical bucket = 2cm

We know that

Volume of water in bucket = π r2h

By substituting the values

Volume of water in bucket = (22/7) × 152 × 20

So we get

Volume of water in bucket = 14142.8571 cm3

It is given that

Radius of spherical ball = 9cm

We know that

Volume of spherical ball = 4/3 πr3

By substituting the values

Volume of spherical ball = 4/3 × (22/7) × 93

So we get

Volume of spherical ball = 3054.8571 cm3 ……. (1)

Consider h cm as the increase in water level

So we get

Volume of increased water level = π r2h

By substituting the values

Volume of increased water level = (22/7) × 152 × h …… (2)

By equating both the equations

3054.8571 = (22/7) × 152 × h

On further calculation

h = 3054.8571/ ((22/7) × 152) = 4.32 cm

Therefore, the increase in the level of water is 4.32cm.

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 21

A cylindrical bucket with base radius 15cm is filled with water up to a height of 20cm. A heavy iron spherical ball of radius 9cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

ANSWER:

It is given that

Radius of the cylindrical bucket = 15cm

Height of the cylindrical bucket = 2cm

We know that

Volume of water in bucket = π r2h

By substituting the values

Volume of water in bucket = (22/7) × 152 × 20

So we get

Volume of water in bucket = 14142.8571 cm3

It is given that

Radius of spherical ball = 9cm

We know that

Volume of spherical ball = 4/3 πr3

By substituting the values

Volume of spherical ball = 4/3 × (22/7) × 93

So we get

Volume of spherical ball = 3054.8571 cm3 ……. (1)

Consider h cm as the increase in water level

So we get

Volume of increased water level = π r2h

By substituting the values

Volume of increased water level = (22/7) × 152 × h …… (2)

By equating both the equations

3054.8571 = (22/7) × 152 × h

On further calculation

h = 3054.8571/ ((22/7) × 152) = 4.32 cm

Therefore, the increase in the level of water is 4.32cm.

Our top 5% students will be awarded a special scholarship to Lido.

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