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Rs aggarwal solutions

CHAPTERS

3. Factorization of Polynomials

4. Linear Equations In Two Variables

6. Introduction To Euclids Geometry

9. Congruence Of Triangles And Inequalities In A Triangle

11. Areas Of Parallelograms And Triangles

14. Areas Of Triangles And Quadrilaterals

15. Volume And Surface Area Of Solids

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.

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.**

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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