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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 tub of radius 12cm contains water to a depth of 20cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75cm. What is the radius of the ball?

ANSWER:

Consider r cm as the radius of ball and R cm as the radius of cylindrical tub

So we get

4/3 πr3 = π R2h

By substituting the values

4/3 × π × r3 = π × 122 × 6.75

On further calculation

r3 = (π × 122 × 6.75)/ 4/3 × π

So we get

r3 = 2916/ 4 = 729

By taking cube root

r = 9cm

**Therefore, the radius of the ball is 9cm.**

ANSWER:

Consider r cm as the radius of ball and R cm as the radius of cylindrical tub

So we get

4/3 πr3 = π R2h

By substituting the values

4/3 × π × r3 = π × 122 × 6.75

On further calculation

r3 = (π × 122 × 6.75)/ 4/3 × π

So we get

r3 = 2916/ 4 = 729

By taking cube root

r = 9cm

**Therefore, the radius of the ball is 9cm.**

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