Question

A ball whose density is $0.4\times {10}^{3}kg/m^3$ falls in to water fro a height of 9 cm the depth the ball will sink is $n\times 3$. Then n= _____

Answer 1

A ball whose density 0.4 × 10³ kg/m³ falls into water . Before falling into the water, velocity of ball at the surface of water , u = √{2gh}

Now, use formula for finding acceleration of ball,

a = apparent weight/mass of ball

Apparent weight = V(ρ - σ)g

Where , V is the volume of ball , ρ is density of ball and σ is density of water.

Now, apprent weight = V(0.4 - 1) × 10³ × g = -0.6gV × 10³

∴ a = -0.6gV × 10³/ρV

= -0.6g/0.4 = -3g/2 m/s²

Use formula, v² = u² + 2aS

⇒0 = {√{2gh})² + 2(-3g/2)S

⇒2gh = 3gS

⇒2 × 9 = 3S ⇒S = 6cm