Question

1. A rubber ball of mass 100 g and radius 5 cm is sub-
merged in water to a depth of 1 m and released. To what height will the ball jump up above the surface of water? ​

Answer 1

ANSWER:-

let the rubber ball reach to the height h above the surface of water. when the ball is taken to the depth of 1 m below the water surface the gravitational potential energy decrease by an amount equal to

mgh = 10^-1 × 10× 1

= 1 J

the work done against the buoyant force is

W = buoyant force × displacement

V ρ gh = 4/3 πr^3 ρgh

where ρ is the density of water

W = 4/3 × (22/7) × ( 5×10^-2)^3. × 1000× 10

= 5.24 J

total energy E = 5.24 - 1

= 4.24 J

E is the potential energy of ball above the water surface.

E = mgh

= 0.1 × 10 × h

0.1 × 10 × h = 4.24

h = 4.24 m

the height reached by the ball above the water surface is 4.24 m

hope it will help you....

Answer 2

Answer:

h=4.24 .........

Explanation:

look above for explanation