Question

A ball of uniform density 2/3rd of that of water is dropped freely into a pond from a height 10m above its surface. The maximum depth of a ball can travel in water is-

(1) 21 m
(2) 10 m
(3) 20 m
(4) 30 m

Answer 1

Answer:

s= 20 m

Explanation:

Buoyant force is given as;

Fb = Vpg;

Where Fb is the force acting on fluid, V is the volume of object immersed, p is the density of fluid and g is the gravitational.

Now volume is the product of mass and density;

V = m / d

Fb = m/d x p x g

Fb = 3/2 mpg

As p = 1 gm/cm^3 for water

So; Fb is 3/2 mg

Resultanat force acting on ball = Fb – Mg

                    = 3/2mg – mg

                    = mg/2 (upward)

Now force = m x a;

Mg/2 = m x a;

A = -g/2  

Now using the formulas;

U = √2gh

U^2 = 2gh = 2 (10) (10)

        = 200

Now;

V^2 – U^2 = 2 as;

Here V=0;

200=2 (10/2)s

So s= 20 m

Hence the maximum depth of a ball can travel in water is 20m

Answer 2

Answer:

The maximum depth of a ball can travel in water is 20m (s= 20 m)

Explanation:

Buoyant force is measured as:

Fb = Vpg;

wherein Fb =force that acts on the liquid;

V = Volume of the object that is immersed in the liquid;

p= Density of the liquid;

g= gravitational force

Volume = mass x density

V = m / d

Therefore, converting in above equation;

Fb = m/d x p x g

It is stated that; Fb = 3/2 mpg

Since, p = 1 gm/cm^3 for water

and Fb is 3/2 mg

The resulting force that acts on the ball is:

                   = Fb – Mg

                   = 3/2mg – mg

                   = mg/2 (upward)

Force = m x a;

As determined above;

Mg/2 = m x a;

A = -g/2  

Based on the formula used for potential energy;

U = √2gh

U^2 = 2gh = 2 (10) (10)

       = 200

Then;

V^2 – U^2 = 2 as;

Here V=0;

200=2 (10/2)s

So s= 20 m

Thus, the maximal depth to which the ball travels in 20 m

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