A ferry boat has internal volume 1 m³ and weight 50 kg neglecting the thickness of wood, if a leak develops in the bottom and water starts coming in the fraction of boat will be filled with water before water starts coming in from sides is

Given that mass of boat = 50 kg

Total volume = 1 m³

When a boat is floating in water

Let mass of boat = M

The volume of boat immersed or volume of water displaced by the body = V

Then we have weight of the boat = Weight of water displaced

$\text{Mg}={\text{V}}_{\text{immersed}}{\text{d}}_{\text{w}}\text{g}\to \left(1\right)$

Now  a leak is developed at the bottom of boat and water started coming in

Let the volume of water entered in to the boat $={\text{V}}^{\text{|}}$

Now boat is completely immersed and water is about to enter into the boat.

$\text{Mg}+{\text{V}}^{\text{|}}{\text{d}}_{\text{w}}{\text{g = V}}_{\text{total}\text{volume}}{\text{d}}_{\text{w}}\text{g}\to \left(2\right)$

$\because \text{Mg}={\text{V}}_{\text{immersed}}{\text{d}}_{\text{w}}\text{g}$  

We can have

${\text{V}}_{\text{immersed}\text{while}\text{floating}}{\text{d}}_{\text{w}}\text{g}+{\text{V}}^{\text{|}}{\text{d}}_{\text{w}}\text{g}={\text{V}}_{\text{total}\text{volume}}{\text{d}}_{\text{w}}\text{g}$

$\Rightarrow {\text{V}}_{\text{immersed}\text{while}\text{floating}}+{\text{V}}_{\text{of}\text{water}\text{entered}}={\text{V}}_{\text{total}\text{volume}}$

$\frac{{\text{V}}_{\text{immersed}\text{while}\text{floating}}}{\text{V}}+\frac{{\text{V}}_{\text{of}\text{water}\text{entered}}}{\text{V}}=1$

$\therefore$  fraction of boat filled with water

$\frac{{\text{V}}_{\text{of}\text{water}\text{entered}}}{\text{V}}=1-\frac{{\text{V}}_{\text{immersed}\text{while}\text{floating}}}{\text{V}}\to \left(3\right)$

Using the values

M = 50 kg

Total volume = 1 m³

${\text{d}}_{\text{w}}=1000{\text{kg/m}}^{\text{3}}$

Volume immersed while floating $=\frac{\text{Mg}}{{\text{d}}_{\text{w}}\times \text{g}}=\frac{\text{50}\text{kg}}{\text{1000}{\text{kg/m}}^{\text{3}}}=\frac{\text{1}}{\text{20}}{\text{m}}^{\text{3}}$

On substituting this in (3)

$\frac{{\text{V}}_{\text{of}\text{water}\text{entered}}}{\text{Total}\text{volume}}=1-\frac{1}{20}=\frac{19}{20}$

$\therefore$ The fraction of water entered to the total volume before water coming in from sides is $=\frac{19}{20}$