Question

Question #11327

Answer 1

Let the mass of the boat floating in the tank be `m_b`.

and it carries total `N` steel balls each of mass `m`.

Hence total mass of boat + ball = `m_b+Nm`.

So by the condition of floatation we can say that initially the mass of displaced water will be `m_b+Nm`

So initially the volume of water displaced will be

`V_w=(m_b+Nm)/d_w...[1]`, where `d_w->"density of water"`

When `n` balls are thrown into the water then it will carry `(N-n)` balls .

At this stage the volume of water displaced by the floating boat carrying `(N-n)` balls will be `(m_b+(N-n)m)/d_w`

The volume water displaced by `n` balls thrown in to the tank will be `(nm)/d_s`, where `d_s` represents the density of steel.

Hence the total volume of water displaced in the 2nd case will be
`V'_w=(m_b+(N-n)m)/d_w+(nm)/d_s.....[2]`

Now subtracting [2] from [1] we get

`V_w-V'_w=(m_b+Nm)/d_w-(m_b+(N-n)m)/d_w-(nm)/d_s`

`V_w-V'_w=(nm)/d_w-(nm)/d_s=nm(1/d_w-1/d_s)....[3]`

Now density of steel `d_s` is much greater than that of water
`d_w`

So

`d_s>d_w`

`=>1/d_s<1/d_w`

`=>1/d_w-1/d_s>0.....[4]`

Hence from relation [3] an [4] we get

`V_w-V'_w=nm(1/d_w-1/d_s)>0.....[5]`

This Eq[5] reveals that the difference (decrease) in volume of displaced water continuously increases with the increase in number of steel balls `(n)` thrown into the tank.

This means that the level of water in the tank will gradually fall with increase number of balls `n` thrown into the tank.