Ncert Exercises & Solutions

Q. 15 A cubical block of density $ρ$ $ρ$ is floating on the surface of the water. Out of its
height L, fraction x is submerged in water. The vessel is in an elevator
accelerating upward with acceleration a. What is the fraction immersed?

Hint: The pseudo acceleration acts on both the block and the water.

Step 1: Find the initial part of the block submerged in water.

Consider the diagram.

Let the density of water be

ρ_{w}

$ρ_{w}$ and a cubical block of ice of side L be floating in water with x of its height (L) submerged in water.

The volume of the block, (V)=

L^{3}

$L^{3}$

Mass of the block, (m) =

V ρ = L^{3} ρ

$V ρ = L^{3} ρ$

Weight of the block, W = mg =

L^{3} ρ g

$L^{3} ρ g$

Case 1:
The volume of the water displaced by the submerged part of the block =

x L^{2}

$x L^{2}$

∴

$∴$ Weight of the water displaced by the block = x

L^{2} ρ_{w} g

$L^{2} ρ_{w} g$
In floating condition,

Weight of the block = Weight of the water displaced by the block

L^{3} ρ g = x L^{2} ρ_{w} g

$L^{3} ρ g = x L^{2} ρ_{w} g$

\frac{x}{L} = \frac{ρ}{ρ_{w}}

$\frac{x}{L} = \frac{ρ}{ρ_{w}}$

Step 2: Find the final fraction of the block submerged in water.

Case 2:
When the elevator is accelerating upward with an acceleration a, then effective acceleration=(g + a)

(∵

$(∵$ Pseudo force is downward

)

$)$

Then, the weight of the block =m(g + a)

L^{3} ρ (g + a)

$L^{3} ρ (g + a)$

Let x' fraction be submerged in water when the elevator is accelerating upwards.
Now, in the floating condition, the weight of the block = weight of the displaced water

L^{3} ρ (g + a) = (x' L^{2}) ρ_{w} (g + a)

$L^{3} ρ (g + a) = (x' L^{2}) ρ_{w} (g + a)$

\frac{x'}{L} = \frac{ρ}{ρ_{w}}

$\frac{x'}{L} = \frac{ρ}{ρ_{w}}$

From the 1st and 2nd case, we see that the fraction of the block submerged in water is independent of the acceleration of the elevator.

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