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A wooden block with a coin placed on its top

Question :- A wooden block with a coin placed on its top, floats in water as shown in the figure. The distance $l$ and $h$ are shown. After some time, the coin falls into the water. Then,

(A) $l$ decreases and $h$ increases

(B) $l$ increases and $h$ decreases

(D) Both $l$ and $h$ decrease

Solution :-

When the coin slips into the water, the wooden block moves up and $l$ decreases, because weight of water displaced by block alone is lesser than weight of water displaced by (block + coin).

When the coin was floating, it displaced water equal to its own weight, so

weight of the water displaced by the coin = weight of coin

⇒ Volume of water displaced × density of water × g = Volume of coin × density of coin × g

$\displaystyle \Rightarrow {{\left( {{V}_{water}} \right)}_{1}}{{\rho }_{water}}g={{V}_{coin}}{{\rho }_{coin}}g$

$\displaystyle \Rightarrow {{\left( {{V}_{water}} \right)}_{1}}=\frac{{{\rho }_{coin}}}{{{\rho }_{water}}}{{V}_{coin}}$ …..(1)

Now when the coin is inside the water, it displaces water equal to its own volume, so

$\displaystyle {{\left( {{V}_{water}} \right)}_{2}}={{V}_{coin}}$ …..(2)

Now as $\displaystyle {{\rho }_{coin}}>{{\rho }_{water}}$ , so from equations (1) and (2), we find that,

$\displaystyle {{\left( {{V}_{water}} \right)}_{2}}<{{\left( {{V}_{water}} \right)}_{1}}$

It means as the cons falls into the water it displaces less water in comparison to when it was floating on the wooden block.

Hence, $h$ decreases when coin falls into the water.

The correct option is (D) Both $l$ and $h$ decrease.

A wooden block with a coin placed on its top

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