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# Suppose a cubical block of wood weighing $200g$ has a lead fastened underneath. What will be the mass of the lead which will just allow the block to float in water's gravity of wood is $0.8$ and that of lead is $11.3$.

Last updated date: 17th Jul 2024
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Hint: In order to answer this question we first calculate the volume of all substances and then the volume of water displacement by the volume of lead and wooden block. Specific gravity of water $=1$.

Given, weight of wooden block $= 200g$
Let us assume that m is the mass of lead,
So, mass of displaced water $= m+200$
Again , we know , the volume of any object is the ratio of its mass and its specific gravity .
$\therefore {V_{lead}}\Rightarrow \dfrac{m}{11.3}$
$\therefore {V_{wooden block}} \Rightarrow \dfrac{200}{0.8}$
$\therefore {V_{water}} \Rightarrow \dfrac{m+200}{1}$ [specific gravity of water is $=1$]
Therefore the volume of water displacement $\Rightarrow$volume of lead $+$volume of block
$\Rightarrow m+200=\dfrac{m}{11.3}+\dfrac{200}{0.8}$
$\Rightarrow m-\dfrac{m}{11.3}=250-200$
$\Rightarrow m\left( \dfrac{10.3}{11.3} \right)=50$
$\Rightarrow m=54.89$
Therefore the mass of the lead is $54.89g$.

Note: The unit of force in this problem is taken as (gm) not the (N) or (dyne) because at the time of calculation ‘g’ is canceled out. If the specific gravity of any substance is less than the fluid , then it will float on that liquid.