Question 230: Properties of Matter

Question

A tub is filled with water and a wooden cube $10\ 	ext{cm} 	imes 10\ 	ext{cm} 	imes 10\ 	ext{cm}$ is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by $3.87$ cm. The mass of the metal coin is ________ gram.
(Take water density as $1\ 	ext{g/cm}^3$ and density of wood as $0.4\ 	ext{g/cm}^3$)

Accepted Answer

A soap bubble has two surfaces (inner and outer). The total surface energy $(U)$ is given by :
$$ U=2 	imes S 	imes A=2 	imes S 	imes\left(4 \pi R^2ight) $$
where $S$ is the surface tension and $A$ is the surface area.
The work done ( $W$ ) to increase the size is equal to the change in surface energy :
$$ \mathrm{W}=\Delta \mathrm{U}=8 \pi \mathrm{~S}\left(\mathrm{R}_2^2-\mathrm{R}_1^2ight) $$
$$ \mathrm{W}=8 \pi \mathrm{~S}\left[\left(\frac{\mathrm{D}_2}{2}ight)^2-\left(\frac{\mathrm{D}_1}{2}ight)^2ight]=2 \pi \mathrm{~S}\left(\mathrm{D}_2^2-\mathrm{D}_1^2ight) $$
The given values are,
Surface tension $\mathrm{N}=0.04 \mathrm{~N} / \mathrm{m}$
Initial diameter $D_1=7 \mathrm{~cm}=0.07 \mathrm{~m}$
Final diameter $\mathrm{D}_2=14 \mathrm{~cm}=0.14 \mathrm{~m}$
So, the work done to increase the size of bubble is,
$$ \mathrm{W}=2 	imes \frac{22}{7} 	imes 0.04 	imes\left[(0.14)^2-(0.07)^2ight] $$
$$\Rightarrow $$ $\mathrm{W}=\frac{44}{7} 	imes 0.04 	imes[0.0196-0.0049]$
$$ \mathrm{W}=0.003696 \mathrm{~J} $$
Now, as we know $1 \mathrm{~J}=10^6 \mu \mathrm{~J}$,
$$ \mathrm{W}=0.003696 	imes 10^6 \mu \mathrm{~J}=3696 \mu \mathrm{~J} $$
$\Rightarrow 15000-x=3696$
$$\Rightarrow $$ $x=15000-3696$
$$\Rightarrow $$ $$ \mathrm{x}=11304 $$
Therefore, the value of x is 11304.
Hence, the correct answer is $\mathbf{1 1 3 0 4}$.

Properties of Matter

Integer / Numeric Answer Type