Q136 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Physics
A cubical block of wood having mass of $160 \mathrm{~g}$ has a metal piece fastened underneath as shown in the figure. Find the maximum mass of the metal piece which will allow the block to float in water. Specific gravity of wood is 0.8 and that metal is 10 and density of water $=1 \mathrm{~g} / \mathrm{cc}$.
- A$55.5 \mathrm{~g}$
- Correct answerB$44.4 \mathrm{~g}$
- C$33.3 \mathrm{~g}$
- D$66.6 \mathrm{~g}$
Solution
Volume of wood block $=\frac{\text { Mass }}{\text { Density }}=\frac{160}{0.8 \times 1}$
$=\frac{1600}{8}=200 \mathrm{~cm}^3$
Let mass of metal piece is $x \mathrm{~g}$.
$\text { Volume of metal piece }=\frac{\text { Mass }}{\text { Density }}=\frac{x}{10 \times 1}=\frac{x}{10} \mathrm{~cm}^3$
Total volume of fluid displaced
$=\left(200+\frac{x}{10}\right) \times 10^{-6} \mathrm{~m}^3$
Weight of fluid displaced $=V \times \rho_f \times g$
$=\left(200+\frac{x}{10}\right) 10^{-6} \times 1000 \times 10$
$=\left(200+\frac{x}{10}\right) 10^{-2} \mathrm{~N}$
If block floats in water, then weight $=$ upthrust
$\Rightarrow(160+x) 10 \times 10^{-3}=\left(200+\frac{x}{10}\right) \times 10^{-2}$
$160+x=200+\frac{x}{10}$
$\frac{9 x}{10}=40$
$x=\frac{400}{9}=44.4 \mathrm{~g}$
Hence, the correct answer is option: 'B'.