Question

A cubical block of wood of edge 3 cm floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting the new weight. Density of wood $=800{\text{kg/m}}^3$ and spring constant of the spring $=50\text{N/m}.(\text{Take g}=10{\text{m/s}}^2)$

• A. $0.268\text{N}$
• B. $0.354\text{N}$
• C. $0.569\text{N}$
• D. $0.486\text{N}$

Answer 1

The correct option is B $0.354\text{N}$

Fraction of volume immersed before putting the new weight $=\frac{{\rho }_{\text{block}}}{{\rho }_{\text{water}}}$
$=\frac{800}{1000}=0.8$
i.e., 20% of $3\text{cm}$ or $0.6\text{cm}$ is above water. Let $W$ be the new weight, then spring will be compressed by $0.6\text{cm}$
$\therefore W$ + weight of block = Upthrust on whole volume of block + spring force
or
$W=(3\times {10}^{-2}{)}^3\times 1000\times 10+50\times (0.6\times {10}^{-2})-(3\times {10}^{-2}{)}^3\times 800\times 10$
$\therefore W=0.354\text{N}$