Question

A cubical block of wood of edge $3\text{cm}$ 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 it.
(Take density of wood $=800{\text{kg/m}}^3$, spring constant of the spring $=50\text{N/m}$ and $g=10{\text{m/s}}^2$.)

• A. $0.35\text{N}$
• B. $0.6\text{N}$
• C. $0.27\text{N}$
• D. $0.78\text{N}$

Answer 1

The correct option is A $0.35\text{N}$


The specific gravity of the block $=0.8$. Hence the height of cubical block inside water $=3\text{cm}\times 0.8=2.4\text{cm}$. The height outside water $=3\text{cm}-2.4=0.6\text{cm}$.
Suppose, the maximum weight that can be put without wetting it, is $W$. The cubical block in this case is completely immersed in the water.
So, the volume of the displaced water
$=$ volume of the cubical block $=27\times {10}^{-6}{\text{m}}^3$

Hence, the force of buoyancy
$=V{\rho }_{w}g=(27\times {10}^{-6}{\text{m}}^3)\times \left(1000{\text{kg/m}}^3\right)\times 10{\text{m/s}}^2$
$=0.27\text{N}$
The spring is compressed by $0.6\text{cm}$ and hence the upward force exerted by the spring
$=50\text{N/m}\times 0.006\text{m}=0.3\text{N}$
The force of buoyancy and the spring force taken together balance the weight of the block plus the weight $W$ put on the block.
$\therefore$ The weight of the block is
$W^{\prime }=(27\times {10}^{-6}\text{m})\times \left(800{\text{kg/m}}^3\right)\times 10{\text{m/s}}^2$
$=0.216\text{N}$
Thus, weight $W=0.27\text{N}+0.3\text{N}-0.216\text{N}$
$=0.35\text{N}$