Question

A cubical block of plastic 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 it. Density of plastic = 800 kg / $m^3$ and spring constant of the spring = 100 N/m. Take g = 10 $m/s^2$.

• A. $0.65N$
• B. $0.59N$
• C. $6.5N$
• D. $5.5N$

Answer 1

The correct option is A $0.65N$

The specific gravity of the block $=$ 0.8. Hence the height inside waterA$=$3 cm $\times$ 0.8 $=$ 2.4 cm. The height outside water $=$3 cm - 2.4$=$ 0.6 cm. Suppose the maximum weight that can be put without wetting it is W. The block in this case is completely immersed in the water. The volume of the displaced water
$=$ volume of the block $=$ 27 $\times {10}^{-6}{m}^3$.
Hence, the force of buoyancy
$=$ $(27\times {10}^{-6}{m}^3)\times 1\left(1000kg/m^3\right)\times \left(10m/s^2\right)=0.27N$.
The spring is compressed by 0.6 cm and hence the upward force exerted by the spring$=$ 100 N/m $\times$ 0.6 cm $=$ 0.6 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. The weight of the block is
$W^{\prime }=(27\times {10}^{-6}m)\times \left(800kg/m^3\right)\times \left(10m/s^2\right)$ $=$ 0.22 N
Thus, W $=$ 0.27 N + 0.6 N - 0.22 N$=$ 0.65 N