Question

10.A cubical block of wood edge 'a' and density 'p' floats in water of density 2P.The Lower surface of the cube just touches the free end of massless spring of force constant K fixed at the bottom of the vessel.The weight W put over the block so that it is completely immersed in water with out wetting the weight is

Answer 1

Final Answer : $\rho a^3g + k\frac{a}{2}$

Steps:
1) Firstly,
Let 'y' be length up-to which cube is immersed initially.
Then, we can clearly say
$\rho a^3 g = 2\rho * a^2 * y \\ \\ => y = \frac{a}{2}$

2) Now,
After putting weight on block,due to compression in spring of 'a/2' units.
we get,
$W + \rho a^3 g = 2\rho a^3 g +k \frac{a}{2}\\ \\ => W = \rho a^3 g + k\frac{a}{2}$

Answer 2

Answer:

Hope you will understand through this image ☺☺

Explanation: