Question

A cubical block of wood of side a and density $\rho $ floats in water of density $2\rho $. The lower surface of the cube just touches the free end of a 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 without wetting the weight is:
(a) $a({a^2}\rho g + K)$
(b) $a(a\rho g + 2K)$
(c) $a(\dfrac{{a\rho g}}{2} + 2K)$
(d) $a({a^2}\rho g + \dfrac{K}{2})$

Answer 1

Hint: Use the law of floatation which states that the weight of the object submerged in a liquid is the weight of the liquid displaced. When the system is in rest, the forces cancel out to give net force $0$.

Formula Used:
1. In case of a free body,
Law of floatation: Weight of block = weight of water displaced = Buoyant force acting …… (1)
2. In any case,
Buoyant force = weight of liquid displaced = ${\rho _{fluid}} \times {V_{submerged}} \times g$ …… (2)
3. Spring force $F$: $F = - Kx$ …… (3)
where,
$K$ is the force constant of spring.
$x$ is the compression of spring.
4. Weight of block = \[{W_{Block}} = Volume(V) \times density(\rho ) \times g\]
5. Buoyant force, ${F_{buoyant}} = Volume(V) \times density({\rho _{fluid}}) \times g$

Complete step by step solution:
Given:
1. Side of wooden block = $a$
2. Density of wooden block = $\rho $
3. Force constant of spring= $K$
To find: The weight put over the block so that it is immersed in the water without wetting the weight.
Rough sketch:

Step 1 of 3:
Initially, in case (a), the spring is in a natural state as the block just touches it.
Let the height of the cube be submerged in water by $x$. Find $x$ using eq (1) and equation (2):
The weight of the block will be equal to mg and the mass of the block is density times volume.
The buoyant force will be equal to the weight of the fluid displaced.
${W_{\text{block}}} = {F_{\text{buoyant}}} $
$ {a^3}\rho g = x \times {a^2} \times (2\rho )g $
$ x = \dfrac{a}{2}$ (used values of buoyant forces and weight) …… eq(4)
where $x$ is the displacement of the water in the upward direction.

Step 2 of 3:
When the block is completely immersed (case (b)), the spring is compressed by:
Using equation (4), we get $a - x = \dfrac{a}{2}$=compressed length of spring.

Step 3 of 3:
As the system is at rest, the net force in the vertical direction is 0.
Net force in downward direction = net force in the upward direction
(Force due to gravity from the block) + (Force due to gravity from the extra weight)
= (Buoyant force on the block) + (Force due to spring)
$ {W_{block}} + W = {F_{buoyant}} + {F_{spring}}$
$ \Rightarrow {a^3}\rho g + W = {a^3}(2\rho )g + K(\dfrac{a}{2})$
$ \Rightarrow W = a({a^2}\rho g + \dfrac{K}{2}) $

The weight W put over the block so that it is completely immersed in water without wetting the weight is: (D) $a({a^2}\rho g + \dfrac{K}{2})$.

Note:
Buoyant force by liquid plays an important role in any suspended body inside the liquid. It is the only force that acts upward by the liquid. Therefore, whenever there is a problem like these, do consider buoyant forces.