Question

The lower surface of the cube floating in water just touches the free end of a vertical spring fixed at the bottom of the vessel. The maximum weight that can be put on the block without wetting it is-

• A. $0.20\text{N}$
• B. $0.30\text{N}$
• C. $0.35\text{N}$
• D. $0.25\text{N}$

Answer 1

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

Let the mass has kept on the block be $m$ .


FBDs of block in initial and final cases:


From initial case :
On applying equilibrium condition along vertical direction,
$\Rightarrow F_b-Mg=0$
$\Rightarrow F_b=Mg$
$\Rightarrow \rho gV={\rho }^{\prime }{V}^{\prime }g$
where $V\to \text{Volume of water displaced};V^{\prime }\to \text{Volume of cube}$
$\Rightarrow 1000\times 10\times (0.03{)}^2\times x=800\times (0.03{)}^3\times 10$
$\Rightarrow x=0.024\text{m}..\left(1\right)$

From final case:
Block will be completely immersed and mass kept over it will not get wet. i.e depth of immersion will be $3\text{cm}$
$\therefore$ compression in spring = Change in depth of immersion
$\Rightarrow y=0.03-0.024\text{m}$
$\therefore y=0.006\text{m}$
On applying equilibrium condition along vertical direction,
$F_s+F_b^{\prime }-(M+m)g=0$
$\Rightarrow ky+\rho g{V}^{\prime }-({\rho }^{\prime }{V}^{\prime }g+mg)=0$
$\Rightarrow (50\times 0.006)+(1000\times 10\times (0.03{)}^3)=(800\times (0.03{)}^3+m)\times 10$
$\Rightarrow 0.57=(0.0216+m)\times 10$
$\Rightarrow m=0.0354\text{kg}$
Now, weight, $W=mg=0.0354\times 10$
$\therefore W\simeq 0.35\text{N}$
Maximum weight that can be put on block without wetting it is $0.35\text{N}$.