Q. A cubical block of wooden edge I and a density $\rho$ floats in water of density$2\rho$. The lower surface of 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) $a(l\ \rho\ g + k)$
B) $a(l^{2}\ \rho\ g + k)$
C) $a\left( \frac{l\ \rho\ g}{2} + 2k \right)$
D) $a\left( l^{2}\ \rho\ g + \frac{k}{2} \right)$
Explanation
(d) Initially the position of wooden block is as shown in figure. Since, the density of block is half than that of water, hence half of its volume is immersed in water
[image:10_image94.jpeg]
When weight $w$ is put on the block, the remaining half of the volume of block is immersed in water, figure (b). Therefore, $w =$additional upthrust + spring force
$= l \times l \times \frac{1}{2} \times 2\rho \times g + k\left( \frac{1}{2} \right) = l\left( l^{2}\rho\ g + \frac{k}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.