Question

$N$ similar slabs of cubical shape of edge $b$ are lying on ground. Density of material of slab is $\rho$. Work done to arrange them one over the other is

• A. $(N^2-1)b^3\rho g$
• B. $(N-1)b^4\rho g$
• C. $\frac{1}{2}(N^2-N)b^4\rho g$
• D. $(N^2-N)b^4\rho g$

Answer 1

Answer: (C)

$\frac{1}{2}(N^2-N)b^4\rho g$

Position of Center of Gravity of first slab is at a height $=\frac{b}{2}$ from the bottom edge of the slab.
Weight of each stab $=\text{volume}\times \text{density}\times g$$=b^3\rho g$

Now, for the column of N no. of slabs, the CG will be at a height $

Center of gravity of column of slabs $=\frac{TotalheightofNslabs}{2}$ $=\frac{Nb}{2}$ from the ground.

So, $P{E}_{final}=$ $\text{Total weight of N slabs}\times \text{height of center of mass}$ $=N\times b^3\rho g\times \frac{Nb}{2}=\frac{N^2{b}^4\rho g}{2}$

And, PE of all the slabs staying on ground, before arranging those $=P{E}_{initial}=N{b}^3\rho g\times \frac{b}{2}=\frac{N{b}^4\rho g}{2}$

So, change in potential energy, $\Delta PE=P{E}_{final}-P{E}_{initial}=Workdone=\frac{N^2{b}^4\rho g}{2}-\frac{N{b}^4\rho g}{2}=\frac{1}{2}(N^2-N)b^4\rho g$