Question

$N$ similar slabs of cubical shape of edge $b$ are lying on the ground. Density of material of the slab is $\rho$. Work done to arrange them one over the other is : (Given acceleration due to gravity $=g$)

• 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)

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

Center of Gravity of first slab $=\frac{b}{2}$
Weight of each stab
$=\text{volume}\times \text{density}\times g$

$=b^3\rho g$

$\text{Center of Gravity of column of slabs}$

$=\frac{\text{Total height of N slabs}}{2}$
$=\frac{Nb}{2}$

$P{E}_{final}=$ $\text{Total weight of N slabs}\times \text{height of center of mass}$

$=N\times b^3\rho g\times \frac{Nb}{2}=\frac{N^2{b}^4\rho g}{2}$

$P{E}_{initial}=N{b}^3\rho g\times \frac{b}{2}=\frac{N{b}^4\rho g}{2}$

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