Question

Similar of cubical shape of edge b are lying on ground. Density of material of slab is δ. Work done to arrange then one over the other is

• A. (N² - 1)b²ρg
• B. (N - 1)b⁴ρg
• C. $\frac { 1 } { 2 }$(N² - N)b⁴ρg
• D. (N² - N)b⁴ρg

Answer 1

Answer: (C)

$\frac { 1 } { 2 }$(N² - N)b⁴ρg

Answer 2

C.G. of first slab = $\frac { \mathrm { b } } { 2 }$

Weight of each stab = volume × density × g = b²ρg

C.G. of column of slabs

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

Height of displacement of force

= $\left( \frac { \mathrm { Nb } } { 2 } - \frac { \mathrm { b } } { 2 } \right) = ( \mathrm { N } - 1 ) \frac { \mathrm { b } } { 2 }$

Work done = N × b²ρG × (N - 1) $\frac { \mathrm { b } } { 2 }$

= $\frac { 1 } { 2 }$(N² - N)b⁴ρ