An 8 kg metal block of dimensions `16 cm xx 8 cm xx 6 cm` is lying on a table with its face of largest area touching the table. If `g = 10 m s^(-2)`, the minimum amount of work done in making it stand with its length vertical is 

Options

A. `0.4 J`
B. `6.4 J`
C. `4J`
D. `12.8 J`

Answer

C

Explanation

When face of the largest area is touching the table, height of centre of gravity above the table `h_1 = 6/2 = 3 cm`
With its length vertical, height of centre of gravity would become `h_2 = 16/2 = 8 cm`
Minimum work required `W = (P.E.)_(2) - (P.E.)_(1) = mg(h_2 - h_1)`
`:. W = (8 xx 10(8 - 3))/(100) = 4J` .