Question

A held is 16 m long and 12 m wide. From two opposite corners of the field, two pits, each
measuring 2 m by 2 m and 1.15 m deep, are dug out. If the earth thus dug out is spread over the
remaining field, find by how much does the level of the field go up?​

Answer 1

Given are the dimensions of a field and two pits, find the height raised by spreading the dug-out dirt over the field.

Explanation:

• We know the volume of a cuboidal stricture having length 'l', width 'w', and height 'h' is given by, $V=lwh$
• Here let the height by which the level of the field goes up be denoted by 'h'.
• Hence for the field we have, $l=16\ m,\ \ w=12\ m$  
• Now given are the dimensions of both pits, $l'=2\ m,\ \ w'=2\ m,\ \ h'=1.15\ m$  
• Then the volume of the raised field is,  
• $V_f=(lw-l'w')h\\->V_f=(16(12)-2(2))h\ \ \ \ \ \\->V_f=184h$  
• Then the volume of the dirt dug out is,
• $V_d=2l'w'h'\ \ \ \ \ \ \ \ \\->V_d=2(2)(2)(1.15)\\->V_d=9.2\ m^3$    
• Now since this dirt is spread over the field which leads to an increase in the level of the field we can say, $V_f=V_d$  
• $184h=9.2\\->h=0.05\ m$  
• The level of the field goes up by $0.05\ m\ (i.e,\ 5\ cm)$.