Question

The filed is 70m ,long and 40m broad. In one corner of the field a pit which is 10m long , 8m broad and 5m deep , has dug out .the earth taken out of it evenly spread oe in the remaining part of the field . find the rise in the level of the field . In the shower ,of rain the falls . find the volume ofwater that falls on 2 hectares of ground .

Answer 1

Solution:-

2nd part of your question seems to be incomplete, so I am solving the first part.

Area of the rectangular field = 70*40
= 2800 sq m
Volume of the earth taken out = L*B*H
= 10*8*5
= 400 cu m
Area of the pit = 10*8 
= 80 sq m
Area of the remaining filed = Area of the rectangular field - Area of the pit  
= 2800 - 80
Area of the remaining field = 2720 sq m
The earth taken out is evenly spread over the field of area 2720 sq m.

Let 'H' be the height of the rise in the level of the field.
∴ 2720 × H = 10*8*5
H = 400/2720
H = 5/34 m

Height in cm
H = (5*100)/34
H = 14.7 cm
So, the rise in the level of field is 14.7 cm
Answer.

Answer 2

Answer:

Area of a field on which earth taken out is to be spared

=70×40m^2_10×8m^2

=2800_80m^2= 2820m^2

Volume of earth dug out= 10×8×5= 400m^3

Rice in level of field = volume of earth dugout

÷

Area on which Earth taken out in the spread

=400÷2720= 0.147m= 14.7cm