Question

In the middle of a rectangular field measuring 30m ×20m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.

Answer 1

Answer:

Vol of well = vol of cuboid-vol of well

As the sand that is of the well gets spread out on the remaining rectangle itself.

πr^2h = (l×b×h)-πr^2h

Height of cylinder on right side will be equal to cuboid as we are subtracting the upper layer.

22÷7×7÷2×7÷2×10=(30×20×h)-22÷7×7÷2×7÷2×h

11×35=60h-77÷2×h

11×35=(120h-77h)÷2. {Take lcm}

11×35=43h÷2

385×2=43h

770÷43=h

h=17.906 (approx)

I hope it was correct

Answer 2

Answer: Height is 0.6857 cm

Pls mark as Brainliest answer

Step-by-step explanation:

Volume of earth dug = Volume of cuboid - volume of cylinder formed with same height

π(r^2)h1= lbh - π(r^2)h

(22/7)3.5*3.5*10= 30*20*h - (22/7)3.5*3.5*h

385= 600h-38.5h

385= 561.5h

h= 385÷561.5

h= 0.68566340160284

h= 0.6857 m (approx)