Given that, length of cuboid pen stand (l) = 10 cm
Breadth of cuboid pen stand (b) = 4 cm
and height of cuboid pen stand (h) = 4 cm
`therefore` Volume of cuboid pen stand `= l xx b xx h xx = 10 xx 5 xx 4 = 200 cm^(3)`
Also, radius of conical depression (r) = 0.5 cm
and height (depth ) of a conical depression `(h_(1))` = 2.1 cm
`therefore` Volume of a conical depression `= (1)/(3) pi r h_(1)`
`= (1)/(3) xx (22)/(7) xx 0.5 xx 0.5 xx 2.1`
`= (22 xx 5xx5)/(1000) = (22)/(40) = (11)/(20) = 0.55 cm ^(3)`
Also given
Edge of cubical depression (a) = 30
`therefore` Volume of cubical depression `= (a)^(3) = (3)^(3) = 27 cm ^(3)`
So, volume of 4 conical depressions
`= 4 xx "Volume of a conical depression "`
`= 4 xx (11)/(20) = (11)/(5) cm^(3)`
Hence , the volume of wood in the entire pen stand
= Volume of cuboid pen stand - Volume of 4 conical depression - Volume of cuboid depressions
`= 200- (11)/(5) - 27 = 200 -(146)/(5)`
`= 200- 29.2 = 170.8 cm^(3)`
So, the required volume of the wood in the entire stand is `170.8 cm^(3)`.
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