A pen stand made of wood is in the shape of a | vedclass

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(D) Given:Length of the cuboid $(l) = 10 \, cm$,breadth $(b) = 5 \, cm$,and height $(h) = 4 \, cm$.Volume of the cuboid $= l 	imes b 	imes h = 10

Vedclass · Accepted Answer

$170.8$

Vedclass · Answer

(D) Given:Length of the cuboid $(l) = 10 \, cm$,breadth $(b) = 5 \, cm$,and height $(h) = 4 \, cm$.Volume of the cuboid $= l 	imes b 	imes h = 10 	imes 5 	imes 4 = 200 \, cm^3$.Radius of each conical depression $(r) = 0.5 \, cm$ and depth $(h_1) = 2.1 \, cm$.Volume of one conical depression $= \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} 	imes \frac{22}{7} 	imes (0.5)^2 	imes 2.1 = \frac{1}{3} 	imes \frac{22}{7} 	imes 0.25 	imes 2.1 = 0.55 \, cm^3$.Volume of $4$ conical depressions $= 4 	imes 0.55 = 2.2 \, cm^3$.Edge of the cubical depression $(a) = 3 \, cm$.Volume of the cubical depression $= a^3 = 3^3 = 27 \, cm^3$.Volume of wood in the stand $=$ Volume of cuboid $-$ (Volume of $4$ conical depressions $+$ Volume of cubical depression)$= 200 - (2.2 + 27) = 200 - 29.2 = 170.8 \, cm^3$.

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