Question

Question 20
A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

Answer 1

Given that, length of cuboid pen stand (l) = 10 cm
Breadth of cuboid pen stand (b) = 5 cm
And height of cuboid pen stand (h) = 4 cm

$\therefore$ Volume of cuboid =$l\times b\times h=10\times 5\times 4=200c{m}^3$
Also, radius of conical depression (r) = 0.5 cm
And height of (depth) of a conical depression $h_1=2.1cm$
$\therefore$ Volume of a conical depression = $\frac{1}{3}\pi r^2{h}_1$

$=\frac{1}{3}\times \frac{22}{7}\times 0.5\times 0.5\times 2.1=\frac{22\times 5\times 5}{1000}=\frac{22}{40}=\frac{11}{20}=0.55c{m}^3$
Also , given Edge of cubical depression (a)= 3cm
volume of cubical depression = $(a{)}^3=(a{)}^3=27c{m}^3$
So, volume of 4 conical depression
$=4\times \frac{11}{20}=\frac{11}{5}c{m}^3$
Hence the volume of wood in the entire pen stand

= Volume of cuboid pen stand – Volume of 4 conical
Depressions – Volume of a cuboid depressions
$=200-\frac{11}{5}-27=200-\frac{146}{5}=200-29.2=170.8c{m}^3$
So, the required volume of the wood in the entire stand is $170.8c{m}^3$.