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A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15cm \times 10cm \times 3.5cm$. The diameter of each of the depressions is 1cm and the depth is 1.4cm. Find the volume of the wood in the entire stand.

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Hint: Make use of the formula of the volume of the cuboid and volume of the cone to solve this question.

Complete step-by-step answer:
Dimension of the cuboid are given as $15cm \times 10cm \times 3.5cm$
Volume of the cuboid is given by $l \times b \times h$ cubic centimetres
So, we can write Volume=$15 \times 10 \times 3.5 = 525c{m^3}$
It is also given that there are four conical depressions and the radius of each depression is given as 0.5cm and the depth is given as 1.4cm
So, the volume of wood taken out to make four conical cavities
=$4 \times \dfrac{1}{3}\pi {r^2}h$
The radius of each depression is given as 0.5 cm and the depression(height) is given as 1.4cm
So, lets substitute these values in the formula, so, we get
$4 \times \dfrac{1}{3} \times \dfrac{{22}}{7} \times 0.5 \times 0.5 \times 1.4$ $c{m^3}$
=$\dfrac{{22}}{{15}}c{m^3}$ =$1.47c{m^3}$
Hence, the volume of the wood in the entire stand=Volume of the cuboid-Volume of the conical cavities
So, the volume of the wood =(525-1.47)$c{m^3}$
So, the volume of wood in the entire stand=$523.53c{m^3}$

Note: In these types of questions it has to be noted that after finding out the volume of the cone and volume of the cuboid, we also have to find out the difference between the volumes which will give us the required volume of the entire stand.
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