Question

A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are 15cm by 10cm by 3.5cm. The radius of each depression is 0.5 cm and the depth is 1.4cm. Find the amount of the wood in the entire stand.

Answer 1

Hint: We use the formula of cuboid to find the volume of pen stand without depressions. Calculate the volume of three depressions using the formula of volume of cone. Subtract volume of three conical depressions from total volume of cuboid to obtain the volume of pen stand.
* Volume of cuboid having length ‘l’, breadth ‘b’ and height ‘h’ is given by \[lbh\]
* Volume of a cone having height ‘h’ and radius of the base ‘r’ is given by \[\dfrac{1}{3}\pi {r^2}h\]

Complete step-by-step solution:
Since we know a pen stands in the shape of a cuboid with three conical depressions made in it to hold the pens.
We calculate the volume of the cuboidal pen stand by calculating the volume of cuboid and subtract the volume of three conical depressions from it.
(i) Volume of cuboid:
Length of cuboid, \[l = 15\]cm
Breadth of cuboid, \[b = 10\]cm
Height of the cuboid, \[h = 3.5\]cm
\[\because \]Volume of the cuboid \[ = lbh\]
Substitute the values of ‘l’, ‘b’ and ‘h’ in the formula
\[ \Rightarrow \]Volume of the cuboid \[ = \left( {15 \times 10 \times 3.5} \right)\] \[c{m^3}\]
 \[ \Rightarrow \]Volume of the cuboid \[ = 525\] \[c{m^3}\]....................… (1)
(ii) Volume of 3 conical depressions:
Height of the conical depression, \[h = 1.4\]cm
Radius of base of conical depression,\[r = 0.5\]cm
\[\because \]Volume of the 1 conical depression \[ = \dfrac{1}{3}\pi {r^2}h\]
Substitute the values of ‘r’ and ‘h’ in the formula
\[ \Rightarrow \] Volume of the 1 conical depression \[ = \left( {\dfrac{1}{3} \times \dfrac{{22}}{7} \times 0.5 \times 0.5 \times 1.4} \right)\] \[c{m^3}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow \] Volume of the 1 conical depression \[ = 0.366\] \[c{m^3}\]
So, volume of 3 conical depressions \[ = 3 \times 0.366\] \[c{m^3}\]
\[ \Rightarrow \]Volume of 3 conical depressions \[ = 1.098\] \[c{m^3}\]...............… (2)
Now we subtract volume from equation (2) from volume in equation (1)
\[ \Rightarrow \]Volume of cuboidal pen stand\[ = \left( {525 - 1.098} \right)\]\[c{m^3}\]
\[ \Rightarrow \]Volume of cuboidal pen stand\[ = 523.902\]\[c{m^3}\]
\[\therefore \]The amount of the wood in the entire stand is 523.902\[c{m^3}\]

Note: Students might make the mistake of subtracting the volume obtained by one conical depression from volume obtained by cuboidal box. They forget to multiply the volume of one conical depression by the number of depressions. Also, always write the SI unit of volume along with the final answer.