Q4:

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 $15$ cm by $10$ cm by $3.5$ cm. The radius of each of the depressions is $0.5$ cm and the depth is $1.4$ cm. Find the volume of wood in the entire stand (see Fig. 12.16).

Given:

Dimensions of the cuboidal pen stand: Length ($l$) = $15$ cm, Breadth ($b$) = $10$ cm, Height ($h_{cuboid}$) = $3.5$ cm.

Number of conical depressions ($n$) = $4$.

Radius of each conical depression ($r$) = $0.5$ cm.

Depth (height) of each conical depression ($h_{cone}$) = $1.4$ cm.

To Find:

The volume of wood remaining in the entire pen stand.

Step 1: Calculate the volume of the cuboidal block.

The formula for the volume of a cuboid is $V_{cuboid} = l \times b \times h$.

$V_{cuboid} = 15 \text{ cm} \times 10 \text{ cm} \times 3.5 \text{ cm}$

$V_{cuboid} = 150 \times 3.5 = 525 \text{ cm}^3$

Step 2: Calculate the volume of one conical depression.

The formula for the volume of a cone is $V_{cone} = \frac{1}{3}\pi r^2 h$.

Using $\pi \approx \frac{22}{7}$:

$V_{cone} = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4$

$V_{cone} = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4$

$V_{cone} = \frac{1}{3} \times 22 \times 0.25 \times 0.2$ [Since $1.4 / 7 = 0.2$]

$V_{cone} = \frac{1}{3} \times 22 \times 0.05 = \frac{1.1}{3} \text{ cm}^3$

Step 3: Calculate the total volume of four conical depressions.

$V_{total\_cones} = 4 \times V_{cone}$

$V_{total\_cones} = 4 \times \frac{1.1}{3} = \frac{4.4}{3} \text{ cm}^3$

$V_{total\_cones} \approx 1.4667 \text{ cm}^3$

Step 4: Calculate the volume of wood in the stand.

The volume of wood is the volume of the cuboid minus the volume of the four conical depressions.

$V_{wood} = V_{cuboid} - V_{total\_cones}$

$V_{wood} = 525 - \frac{4.4}{3}$

$V_{wood} = \frac{1575 - 4.4}{3}$

$V_{wood} = \frac{1570.6}{3}$

$V_{wood} = 523.5333... \text{ cm}^3$

Final Answer: The volume of wood in the entire stand is approximately $523.53 \text{ cm}^3$.