The volume $V$ of a cylindrical wire is given by:

$V = \pi r^2 l$

where:

• $r$ is the radius of the wire,
• $l$ is the length of the wire.

Now, let the percentage change in length be $\Delta l = 0.02\%$ and the percentage change in radius be $\Delta r = -0.005\%$.

Step 1: Percentage change in volume formula

The total change in volume can be approximated using the formula for small percentage changes:

$\frac{\Delta V}{V} \approx 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}$

This equation is derived from differentiating the volume formula with respect to $r$ and $l$.

Step 2: Applying the values

• $\frac{\Delta r}{r} = -0.005\%$
• $\frac{\Delta l}{l} = 0.02\%$

Substitute these values into the formula:

$\frac{\Delta V}{V} \approx 2(-0.005\%) + 0.02\%$

$\frac{\Delta V}{V} \approx -0.01\% + 0.02\% = 0.01\%$

Final Answer:

The percentage change in volume is 0.01%.

Would you like more details or further clarification?

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Here are five related questions for further exploration:

1. How does the volume change if the radius decreases by a larger percentage?
2. What happens to the volume if both the radius and length increase?
3. How does the volume change when only the radius changes?
4. Can this method be used for other shapes, such as spheres?
5. What is the significance of small percentage changes in engineering calculations?

Tip: When dealing with small changes, using approximations like linearization makes calculations easier and faster!