Question

The dimensions of a cone are measured using a scale with a least count of 2 mm. The diameter of the base and the height are both measured to be 20.0 cm. The maximum percentage error in thedetermination of the volume is ______.

Answer 1

Correct Answer: 3

To solve the problem, we need to calculate the maximum percentage error in the volume of a cone when both the diameter and height are measured using a scale with a least count of 2 mm.

1. Formula for Volume of a Cone:
The volume of a cone is given by:

$ V = \frac{1}{3} \pi r^2 h $
where $r$ is the radius and $h$ is the height.

2. Error Propagation in Volume:
Since $V \propto r^2 h$, the relative error in volume is:

$ \frac{\Delta V}{V} = 2 \cdot \frac{\Delta r}{r} + \frac{\Delta h}{h} $

3. Converting Measurements:
Measured diameter = 20.0 cm ⇒ Radius $r = 10.0$ cm
Height $h = 20.0$ cm
Least count = 2 mm = 0.2 cm

4. Calculating Maximum Errors:
$ \frac{\Delta r}{r} = \frac{0.2}{10.0} = 0.02 = 2\% $
$ \frac{\Delta h}{h} = \frac{0.2}{20.0} = 0.01 = 1\% $

5. Calculating Total Percentage Error:
$ \text{Total error} = 2 \cdot 2\% + 1\% = 4\% + 1\% = 5\% $

6. Correcting for Diameter vs Radius Measurement:
The question specifies the diameter is measured using the scale, not the radius. Thus, the percentage error in radius is half that of diameter.
$ \Delta d = 0.2 \Rightarrow \Delta r = \frac{0.2}{2} = 0.1 \Rightarrow \frac{\Delta r}{r} = \frac{0.1}{10} = 0.01 = 1\% $
So revised total error is:
$ \text{Total error} = 2 \cdot 1\% + 1\% = 2\% + 1\% = 3\% $

Final Answer:
The maximum percentage error in the volume of the cone is 3%.

Answer 2

To solve the problem, we need to calculate the percentage error in the volume of a cone based on the absolute errors in its diameter and height measurements.

1. Formula for Volume of a Cone:
Let the diameter of the base be $d$ and the height be $h$. The radius of the base is $r = \frac{d}{2}$. The volume of the cone is:

$$ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{d}{2}\right)^2 h = \frac{\pi}{12} d^2 h $$

2. Given Data:
The least count of the scale is 2 mm = 0.2 cm. The diameter and height are both measured to be 20.0 cm, so $d = h = 20.0$ cm.

The absolute errors in the measurements of the diameter and height are given by $\Delta d = \Delta h = 0.2$ cm.

3. Formula for Percentage Error in Volume:
The percentage error in the volume is given by the formula:

$$ \frac{\Delta V}{V} \times 100 = \left| \frac{\partial V}{\partial d} \frac{\Delta d}{V} \right| \times 100 + \left| \frac{\partial V}{\partial h} \frac{\Delta h}{V} \right| \times 100 $$

4. Calculating Partial Derivatives:
We calculate the partial derivatives of $V$ with respect to $d$ and $h$:

$$ \frac{\partial V}{\partial d} = \frac{2 \pi}{12} d h = \frac{\pi}{6} d h $$

$$ \frac{\partial V}{\partial h} = \frac{\pi}{12} d^2 $$

5. Applying the Values:
Substituting the partial derivatives into the percentage error formula:

$$ \frac{\Delta V}{V} \times 100 = \left| \frac{2}{d} \Delta d \right| \times 100 + \left| \frac{1}{h} \Delta h \right| \times 100 $$

Since $\Delta d = \Delta h$ and $d = h$, we get:

$$ \frac{\Delta V}{V} \times 100 = \left( \frac{2}{d} + \frac{1}{h} \right) \Delta d \times 100 $$

Substitute the values $d = 20$, $h = 20$, and $\Delta d = 0.2$:

$$ \frac{\Delta V}{V} \times 100 = \left( \frac{2}{20} + \frac{1}{20} \right) \times 0.2 \times 100 = \frac{3}{20} \times 0.2 \times 100 = \frac{3}{20} \times 20 = 3 $$

6. Verifying the Result:
The relative errors in the diameter and height are:

$$ \frac{\Delta d}{d} = \frac{0.2}{20} = 0.01 $$

$$ \frac{\Delta h}{h} = \frac{0.2}{20} = 0.01 $$

Thus, the total relative error in volume is:

$$ \frac{\Delta V}{V} = 2 \times \frac{\Delta d}{d} + \frac{\Delta h}{h} = 2 \times 0.01 + 0.01 = 0.03 = 3\% $$

Final Answer:
The final answer is $\boxed{3}$.