Maximum Percentage Error in Volume of Cone

To determine the maximum percentage error in the determination of the volume of the cone, we use the formula for the volume $V$ of a cone in terms of its base diameter $D$ and height $h$:

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

Taking the natural logarithm on both sides: $\ln V = \ln\left(\frac{\pi}{12}\right) + 2\ln D + \ln h$

Differentiating to find the maximum fractional error: $\frac{\Delta V}{V} = 2\left(\frac{\Delta D}{D}\right) + \frac{\Delta h}{h}$

Given data:

• Least count of the measuring scale, $\Delta D = \Delta h = 2\text{ mm} = 0.2\text{ cm}$
• Diameter of the base, $D = 20.0\text{ cm}$
• Height of the cone, $h = 20.0\text{ cm}$

Substitute these values into the error formula: $\frac{\Delta V}{V} = 2\left(\frac{0.2\text{ cm}}{20.0\text{ cm}}\right) + \left(\frac{0.2\text{ cm}}{20.0\text{ cm}}\right)$ $\frac{\Delta V}{V} = 2(0.01) + 0.01 = 0.03$

Thus, the maximum percentage error in the volume is: $\text{Percentage Error} = \frac{\Delta V}{V} \times 100\% = 0.03 \times 100\% = 3\%$

3