The length of a cylinder is measured with a meter rod having least count $0.1$ cm. Its diameter is measured with vernier calipers having least count $0.01$ cm. Length of cylinder is $8.0$ cm and radius is $4.0$ cm. Find the percentage error in the calculated value of the volume.

The volume of a cylinder is given by $V = \pi R^2 L$.

Since the diameter $D$ is measured, we can write the volume in terms of diameter as $V = \dfrac{\pi D^2 L}{4}$.

The maximum percentage error in the volume is given by:
$\dfrac{\Delta V}{V} \times 100 = \left( 2 \dfrac{\Delta D}{D} + \dfrac{\Delta L}{L} \right) \times 100$

Given values:
Length of the cylinder, $L = 8.0$ cm
Least count of meter rod, $\Delta L = 0.1$ cm
Radius of the cylinder, $R = 4.0$ cm $\Rightarrow$ Diameter, $D = 2R = 8.0$ cm
Least count of vernier calipers, $\Delta D = 0.01$ cm

Substituting these values into the percentage error formula:
$\dfrac{\Delta V}{V} \times 100 = \left( 2 \times \dfrac{0.01}{8.0} + \dfrac{0.1}{8.0} \right) \times 100$

$\dfrac{\Delta V}{V} \times 100 = \left( \dfrac{0.02}{8.0} + \dfrac{0.1}{8.0} \right) \times 100$

$\dfrac{\Delta V}{V} \times 100 = \left( \dfrac{0.12}{8.0} \right) \times 100$

$\dfrac{\Delta V}{V} \times 100 = \dfrac{12}{8.0} = 1.5\%$

Answer: $1.5\%$