The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its diameter is measured with venier calipers having least count 0.01 cm. Given that length is 5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume will be

The correct option is 3%.

Step-by-step Explanation:

The formula for the volume ($V$) of a cylinder of radius $r$ and length $l$ is given by:
$V=\pi r^{2}l$
Here, $\pi$ is a constant, so it does not contribute to the error in the calculation.

The relative or fractional error in the volume is the sum of the fractional errors in the measured quantities, multiplied by their respective powers:
$\frac{\Delta V}{V}=2\frac{\Delta r}{r}+\frac{\Delta l}{l}$
Multiplying by 100 on both sides gives the percentage error in the volume:
$\frac{\Delta V}{V}\times 100=\left(2\frac{\Delta r}{r}+\frac{\Delta l}{l}\right)\times 100$

Given data from the problem:
• Length of the cylinder, $l=5.0\text{cm}$
• Least count of the meter rod (error in length), $\Delta l=0.1\text{cm}$
• Radius of the cylinder, $r=2.0\text{cm}$
• Least count of the vernier calipers (error in radius/diameter measurement), $\Delta r=0.01\text{cm}$

Now, substitute these values into the percentage error formula:
$\frac{\Delta V}{V}\times 100=\left(2\times \frac{0.01}{2.0}+\frac{0.1}{5.0}\right)\times 100$

Calculate the terms inside the parentheses:
$2\times \frac{0.01}{2.0}=0.01$
and
$\frac{0.1}{5.0}=0.02$

Adding the two fractional errors:
$0.01+0.02=0.03$

Calculating the percentage error:
$\text{Percentage Error}=0.03\times 100=3\%$

Therefore, the percentage error in the calculated value of the volume is 3%.