Question

The length of a cylinder is measured with a metre scale having least count $0.1cm$. Its diameter is measured with Vernier Callipers having least count $0.01cm$. Given the length is $5.0cm$ and the diameter is $2.00cm$. The percentage error in the calculated value of volume will be:
(A) $2\%$
(B) $1\%$
(C) $3\%$
(D) $4\%$

Answer 1

To find the percentage error for any quantity using other physical quantities, we need to first establish a relationship among the quantities that we’ve been given. After we have done that, we can proceed to find the relative errors of the different quantities whose information has been given to us.

Formula Used: $\text{percentage error = }\dfrac{\text{(observed value - true value)}}{\text{true value}}\times 100$

Complete step by step answer:
We know that volume of a cylinder $(V)=\pi {{r}^{2}}h$
Since we know that the value of the radius is half the value of diameter, we have

& V=\pi {{\left( \dfrac{d}{2} \right)}^{2}}h \\\ & \Rightarrow V=\dfrac{\pi }{4}\times {{d}^{2}}h \\\ \end{aligned}$$ Now relative error in volume will be equal to the sum of relative errors of both the diameter and the length or height, times the product of their respective exponential coefficients. The relative error is nothing but the ratio of the least count of the instrument used to measure the value of the quantity measured. In cases when the least count is not known to us, we take the difference of the observed value and the true value and use it instead of the least count. Mathematically, we can say that $$\dfrac{\Delta V}{V}=2\dfrac{\Delta d}{d}+\dfrac{\Delta h}{h}$$ Substituting the values in the above equation, we get $$\begin{aligned} & \dfrac{\Delta V}{V}=2\dfrac{0.01}{2}+\dfrac{0.1}{5} \\\ & \Rightarrow \dfrac{\Delta V}{V}=0.01+0.02=0.03 \\\ \end{aligned}$$ Now we know that percentage error is just relative error multiplied by $$100$$ Hence percentage error in volume is $$\dfrac{\Delta V}{V}\times 100=0.03\times 100=3\%$$ Hence option (C) is the correct answer. **Note:** One very common error made while calculating relative error or percentage error is that students consider the constants as well. Error is found in only those values which are measured by us. Since constants have predefined fixed values, their presence will not change in any way the error in the observation. This is why we neglected the $$\dfrac{\pi }{4}$$ portion of the volume’s formula.