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# 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) $4\%$

Last updated date: 17th Apr 2024
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Hint: 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: $\text{percentage error = }\dfrac{\text{(observed value - true value)}}{\text{true value}}\times 100$

We know that volume of a cylinder $(V)=\pi {{r}^{2}}h$
\begin{align} & V=\pi {{\left( \dfrac{d}{2} \right)}^{2}}h \\ & \Rightarrow V=\dfrac{\pi }{4}\times {{d}^{2}}h \\ \end{align}
Mathematically, we can say that $\dfrac{\Delta V}{V}=2\dfrac{\Delta d}{d}+\dfrac{\Delta h}{h}$
\begin{align} & \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{align}
Now we know that percentage error is just relative error multiplied by $100$
Hence percentage error in volume is $\dfrac{\Delta V}{V}\times 100=0.03\times 100=3\%$
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.