Question

The following observation were taken for determining surface tension of water by capillary tube method: Diameter, \[D=1.25\times {{10}^{-2}}m\], Rise in capillary, \[h=1.45\times {{10}^{-2}}m\]. Taking \[g=9.80m{{s}^{-2}}\] using the relation \[T=\dfrac{Rhg}{2}\times {{10}^{3}}N{{m}^{-1}}\]. What is the possible error in Surface tension, T?
\[\begin{align}
  & A)\text{ }0.15\% \\
 & B)\text{ }1.50\% \\
 & C)\text{ }15\% \\
 & D)\text{ }2.4\% \\
\end{align}\]

Answer 1

Hint: We can find the error percentage of a physical formula. The relation between the quantities involved in the formula gives a direct approach to the error that can be involved while conducting a laboratory experiment which is mainly due to imperfect scale readings mostly.

Complete answer:
 We have learnt earlier that the error that can occur in the final answer of an experiment that is either added or subtracted from the result can be derived from the formula itself.
We can either find the total error by computing the individual errors of physical quantities involved, i.e., For example, let
\[\begin{align}
  & z=\dfrac{ab}{c} \\
 & \text{The percentage error involved will be - } \\
 & \dfrac{\Delta z}{z}\times 100=\dfrac{\Delta a}{a}\times 100+\dfrac{\Delta b}{b}\times 100+\dfrac{\Delta c}{c}\times 100\text{ --(1)} \\
\end{align}\]
Here, in our question, the value for each of the physical quantities is given. We need to find the error involved in each physical quantity by knowing the least count of the apparatus used. Also, in theoretical cases, we can use precision of the values given.
Here, the diameter is given as \[D=1.25\times {{10}^{-2}}m\], this means that the change in diameter which accounts for the error can be given as -
\[\Delta D=0.01m\]
Similarly,
\[\begin{align}
  & h=1.45\times {{10}^{-2}}m \\
 & \Rightarrow \text{ }\Delta h=0.01m \\
 & \\
\end{align}\]
Now we can find the percentage error involved by substituting in the (1) as –
\[\begin{align}
  & T=\dfrac{Dhg}{4}\times {{10}^{3}}N{{m}^{-1}} \\
 & \text{The percentage error involved will be - } \\
 & \dfrac{\Delta T}{T}\times 100=\dfrac{\Delta D}{D}\times 100+\dfrac{\Delta h}{h}\times 100 \\
\end{align}\]
Remember that,
\[R=\dfrac{D}{2}\]
 Substituting the values for D, h and g,
\[\begin{align}
  & \dfrac{\Delta T}{T}\times 100=\dfrac{0.01}{1.25}\times 100+\dfrac{0.01}{1.45}\times 100 \\
 & \Rightarrow \text{ }\dfrac{\Delta T}{T}\times 100=0.8+0.69 \\
 & \Rightarrow \text{ }\dfrac{\Delta T}{T}\times 100=1.5\% \\
\end{align}\]
The error percent involved is \[1.5\%\]

The correct answer is option B.

Note:
- Error analysis is the most important part of an experiment. It allows to see the possible variations in outcome as compared to an actual value. Also, these values give an insight of reducing error and making an accurate observation and inference with the possible precision.
- The constant g is not measured in the experiment, so needs to be added for error calculation.