Question

In an experiment to determine the acceleration due to gravity $g$, the formula used for the time period of a periodic motion is $T = 2\pi \sqrt {\dfrac{{7\left( {R - r} \right)}}{{5g}}} $. The values of $R$ and $r$ are measured to be $\left( {60 \pm 1} \right)mm$ and $\left( {10 \pm 1} \right)mm$, respectively. In five successive measurements, the time period is found to be \[0.52s\], $0.56s$, $0.57s$, $0.54s$ and $0.59s$. The least count of the watch used for the measurement of time period is $0.01s$. Which of the following statement(s) is (are) true?
This question has multiple correct options.
(A) the error in the measurement of $r$ is \[10\% \]
(B) the error in the measurement of $T$ is \[3.57\% \]
(C) the error in the measurement of $T$ is \[2\% \]
(D) The error in the determined value of $g$ is \[11\% \]

Answer 1

Hint: The amount of uncertainty that is present in the measurement made with a measuring instrument Is known as the Error. The difference between the measured value and the true value can be observed. The smallest value that can be measured by an instrument is called its least count.

Complete step by step answer:
The magnitude of the difference between the true value of the measured physical quantity and the value of individual measurement is called absolute error of the given measurements.
If ${x_{mean}} = {\text{actual value}}$, and ${x_i} = {i^{th}}$ observed value, then the absolute error$\left| {\Delta x} \right|$ in the ${i^{th}}$ observed value is defined as;
$\left| {\Delta x} \right| = \left| {{x_i} - {x_{mean}}} \right|$ ……………………..(1)
Always remember that the absolute error is always positive and has the same units as that of the quantity being used.
Given, in five successive measurements, the observed values of time period ${T_i}$ are \[0.52s\],$0.56s$, $0.57s$, $0.54s$, and $0.59s$ .
Sum of all these values, $\sum\limits_{i = 0}^5 {{T_i}} = 0.52 + 0.56 + 0.57 + 0.54 + 0.59 = 2.78s$
The mean value of time period is given by, ${T_{mean}} = \dfrac{{\sum\limits_{i = 0}^5 {{T_i}} }}{5}$
${T_{mean}} = \dfrac{{2.78}}{5}0.56s$
Magnitude of absolute error in each observation is calculated as follows using equation (1),
We can substitute the values of ${T_i}$ and ${T_{mean}}$ for respective $\Delta T$ values we get:
$\left| {\Delta {T_1}} \right| = \left| {{T_i} - {T_{mean}}} \right|$
$ \Rightarrow \left| {0.56 - 0.52} \right|$
$ \Rightarrow 0.04s$
$\left| {\Delta {T_2}} \right| = \left| {{T_i} - {T_{mean}}} \right|$
$ \Rightarrow \left| {0.56 - 0.56} \right|$
$ \Rightarrow 0.00s$
$\left| {\Delta {T_3}} \right| = \left| {{T_i} - {T_{mean}}} \right|$
$ \Rightarrow \left| {0.57 - 0.56} \right|$
$ \Rightarrow 0.01s$
$\left| {\Delta {T_4}} \right| = \left| {{T_i} - {T_{mean}}} \right|$
$ \Rightarrow \left| {0.56 - 0.54} \right|$
$ \Rightarrow 0.02s$
$\left| {\Delta {T_5}} \right| = \left| {{T_i} - {T_{mean}}} \right|$
$ \Rightarrow \left| {0.59 - 0.56} \right|$
$ \Rightarrow 0.03s$
The arithmetic mean of all the absolute errors is considered as the mean absolute error or final absolute error of the value of the physical quantity concerned.
Mean absolute error in time period, $\Delta {T_{mean}} = \dfrac{{0.04 + 0.00 + 0.01 + 0.02 + 0.03}}{5}$
$\Delta {T_{mean}} = \dfrac{{0.1}}{5}$
$ \Rightarrow 0.02s$
Now find the certain values so that we can select correct options.
The error in the measurement of $r$:
\[percentage{\text{ }}error = relative{\text{ }}error \times 100\]
Given, the value of r measure, $\left( {10 \pm 1} \right)mm = r \pm \Delta r$
Percentage error in measurement of $r$ is given by,
 $\dfrac{{\Delta r}}{r} \times 100 = \dfrac{1}{{10}} \times 100$
$ \Rightarrow 10\% $
The error in the measurement of $t$:
We have,
${T_{mean}} = T$
$ \Rightarrow 0.56s$ and
  $\Delta {T_{mean}} = \Delta T$
$ \Rightarrow 0.02s$
Now let us calculate the Percentage error in measurement and that is given by,
$\dfrac{{\Delta T}}{T} \times 100 = \dfrac{{0.02}}{{0.56}} \times 100$
$ \Rightarrow 3.57\% $
The error in the determined value of $g$ :
From given equation, $T = 2\pi \sqrt {\dfrac{{7\left( {R - r} \right)}}{{5g}}} $
g value can be written as, $g = \dfrac{{28{\pi ^2}\left( {R - r} \right)}}{{5{T^2}}}$
Percentage error in measurement of $g$ is given by,
$\dfrac{{\Delta g}}{g} \times 100 = \dfrac{{\Delta R + \Delta r}}{{\left( {R - r} \right)}} \times 100 + 2\dfrac{{\Delta T}}{T} \times 100$
Let us substitute the respective values in the equation
$ \Rightarrow \dfrac{{\Delta g}}{g} \times 100 = \dfrac{{1 + 1}}{{\left( {60 - 10} \right)}} \times 100 + 2\left( {3.57\% } \right)$
$\therefore \dfrac{{\Delta g}}{g} \times 100 = 11.14\% $

From these calculations we can say, options (A), (B) and (D) are correct. Option (C) is not possible.

Note:
Error can be positive or negative.
Relative error is a pure number having no units.
The maximum possible error in the final results is the sum of absolute error in the individual quantities when we add or subtract the values.