
In an experiment, the percentage of error occurred in the measurement of physical quantities $A,B,C$ and $D$ are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement $X$, where $X=\dfrac{{{A}^{2}}{{B}^{{1}/{2}\;}}}{{{C}^{{1}/{3}\;}}{{D}^{3}}}$, will be
$A.\,\,\,(\dfrac{3}{13})$%
$B.\,\,\,16$%
$C.\,\,\,-10$%
$D.\,\,\,10$%
Answer
543.3k+ views
Hint: To get the percentage of error in the measurement $X$, we need to differentiate the equation given to get the equation in terms of error that occurred in all the physical quantities, from which we can get the final percentage error.
Complete answer:
The formula to be used is given in the question, that is,
$X=\dfrac{{{A}^{2}}{{B}^{{1}/{2}\;}}}{{{C}^{{1}/{3}\;}}{{D}^{3}}}$
where, $A,B,C$ and $D$ are the physical quantities for which the percentage error occurred in measurement are 1%, 2%, 3% and 4%.
Now,
Differentiating the above equation to get the equation in terms of error in physical quantities, we get:
Error,$\dfrac{\Delta X}{X}=2\times \dfrac{\Delta A}{A}+\dfrac{1}{2}\times \dfrac{\Delta B}{B}+\dfrac{1}{3}\times \dfrac{\Delta C}{C}+3\times \dfrac{\Delta D}{D}$
So, Percentage error can be written as:
% error,$\dfrac{\Delta X}{X}\times 100=2\times \dfrac{\Delta A}{A}\times 100+\dfrac{1}{2}\times \dfrac{\Delta B}{B}\times 100+\dfrac{1}{3}\times \dfrac{\Delta C}{C}\times 100+3\times \dfrac{\Delta D}{D}\times 100$
$=\,\,(2\times 1\,\,+\,\,\dfrac{1}{2}\times 2\,\,+\,\,\dfrac{1}{3}\times 3\,\,+\,\,3\times 4)$%
$=\,(2\,\,+\,\,1\,\,+\,\,1\,\,+\,\,12)$%
$=\,16$%
Maximum percentage of error in the measurement $X$ is $16$%.
Therefore, the correct answer is Option (B).
Note:
Error in a quantity can be identified by differentiating the quantity so that we can get the change in the quantity. Remember the way of solving this type of problem, as this type of questions are common in the paper where we need to find the percentage error and if you don’t have any formula for finding that, then differentiating the given equation to get the equation in terms of error in each quantity is the best option.
Complete answer:
The formula to be used is given in the question, that is,
$X=\dfrac{{{A}^{2}}{{B}^{{1}/{2}\;}}}{{{C}^{{1}/{3}\;}}{{D}^{3}}}$
where, $A,B,C$ and $D$ are the physical quantities for which the percentage error occurred in measurement are 1%, 2%, 3% and 4%.
Now,
Differentiating the above equation to get the equation in terms of error in physical quantities, we get:
Error,$\dfrac{\Delta X}{X}=2\times \dfrac{\Delta A}{A}+\dfrac{1}{2}\times \dfrac{\Delta B}{B}+\dfrac{1}{3}\times \dfrac{\Delta C}{C}+3\times \dfrac{\Delta D}{D}$
So, Percentage error can be written as:
% error,$\dfrac{\Delta X}{X}\times 100=2\times \dfrac{\Delta A}{A}\times 100+\dfrac{1}{2}\times \dfrac{\Delta B}{B}\times 100+\dfrac{1}{3}\times \dfrac{\Delta C}{C}\times 100+3\times \dfrac{\Delta D}{D}\times 100$
$=\,\,(2\times 1\,\,+\,\,\dfrac{1}{2}\times 2\,\,+\,\,\dfrac{1}{3}\times 3\,\,+\,\,3\times 4)$%
$=\,(2\,\,+\,\,1\,\,+\,\,1\,\,+\,\,12)$%
$=\,16$%
Maximum percentage of error in the measurement $X$ is $16$%.
Therefore, the correct answer is Option (B).
Note:
Error in a quantity can be identified by differentiating the quantity so that we can get the change in the quantity. Remember the way of solving this type of problem, as this type of questions are common in the paper where we need to find the percentage error and if you don’t have any formula for finding that, then differentiating the given equation to get the equation in terms of error in each quantity is the best option.
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