Question

The percentage error in the measurement of mass and speed are $2\%$ and $3\%$ respectively. The error in the estimate of kinetic energy obtained by measuring mass and speed will be
(A). $12\%$
(B). $10\%$
(C). $8\%$
(D). $2\%$

Answer 1

Hint: This problem can be solved by finding out the percentage error in the measurement of the kinetic energy by using the formula for the relative error in a quantity that is the product of two other quantities using the percentage errors of the individual quantities. By writing the formula for kinetic energy and using it to find a relation for the relative percentage errors of mass, speed and the kinetic energy, we can arrive at the required answer.
Formula used:
For a quantity $Z$ defined as $Z={{X}^{n}}{{Y}^{m}}$ where $X,Y$ are physical quantities and $n.m$ are real numbers, the percentage error in $Z$, that is, $\left( \dfrac{\Delta Z}{Z}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$ is given by
$\left( \dfrac{\Delta Z}{Z}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)=\left( \left| n \right|\times \dfrac{\Delta X}{X}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)+\left( \left| m \right|\times \dfrac{\Delta Y}{Y}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$
Where $\left( \dfrac{\Delta X}{X}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$, $\left( \dfrac{\Delta Y}{Y}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$ are the percentage errors in $X,Y$ respectively.
Kinetic energy $KE$ of a body of mass $m$ moving at a speed $v$ is given by
$KE=\dfrac{1}{2}m{{v}^{2}}$

Complete step by step answer:
As explained in the hint, we will first find the relation of kinetic energy with the mass and speed of the body and then use this relation to find a relation for the percentage error in the kinetic energy in terms of the percentage errors in the mass and speed of the body.
Hence, let us proceed to do that.
Kinetic energy $KE$ of a body of mass $m$ moving at a speed $v$ is given by
$KE=\dfrac{1}{2}m{{v}^{2}}$ --(1)
Also,
for a quantity $Z$ defined as $Z={{X}^{n}}{{Y}^{m}}$ where $X,Y$ are physical quantities and $n.m$ are real numbers, the percentage error in $Z$, that is, $\left( \dfrac{\Delta Z}{Z}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$ is given by
$\left( \dfrac{\Delta Z}{Z}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)=\left( \left| n \right|\times \dfrac{\Delta X}{X}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)+\left( \left| m \right|\times \dfrac{\Delta Y}{Y}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$ --(2)
Where $\left( \dfrac{\Delta X}{X}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$, $\left( \dfrac{\Delta Y}{Y}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$ are the percentage errors in $X,Y$ respectively.
Hence, using (1) and (2), we get,
$\dfrac{\Delta KE}{KE}\left( \text{in }\!\!\%\!\!\text{ } \right)=\left( 1\times \dfrac{\Delta \dfrac{1}{2}}{\dfrac{1}{2}}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)+\left( 1\times \dfrac{\Delta m}{m}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)+\left( 2\times \dfrac{\Delta v}{v}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$ --(3)
where $\left( \dfrac{\Delta KE}{KE}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$,$\left( \dfrac{\Delta m}{m}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$,$\left( \dfrac{\Delta v}{v}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)$ are the percentage errors in the kinetic energy, mass and speed of the body respectively.
Now, according to the question, the percentage error in the mass and speed of the body are $2\%$ and $3\%$ respectively.
$\therefore \left( \dfrac{\Delta m}{m}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)=2\%$
$\therefore \left( \dfrac{\Delta v}{v}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)=3\%$
Putting these two values in (3), we get,
$\dfrac{\Delta KE}{KE}\left( \text{in }\!\!\%\!\!\text{ } \right)=\left( 1\times \dfrac{0}{\dfrac{1}{2}}\left( \text{in }\!\!\%\!\!\text{ } \right) \right)+\left( 1\times 2\% \right)+\left( 2\times 3\% \right)$ $\left( \Delta \dfrac{1}{2}=0,\text{ since it is a constant and does not change} \right)$
$\dfrac{\Delta KE}{KE}\left( \text{in }\!\!\%\!\!\text{ } \right)=0+\left( 2\% \right)+\left( 6\% \right)=8\%$
Hence, the required percentage error in the kinetic energy of the body is $8\%$.
Therefore the correct answer is C) $8\%$.

Note: This is the most general approach to solving problems related to finding the percentage or relative errors of complex quantities that are defined as a combination of other quantities. However, students must keep in mind that the percentage or relative errors in the component quantities must always be added regardless of their power (which is indicated by the modulus signs on the powers in equation (2)). This is very important as students often get confused and think that if a quantity is in the denominator, its relative or percentage error must be subtracted. However, that is completely wrong. Questions are purposefully set to confuse students who do not have a thorough understanding of this concept.