Q. A physical parameter $a\$can be determined by measuring the parameters b, c, d and e using the relation $a = b^{\alpha}c^{\beta}/d^{\gamma}e^{\delta}$. If the maximum errors in the measurement of b, c, d and e are $b_{1}\%,c_{1}\%,d_{1}\%\$and $e_{1}\%,$ then the maximum error in the value of $a$ determined by the experiment is
A) $\left( b_{1} + c_{1} + d_{1} + e_{1} \right)\%$
B) $\left( b_{1} + c_{1}\ –d_{1}\ –e_{1}\ \right)\%$
C) $\left( \alpha b_{1} + \beta c_{1} - \gamma d_{1} - \delta e_{1} \right)\%$
D) $\left( \alpha b_{1} + \beta c_{1} + \gamma d_{1} + \delta e_{1} \right)\%$
Explanation
$a = b^{\alpha}c^{\beta}/d^{\gamma}e^{\delta}$ So maximum error in $a$ is given by $\left( \begin{array}{r} \frac{\Delta a}{a} \times 100 \\ \ \end{array} \right)_{\max} = \alpha.\frac{\mathrm{\Delta}b}{b} \times 100 + \beta.\frac{\Delta c}{c} \times 100 + \gamma.\frac{\Delta d}{d} \times 100 + \delta.\frac{\Delta e}{e} \times 100$ $= (\alpha b_{1} + \beta c_{1} + \gamma d_{1} + \delta e_{1})\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.