A physical parameter a can be determined by m | vedclass

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(D) Given the relation: $a = \frac{b^\alpha c^\beta}{d^\gamma e^\delta}$.According to the theory of propagation of errors,for a quantity $a = b^\alpha

Vedclass · Accepted Answer

$(\alpha {b_1} + \beta {c_1} + \gamma {d_1} + \delta {e_1})\%$

Vedclass · Answer

(D) Given the relation: $a = \frac{b^\alpha c^\beta}{d^\gamma e^\delta}$.According to the theory of propagation of errors,for a quantity $a = b^\alpha c^\beta d^{-\gamma} e^{-\delta}$,the maximum relative error is given by the sum of the relative errors of the individual components multiplied by their respective powers.The maximum percentage error is calculated as:$\left( \frac{\Delta a}{a} 	imes 100 ight)_{\max} = |\alpha| \left( \frac{\Delta b}{b} 	imes 100 ight) + |\beta| \left( \frac{\Delta c}{c} 	imes 100 ight) + |-\gamma| \left( \frac{\Delta d}{d} 	imes 100 ight) + |-\delta| \left( \frac{\Delta e}{e} 	imes 100 ight)$.Substituting the given percentage errors $b_1\%, c_1\%, d_1\%, e_1\%$:$\left( \frac{\Delta a}{a} 	imes 100 ight)_{\max} = (\alpha b_1 + \beta c_1 + \gamma d_1 + \delta e_1)\%$.Thus,the correct option is $D$.

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