A physical parameter a can be determined by measuring the paramet… - Vidyadip

MCQ

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
(${b_1}\, + \,{c_1}\, + \,{d_1}\, + \,{e_1}$)$\%$
B
(${b_{1\,}}\, + \,{c_1}\, - \,{d_1}\, - \,{e_1}$)$\%$
C
($\alpha {b_1}\, + \,\beta {c_1}\, - \,\gamma {d_1}\, - \delta {e_1}$)$\%$
✓
($\alpha {b_1} + \,\beta {c_1}\, + \,\gamma {d_1}\, + \,\delta {e_1}$)$\%$

✓

Answer

Correct option: D.

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

d
(d) $a = {b^\alpha }\,{c^\beta }/{d^\gamma }\,{e^\delta }$

So maximum error in a is given by

${\left( {\frac{{\Delta a}}{a} \times 100} \right)_{\max }} = \alpha \,.\,\frac{{\Delta b}}{b} \times 100 + \beta \,.\,\frac{{\Delta c}}{c} \times 100$

$ + \gamma \,.\,\frac{{\Delta d}}{d} \times 100 + \delta \,.\,\frac{{\Delta e}}{e} \times 100$

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

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