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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
(${b_1}\, + \,{c_1}\, + \,{d_1}\, + \,{e_1}$)$\%$
(${b_{1\,}}\, + \,{c_1}\, - \,{d_1}\, - \,{e_1}$)$\%$
($\alpha {b_1}\, + \,\beta {c_1}\, - \,\gamma {d_1}\, - \delta {e_1}$)$\%$
($\alpha {b_1} + \,\beta {c_1}\, + \,\gamma {d_1}\, + \,\delta {e_1}$)$\%$
Solution
(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)\% $
Similar Questions
Three students $S_{1}, S_{2}$ and $S_{3}$ perform an experiment for determining the acceleration due to gravity $(g)$ using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.
Student No. | Length of pendulum $(cm)$ | No. of oscillations $(n)$ | Total time for oscillations | Time period $(s)$ |
$1.$ | $64.0$ | $8$ | $128.0$ | $16.0$ |
$2.$ | $64.0$ | $4$ | $64.0$ | $16.0$ |
$3.$ | $20.0$ | $4$ | $36.0$ | $9.0$ |
(Least count of length $=0.1 \,{m}$, least count for time $=0.1\, {s}$ )
If $E_{1}, E_{2}$ and $E_{3}$ are the percentage errors in $'g'$ for students $1,2$ and $3$ respectively, then the minimum percentage error is obtained by student no. ……. .