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(B) The dimensional formula for torque $(	au)$ is given by $[	au] = [M^1 L^2 T^{-2}]$.According to the theory of propagation of errors,if a physical

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$25$

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(B) The dimensional formula for torque $(	au)$ is given by $[	au] = [M^1 L^2 T^{-2}]$.According to the theory of propagation of errors,if a physical quantity $X$ is given by $X = M^a L^b T^c$,the relative error is $\frac{\Delta X}{X} = a \frac{\Delta M}{M} + b \frac{\Delta L}{L} + c \frac{\Delta T}{T}$.Given that the percentage error in mass,length,and time is $5 \%$ each,i.e.,$\frac{\Delta M}{M} 	imes 100 = 5 \%$,$\frac{\Delta L}{L} 	imes 100 = 5 \%$,and $\frac{\Delta T}{T} 	imes 100 = 5 \%$.The percentage error in torque is calculated as:$\frac{\Delta 	au}{	au} 	imes 100 = (1 	imes \% M) + (2 	imes \% L) + (2 	imes \% T)$Substituting the given values:$\frac{\Delta 	au}{	au} 	imes 100 = 1(5 \%) + 2(5 \%) + 2(5 \%)$$\frac{\Delta 	au}{	au} 	imes 100 = 5 \% + 10 \% + 10 \% = 25 \%$.

$A$ torque meter is calibrated to reference standards of mass,length,and time,each with $5 \%$ accuracy. After calibration,the measured torque with this torque meter will have a net accuracy of $............\%$.

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