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(B) The formula for linear thermal expansion is given by $\Delta L = L_0 \alpha \Delta 	heta$.Here,$\Delta L$ is the allowed error in length,$L_0$ is

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(B) The formula for linear thermal expansion is given by $\Delta L = L_0 \alpha \Delta 	heta$.Here,$\Delta L$ is the allowed error in length,$L_0$ is the original length,$\alpha$ is the coefficient of linear expansion,and $\Delta 	heta$ is the change in temperature.Given: $\Delta L = 5 	imes 10^{-5} \ mm$,$L_0 = 1 \ mm$ (as we are considering millimeter intervals),and $\alpha = 10 	imes 10^{-6} \ K^{-1}$.Rearranging the formula to solve for $\Delta 	heta$:$\Delta 	heta = \frac{\Delta L}{\alpha L_0} = \frac{5 	imes 10^{-5}}{10 	imes 10^{-6} 	imes 1} = \frac{5 	imes 10^{-5}}{10^{-5}} = 5 \ ^oC$.Thus,the maximum temperature variation allowable is $5 \ ^oC$.

List-$I$	List-$II$
$A$. Figure of merit $(k)$	$P$. in series
$B$. Current sensitivity $(S_I)$	$Q$. to get power loss minimized
$C$. Deflection is made half by using resistance	$R$. in shunt/parallel
$D$. High resistance box connected	$S$. current per unit deflection
	$T$. deflection per unit current

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