In the context of accuracy of measurement and | vedclass

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(C) Statement $(1)$ is correct because accuracy is related to the number of significant figures and the precision of the measurement. $50.14 \, cm$ ha

Vedclass · Accepted Answer

Both are correct

Vedclass · Answer

(C) Statement $(1)$ is correct because accuracy is related to the number of significant figures and the precision of the measurement. $50.14 \, cm$ has $4$ significant figures,while $0.00025 \, A$ has only $2$ significant figures. Thus,the first measurement is more accurate.Statement $(2)$ is correct because when adding quantities with different units,we must consider the precision. $478 \, km = 478000 \, m$. Adding $397 \, m$ gives $478397 \, m$. However,in terms of significant figures and precision,the result is limited by the least precise measurement. Since $478 \, km$ has precision up to the units place $(1 \, km)$,the addition of $397 \, m$ $(0.397 \, km)$ is insignificant relative to the scale of $478 \, km$. Thus,the total distance is rounded to $478 \, km$.

List-$I$	List-$II$
$(a)$ Torque	$(i)$ $[M^{-1}L^{-2}T^4A^2]$
$(b)$ Gravitational constant	(ii) $[M^1L^2T^{-1}]$
$(c)$ Capacitance	(iii) $[M^{-1}L^3T^{-2}]$
$(d)$ Planck's constant	(iv) $[M^1L^2T^{-2}]$

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