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 $478 \, km$.

$A$. $Pa \cdot s$	$(i)$. $[L^2 T^{-2} K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$. $[MLT^{-3} K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$. $[ML^{-1} T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$. $[ML^2 T^{-2} K^{-1}]$

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