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(A) For $74.083$,all non-zero digits and the zero between them are significant,so there are $5$ significant figures $(IV)$.For $0.029$,leading zeros a

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$IV, III, II, I$

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(A) For $74.083$,all non-zero digits and the zero between them are significant,so there are $5$ significant figures $(IV)$.For $0.029$,leading zeros are not significant,so there are $2$ significant figures $(III)$.For $0.002407$,leading zeros are not significant,but the zero between $2$ and $4$ is significant,so there are $4$ significant figures $(II)$.For $2.74 	imes 10^7$,the exponential term does not contribute to significant figures,so there are $3$ significant figures $(I)$.Thus,the correct matching is $A-IV, B-III, C-II, D-I$.

$A$. $74.083$	$I$. $3$
$B$. $0.029$	$II$. $4$
$C$. $0.002407$	$III$. $2$
$D$. $2.74 \times 10^7$	$IV$. $5$

Match the measurements given in List-$I$ with the number of significant figures given in List-$II$.
$A$. $74.083$ $I$. $3$
$B$. $0.029$ $II$. $4$
$C$. $0.002407$ $III$. $2$
$D$. $2.74 \times 10^7$ $IV$. $5$

The correct answer is:

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Match the measurements given in List-$I$ with the number of significant figures given in List-$II$.$A$. $74.083$$I$. $3$$B$. $0.029$$II$. $4$$C$. $0.002407$$III$. $2$$D$. $2.74 \times 10^7$$IV$. $5$The correct answer is: