If the value of resistance is 10.845 , Omega | vedclass

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Question

(A) According to Ohm's law,the potential difference $V$ is given by $V = I 	imes R$.Given,$R = 10.845 \, \Omega$ (which has $5$ significant figures)

Vedclass · Accepted Answer

$35.0$

Vedclass · Answer

(A) According to Ohm's law,the potential difference $V$ is given by $V = I 	imes R$.Given,$R = 10.845 \, \Omega$ (which has $5$ significant figures) and $I = 3.23 \, 	ext{A}$ (which has $3$ significant figures).Calculating the product: $V = 10.845 	imes 3.23 = 35.02935 \, V$.According to the rules of significant figures in multiplication,the final result should have the same number of significant figures as the measurement with the least number of significant figures.Here,the least number of significant figures is $3$ (from $3.23$).Rounding $35.02935$ to $3$ significant figures,we get $35.0 \, V$.

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

If the value of resistance is $10.845 \, \Omega$ and the value of current is $3.23 \, \text{A}$,the value of potential with significant figures would be ........... $V$.

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Keeping the significant figures in view,the sum of the physical quantities $52.01 \ m$,$153.2 \ m$,and $0.123 \ m$ is: (in $m$)

The respective number of significant figures for the numbers $23.023$,$0.0003$,and $2.1 \times 10^3$ are:

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

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