Question

If \(N_A\), \(N_B\) and \(N_C\) are the number of significant figures in \(A=0.001204\text{ m}\), \(B=43120000\text{ m}\) and \(C=1.200\text{ m}\) respectively then

• A. \(N_A=N_B=N_C\)
• B. \(N_A\gt N_B\gt N_C\)
• C. \(N_A\lt N_B\lt N_C\)
• D. \(N_A\gt N_B\lt N_C\)

Hint:

Leading zeros are never significant, while zeros between non-zero digits and trailing zeros after a decimal point are significant.

Answer 1

The Correct Option is A

Step 1: Recall the rule for leading zeros.
Any zeros that appear before the first non zero digit are only place holders, so they are never counted as significant. We will use this idea for the first number.
Step 2: Count the significant figures of $A=0.001204$.
The leading zeros $0.00$ are dropped. The remaining digits are $1,2,0,4$, and the middle zero sits between two non zero digits so it counts. Hence $N_A=4$.
Step 3: Recall the rule for trailing zeros with no decimal point.
When a whole number ends in zeros and has no decimal point, those final zeros are taken as not significant. We apply this to the second number.
Step 4: Count the significant figures of $B=43120000$.
The trailing zeros are dropped, leaving the digits $4,3,1,2$. Hence $N_B=4$.
Step 5: Count the significant figures of $C=1.200$.
Here a decimal point is present, so the zeros after $2$ are deliberately written and are significant. The digits $1,2,0,0$ give $N_C=4$.
Step 6: Compare and conclude.
All three counts came out equal, $N_A=N_B=N_C=4$, so the matching option is the one stating equality.
\[ \boxed{N_A=N_B=N_C} \]