Numerical Problems Based on Significant Figures for Class 11 Physics

• (i) $0.009$: Leading zeros are not significant. Only the ‘9’ is significant. (1 sig fig)
• (ii) $5.049$: All non-zero digits are significant. Zeros between non-zero digits are significant. (4 sig figs)
• (iii) $0.1890$: Leading zeros are not significant. Trailing zeros after a decimal point are significant. (4 sig figs)
• (iv) $1.90 \times 10^{11}$: The power of 10 is irrelevant. In ‘1.90’, trailing zeros after a decimal point are significant. (3 sig figs)
• (v) $0.020800$: Leading zeros (0.0…) are not significant. Zeros between non-zero digits (208) are significant. Trailing zeros after a decimal point (…800) are significant. (5 sig figs)
• (vi) $5.308$: Zeros between non-zero digits are significant. (4 sig figs)

Before adding or subtracting numbers in scientific notation, we must align the powers of 10.

Perform the subtraction on the coefficients:

The rule for addition/subtraction states the final result should have the same number of decimal places as the number with the fewest decimal places in the original aligned terms.
$3.9$ has one decimal place.
$0.25$ has two decimal places.
Therefore, our result must be rounded to one decimal place.

Rounding $3.65$ to one decimal place: Since the digit to be dropped is exactly 5, and the preceding digit (6) is even, we leave it as 6. Thus, $3.65$ becomes $3.6$.

(Note: The provided answer key states $3.7 \times 10^5$. This implies standard rounding where a 5 always rounds up, rather than the “round to even” rule standard in physics. If using standard rounding up, $3.65 \rightarrow 3.7$).

• (i) $15.654$ to 3 digits: Look at the 4th digit (5). It is followed by a non-zero digit (4), so we round up. Result: $15.7$
• (ii) $15.75$ to 3 digits: The 4th digit is exactly 5. The preceding digit (7) is odd, so we round up to make it even. Result: $15.8$
• (iii) $15.654$ to 4 digits: Look at the 5th digit (4). It is less than 5, so we leave the 4th digit as is. Result: $15.65$
• (iv) $15.65$ to 3 digits: The 4th digit is exactly 5. The preceding digit (6) is even, so it remains unchanged. Result: $15.6$
• (v) $142667$ to 5 digits: Look at the 6th digit (7). It is greater than 5, so round up the 5th digit. Result: $142670$ (Note: zero is added as a placeholder).
• (vi) $5.996 \times 10^5$ to 3 digits: Look at the 4th digit (6). It is greater than 5, so round up the 9. This carries over: $5.99 \rightarrow 6.00$. Result: $6.00 \times 10^5$
• (vii) $0.7995$ to 1 digit: The 2nd significant digit is 9. Rounding up the 7 gives 8. Result: $0.8$
• (viii) $2.5946 \times 10^{-4}$ to 2 digits: The 3rd digit is 9. Rounding up the 5 gives 6. Result: $2.6 \times 10^{-4}$

First, convert the weights to the same units. Let’s use kg.

Now, perform the addition:

According to the rules for addition and subtraction, the final result must be reported to the same number of decimal places as the measurement with the fewest decimal places.

$1.2$ has 1 decimal place.
$0.00542$ has 5 decimal places.

Therefore, our result must be rounded to 1 decimal place.

Given the diameter $D = 1.06 \text{ m}$.

The radius $r = \frac{D}{2} = \frac{1.06}{2} = 0.53 \text{ m}$.

The area of a circle is $A = \pi r^2$.

In multiplication, the result must be reported with the same number of significant figures as the measurement with the fewest significant figures.

Diameter $= 1.06$ (3 significant figures).
$\pi = 3.14$ (3 significant figures).

Therefore, our answer must be rounded to 3 significant figures.