Questions & Answers: "significant figures"

For $0.007\text{ m}^2$, the leading zeros are not significant; only '7' is significant (1 significant figure). For $2.64 \times 10^{24}\text{ kg}$, the exponential term does not contribute to the significant figures; only '2', '6', and '4' are significant (3 significant figures). For $0.2370\text{ g cm}^{-3}$, the trailing zero after the decimal point is significant, meaning '2', '3', '7', and '0' are all significant (4 significant figures).

In addition or subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places. Among the given values, $227.2\text{ g}$ has the least number of decimal places (only 1). The raw sum is $436.32 + 227.2 + 0.301 = 663.821\text{ g}$. Rounding this off to 1 decimal place gives $663.8\text{ g}$.

Density is calculated as $\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{4.237\text{ g}}{2.51\text{ cm}^3} \approx 1.688047\text{ g cm}^{-3}$. In multiplication or division, the final result must be rounded off to the same number of significant figures as the term with the least number of significant figures. Here, the volume ($2.51$) has 3 significant figures, while the mass ($4.237$) has 4. Therefore, the result must be rounded off to 3 significant figures, yielding $1.69\text{ g cm}^{-3}$.

Under NCERT/CBSE guidelines, if the digit to be dropped is exactly 5, we look at the preceding digit. If the preceding digit is even, it is left unchanged; hence, $2.745$ becomes $2.74$ (since 4 is even). If the preceding digit is odd, it is increased by 1; hence, $2.735$ becomes $2.74$ (since 3 is odd).

A change of units does not change the number of significant figures in a measurement. The original value $2.308\text{ cm}$ has 4 significant figures. In $0.02308\text{ m}$, the leading zeros are not significant, keeping the significant figures at 4. The other options change the precision and have 5 significant figures (e.g., $23.080\text{ mm}$ and $0.000023080\text{ km}$ contain an extra trailing zero, which is significant).

Exact numbers, such as pure mathematical constants ($\pi$) or exact counts of objects (20 bananas), have an infinite number of significant figures because they are exact values and can be represented as having an infinite number of decimal zeros (e.g., $20.0000...$).

The mathematical subtraction gives: $9.99 - 0.0099 = 9.9801\text{ m}$. In subtraction, the answer must contain the same number of decimal places as the number with the fewest decimal places. $9.99\text{ m}$ has 2 decimal places, whereas $0.0099\text{ m}$ has 4 decimal places. Rounding $9.9801$ to 2 decimal places yields $9.98\text{ m}$.

To show that a measurement of $4700\text{ m}$ is accurate to the nearest meter, all four digits must be significant (implying an uncertainty of $\pm 1\text{ m}$). Scientific notation resolves the ambiguity of trailing zeros. The expression $4.700 \times 10^3\text{ m}$ has 4 significant figures. In contrast, $4.7 \times 10^3$ has 2 significant figures, and $4700$ is ambiguous because trailing zeros without a decimal point are generally not considered significant.

First, perform the multiplication: $1.25 \times 3.2 = 4.0$. Since $3.2$ has 2 significant figures, the result of the multiplication must also have 2 significant figures, which makes it $4.0$ (having 1 decimal place). Now perform the subtraction: $4.0 - 0.12 = 3.88$. Since $4.0$ has only 1 decimal place, the final result must be rounded to 1 decimal place. Rounding $3.88$ to 1 decimal place gives $3.9$.

Let us count the significant figures for each: Student A: $0.06900$ has 4 significant figures (leading zeros are not significant; trailing zeros in a decimal are). Student B: $0.0069$ has 2 significant figures. Student C: $6.90 \times 10^{-2}$ has 3 significant figures. Student D: $6.9 \times 10^{-3}$ has 2 significant figures. Thus, Student A's measurement has the maximum number of significant figures.

The raw difference is $12.12 - 11.0 = 1.12\text{ cm}$. For subtraction, the final result must match the least number of decimal places of the inputs. $11.0\text{ cm}$ has 1 decimal place, whereas $12.12\text{ cm}$ has 2. Rounding $1.12\text{ cm}$ to 1 decimal place gives $1.1\text{ cm}$.

Calculating the value: $X = \frac{(2.0)^2 \times 3.00}{(1.000)^3} = \frac{4.0 \times 3.00}{1.000} = 12$. For multiplication and division, the result must possess the same number of significant figures as the term with the fewest significant figures. The component 'a' ($2.0$) has 2 significant figures, 'b' ($3.00$) has 3, and 'c' ($1.000$) has 4. Therefore, the result must have exactly 2 significant figures, which is represented by $12$.

To round off to four significant figures: For $143.45$, the digit to be dropped is $5$. The preceding digit $4$ is even, so it remains unchanged, yielding $143.4$. For $143.55$, the digit to be dropped is $5$. The preceding digit $5$ is odd, so it is rounded up by 1, yielding $143.6$.

The original measurement $5.00\text{ g}$ has 3 significant figures. When converting units, the number of significant figures must remain 3. Writing $5000\text{ mg}$ is ambiguous and generally implies 1 significant figure. Writing $5.00 \times 10^3\text{ mg}$ explicitly retains exactly 3 significant figures, maintaining the original precision.

Let's analyze each measurement: $0.0007\text{ s}$ has only 1 significant figure (leading zeros are not significant). $7.000\text{ s}$ has 4 significant figures (trailing zeros after a decimal point are significant). $70.00\text{ s}$ has 4 significant figures. $7.007\text{ s}$ has 4 significant figures (zeros between non-zero digits are significant). Therefore, $0.0007\text{ s}$ has the least number of significant figures.