Class 11 NCERT PHYSICS • Chapter 1: Units and Measurement • Question 10

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We need to determine the number of significant figures for the provided measurements by applying standard scientific precision rules.

Concept: Significant figures

Significant figures are the digits in a measurement that carry meaning contributing to its precision. The following rules apply:

All non-zero digits are significant.
Zeros between non-zero digits (captive zeros) are significant.
Leading zeros (to the left of the first non-zero digit) are not significant.
Trailing zeros are significant only if a decimal point is explicitly shown.
In scientific notation ( $a \times 10^{b}$ ), all digits in the coefficient $a$ are significant.

Given:

(a) $0.007\,\text{m}^2$
(b) $2.64 \times 10^{24}\,\text{kg}$
(c) $0.2370\,\text{g}\cdotp\text{cm}^{-3}$
(d) $6.320\,\text{J}$
(e) $6.032\,\text{N}\cdotp\text{m}^{-2}$
(f) $0.0006032\,\text{m}^2$

Solving:

Analysis of (a): $0.007\,\text{m}^2$
- The zeros before the '7' are leading zeros acting as placeholders.
- Only the digit $7$ is significant.
- Count = 1.
Analysis of (b): $2.64 \times 10^{24}\,\text{kg}$
- The measurement is in scientific notation.
- The coefficient $2.64$ contains three non-zero digits $2, 6, 4$ .
- Count = 3.
Analysis of (c): $0.2370\,\text{g}\cdotp\text{cm}^{-3}$
- The leading zero is not significant.
- The digits $2, 3, 7$ are non-zero and significant.
- The trailing zero after the decimal point is significant.
- Count = 4.
Analysis of (d): $6.320\,\text{J}$
- The digits $6, 3, 2$ are non-zero and significant.
- The trailing zero after the decimal point is significant.
- Count = 4.
Analysis of (e): $6.032\,\text{N}\cdotp\text{m}^{-2}$
- The digits $6, 3, 2$ are non-zero and significant.
- The zero between 6 and 3 is a captive zero and is significant.
- Count = 4.
Analysis of (f): $0.0006032\,\text{m}^2$
- The zeros before the 6 are leading zeros and are not significant.
- The digits $6, 0, 3, 2$ are significant.
- The zero between 6 and 3 is a captive zero and is significant.
- Count = 4.

Answer

(a) 1
(b) 3
(c) 4
(d) 4
(e) 4
(f) 4

NCERT Solutions

Units and Measurement

State the number of significant figures in the following :

(a) $0.007 m^2$

(b) $2.64 × {10}^{24} kg$

(c) $0.2370 g {cm}^{−3}$

(d) $6.320 J$

(e) $ 6.032 N {m}^{−2}

(f) $0.0006032 m^2$

SOLUTION

We need to determine the number of significant figures for the provided measurements by applying standard scientific precision rules.

Concept: Significant figures

Given:

Solving:

Analysis of (a): $0.007\,\text{m}^2$

Analysis of (b): $2.64 \times 10^{24}\,\text{kg}$

Analysis of (c): $0.2370\,\text{g}\cdotp\text{cm}^{-3}$

Analysis of (d): $6.320\,\text{J}$

Analysis of (e): $6.032\,\text{N}\cdotp\text{m}^{-2}$

Analysis of (f): $0.0006032\,\text{m}^2$

Answer

Units and Measurement

State the number of significant figures in the following :

(a) 0.007m20.007 m^20.007m2

(b) 2.64×1024kg2.64 × {10}^{24} kg2.64×1024kg

(c) 0.2370gcm−30.2370 g {cm}^{−3}0.2370gcm−3

(d) 6.320J6.320 J6.320J

(e) $ 6.032 N {m}^{−2}

(f) 0.0006032m20.0006032 m^20.0006032m2

SOLUTION

We need to determine the number of significant figures for the provided measurements by applying standard scientific precision rules.

Concept: Significant figures

Given:

Solving:

Analysis of (a): 0.007 m20.007\,\text{m}^20.007m2

Analysis of (b): 2.64×1024 kg2.64 \times 10^{24}\,\text{kg}2.64×1024kg

Analysis of (c): 0.2370 g⋅cm−30.2370\,\text{g}\cdotp\text{cm}^{-3}0.2370g⋅cm−3

Analysis of (d): 6.320 J6.320\,\text{J}6.320J

Analysis of (e): 6.032 N⋅m−26.032\,\text{N}\cdotp\text{m}^{-2}6.032N⋅m−2

Analysis of (f): 0.0006032 m20.0006032\,\text{m}^20.0006032m2

Answer

(a) $0.007 m^2$

(b) $2.64 × {10}^{24} kg$

(c) $0.2370 g {cm}^{−3}$

(d) $6.320 J$

(f) $0.0006032 m^2$

Analysis of (a): $0.007\,\text{m}^2$

Analysis of (b): $2.64 \times 10^{24}\,\text{kg}$

Analysis of (c): $0.2370\,\text{g}\cdotp\text{cm}^{-3}$

Analysis of (d): $6.320\,\text{J}$

Analysis of (e): $6.032\,\text{N}\cdotp\text{m}^{-2}$

Analysis of (f): $0.0006032\,\text{m}^2$