Significant Figures Worksheet for Class 11 Physics | PDF

Significant Figures Worksheet for Class 11 Physics | PDF

Significant Figures Worksheet Class 11 PhysicsDownload

Answer Key — Significant Figures Worksheet

Section A — Identify Significant Figures

  • (a) 0.00450 → 3 s.f.
  • (b) 3.020 × 10⁵ → 4 s.f.
  • (c) 700.0 → 4 s.f.
  • (d) 0.000708 → 3 s.f.
  • (e) 6.00 × 10⁻³ → 3 s.f.
  • (f) 5002 → 4 s.f.
  1. Exactly three significant figures:
  • 0.00340
  • 3.40 × 10²
  • 304
  1. Examples:
  • (a) 1 s.f. → 5
  • (b) 2 s.f. → 5.2
  • (c) 4 s.f. → 5.234
  • (d) 5 s.f. → 5.2346

Section B — Rounding Off

  1. To 3 s.f.:
  • (a) 47.386 → 47.4
  • (b) 0.006784 → 0.00678
  • (c) 999.6 → 1.00 × 10³
  • (d) 12.049 → 12.0
  • (e) 5.995 → 6.00
  1. To 2 s.f.:
  • (a) 0.004967 → 0.0050
  • (b) 735.8 → 740
  • (c) 9.049 → 9.0
  • (d) 0.0996 → 0.10
  • (a) 0.0005608 → 5.61 × 10⁻⁴
  • (b) 48,730 → 4.873 × 10⁴
  • (c) 0.009995 → 0.0100
  • (d) 602,400 → 6.02 × 10⁵

Section C — Calculations

  • (a) 12.5×3.42=12.5\times3.42= 42.8
  • (b) 0.0045×2.31=0.0045\times2.31= 0.010
  • (c) 48.6÷2.4=48.6\div2.4= 20.
  • (d) 7.250÷0.25=7.250\div0.25= 29
  • (a) 12.35+4.6+0.789=12.35+4.6+0.789= 17.7
  • (b) 105.72−8.4=105.72-8.4= 97.3
  • (c) 0.0567+0.004+0.12=0.0567+0.004+0.12= 0.18
  • (d) 18.00−3.276=18.00-3.276= 14.72
  • (a) (4.52×2.1)+3.15=(4.52\times2.1)+3.15= 12.6
  • (b) (15.6−2.34)×4.2=(15.6-2.34)\times4.2= 56
  • (c) 25.0÷(3.2+1.84)=25.0\div(3.2+1.84)= 5.0

Section D — Conceptual

  1. 0.0050 and 0.050 each have two significant figures because leading zeros are not significant, while the final zero after the decimal is significant.
  2. No. 3000 is ambiguous when written without a decimal point or scientific notation. It could represent 1, 2, 3, or 4 significant figures depending on the intended precision. Scientific notation removes the ambiguity.
  3. Zeros between non-zero digits are significant because they represent measured digits.
    Examples: 1002 → 4 s.f.; 3.05 → 3 s.f.
  4. A calculated result cannot legitimately claim greater precision than the least precise measurement because the uncertainty of that measurement limits the reliability of the result.
  5. Both have the same numerical value, but 2.00 m indicates greater precision than 2.0 m.

Section E — Measurement

A=12.46×2.3=28.658 cm2A=12.46\times2.3=28.658\text{ cm}^2

Least precise measurement = 2 s.f.

Answer = 29 cm²

A=25.0×12.5=312.5A=25.0\times12.5=312.5

Both measurements have 3 s.f.

Answer = 313 cm²

V=43(3.14)(2.50)3V=\frac43(3.14)(2.50)^3

V=65.4167 cm3V=65.4167\text{ cm}^3

Correct to 3 s.f.:

Answer = 65.4 cm³

ρ=15.204.0=3.8 g cm−3\rho=\frac{15.20}{4.0}=3.8\text{ g cm}^{-3}

Answer = 3.8 g/cm³

Section F — Challenge

Q=(6.25)(0.0400)2.5=0.100Q=\frac{(6.25)(0.0400)}{2.5}=0.100

Least number of significant figures = 2.

Answer = 0.10

12.0+0.56=12.56→12.612.0+0.56=12.56\rightarrow12.6

12.6×3.254.0=10.2375\frac{12.6\times3.25}{4.0}=10.2375

Correct to 2 s.f.:

Answer = 10.

  1. 4.700 cm is incorrect. It claims four significant figures when the least precise measurement has only two.

The answer should be reported to 2 significant figures.

  1. 5.00 kg is more precise than 5.0 kg.
  • 5.0 → 2 s.f.
  • 5.00 → 3 s.f.

The stated precision increases by a factor of 10 in the decimal-place resolution.

  1. The result should be reported to the least number of significant figures among the inputs, which is 3.

0.003846271→0.003850.003846271\rightarrow\boxed{0.00385}

  1. (A) Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

2.35×4.2=9.872.35\times4.2=9.87

The answer 9.870 is incorrect because the least precise factor, 4.2, has only 2 significant figures.

Correct answer = 9.9.