Q1. Define measurement. Why is the choice of a standard unit important in physics?

Ans: Measurement is the comparison of an unknown physical quantity with a predefined, internationally accepted standard of the same quantity.

The choice of a standard unit is important because:

  • It ensures uniformity and reproducibility.
  • It allows comparison across experiments.
  • It forms the basis of standard scientific laws.

Q2. List the seven base units of the SI system and write one derived unit.

Ans: Base units:

  • Length (m), Mass (kg), Time (s), Current (A), Temperature (K), Amount of substance (mol), Luminous intensity (cd)

Derived unit example: Force → N=kg⋅m/s2\text{N} = \text{kg} \cdot \text{m/s}^2

Q3. Explain the term ‘significant figures’. A quantity is measured as 2.350 m2.350\ \text{m}. How many significant figures does it have?

Ans: Significant figures are digits in a measurement known with certainty, plus the first doubtful digit. 2.350 m2.350\ \text{m} has 4 significant figures — 2, 3, 5 are significant and the trailing 0 is significant because it's after a decimal.

Q4. What is dimensional analysis? State any two uses.

Ans: Dimensional analysis uses the dimensions of quantities to:

  1. Check dimensional consistency
  2. Derive relations between quantities

It is based on the principle that all terms in an equation must have the same dimensions.

Q5. Using dimensional analysis, derive the formula for time period TT of a simple pendulum.

Ans: Let T∝lagbT \propto l^a g^b [T]=[L]a[LT−2]b=[La+bT−2b][T] = [L]^a [L T^{-2}]^b = [L^{a + b} T^{-2b}] Equating powers: a+b=0a + b = 0, −2b=1-2b = 1 ⇒ b=−1/2b = -1/2, a=1/2a = 1/2 So: T=klgT = k \sqrt{\frac{l}{g}}

Q6. A force of 1 dyne acts on a mass of 1 g. Calculate the acceleration in SI units.

Ans: (F = 1\,\text{dyne},\ m = 1\,\text{g})

In CGS: a=F/m=1 cm/s2a = F/m = 1\ \text{cm/s}^2 Convert to SI: 1 cm/s2=0.01 m/s21\ \text{cm/s}^2 = 0.01\ \text{m/s}^2

Ans: a=0.01 m/s2a = 0.01\ \text{m/s}^2

Q7. Differentiate between accuracy and precision with examples.

Ans:

  • Accuracy = closeness to true value
  • Precision = repeatability

Example:

  • Readings 1.99, 2.01, 2.00 → accurate & precise
  • Readings 1.81, 1.80, 1.79 → precise but not accurate

Q8. Show that the equation v2=u2+2asv^2 = u^2 + 2as is dimensionally correct.

Ans:

  • v2v^2: [L2T−2][L^2 T^{-2}]
  • u2u^2: [L2T−2][L^2 T^{-2}]
  • 2as2as: [LT−2][L]=[L2T−2][L T^{-2}] [L] = [L^2 T^{-2}]

✅ All terms have same dimensions. Hence, the equation is dimensionally homogeneous.