JEE Physics Errors in Measurement MCQs Set 01

Answer: (b) 8%

Explanation:
1. Kinetic energy \( K = \tfrac12 m v^2 \), so \( \frac{\Delta K}{K} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v} \).
2. Maximum error = 2% + 2 × 3% = 8%.

Teacher's Note:
1. Powers multiply the percentage error of that quantity, and different quantities always add (never subtract) for the maximum error.
2. Constants such as 1/2 carry no error.

Answer: (c) unit less

Explanation:
1. Percentage error \( = \frac{\Delta a}{a} \times 100 \), a ratio of two quantities with the same unit.
2. The unit cancels, so percentage error is a pure number.

Teacher's Note:
1. Relative and percentage errors are dimensionless; only the absolute error carries the unit of the quantity.
2. That is why they can be added for different quantities.

Answer: (c) 0.94 ± 0.02 cm

Explanation:
1. Difference = 4.19 − 3.25 = 0.94 cm.
2. For a difference the absolute errors add: 0.01 + 0.01 = 0.02 cm.
3. So the answer is 0.94 ± 0.02 cm.

Teacher's Note:
1. Absolute errors add for both sum and difference.
2. Subtraction never reduces the error.

Answer: (b) \( (3.45 \pm 0.3)\ \text{m s}^{-1} \)

Explanation:
1. \( v = 13.8/4.0 = 3.45 \) m/s.
2. \( \frac{\Delta v}{v} = \frac{0.2}{13.8} + \frac{0.3}{4.0} = 0.0145 + 0.075 = 0.0895 \).
3. \( \Delta v = 3.45 \times 0.0895 \approx 0.31 \), which is quoted as 0.3 m/s.

Teacher's Note:
1. Relative errors add in a quotient; convert back to an absolute error by multiplying with the result.
2. Quote the error to one significant figure.

Answer: (c) 8.95%

Explanation:
1. \( \frac{\Delta v}{v} \times 100 = \left(\frac{0.2}{13.8} + \frac{0.3}{4.0}\right) \times 100 \).
2. = 1.45% + 7.5% = 8.95%.

Teacher's Note:
1. Add the percentage errors of the quantities in a quotient.
2. Keep intermediate values unrounded.

Answer: (d) Δl = 1 mm, ΔT = 0.1 s, n = 50, amplitude 1 mm

Explanation:
1. \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \). The time of n oscillations carries error \( \Delta T + 0.1 \), so \( \frac{\Delta T_{period}}{T} = \frac{\Delta T + 0.1}{nT} \).
2. A smaller Δl, a smaller least count and a larger n all reduce the error. Set (d) is best on each of the three counts.
3. Amplitude does not enter the formula, so it does not matter for small oscillations.

Teacher's Note:
1. Timing many oscillations divides the stop watch error by n.
2. Check each term of the error formula separately; here one option is smallest in all of them.

Answer: (a) high resistivity and low temperature coefficient

Explanation:
1. High resistivity lets a long, thin enough wire give a measurable resistance, and gives a sensitive balance.
2. A low temperature coefficient keeps the resistance per unit length constant when the wire warms up, so the length ratio stays reliable.

Teacher's Note:
1. Materials such as manganin and constantan are used for this reason.
2. Heating by current is a standard source of error in bridge experiments.

Answer: (a) 5.0 ± 11%

Explanation:
1. Relative density \( = \frac{W_{air}}{W_{air} - W_{water}} = \frac{5.00}{1.00} = 5.0 \).
2. Loss of weight = 1.00 ± 0.10 N (absolute errors add in the difference), i.e. 10%.
3. Maximum error = 0.05/5.00 × 100 + 10 = 1% + 10% = 11%.

Teacher's Note:
1. Take the error of the difference first, then treat the ratio.
2. A small difference of two nearly equal readings has a large relative error.

Answer: (b) 0.11 s

Explanation:
1. Mean = (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/5 = 2.624 s.
2. Absolute deviations: 0.006, 0.064, 0.204, 0.086, 0.176 s.
3. Their mean = 0.536/5 = 0.107 s, i.e. 0.11 s.

Teacher's Note:
1. Average absolute error is the mean of the magnitudes of deviations from the mean.
2. Quote it with the same number of decimal places as the readings.

Answer: (b) ± 16%

Explanation:
1. Maximum error = 2(3%) + 4% + 6% = 16%.

Teacher's Note:
1. The power of I is 2, so its percentage error is doubled.
2. Add all contributions.

Answer: (b) \( 9.378\ \text{m}^2 \)

Explanation:
1. Area = 3.124 × 3.002 = 9.378248 m².
2. Rounding to four significant figures: the fifth digit is 2, so it is dropped, giving 9.378 m².

Teacher's Note:
1. Count significant figures from the first non-zero digit.
2. Round only at the end of the calculation.

Answer: (c) 3%

Explanation:
1. \( V = \pi r^2 l \), so \( \frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \).
2. Error in r = least count of the vernier = 0.01 cm, so 2 × 0.01/2.0 = 1%.
3. Error in l = 0.1/5.0 = 2%.
4. Total = 1% + 2% = 3%.

Teacher's Note:
1. The radius enters squared, so its percentage error is doubled.
2. The least count of the instrument is taken as the absolute error.

Answer: (c) 6.40%

Explanation:
1. \( \frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + 2\frac{\Delta D}{D} + \frac{\Delta l}{l} \). The least count in each reading is its last digit.
2. ΔM/M = 0.01/3.00 = 0.33%; ΔL/L = 0.001/2.820 = 0.035%; 2ΔD/D = 2 × 0.001/0.041 = 4.88%; Δl/l = 0.001/0.087 = 1.15%.
3. Sum = 0.33 + 0.035 + 4.88 + 1.15 = 6.40%.

Teacher's Note:
1. g is taken as exact. The smallest measurement (l and D) contributes the most.
2. Squared quantities double their percentage error.

Answer: (c) n = 10, m = 0.8 cm

Explanation:
1. Least count = 1 MSD − 1 VSD, so 1 VSD = 0.1 − 0.02 = 0.08 cm.
2. For n = 10 divisions the vernier scale length m = 10 × 0.08 = 0.8 cm.
3. Check the others: n = 9 would need m = 0.72 cm, and 0.5, 0.2 give VSD of 0.05, 0.02 cm, none with L.C. 0.02 cm.

Teacher's Note:
1. Least count = value of one main scale division − value of one vernier division.
2. Also equals MSD divided by the number of vernier divisions only when n VSD = (n−1) MSD.

Answer: (a) 1

Explanation:
1. \( N \) MSD = \( (N+m) \) VSD, so 1 VSD = \( \frac{N}{N+m} \) MSD.
2. Least count = 1 MSD − 1 VSD = \( \frac{m}{N+m} \) MSD (in magnitude).
3. \( \frac{m}{N+m} \) increases with m, so it is smallest for the smallest allowed m = 1.

Teacher's Note:
1. Write the vernier division in terms of the main scale division first.
2. A function m/(N+m) is increasing in m.

Answer: (a) So that the wire can have its desired change in length

Explanation:
1. A loaded wire stretches slowly (elastic after effect), so the full extension is reached only after a short time.
2. Reading immediately would give a smaller extension than the true one.

Teacher's Note:
1. Waiting is about letting the wire settle to the equilibrium extension.
2. The wire is in a closed loop of readings: load, wait, read.

Answer: (a) 4%

Explanation:
1. Area \( = l^2 \), so \( \frac{\Delta A}{A} = 2\frac{\Delta l}{l} = 2 \times 2\% = 4\% \) (small fractional change, first order).

Teacher's Note:
1. Areal expansion is twice the linear expansion for small changes.
2. This is the power rule for errors applied to expansion.

Answer: (a) 5%

Explanation:
1. Volume = l × b × h, so the percentage errors add.
2. 0.1/5 + 0.1/10 + 0.1/5 = 2% + 1% + 2% = 5%.

Teacher's Note:
1. Take the least count as the absolute error in each dimension.
2. Smaller dimensions contribute more percentage error.

Answer: (a) 15 Ω ± 2%, 3.3 Ω ± 3%

Explanation:
1. Series: R = 15 Ω, ΔR = 0.2 + 0.1 = 0.3 Ω, so 0.3/15 = 2%.
2. Parallel: \( R = \frac{5 \times 10}{15} = 3.33\ \Omega \) and \( \frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \).
3. \( \Delta R = 11.11\left(\frac{0.2}{25} + \frac{0.1}{100}\right) = 0.1\ \Omega \), i.e. 0.1/3.33 = 3%.

Teacher's Note:
1. For parallel resistors differentiate 1/R = 1/R₁ + 1/R₂ rather than using the quotient rule.
2. Series errors add directly.

Answer: (c) Both are correct

Explanation:
1. 50.14 has four significant figures while 0.00025 has two, so (I) is correct.
2. Total = 478 km + 0.397 km = 478.397 km. The least precise number, 478, has no decimal place, so the sum is 478 km: (II) is correct.

Teacher's Note:
1. Leading zeros are not significant.
2. In addition and subtraction, keep as many decimals as the least precise term.