JEE Physics Errors in Measurement MCQs Set 03

Answer: (d) 204.8 cm/s

Explanation:
1. λ = 2(l₂ − l₁), so Δλ = 2(Δl₁ + Δl₂) = 2(0.1 + 0.1) = 0.4 cm.
2. f is exact, so Δv = f Δλ = 512 × 0.4 = 204.8 cm/s.

Teacher's Note:
1. The error in a difference is the sum of the errors of the two readings.
2. A tuning fork frequency is taken as exact here.

Answer: (b) \( 2.66\ \text{g/cm}^3 \)

Explanation:
1. Least count = 1 mm − 0.9 mm = 0.1 mm. Side = 10 mm + 1 × 0.1 mm = 10.1 mm = 1.01 cm.
2. Volume = 1.01³ = 1.0303 cm³, which has three significant figures: 1.03 cm³.
3. Density = 2.736/1.0303 = 2.656 g/cm³, quoted to three significant figures: 2.66 g/cm³.

Teacher's Note:
1. The result cannot have more significant figures than the least precise measurement.
2. Mass has four significant figures but the side has only three.

Answer: (b) f < x < 2f

Explanation:
1. In parallax, the farther object appears to move in the direction opposite to the eye's movement, relative to the nearer one.
2. The eye moves left and the image appears to move right, so the image is farther from the eye (further from the mirror) than the object pin.
3. For a real inverted image, v > u means the object is between f and 2f.

Teacher's Note:
1. Parallax is used to place the pin at the position of the image: no relative shift means they coincide.
2. For x between f and 2f the image forms beyond 2f.

Answer: (b) \( (2.0 \pm 0.2) \times 10^{11}\ \text{N/m}^2 \)

Explanation:
1. \( Y = \frac{4MgL}{\pi d^2 l} = \frac{4 \times 1.0 \times 9.8 \times 2}{\pi(0.4\times10^{-3})^2(0.8\times10^{-3})} \approx 1.95 \times 10^{11} \approx 2.0 \times 10^{11}\ \text{N/m}^2 \).
2. \( \frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta l}{l} = 2\times\frac{0.01}{0.4} + \frac{0.05}{0.8} = 5\% + 6.25\% = 11.25\% \).
3. \( \Delta Y \approx 0.1125 \times 2.0\times10^{11} = 0.2\times10^{11} \).

Teacher's Note:
1. Mass, g and length are exact, so only d and the extension contribute.
2. Diameter is squared in Y.

Answer: (d) 3.38 mm

Explanation:
1. Pitch = 1 mm/2 = 0.5 mm, so least count = 0.5/50 = 0.01 mm.
2. Observed reading = 3 + 35 × 0.01 = 3.35 mm.
3. Zero error is − 0.03 mm, so corrected diameter = 3.35 − (− 0.03) = 3.38 mm.

Teacher's Note:
1. Correction = − (zero error).
2. The pitch is the distance moved in ONE rotation, not two.

Answer: (a) one minute

Explanation:
1. 30 VSD = 29 MSD, so 1 VSD = (29/30) MSD.
2. Least count = 1 MSD − 1 VSD = MSD/30 = 0.5°/30 = (1/60)° = 1 minute.

Teacher's Note:
1. Least count = smallest main scale division / number of vernier divisions.
2. 1° = 60 minutes.

Answer: (a) 0.052 cm

Explanation:
1. Least count = 1 mm/100 = 0.01 mm.
2. Diameter = 0 + 52 × 0.01 mm = 0.52 mm = 0.052 cm.

Teacher's Note:
1. Convert to the unit asked in the options at the end.
2. 1 mm = 0.1 cm.

Answer: (a) 58.65°

Explanation:
1. Least count = 0.5°/30 = 1/60° (1 minute).
2. Reading = 58.5° + 9 × (1/60)° = 58.5° + 0.15° = 58.65°.

Teacher's Note:
1. Vernier reading = coinciding division × least count.
2. Convert minutes to degrees by dividing by 60.

Answer: (c) 6%

Explanation:
1. R = V/I, so the maximum percentage error is 3% + 3% = 6%.

Teacher's Note:
1. Errors never cancel in the maximum error of a quotient.
2. Use the zero answer only if the errors were known to be correlated.

Answer: (d) A vernier callipers where 10 divisions in the vernier scale match 9 divisions in the main scale and the main scale has 10 divisions in 1 cm

Explanation:
1. 3.50 cm is quoted to the second decimal place, so the least count must be 0.01 cm = 0.1 mm.
2. Least counts: (a) 0.01 mm = 0.001 cm; (b) 0.02 mm = 0.002 cm; (c) 0.1 cm; (d) MSD = 0.1 cm and LC = 0.1/10 = 0.01 cm.
3. Only option (d) has a least count of 0.01 cm.

Teacher's Note:
1. The number of decimal places in a reading tells you the least count.
2. Screw gauges measure small thicknesses, not rod lengths like 3.5 cm.

Answer: (d) 3%

Explanation:
1. \( \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \).
2. ΔL/L = 0.1/20.0 = 0.5%; ΔT/T = 1/90 = 1.11%.
3. Total = 0.5% + 2(1.11%) = 2.7%, which is about 3%.

Teacher's Note:
1. Total time for 100 oscillations has the same relative error as one period.
2. Quote the answer to the nearest whole percent.

Answer: (d) 92 ± 2 s

Explanation:
1. Mean = (90 + 91 + 95 + 92)/4 = 92 s.
2. Absolute deviations: 2, 1, 3, 0; their mean = 1.5 s.
3. The error cannot be smaller than the least count of 1 s; 1.5 s is quoted as 2 s (same decimal places as the readings).

Teacher's Note:
1. The reported error should have the same decimal places as the readings.
2. Mean absolute deviation, not the largest deviation, is used.

Answer: (a) 0.80 mm

Explanation:
1. Least count = 0.5/50 = 0.01 mm.
2. Zero error = − (50 − 45) × 0.01 = − 0.05 mm.
3. Observed = 0.5 + 25 × 0.01 = 0.75 mm.
4. Corrected thickness = 0.75 − (− 0.05) = 0.80 mm.

Teacher's Note:
1. When the jaws are closed and the reading is above 25 divisions, the zero error is negative.
2. Subtract the zero error with its sign.

Answer: (b) E_I is minimum

Explanation:
1. \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t} \) (the time error for the total time t is 0.1 s; the factor n cancels).
2. I: 0.1/64.0 + 2(0.1/128.0) = 0.0016 + 0.0016 = 0.31%.
3. II: 0.1/64.0 + 2(0.1/64.0) = 0.0016 + 0.0031 = 0.47%.
4. III: 0.1/20.0 + 2(0.1/36.0) = 0.0050 + 0.0056 = 1.06%.
5. So E_I is the minimum (and E_III is the maximum).

Teacher's Note:
1. A longer total time reduces the relative time error.
2. Compare numbers, not guesses.

Answer: (c) (66, 33) and (78, 39)

Explanation:
1. Mirror formula: \( v = \frac{uf}{u - f} \) with f = 24 cm (magnitudes).
2. u = 42: v = 56.0; u = 48: v = 48.0; u = 60: v = 40.0 (all match).
3. u = 66: v = 37.7 cm but 33 is recorded (difference 4.7 cm, far more than 0.2 cm).
4. u = 78: v = 34.7 cm but 39 is recorded (difference 4.3 cm).

Teacher's Note:
1. Compute the expected v for each u, then compare with the maximum error 0.2 cm.
2. Check (u, v) pairs with 1/u + 1/v = 1/f.

Answer: (d) 0.2 mm

Explanation:
1. 20 VSD = 16 MSD = 16 mm, so 1 VSD = 0.8 mm.
2. Least count = 1 MSD − 1 VSD = 1 − 0.8 = 0.2 mm.

Teacher's Note:
1. Least count is the difference of one main and one vernier division.
2. Here the vernier division is shorter than the main scale division.

Answer: (c) 3.1%

Explanation:
1. Least count = 0.5/50 = 0.01 mm; diameter = 2.5 + 20 × 0.01 = 2.70 mm.
2. Density \( \rho = \frac{m}{\frac{\pi}{6}d^3} \), so \( \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta d}{d} \).
3. 3 × 0.01/2.70 = 1.11%; total = 2% + 1.11% = 3.1%.

Teacher's Note:
1. The diameter is cubed in the volume of a sphere.
2. Take the least count as the error in d.

Answer: (a) due to the errors in the measurements of d and l are the same

Explanation:
1. Both instruments have least count = 0.5 mm/100 = 0.005 mm.
2. Contribution of d = 2 × 0.005/0.5 = 2%.
3. Contribution of l = 0.005/0.25 = 2%.
4. So the two contributions are equal.

Teacher's Note:
1. Because d is squared its contribution is doubled, but it is also twice as large as l is small.
2. Compute both percentages before comparing.

Answer: (b) 5.124 cm

Explanation:
1. Main scale divisions are 0.05 cm apart (5.10, 5.15), so 49 MSD = 2.45 cm and 50 VSD = 49 MSD.
2. 1 VSD = 2.45/50 = 0.049 cm; least count = 0.05 − 0.049 = 0.001 cm.
3. Diameter = 5.10 + 24 × 0.001 = 5.124 cm.

Teacher's Note:
1. Find the main scale division from the readings between which the zero lies.
2. Least count = MSD − VSD.

Answer: (d) 4.5%

Explanation:
1. \( \rho = \frac{m}{a^3} \), so \( \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta a}{a} \).
2. = 1.5% + 3(1%) = 4.5%.

Teacher's Note:
1. Volume of a cube is a³, so the length error counts three times.
2. Add the mass error.