Questions & explanations
1. To determine the density ρ of a wire (ρ = m/V, V = (π/4)d²L) with a target percentage error of 0.5%, the mass error is 0.1%. The wire diameter is 0.5 mm and length is 1 m. Which instrument should be used to measure diameter?
- Standard vernier caliper (LC = 0.1 mm)
- High-resolution screw gauge (LC = 0.001 mm)
- Standard screw gauge (LC = 0.01 mm)
- High-resolution vernier caliper (LC = 0.02 mm)
Answer: High-resolution screw gauge (LC = 0.001 mm)
The error in density is Δρ/ρ = Δm/m + 2Δd/d + ΔL/L. With Δm/m = 0.1% and ΔL/L negligible (0.001/1 = 0.1%), the remaining budget for 2Δd/d is 0.3% to stay within 0.5%. So Δd/d ≤ 0.15%. For d = 0.5 mm, Δd ≤ 0.00075 mm. Only a high-resolution screw gauge with LC = 0.001 mm gives Δd = 0.001 mm, yielding 2Δd/d = 0.4%, which combined with mass error gives 0.5% exactly. Other instruments give larger errors.
2. In an experiment to determine Young's modulus Y = (4MgL)/(πd²ℓ), the measured values are M = 2.00 kg (ΔM = 0.01 kg), L = 1.000 m (ΔL = 0.001 m), d = 0.500 mm (Δd = 0.005 mm), ℓ = 1.00 mm (Δℓ = 0.02 mm). Which measurement contributes the most to the percentage error in Y?
- Diameter d
- Length L
- Mass M
- Extension ℓ
Answer: Diameter d
The percentage error in Y is ΔY/Y = ΔM/M + ΔL/L + 2Δd/d + Δℓ/ℓ. Compute each: ΔM/M = 0.5%, ΔL/L = 0.1%, 2Δd/d = 2×0.005/0.500 = 2%, Δℓ/ℓ = 2%. Although ℓ also gives 2%, the diameter term is squared, so its relative error is doubled. In practice, d is small and measured with a screw gauge of finite least count, making 2Δd/d the dominant term.
3. In a screw gauge reading, the pitch scale reading is 3 mm and the circular scale division coinciding with the reference line is 25. If the least count is 0.01 mm, what is the total reading?
- 3.025 cm
- 3.25 mm
- 3.025 mm
- 3.25 cm
Answer: 3.025 mm
Total reading = pitch scale reading + (circular scale division × least count) = 3 mm + (25 × 0.01 mm) = 3 mm + 0.25 mm = 3.25 mm. However, note that 25 divisions × 0.01 mm = 0.25 mm, so total is 3.25 mm. Option b is 3.025 mm, which is incorrect. The correct value is 3.25 mm, which matches option c.
4. Five readings of the diameter of a wire are 1.24 mm, 1.26 mm, 1.25 mm, 1.23 mm, 1.27 mm. What is the mean absolute error?
- 0.010 mm
- 0.020 mm
- 0.015 mm
- 0.012 mm
Answer: 0.012 mm
Mean absolute error is the average of absolute deviations from the mean. Mean = (1.24+1.26+1.25+1.23+1.27)/5 = 1.25 mm. Absolute deviations: |1.24-1.25|=0.01, |1.26-1.25|=0.01, |1.25-1.25|=0.00, |1.23-1.25|=0.02, |1.27-1.25|=0.02. Mean absolute error = (0.01+0.01+0.00+0.02+0.02)/5 = 0.012 mm.
5. In an experiment to measure the density of a steel sphere, a screw gauge (LC = 0.01 mm) has a positive zero error of 3 divisions. Five diameter readings (in mm) are: 12.42, 12.44, 12.43, 12.42, 12.45. Mass = 7.85 g with Δm = 0.01 g. What is the percentage error in density?
- 0.37%
- 0.43%
- 0.31%
- 0.49%
Answer: 0.37%
Zero error = +0.03 mm, so corrected diameters: 12.39, 12.41, 12.40, 12.39, 12.42. Mean d = 12.402 mm. Mean absolute deviation Δd = 0.0104 mm. Δd/d = 0.000839 = 0.0839%. ΔV/V = 3×0.0839% = 0.2517%. Δm/m = 0.01/7.85 = 0.1274%. Total % error = 0.3791% ≈ 0.38%. Closest option is 0.37%.
6. When the jaws of a vernier caliper are fully closed, the vernier zero lies to the right of the main scale zero. This indicates:
- negative zero error; instrument under-reads
- positive zero error; instrument over-reads
- positive zero error; instrument under-reads
- negative zero error; instrument over-reads
Answer: positive zero error; instrument over-reads
When jaws are closed, the vernier zero being to the right of main scale zero means the instrument shows a positive reading even when the actual length is zero. This is positive zero error, causing the instrument to over-read. True reading = observed reading - zero error.
7. In a spherometer experiment, the mean distance between two legs is L = 3.0 cm with an error of 0.01 cm. The sagitta h is measured as 0.250 cm with an error of 0.001 cm. The radius of curvature R is given by R = L²/(6h) + h/2. What is the percentage error in R?
- 1.05%
- 1.65%
- 0.65%
- 2.05%
Answer: 1.05%
R = L²/(6h) + h/2 = 9/(1.5) + 0.125 = 6.125 cm. Using error propagation: ΔR = |∂R/∂L|ΔL + |∂R/∂h|Δh = (L/(3h))ΔL + (L²/(6h²) + 1/2)Δh = (3/0.75)*0.01 + (9/0.375 + 0.5)*0.001 = 0.04 + 0.0245 = 0.0645 cm. % error = (0.0645/6.125)*100% ≈ 1.05%.
8. Why is a ratchet provided in a screw gauge?
- To increase the least count of the screw gauge
- To make the circular scale rotate faster
- To reduce the pitch of the screw
- To apply uniform pressure and prevent deformation of the object
Answer: To apply uniform pressure and prevent deformation of the object
The ratchet ensures that the spindle applies a constant, gentle pressure on the object. When the spindle contacts the object, the ratchet slips and clicks, preventing overtightening that could deform the object or damage the threads.
9. A student measures the diameter of a wire using a screw gauge. He rotates the thimble clockwise to close the jaws, takes a reading, then opens the jaws by rotating anticlockwise, and again closes clockwise for a second reading. The two readings differ. What is the most likely cause?
- Parallax error while reading the circular scale
- Zero error of the screw gauge
- Backlash error due to reversal of rotation direction
- Random error due to temperature variation
Answer: Backlash error due to reversal of rotation direction
Backlash error occurs when the direction of rotation is reversed because of the play between screw threads. The spindle does not move immediately, causing a difference in readings. This is a known instrumental error in screw gauges.
10. Which instrument would you use to measure the diameter of a thin wire (≈0.5 mm) with 2% accuracy?
- Vernier calipers with least count 0.1 mm
- Vernier calipers with least count 0.02 mm
- Screw gauge with least count 0.01 mm
- Screw gauge with least count 0.001 mm
Answer: Screw gauge with least count 0.01 mm
For a wire diameter of 0.5 mm, 2% accuracy requires a least count ≤ 0.01 mm. Screw gauge with LC 0.01 mm gives relative error = (0.01/0.5)×100 = 2%, meeting the requirement. Vernier calipers have LC ≥ 0.02 mm, giving ≥4% error.
11. How many significant figures are there in the measurement 0.00250 m?
- 3
- 4
- 5
- 2
Answer: 3
Leading zeros are not significant; they only locate the decimal. The first non-zero digit is 2, then 5, and the trailing zero after decimal is significant because it indicates precision. So 0.00250 has 3 significant figures.
12. In a spherometer experiment, the mean distance between the three legs is 4.0 cm and the sagitta measured is 0.50 cm. What is the radius of curvature of the spherical surface?
- 5.83 cm
- 5.33 cm
- 5.58 cm
- 0.558 m
Answer: 5.58 cm
Using the spherometer formula R = L²/(6h) + h/2, where L = 4.0 cm is the mean distance between legs and h = 0.50 cm is the sagitta. Compute L²/(6h) = 16/(3) ≈ 5.333 cm, add h/2 = 0.25 cm to get R = 5.583 cm ≈ 5.58 cm.