3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
The main scale of a vernier calliper is graduated in divisions of 0.5 mm. Its vernier scale has 25 divisions whose total length equals that of 49 main-scale divisions (an extended vernier). With the jaws closed, the zero of the vernier lies slightly to the left of the zero of the main scale and the 20th vernier mark coincides with a main-scale mark. When a rod is held between the jaws, the vernier zero lies between the main-scale marks 32.5 mm and 33.0 mm, and the 12th vernier mark coincides with a main-scale mark. What is the corrected diameter of the rod? Exactly one option is correct.
A33.14 mm
B32.64 mm
C32.84 mm
D32.98 mm
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Answer: C
One vernier division =2549×0.5=0.98 mm. The nearest whole number of main divisions to one vernier division is two (1.00 mm), so the least count is 2(0.5)−0.98=0.02 mm.
Zero error: vernier zero at −δ. The kth vernier mark is at −δ+0.98k; for k=20 this is 19.6−δ, which must equal a main mark, 19.5 mm, so δ=0.10 mm, i.e. the zero error is −0.10 mm (equivalently −(25−20)×0.02 mm).
Observed reading =32.5+12×0.02=32.74 mm. Corrected reading = observed − zero error =32.74+0.10=32.84 mm.
Traps: taking the zero error as −20×0.02=−0.40 mm gives 33.14 mm; subtracting 0.10 gives 32.64 mm; using a least count of 1/25 mm gives 32.98 mm.
Difficulty: HardQuestion 2Level 2 · ExamNumerical answer
A screw gauge has a main scale graduated in 0.5 mm divisions, a pitch of 0.5 mm and 50 divisions on the circular scale. With the studs in contact, the main-scale zero is just visible (the thimble edge has not reached it) and the 3rd circular division is on the reference line. A small steel ball is placed between the studs three times, in different orientations. Each time the thimble edge has uncovered 6 main-scale divisions, and the circular-scale readings are 24, 26 and 25. The mass of the ball is 0.1370 g. Using the mean corrected diameter, find the density of the ball in g cm−3. Give the answer correct to two decimal places.
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Answer: 7.84
Least count =0.5/50=0.01 mm. Zero error: thimble edge not yet at zero, circular reading 3, so the gauge reads +0.03 mm with nothing between the studs; zero error =+0.03 mm.
Mean circular reading =25. Observed reading =6×0.5+25×0.01=3.25 mm. Corrected diameter d=3.25−0.03=3.22 mm =0.322 cm.
Volume V=6πd3=6π(0.322)3=0.017481 cm3.
Density =0.0174810.1370=7.837≈7.84 g cm−3.
Difficulty: HardQuestion 3Level 2 · ExamOne or more correct
The diameter of a solid cylinder is measured with a vernier calliper of least count 0.01 cm as d=2.54 cm, and its length with a metre scale of least count 0.1 cm as L=12.3 cm. Take the uncertainty of each reading equal to the least count of the instrument used and combine errors by adding fractional errors (maximum error). Which of the following statements is (are) correct? One or more options may be correct.
AThe volume 4πd2L, reported with the correct number of significant figures, is 62.3 cm3.
BReplacing the vernier calliper by a screw gauge of least count 0.001 cm (the diameter reading becoming 2.540 cm) reduces the maximum percentage error in the volume to less than half of its earlier value.
CThe recorded value 2.54 cm shows that the vernier calliper can resolve 0.001 cm.
DThe maximum percentage error in the volume is about 1.6%.
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Answer: A, D
V=4π(2.54)2(12.3)=62.325 cm3. The least precise factors (2.54, 12.3) have 3 significant figures, so V=62.3 cm3: (A) correct.
Maximum fractional error: 22.540.01+12.30.1=0.787%+0.813%=1.60%: (D) correct.
With the screw gauge: 22.5400.001+0.813%=0.079%+0.813%=0.89%, which is more than half of 1.60% because the length term dominates: (B) incorrect.
A reading of 2.54 cm has its last digit in the 0.01 cm place, matching the least count 0.01 cm; nothing finer is resolved: (C) incorrect.
Difficulty: Hard