The Workbook

1. Relative error \(= \dfrac{\Delta L}{L} = \dfrac{0.01}{2.50} = 0.004\).

2. Percentage error \(= 0.004 \times 100 = 0.4\%\).

1. The rod runs from \(0\) to \(12.3\) cm. The smallest division is \(1\) mm \(= 0.1\) cm, so the least count is \(0.1\) cm.

2. The error of a single reading is taken as the least count: \(L = 12.3 \pm 0.1\) cm.

3. \(\dfrac{0.1}{12.3} \times 100 \approx 0.8\%\).

1. \(\bar T = \dfrac{2.1 + 2.0 + 2.2 + 1.9 + 2.3}{5} = \dfrac{10.5}{5} = 2.1\) s.

2. \(|T_i - \bar T|\): \(0\), \(0.1\), \(0.1\), \(0.2\), \(0.2\) s. Mean absolute error \(= \dfrac{0.6}{5} = 0.12\) s, rounded to \(0.1\) s to match the readings.

3. \(T = 2.1 \pm 0.1\) s. Relative error \(= \dfrac{0.1}{2.1} \approx 0.05\).

Diameter: \(\dfrac{0.01}{0.4} \times 100 = 2.5\%\). Mass: \(\dfrac{1}{10} \times 100 = 10\%\).

The diameter has the smaller percentage error, so it is the more precise measurement. The size of the least count alone does not decide precision; the least count compared with the reading does.

\(m = 25.30 \pm 0.01\) g. Percentage error \(= \dfrac{0.01}{25.30} \times 100 \approx 0.04\%\).

1. Absolute errors add: \(L_1 + L_2 = 14.0 \pm 0.2\) cm.

2. Absolute errors add here too: \(L_1 - L_2 = 6.0 \pm 0.2\) cm.

3. Sum: \(\dfrac{0.2}{14.0} \approx 1.4\%\). Difference: \(\dfrac{0.2}{6.0} \approx 3.3\%\). The difference has the larger percentage error, because the same absolute error sits on a smaller number.

1. \(A = 4.0 \times 2.0 = 8.0\) cm\(^2\).

2. \(\dfrac{0.1}{4.0} = 2.5\%\) and \(\dfrac{0.1}{2.0} = 5\%\). For a product the percentage errors add: \(7.5\%\).

3. \(\Delta A = 0.075 \times 8.0 = 0.6\) cm\(^2\), so \(A = 8.0 \pm 0.6\) cm\(^2\).

For a quotient the percentage errors add: \(\dfrac{\Delta\rho}{\rho} = \dfrac{\Delta m}{m} + \dfrac{\Delta V}{V} = 2\% + 3\% = 5\%\).

\(\dfrac{\Delta K}{K} = \dfrac{\Delta m}{m} + 2\dfrac{\Delta v}{v} = 1\% + 2 \times 2\% = 5\%\). The power \(2\) doubles the error of \(v\). The factor \(\dfrac{1}{2}\) has no error.

\(\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T} = 1\% + 2 \times 2\% = 5\%\).

1. \(R = \dfrac{10.0}{2.0} = 5.0\ \Omega\).

2. \(V\): \(\dfrac{0.2}{10.0} = 2\%\). \(I\): \(\dfrac{0.1}{2.0} = 5\%\). \(R\): \(2\% + 5\% = 7\%\).

3. \(\Delta R = 0.07 \times 5.0 = 0.35 \approx 0.4\ \Omega\), so \(R = 5.0 \pm 0.4\ \Omega\).

\(\dfrac{\Delta Z}{Z} = 2\dfrac{\Delta a}{a} + \dfrac{\Delta b}{b} + \dfrac{1}{2}\dfrac{\Delta c}{c} = 2 \times 1\% + 2\% + \dfrac{1}{2} \times 4\% = 6\%\).

1. \(0.0250\): the leading zeros only fix the decimal point; \(2\), \(5\) and the trailing \(0\) count. Three.

2. \(4.00\): trailing zeros after a decimal point count. Three.

3. \(305\): a zero between non-zero digits counts. Three.

4. \(1.20 \times 10^{3}\): the power of ten does not count. Three.

5. \(0.7\): one.

1. \(3.14\) (the dropped digit \(1\) is less than \(5\)).

2. \(2.72\) (the dropped digit \(8\) is more than \(5\)).

3. \(0.0123\).

4. \(12.65\): the dropped digit is exactly \(5\) and the preceding digit \(6\) is even, so it stays: \(12.6\).

5. \(12.75\): the preceding digit \(7\) is odd, so it is raised: \(12.8\).

\(12.3 + 1.234 + 0.05 = 13.584\). In addition the answer keeps as many decimal places as the term with the fewest, here one (\(12.3\)). So the sum is \(13.6\).

1. \(2.5 \times 3.14 = 7.85\). The answer keeps the fewer significant figures, two: \(7.9\).

2. \(10.0 \div 3.0 = 3.33\ldots\), to two significant figures: \(3.3\).

3. \(20.46 - 3.2 = 17.26\), to one decimal place: \(17.3\).

\(V = 2.5^3 = 15.625\) cm\(^3\). The side has two significant figures, so \(V = 16\) cm\(^3\).

\(\rho = \dfrac{12.4}{3.2} = 3.875\) g/cm\(^3\). The volume has two significant figures, so \(\rho = 3.9\) g/cm\(^3\).

1. \(T = \dfrac{40.0}{20} = 2.00\) s.

2. The whole timing carries the error \(0.1\) s, so \(\Delta T = \dfrac{0.1}{20} = 0.005\) s.

3. Timing one oscillation would give \(\Delta T = 0.1\) s, twenty times worse. Dividing a long timed interval by the number of oscillations divides the error by that number.

1. \(g = \dfrac{4\pi^2 \times 1.000}{2.00^2} = \pi^2 \approx 9.87\) m/s\(^2\).

2. \(\dfrac{\Delta g}{g} = \dfrac{0.1}{100.0} + 2 \times \dfrac{0.02}{2.00} = 0.1\% + 2\% = 2.1\%\). Almost all of the error comes from the timing.

\(A = \pi r^2\), so \(\dfrac{\Delta A}{A} = 2\dfrac{\Delta r}{r} = 2 \times 1.5\% = 3\%\).

First: \(\dfrac{1}{100} = 1\%\). Second: \(\dfrac{5}{1000} = 0.5\%\). The second measurement is more accurate, because accuracy is judged by the relative error, not by the absolute error.

\(R = R_1 + R_2 = 300\ \Omega\). Absolute errors add in a sum: \(\Delta R = 3 + 4 = 7\ \Omega\). So \(R = 300 \pm 7\ \Omega\).

Mean \(= \dfrac{2.01 + 2.03 + 1.99 + 2.01}{4} = \dfrac{8.04}{4} = 2.01\) mm.

Deviations: \(0\), \(0.02\), \(0.02\), \(0\) mm. Mean absolute error \(= \dfrac{0.04}{4} = 0.01\) mm. Percentage error \(= \dfrac{0.01}{2.01} \times 100 \approx 0.5\%\).

1. \(1\) VSD \(= \dfrac{9}{10}\) MSD \(= 0.9\) mm.

2. LC \(= 1\) MSD \(- 1\) VSD \(= 1 - 0.9 = 0.1\) mm \(= 0.01\) cm.

\(1\) VSD \(= \dfrac{19}{20} \times 0.5 = 0.475\) mm. LC \(= 0.5 - 0.475 = 0.025\) mm. Equivalently LC \(= \dfrac{\text{1 MSD}}{20} = \dfrac{0.5}{20} = 0.025\) mm.

LC \(= \dfrac{\text{pitch}}{\text{number of circular divisions}}\).

1. \(\dfrac{1}{100} = 0.01\) mm.

2. \(\dfrac{0.5}{50} = 0.01\) mm.

Pitch \(=\) distance per rotation \(= \dfrac{2}{4} = 0.5\) mm. LC \(= \dfrac{0.5}{100} = 0.005\) mm.

1. The vernier zero lies between \(1.2\) and \(1.3\) cm, so the main-scale reading is \(1.2\) cm.

2. The \(3\)rd vernier line is exactly in line with a main-scale line.

3. Reading \(=\) MSR \(+\) VC \(\times\) LC \(= 1.2 + 3 \times 0.01 = 1.23\) cm.

Observed \(= 2.4 + 7 \times 0.01 = 2.47\) cm. True \(=\) observed \(-\) zero error \(= 2.47 - 0.02 = 2.45\) cm.

1. The edge of the thimble is past the \(5\) mm mark and before any further mark: sleeve \(= 5\) mm. The reference line meets the circular scale at \(32\).

2. Reading \(= 5 + 32 \times 0.01 = 5.32\) mm.

Observed \(= 3 + 45 \times 0.01 = 3.45\) mm. True \(= 3.45 - (-0.03) = 3.48\) mm. A negative zero error makes readings too small, so it is added back.

1. Positive. With nothing between the jaws the instrument already shows a reading, because the vernier zero is to the right of the main zero.

2. The \(4\)th division coincides: zero error \(= +4 \times 0.01 = +0.04\) cm.

3. True \(= 3.26 - 0.04 = 3.22\) cm.

1. The vernier zero is to the left, so the error is negative. Its size is \((10 - 6) \times 0.01 = 0.04\) cm: zero error \(= -0.04\) cm.

2. True \(= 3.26 - (-0.04) = 3.30\) cm.

1. The zero has not reached the line, so the error is negative: \(-5 \times 0.01 = -0.05\) mm. True \(= 2.40 + 0.05 = 2.45\) mm.

2. The zero has gone past the line, so the error is positive: \(+0.03\) mm. True \(= 2.40 - 0.03 = 2.37\) mm.

A screw gauge has a least count of \(0.01\) mm, ten times smaller than the usual vernier's \(0.1\) mm. For a wire of diameter about \(0.5\) mm the vernier would give an error of \(20\%\), the screw gauge \(2\%\).

Backlash error: the screw has a little play in its nut. On reversing the direction of rotation the thimble turns a little before the screw moves. It is avoided by always turning the screw in the same direction when taking a reading.

Pitch \(= \dfrac{5}{5} = 1\) mm. LC \(= \dfrac{1}{100} = 0.01\) mm.

1. The centre \(O\), a leg and the foot of the screw form a right triangle. The hypotenuse is \(R\) (centre to leg), one side is \(a\), and the other is \(R - h\) (centre to the plane of the legs). Pythagoras gives \(R^2 = a^2 + (R - h)^2\).

2. \(R^2 = a^2 + R^2 - 2Rh + h^2\), so \(2Rh = a^2 + h^2\) and \(R = \dfrac{a^2}{2h} + \dfrac{h}{2}\).

3. For an equilateral triangle of side \(l\) the distance from the centre to a vertex is \(\dfrac{l}{\sqrt{3}}\), so \(a^2 = \dfrac{l^2}{3}\). Then \(R = \dfrac{l^2}{6h} + \dfrac{h}{2}\).

\(R = \dfrac{9}{6 \times 0.05} + \dfrac{0.05}{2} = 30 + 0.025 = 30.025 \approx 30\) cm. The \(\dfrac{h}{2}\) term is negligible because \(h\) is far smaller than \(l\).

1. \(h = 1 \times 1\) mm \(+ 50 \times 0.01\) mm \(= 1.50\) mm.

2. \(R = \dfrac{30^2}{6 \times 1.5} + \dfrac{1.5}{2} = 100 + 0.75 = 100.75 \approx 101\) mm, about \(10\) cm.

1. Down. The surface dips below the plane of the legs, so the screw must be lowered to touch it.

2. No. The same right triangle with sides \(a\) and \(R - h\) and hypotenuse \(R\) still holds, now with the centre of curvature above the surface. \(h\) is simply the size of the screw's movement.

3. \(R = \dfrac{9}{6 \times 0.15} + \dfrac{0.15}{2} = 10 + 0.075 = 10.075 \approx 10.1\) cm.

LC \(= \dfrac{\text{1 MSD}}{50} = \dfrac{0.5}{50} = 0.01\) mm.

\(n\) VSD \(= (n - 1)s\), so \(1\) VSD \(= \dfrac{(n - 1)s}{n}\). LC \(= 1\) MSD \(- 1\) VSD \(= s - \dfrac{(n - 1)s}{n} = \dfrac{s}{n}\).

\(r = 0.60\) cm. \(V = \pi r^2 h = \pi \times 0.36 \times 5.00 = 1.8\pi \approx 5.65\) cm\(^3\).

\(\dfrac{\Delta d}{d} = \dfrac{0.01}{0.50} = 2\%\). The area is \(\dfrac{\pi d^2}{4}\), so its percentage error is \(2 \times 2\% = 4\%\).

1. Vernier caliper, using its inside jaws.

2. Screw gauge.

3. Spherometer.

4. Metre scale; a least count of \(1\) mm is enough.

LC \(= \dfrac{0.5}{50} = 0.01\) mm. Reading \(= 2.5 + 24 \times 0.01 = 2.74\) mm.

\(1\) VSD \(= \dfrac{24}{25} = 0.96\) mm, so LC \(= 1 - 0.96 = 0.04\) mm. Reading \(= 10 + 12 \times 0.04 = 10.48\) mm.