Measurement Error & Experiment Solved Examples

MEASUREMENT ERROR & EXPERIMENT SOLVED EXAMPLES Ex 1. A rectangular plate has a length of (21. 3  0.2) cm and a width of ( 9.8  0.1) cm. Calculate the area of the plate, including its uncertainty. Sol. Given L = (21.3  0.2) cm, b = (9.8  0.1) cm L = 0.2 cm, b = 0.1 cm Area = Lb = (21.3  0.2) × (9.8  0.1) = 21.3 × 9.8  0.2 × 9.8  0.1 × 21.3 = 208.74  1.96  2.13 Least number of significant figures in the individual quantities is 2. Hence, the final answer should have only two significant figures.  A = 208.74 = 209 cm2 on rounding = 210 cm2 on further rounding to two significant figures Relative error = = L b  L b 0.2 0.1  = 9.39 × 10–3 + 1.02 × 10–2 = 19.59 × 10–3 21.3 9.8 Percentage error = 19.59 × 10–3 × 100 = 1.959 % ; 2 %  actual area = 210 cm2  2 % Ex 2. Sol. How many significant figures are there in each of the following numbers ? (A) 78.9  0.2 (B) 3.788 × 109 (C) 2.46 × 10–6 (E) 201 (F) 2003 (G) 0.02 (D) 0.0053 (A) 78.9  0.2 has 3 significant figures. (B) 3.788 × 109 has 4 significant figures. (C) 2.46 × 10–6 has 3 significant figures (D) 0.0053 has 2 significant figures as leading zeros are not significant. (E) 3 significant figures (F) 4 significant figures (G) 1 significant figures Ex 3. A jeweller inserts a gem weighing 3.47 g into a box weighing 1.8 kg. Find the total weight of the box and the gem to correct number of significant figures. Sol. Total weight = 1.8 kg + 0.00347 kg = 1.80347 kg In finding the sum or difference, the result should have the number of decimal places equal to the smallest number of decimal places in individual quantities. Therefore, total weight = 1.8 kg since only one decimal places should be used. Ex 4. A long thread of length 1.23 m is cut to obtain a small thread of length 12.3 mm. What is the new length of the long thread ? Sol. New length = 1.23 m – 12.3 × 10–3 = 1.2177 m Least number of decimal places in any of the individual quantities is 2. Hence, new length = 1.22 m. Ex 5. A rectangular piece of copper is 4.10 cm  0.01cm long and 0.91 cm  0.01 cm wide. Find the area of the rectangle and the percentage uncertainty in the area. Sol. Given L = 4.10 cm  0.01 cm, b = 0.91 cm  0.01 cm Area = (L  L) (b  b) = (4.10  0.01) ( 0.91  0.01) = 4.10 × 0.91  percentage error = 3.731 cm2  percentage error  L b   0.01 0.01    Percentage error =    100     100 b   L  4.1 0.91  Percentage error = (2.439 × 10–3 + 10.99 × 10–3) × 100 = 13.429 × 10–1 = 1.343 % The least number of significant figures in any of the individual quantities is 2. Hence, Area = 3.7 cm2  1.3 % Ex 6. The sides of a rectangle are ( 18.52  0.2) m and (3.62  0.5) m. Find the perimeter of the rectangle with error limits. Sol. L = (1.852  0.2) m and b = (3.62  0.5) m Perimeter = 2(L + b) = 2(1.852 + 3.62) = 10.944 m Error in measurement of perimeter, P = (L + b) = 2 (0.2 + 0.5) = 1.4 m  actual perimeter = (10.94  1.4) m Ex 7. The length of suspension string in a simple pendulum is (0.87 m  0.01 m) and the radius of the bob of the pendulum is (3.47 cm  0.01 cm). Find the effective length of the pendulum with error limits and due regard to significant figures. Sol. L = 0.87 m  0.01 m r = 3.47 cm  0.01 cm = (3.47  0.01) 5 10–2 m Effective length of the pendulum, l = L + r = 0.87  + 0.0347 Error in measurement of effective length, l  L  r = 0.01 + 0.0001 = 0.01 m Hence, l = (0.9 m  0.01 m) = 0.9047 m  = 0.91 m MEASUREMENT ERROR & EXPERIMENT Ex 8. The radius of a uniform solid sphere is measured to be (5.60  0.20) cm and its mass is measured to be (8.51  0.02) kg. Determine the density of the sphere in kg m–3 and the uncertainity in density. Sol. Radius, r = (5.60  0.20) cm ; mass, M = (8.51  0.02) kg  Density, Mass M  4 Volume r 3 3   3  8.51kg 4     5.60  102 m  3 = 1.1569 × 104 kg m–3 Uncertainty in density,  M r  3  M r 0.02 0.20  3 8.51 5.60 = 2.35 × 10–3 + 107.1 × 10–3 = 109.45 × 10–3 = 109.45 × 10–3 × 100 % = 10. 945 % Least number of significant figures in any of the individual quantities is 3. Density,  = 1.16 × 104 kg m–3  11 % Ex 9. A circular park has a radius of (12.1 m  0.1m). Find the area of the park with percentage error.. Sol. r = 12.1 m  0.1 m Area =  r2 =  (12.1)2 = 4.6 × 1.65 % Percentage error,  A r 0.1 2  2  100  1.65 % A r 12.1 Area = 460 m2  1.65 % Ex.10 The circular scale of a micrometer has 200 divisions and pitch of 2mm. Find the measured value of thickness of a thin sheet. 10 0 0 (A) 3.41 mm Sol. (B) 6.41 mm 0 1 2 3 50 40 (C) 3.46 mm (D) 3.51 mm Ans. (B) Least count = pitch 2   0.01mm ; Reading = 3× 2 + (46–5) (0.01) = 6.41 mm No. of divisions 200 Ex 11. Estimate the percentage error in the determination of density of a material taken in the form of a cylinder with 2 % error in mass measurement, 1 % error in radius measurement and 3 % error in height measurement. Sol. Density of material of a cylinder,   Mass M  2 Volume r h  M r h  2  =2%+2×1%+3%=7%  M r h Ex 12. The radius of earth is estimated with an error of 1 %. Estimate the error in the measurement of its volume. Sol. Volume, V = 4 3 r 3  error in estimation of volume, Ex.13 V r 3 =3% V r A brilliant student of Class XII constructed a vernier calipers as shown. He used two identical inclines of inclination 37° and tried to measure the length of line AB. The length of line AB is A B 1mm (A) Sol. 21 mm 4 Least count = (B) 25 mm 4 (C) Sol. (D) None of these l  1  5  1  l 1  1     1   mm cos  4/ 5  4  4  l   l  5  (5 ) l4 Length AB = (4)    cos    cos   4 Ex.14 18 mm 4  1  5 25 mm 54  5      4 4 The length of the string of a simple pendulum is measured with a meter scale to be 63.5 cm, the radius of the bob plus the hook is measured with the help of vernier caliper to be 1.55 cm. Select the incorrect statement :(A) Least count of meter scale is 0.1 cm (B) Least count of vernier caliper is 0.01 cm (C) Effective length of pendulum is 65.1 cm (D) Effective length of pendulum is 65.2 cm From measurements least count of meter scale is 0.1 cm and least count of vernier calliper is 0.01 cm. Effective length of simple pendulum = 63.5 + 1.55 = 65.15 = 65.2 cm Ex.15 Two clocks A and B are being tested against a standard clock located in the national laboratory At 10:00 AM by the standard clock, the readings of the two clocks are shown in following table Da y Ist ClockA 10:00 : 06 Clock B 8:15:00 IInd III rd 10:01:13 8:15:01 9:59:08 8:15:04 IVth V th 10:02:15 9:58:10 8:14:58 8:15:02 If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer? (A) clock A (B) clock B (C) either clock A or clock B (D) Neither clock A nor clock B MEASUREMENT ERROR & EXPERIMENT Sol. The average reading of clock A is, closure to the standard time and the variation in time is smaller for clock B. As clock's is zero error is not significant for precision work because a zero error can always be easily corrected. Hence clock B is to be preferred. Ex.16 The side of a cube is (2.00 ± 0.01) cm. The volume and surface area of cube are respectively :– Sol. (A) (8.00 ± 0.12) cm3, (24.0 ± 0.24) cm2 (B) (8.00 ± 0.01) cm3, (24.0 ± 0.01) cm2 (C) (8.00 ± 0.04) cm3, (24.0 ± 0.06) cm2 (D) (8.00 ± 0.03) cm3, (24.0 ± 0.02) cm2 V V aa aa    0.01   33   V V  33V V    3  8    0.12 cm 3  aa   2.00  VV aa Volume V=a3=8 cm3 . Also Therefore V = (8.00 ± 0.12) cm3; Surface Area A = 6a2 = 6 (2.00)2 = 24.0 cm2. Also Ex.17 A A aa 0.01  aa  0.01  22   A A  22A A   2224.0 24.0  0.24. Therefore A = (24.0 ± 0.24) cm2   0.24  aa   2.00 A A aa 2.00 The main scale of a vernier callipers reads in millimeter and its vernier is divided into 10 divisions which coincides with 9 divisions of the main scale. The reading for shown situation is found to be (x/10) mm. Find the value of x. cm 0 0 Sol. Least count = 1 2 3 cm 0 10 1 0 2 3 10 1mm = 0.1 mm; Zero error = –(10–6) × 0.1 = – 0.4 mm 10 Reading = 6 + 5 × (0.1) – (–0.4) = 6.9 mm Ex.18 Round off the following numbers to 3 significant digits(i) 899.68 (ii) 987.52 (iii) 2.0082 Sol. (i) 900 (ii) 988 (iii) 2.01 (iv) 336 (v) 336 Ex.19 Write number of significant digits (iv) 336.5 (v) 335.5 (i) 62.3 cm (ii) 6.23 × 101 cm (iii) 20.000 (iv) 0.02 × 10–19 (v) 500.000 (vi) 0.5210 (vii) 896.80 (ix) 1200 (x) 1200 N (viii) 201 Sol. Ans. (i) 3 (ii) 3 (iii) 5 (iv) 1 (v) 6 (vi) 4 (vii) 5 (viii) 3 (ix) 2 (x) 4 Ex.20 A scale is calibrated to centimeters and the following measurements are estimated by the scale. Find out the significant digits. (i) 200 m (ii) 92.80 m (v) 8.921 mm (vi) 6.001 m (iii) 80.26 m Sol. Ans. (i) 3 (ii) 4 (iii) 4 (iv) 2 (v) 1 (vi) 4 Ex.21 Solve with regards to significant figure (i) 908 + 2.76 (ii) 999 – 989 (iv) 4.0 × 10–4 – 2.5 × 10–5 Sol. Ans. (i) 911 (ii) 10.0 (iii) 4.0 × 10–4 (iv) 8.23 cm (iii) 4.0 × 10–4 – 2.5 × 10–6 (v) 6.75 × 103 + 4.52 × 102 (iv) 3.8 × 10–4 (v) 7.20 × 103 (vi) 5.00 (vi) 625 ÷ 125 Ex.22 Students I1, J1,J3 and I2 perform an experiment for measuring the acceleration due to gravity (G) using a simple pendulum. they use different lengths of the pendulum and record time for different number of oscillations. The observations are shown in the table. Least count for length = 0.1 cm, Least count for time = 1s S tude nts I J J I 1 1 3 2 Le ngth o f the pendulum (cm) No . o f oscillations (n) Tim e pe riod of pendulum (s) 1 00 .0 4 00 .0 1 00 .0 4 00 .0 20 10 10 20 20 40 20 40 If P1,P2,P3 and P4 are the % error in g for students I1,J1,J3 and I2 respectively then(A) P1 = P3 Sol. T  2 (B) P3 is maximum (C) P4 is minimum (D) P2 = P4 l g l 2 T   g  T–2 = . g g l T  l 2  T   0 .1 2 (1)   0.1 2(1)   100   Therefore P   × 100 = 0.6%, P2   × 100 = 0.42%  P1       l  4 00 40 0   100 400  T   0.1 2(1)   0.1 2(1)  P3    P    100 200  × 100 = 1.1%, 4  400 800  × 100 = 0.28% Ex.23 An object covers (16.0 ± 0.4) m distance in (4.0 ± 0.2) s. Find out its speed. Sol. Speed v = Ex.24 The length of a cylinder is measured with a metre rod having least count 0.1 cm. Its diameter is measured with vernier callipers having least count 0.01 cm. Given the length is 5.0 cm and diameter is 2.00 cm. Find the percentage error in the calculated value of volume. Sol. V = r2h  distance 16.0 0.2   s t   0.4  v  = = 4.0 m/s; Error in speed v      4.0   0.3 m/ s    time 4.0 s t 16.0 4.0  DD2 hh VV 22D D hh V   0 .0 1   0 .1        10 0  2     1 00 = 3% 4 VV D D hh V  2 .0 0   5.0   