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Unit # 1 - Part # 2 Error Analysis

(Error Analysis, Order of Magnitude, Significant Figures, Measuring Instruments, Accuracy & Precision)

Prepared By: Er. Veenus Girdhar

(Solutions to all the assignments will be available on the website www.physicswithveenus.com) These Notes Belongs to: ______________________

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Error in measurements ♦Every physical quantity is measured with the help of an instrument or with the help of certain sets of instruments. The measurement of a physical quantity is never free from any errors. ♦The measure of degree of uncertainty in the measurement of physical quantity is called error. ♦Mathematically, an error in the measurement of a physical quantity is defined as the difference between the measured value and actual value of the physical quantity i.e. Error = True value – Measured value. ♦Error can be classified into various categories such as; ♦Systematic Error

Ø These are the errors whose causes are known and hence these errors can be eliminated.

Ø These are the errors that arise due to the system. Ø These are further of 4 types.

v Instrumental Error: due to faulty instruments having faculty design or calibration.

v Personal Error: due to inexperience of performer. It involves lack of proper setting and lack of certain minute steps.

v Error due to imperfection: due to ignoring of certain facts. v Error due to external causes.

♦Random error: Ø These are also known as chance errors or probable errors. Ø They can creep into the experiment during any stage of the experiment. Ø Their cause is not completely known and hence they cannot be completely

eliminated. However; they can be reduced by taking large set of readings. If a scientist doubles the number of observation taken; then he will reduce the random error by half.

♦Gross Error: Ø These errors arise due to simple carelessness of the observer.

Methods of Expressing Errors ♦Absolute Error: It is defined as the actual error present in the experiment and is defined as the difference of measured value and standard or actual value i.e. Absolute Error = Actual value Observed value± −

In case; actual value of a quantity is not known; then average value is taken as actual value i.e. Absolute error = Average value Observed value± −

♦It is represented by aΔ ♦Relative Error: It is defined as the ratio of absolute error present in the measurement to the actual or mean value of the quantity i.e.

Relative error = Absolute ErrorActual Value

±

⇒ Actual value Observed value '

Actual Valuea a aa a

± −Δ −= =

♦The relative error when expressed in percentage is called percentage relative error i.e.


Actual value Observed value% Relative error 100

Actual Valuex

± −=

♦Percentage relative error is considered as best method for expressing error.

Assignment # 1 Q1. Density of a body was measured in CGS unit (gm/cm3) as 2.37; 2.39; 2.42; 2.38 and 2.40. Calculate actual error, relative error and percentage relative error. Q1. The time period of oscillation of a simple pendulum was recorded as 2.56s, 2.62s, 2.70s, 2.58s and 2.45s. Find the absolute error, relative error and percentage relative error. Q3. The refractive index of water is found to have the values 1.29; 1.33; 1.34; 1.35; 1.32; 1.36; 1.30 and 1.33. Calculate the mean value, absolute error, the relative error and the percentage error. Q4. A research worker takes 100 observations in an experiment. If he repeats the same experiment by taking 500 observations, how is the probable error affected? Q5. A student measured the length of a pendulum 1.351 m and time for 30 vibrations is 2 minute 10sec using his wrist watch. The % error in g is therefore;

(a) 1.72% (b) 1.813% (c) 1.63 % (d) 1.513%

Q6. A scientist performs an experiment and takes 100 readings. He repeats the same experiment and now takes 400 readings. By doing so,

(a) The probable error remains the same. (b) The probable error is halved. (c) The probable error is reduced to ¼ (d) The probable error is increased by 4 times. (c)

Q7. The least count of a stopwatch is 1/5 sec. The time of 20 oscillations of the pendulum is measured to be 25 sec. The minimum % error in the measurement of time will be:

(a) 0.1% (b) 8% (c) 1.8% (d) 0.8%

Q8. The period of oscillations of a simple pendulum in the experiment are recorded as 2.63s, 2.56 s, 2.42 s, 2.71s and 2.80 s respectively. The average absolute error is

(a) 0.1 s (b) 0.11s (c) 0.01s (d) 1.0s

Propagation of Errors during Summation ♦Let; z = x + y Let the error in measurement of x & y and hence in z be Δx, Δy and Δz respectively. Thus; Measured value of x = ( )x x±Δ

Measured value of y = ( )y y± Δ Measured value of z = ( )z z±Δ

Hence; ( ) ( ) ( )z z x x y y±Δ = ±Δ + ±Δ


⇒ ( )x y x y= + ±Δ ±Δ z x y= ±Δ ±Δ ⇒ z x y±Δ = ±Δ ±Δ

There are four possibilities;

x yx y

zx yx y

Δ + Δ⎧⎪Δ −Δ⎪

±Δ = ⎨−Δ + Δ⎪⎪−Δ −Δ⎩

Hence; maximum probable error in z is

( )z x yΔ = ± Δ +Δ

This is the required error in measurement of z. Propagation of Errors during Subtraction ♦Let; z = x – y Let the error in measurement of x & y and hence in z be Δx, Δy and Δz respectively. Thus; Measured value of x = ( )x x±Δ


Hence; ( ) ( ) ( )z z x x y y±Δ = ±Δ − ±Δ

⇒ ( )x y x y= − ±Δ Δm z x y= ±Δ Δm ⇒ z x y±Δ = ±Δ Δm


x yx y

zx yx y

Δ + Δ⎧⎪Δ −Δ⎪

±Δ = ⎨−Δ + Δ⎪⎪−Δ −Δ⎩

Hence; maximum probable error in z is

( )z x yΔ = ± Δ +Δ

This is the required error in measurement of z. Important Note ♦If z = 3x + 4y; Then; (3 4 )z x yΔ = ± Δ + Δ ♦If z = 3x – 4y; Then; (3 4 )z x yΔ = ± Δ + Δ

Assignment # 2 Q1. The length and breadth of a rectangular filed are l = (30 ± 1.0) cm; b = (20 ± 3.0) cm. Find the perimeter of the field. Q2. The inner and outer radii of a hollow cylinder are R1 = (3.0 ± 0.1) cm and R2 = (3.5 ± 0.2) cm. Find the thickness of cylinder. Q3. In an experiment two capacitors are measured to have capacitance as (1.3 ± 0.1)µF and (2.4 ± 0.2)µF. Calculate the total capacitance if these capacitors are connected in parallel. Q4. The lengths of two cylinders are measured to be L1 = (5.62 ± 0.01) cm and L2 = (4.34 ± 0.02) cm. Calculate the difference in length with error limits. Q5. The initial and final temperature of water as recorded by an observer are (40.6 ± 0.2)0C and (78.3 ± 0.3)0C. Calculate the rise in temperature with proper limits.


Q6. The original length of the wire is (153.7 ± 0.6) cm. It is stretched to (155.3 ± 0.2) cm. Calculate the elongation in the wire. Q7. Two resistances R1 = (100.0 ± 0.3) ohm and R2 = (150.0 ± 0.5) ohm are connected in series. Calculate the equivalent resistance of combination with error limit. Q8. The external and internal diameter of a hollow cylinder are measured to be (4.13 ± 0.01) cm and (3.8 ± 0.01) cm. The thickness of the wall of the cylinder is

(a) (0.34 ± 0.02) cm (b) (0.17 ± 0.02) cm (c) (0.17 ± 0.01) cm (d) (0.34 ± 0.01) cm

Q9. The measured values of the resistance of two resistors are (8 ± 0.3) ohm and (24 ± 0.5) ohm. The maximum error in the measurement of equivalent resistance when connected in series is

(a) 5% (b) 2.5% (c) 10% (d) 7.5%. Q10. In an experiment, the external diameter d1 and internal diameter d2 of a

metal tube are found to be (64 ± 2) mm and (47 ± 1) mm respectively. The percentage error in (d1-d2) expected from these readings is at most

(a) 18% (b) 1% (c) 5% (d) s6% Propagation of Errors during Multiplication ♦Let; z = x.y Let the error in measurement of x & y and hence in z be Δx, Δy and Δz respectively. Thus; Measured value of x = ( )x x±Δ


Then; ( ) ( ).( )z z x x y y±Δ = ±Δ ±Δ

⇒ 1 1 1z x yz x yz x y

⎛ ⎞Δ Δ Δ⎛ ⎞ ⎛ ⎞± = ± ±⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 .x y x yzx y x y

⎛ ⎞Δ Δ Δ Δ= ± ± ±⎜ ⎟

⎝ ⎠

But .x yx yΔ Δ

≈ zero.

Therefore, z x yz x yΔ Δ Δ

± = ± ±


x yx yx yx yzx yzx yx yx y

Δ Δ⎧ +⎪⎪Δ Δ⎪

−⎪Δ ⎪± = ⎨

Δ Δ⎪− +⎪⎪

Δ Δ⎪− −⎪⎩

Hence; maximum probable relative error in z is

z x yz x y

⎛ ⎞Δ Δ Δ= ± +⎜ ⎟

⎝ ⎠

This is the required error in measurement of z.


Binomial Expansion ♦According to binomial expansion;

2 3( 1) ( 1)( 2)(1 ) 1 ...2.1 3.2.1

n n n n n nx nx x x− − −+ = + + + +

♦If in the binomial expansion 1x << ; then square and higher powers of x can be

neglected. In that case; (1 ) 1 ....(1 ) 1 ....(1 ) 1 ....(1 ) 1 ....

n

n

n

n

x nxx nxx nxx nx

−

−

+ = + +

+ = − +

− = − +

− = + +

Propagation of Errors during Division

♦Let; 1xz xyy

−= =

Let the error in measurement of x & y and hence in z be Δx, Δy and Δz respectively. Thus; Measured value of x = ( )x x±Δ


Then; 1( ) ( ).( )z z x x y y −±Δ = ±Δ ±Δ

⇒ z 1± Δzz

⎛

⎝⎜

⎞

⎠⎟= x 1±

Δxx

⎛

⎝⎜

⎞

⎠⎟ y−1 1±

Δyy

⎛

⎝⎜

⎞

⎠⎟

−1

Expanding binomially and neglecting higher powers, we get

z 1± Δzz

⎛

⎝⎜

⎞

⎠⎟= x 1±

Δxx

⎛

⎝⎜

⎞

⎠⎟ y−1 1∓

Δyy+ ...

⎛

⎝⎜

⎞

⎠⎟ 1 .x y x yz

x y x y⎛ ⎞Δ Δ Δ Δ

= ±⎜ ⎟⎝ ⎠

m m

But .x yx yΔ Δ

≈ zero.

Therefore, z x yz x yΔ Δ Δ

± = ± m


x yx yx yx yzx yzx yx yx y

Δ Δ⎧ +⎪⎪Δ Δ⎪

−⎪Δ ⎪± = ⎨

Δ Δ⎪− +⎪⎪

Δ Δ⎪− −⎪⎩

Hence; maximum probable relative error in z is

z x yz x y

⎛ ⎞Δ Δ Δ= ± +⎜ ⎟

⎝ ⎠

This is the required error in measurement of z.


Important Points ♦Further; if z = 3x2y4

Then; 2 4z x yz x y

⎛ ⎞Δ Δ Δ= ± +⎜ ⎟

⎝ ⎠

♦Further; if z = 3x2y–4

Then; 2 4z x yz x y

⎛ ⎞Δ Δ Δ= ± +⎜ ⎟

⎝ ⎠

♦In general if there is a quantity P such that;3 24a bP

c d= ; then maximum probable

error in P is given by;

13 22

P a b c dP a b c dΔ Δ Δ Δ Δ⎡ ⎤= ± + + +⎢ ⎥⎣ ⎦

♦In case 1z=1x+1y

Then the error is calculated using formula:

Δzz2=Δxx2+Δyy2

Assignment # 3 Q1. The length and breadth of a rectangular filed are l = (30 ± 1.0) cm; b = (20 ± 3.0) cm. Find the area of the field. Q2. In an experiment two capacitors are measured to have capacitance as (1.3 ± 0.1)µF and (2.4 ± 0.2)µF. Calculate the total capacitance if these capacitors are connected in series.

Q3. If 3 24a bP

c d= ; find the error in P if error in a, b, c and d are ± 1%, ± 2%, ± 3%

and ± 4%.

Q4. If3 5

7

4a bPc d

= ; which quantity should be measured accurately to minimize error

in P.

Q5. If3

2

4abPcd

= and error in a, b, c and d are + 1%, -2%, - 3% and 4% respectively.

Find maximum error in P. Q6. The length, breadth and height of a cuboid are measured are l = (5.25 ± 0.04) cm; b = (3.45 ± 0.03) cm and c = (1.15 ± 0.07) cm; find the volume of cuboid. Q7. To study flow of liquid through a narrow tube, the following formula is used;

4Pr

Vlπ

η =

The value of P, r, V and l are measured to be 76cm of Hg, 0.28 cm; 1.2 cm3/sec and 18.2 cm. If these quantities are measured to the accuracy of 0.5cm of Hg, 0.01cm, 0.1cm3/sec and 0.1cm respectively. Find the percentage error in the value of η.

Q8. In the measurement of a physical quantity X =2

1/ 3 3

A BC D

. The percentage errors

introduced in the measurements of the quantities A, B, C and D are 2%, 2%, 4%


and 5% respectively. Then the minimum amount of percentage of error in the measurement of X is contributed by:

(a) A (b) B (c) C (d) D

Q9. A chocolate cookie is a circular disc of diameter (8.50 ± 0.02) cm and thickness (0.050 ± 0.005) cm. The average volume in cm3 is

(a) 2.83 ± 0.3 (b) 2.38 ± 0.27 (c) 11.35 ± 1.2 (d) 9.31 ± 1.12 (a)

Q10. Two capacitors have value of (5.2 ± 0.1) µF and (12.2 ± 0.1) µF. They are connected first in parallel and then in series. Find the percentage error in the two cases.

(a) 2.8% and 1.23% (b) 3.6% and 1.31% (c) 3.4% and 1.3% (d) 3.9% and 1.15% (d)

Q11. A student measured the length of a pendulum 1.351 m and time for 30 vibrations is 2 minute 10sec using his wrist watch. The % error in g is therefore;

(a) 1.72% (b) 1.813% (c) 1.63 % (d) 1.513%

Q12. The heat generated in a circuit is dependent upon the resistance, current and time for which current is passed. If the errors in measurement of above are 1%, 2% and 3% respectively, the maximum error in calculating heat will be

(a) 2% (b) 3% (c) 6% (d) 1%

Q13. The relative density of metal may be found out by hanging a block of metal from a spring balance. Its reading in air is (5.00 ± 0.05) N and in water is (4.00 ± 0.05) N. its relative density is

(a) (5.00 ± 0.05) (b) (5.00 ± 0.10) (c) (5.00 ± 6%) (d) (5.00 ± 11%)

Q14. In an experiment to find the Young’s modules for a brass wire. A student used the data below length of wire = (1.50 ± 0.01)m; diameter of wire = (1.56 ± 0.01)mm mass of load = (6.00 ± 0.01)kg; extension = (4.0 ± 0.1)mm

acceleration of free fall = (9.8 ± 0.1)ms-2 Which of the following leads to the greatest uncertainty in the calculated value of

the Young’s modulus, given that the formula is Y = FLA ⋅ ΔL

?

(a) Measurement of length (b) Measurement of diameter (c) Measurement of load (d) Measurement of extension


Q15. Error in the measurement of radius of a sphere is 1%. Then error in the measurement of volume is

(a) 1% (b) 5% (c) 3% (d) 8%

Q16. Given that n m

r

a bXp

= . The % error in the measurement of a, b and p is 1%,

0.5% and 0.75% respectively. If n = 2, m = 2 and r = 4, then percent error in X is (a) 6% (b) 5.25% (c) 0.75% (d) zero

Q17. An experimentalist measures quantities a, b and c to find the value of x, given by the relation x = a– 1 b2/c3/2. If the % error in the measurement of a, b and c are ±1%, ±3% and ±2% respectively, then the % error in the measurement of x can be

(a) ± 2% (b) ± 10% (c) ± 6% (d) ± 8%

Q18. The length of simple pendulum is about 100cm and is known to an accuracy of 1mm. Its period of oscillations is 2s determined by measuring the time of 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in determination of g?

(a) 0.2% (b) 0.5% (c) 0.1% (d) 2%

Q19. Find the percentage error in specific resistance given by 2r Rl

πρ = where r is

radius having value (0.2 ± 0.02) cm, R is the resistance of (60 ± 2) Ω and l is the length of (150 ± 0.1) cm.

Q20. A physical quantity ρ is related to four variables α, β, γ and η as; 3 2α β

ργη

= .

The percentage error in them is 1%, 3%, 4% and 2% respectively. Find percentage error in ρ.

Q21. The time period of simple pendulum is given by 2 LTg

π= . This equation can

be used to determine the value of g. If we desire to have minimum error in g, which of the two quantities L or t should be measured most accurately and why? Q22. A wire has a mass of (0.3 ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is

(a) 1 (b) 2 (c) 3 (d) 4


Significant Figures ♦Significant figures in the measured value of a physical quantity tell the number of digits in the measurement in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy in the measurement and vice-versa. ♦Following are the rules for counting the number of significant figures in a given number

Ø All non-zero digits are significant i.e. in 1875 there are four significant figures.

Ø All zeroes occurring between two non-zero digit are significant i.e. in 12004508 there are 8 significant figures.

Ø In a number less than one, all zeroes to the right of decimal point and to the left of a non-zero digit are not significant i.e. in 0.0087 have only two significant figures.

Ø All zeroes to the right of the last non-zero digit in a decimal part are significant i.e. in 0.0000800 have three significant figures. Also in 1.00 there are three significant figures.

Ø All zeroes on the right of a non-zero digit are not significant i.e. in 8000 there is only one significant figure.

Ø All zeroes to the right of the last non-zero digit become significant; when they come from a measurement i.e. in 3050 there are only three significant figures but in 3050m; there are four significant figures.

Ø Any number in the power of ten is not significant i.e. in 3.230 x 1078; there are four significant figures.

Arithmetic Operations with Significant figures ♦During addition and subtraction; the number of decimal places in the result should equal the smallest number of decimal places of terms in the operation. ♦During multiplication and division; the number of significant figures in the product or in the quotient is the same as the smallest number of significant figures in any of the factors. Rules for rounding off the last digit ♦The following rules should be taken care of

Ø If the digit to be dropped is less than 5; then the preceding digit is left unchanged i.e. 4.62 is rounded off to 4.6

Ø If the digit to be dropped is more than 5; then the preceding digit is raised by one i.e. 4.57 is rounded off to 4.6

Ø If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one i.e. 2.3452 is rounded off to 2.35

Ø If the digit to be dropped is 5 or 5 followed by zero, then the preceding digit is left unchanged, if it is even i.e. 78.250 or 78.25 is rounded off to 78.2

Ø If the digit to be dropped is 5 or 5 followed by zero, then the preceding digit is raised by one, if it is odd i.e. 78.350 or 78.35 is rounded off to 78.4

♦It must be understood that in any measurement, the number of decimal places are determined by least count of device used. The reading cannot have higher or lesser decimal places than the least count of device.


Assignment # 4 Q1. Write down the number of significant figures in the following:

(a) 5238N (b) 4200 kg (c) 34.000 m (d) 0.012340 N/m (e) 23.4560 x 1045

Q2. Write the result of the following with respect to significant figures (a) 0.080 + 12.34 + 2.678 + 1.9 (b) (4.6 x 0.13) x (2.345) (c) (2.345 x 4.6) x (0.13) (d) (2.345 x 0.13) x (4.6)

(e) 0.9995 1.531.592x

Q3. Subtract 2.5 x 104 from 3.9 x 105 with due regard to significant figures. Q4. Calculate area enclosed by a circle of diameter 1.06 m to correct number of significant figures. Q5. What is the difference between 4.0 and 4.00? Q6. Round off the following

(a) 0.678982 upto four significant figures (b) 36.879 upto four significant figures (c) 1.0084 upto four significant figures (d) 2.3455 upto four significant figures (e) 25.653 upto three significant figures (f) 0.6995 upto one digit.

Q7. Out of 5.0; 5.00 and 5.000 which is most accurate and why? Q8 Mass of a body is measured by two persons as 10.2 kg and 10.23 kg. Which has measured mass more accurately and why? Q9. Subtraction of two nearly equal quantities destroys the accuracy. Comment. Q10. Add 6.75 x 103 cm to 4.52 x 103cm. Q11. The significant figures in the number 6.0023 is

(a) 2 (b) 5 (c) 4 (d) 1

Q12. The length, breadth of a metal sheet is 3.124m and 3.002m respectively. The area of this sheet upto correct significant figure is

(a) 9.378 m2. (b) 9.37 m2. (c) 9.378248 m2. (d) 9.3782 m2.

Q13. The length, breadth and thickness of a block are given by 12cm, 6cm and 2.45cm respectively. The volume of block according to the idea of significant figures should be:

(a) 1 x 102 cm3

(b) 2 x 102 cm3 (c) 1.763 x 102 cm3 (d) None of these


Q14. In arithmetic 17.8 x 3.1143 = 55.4354, but as a result of experimental measurement the best way to express it is

(a) 55.4354 (b) 55.4 (c) 55.44 (d) 55.435

Q15. Which of the following numerical values have three significant figures? (a) 3.033 (b) 0.030 (c) 30.30 (d) 0.300

Q16. In which of the following numerical vales all zeroes are significant? (a) 0.2020 (b) 20.2 x 103 (c) 20.20 (d) None of these

Q17. A student measured the diameter of a wire using screw gauge with least count 0.001cm. The correct measurement is

(a) 5.320 cm (b) 5.32 cm (c) 5.3200 cm (d) 5.3 cm

Q18. The length, breadth and thickness of a rectangular sheet of metal is 4.234m, 1.005m and 2.01 cm. The volume of sheet to correct significant figures is

(a) 0.0855 m3. (b) 0.086 m3. (c) 0.08556 m3. (d) 0.08 m3.

Q19. The number of significant figures in 23.023, 0.0003 and 2.1 x 10 – 3 are (a) 4, 4, 2 (b) 5, 1, 2 (c) 5, 1 ,5 (d) 5, 5, 2

Q20. In the context of accuracy of measurement and significant figures in expressing results of experiment, which of the following are correct Statement # 1: Out of the two measurement 50.14 cm and 0.00025 ampere, the first one has greater accuracy. Statement # 2: If one travel by 478 km by rail and 397m by road, the total distance travelled is 478 km.

(a) Only first statement is correct. (b) Only second statement is correct. (c) Both are correct. (d) Neither of the two is correct.

Q21. The body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 m/s. The magnitude of its momentum is recorded as

(a) 17.565 kg m/s (b) 17.56 kg m/s (c) 17.57 kg m/s (d) 17.6 kg m/s

Q22. If L = 2.331 cm and B = 2.1 cm then L + B = (a) 4.431 cm (b) 4.43 cm


(c) 4.4 cm (d) 4cm

Q23. The length of the string of a simple pendulum is measured with a meter scale to be 90.0 cm. The radius of the bob plus the length of the hook is calculated to be 2.13 cm using measurement of slide callipers. What is the effective length of the pendulum?

(a) 92 cm (b) 92.13 cm (c) 92.1 cm (d) Any of them

Q24. The mass of the box measured by a grocer’s balance is 2.3kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) total mass of the box and (b) the difference in masses of gold pieces to correct significant figures? Q25. A chocolate cookie is a circular disc of diameter (8.50 ± 0.02) cm and thickness (0.050 ± 0.005) cm. The average volume in cm3 is

(a) 2.83 ± 0.3 (b) 2.38 ± 0.27 (c) 11.35 ± 1.2 (d) 9.31 ± 1.12 (a)

Order of Magnitude ♦A number rounded to the nearest power of 10 is called an order of magnitude. It gives an approximate idea of size or magnitude of that object or quantity. The order of magnitude enables us to estimate that how small or large that physical quantity is. ♦Order of magnitude of a numerical quantity N is x and is calculated using the following formula:

0.5 < N10x

≤ 5

♦Few examples of order of magnitude are: Numerical Value Calculations to Express

them in nearest power of 10 Order of magnitude

8 810

= 8101 = 0.8 > 0.5( ) x = 1

12 1210

= 12101 =1.2 > 0.5( ) x = 1

47 4710

= 47101 = 4.7 > 0.5( ) x = 1

54 54100

= 54102 = 0.54 > 0.5( ) x = 2

697 6971000

= 697103 = 0.697 > 0.5( )

x = 3

702 7021000

= 702103 = 0.702 > 0.5( ) x = 3

1020 10201000

= 1020103 =1.020 > 0.5( ) x = 3

2040 20401000

= 2040103 = 2.040 > 0.5( ) x = 3


43000 4300010000

= 43000104 = 4.3000 > 0.5( ) x = 4

0.8 0.81

= 0.8100 = 0.8 > 0.5( ) x = 0

0.05 0.050.01

= 0.0510−2 = 5 = 5 but not ≠ 0.5( ) x = – 2

0.007 0.0070.01

= 0.00710−2 = 0.7 > 0.5( ) x = – 2

Assignment # 5 Q1. Define order of magnitude. What is its significance? Q2. Write the order of magnitude of following measurement:

(a) 25,710,000m (b) 0.00000521 kg [7, – 5]

Q3. Write the order of magnitude of the following: (a) 8 (b) 49 (c) 52 (d) 999

(e) 1001 (f) 75300 (g) 0.05 (h) 0.99 [1, 1, 2, 3, 3, 5, – 2, 0]

Q4. What is order of magnitude of number of seconds in a day? (a) 3 (b) 4

(c) 5 (d) 6

Q5. What is order of magnitude of number of molecules in one mole of substance? (a) 22 (b) 23

(c) 24 (d) 25

Q6. What is the order of magnitude of capacitance of earth, given that capacitance of earth is 711µF.

(a) – 2 (b) – 3

(c) – 4 (d) – 5

Q7. The wavelength of red color photon is approximately 8000Å. What is order of magnitude of this wavelength?

(a) – 4 (b) – 5

(c) – 6 (d) – 7

Q8. The mass of earth is 6 x 1024kg. What is order of magnitude of this value? (a) 24 (b) 23

(c) 25 (d) 26

Q9. The order of magnitude of height of Mount Everest in m is: (a) 4 (b) 6

(c) 5 (d) 3

Q10. The radius of earth is of the order of (a) 107 m (b) 109 m (c) 1015 m (d) 1019 m

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Vernier Calliper ♦It has three parts

(a) Main scale. (b) Vernier Scale. (c) Metallic Strip.

♦Zero error Zero correction in Vernier Calliper: There are two kind of zero error in the caliper as shown.

(a) Positive Zero error: It occurs when zero of vernier scale lies to the left of main scale.

(b) Negative zero error: It occurs when zero of vernier scale lies to the left of main scale.

♦Least Count of Vernier Calliper: It is also called Vernier constant and is defined as the difference between 1 main scale division and 1 vernier scale division i.e.

LC = 1MSD – 1VSD


Assignment # 6 Q1. 25 Vernier scale divisions coincide with twenty four main scale division. If the main scale division is half milli-meter, then the least count of the vernier is

(a) 0.02 cm (b) 0.002 cm (c) 0.002 mm (d) 0.001 cm

Q2. N-division on the main scale of a Vernier calliper coincides with N + 1 divisions on the Vernier scale. If each divisions on the main scale is of ‘a’ units, determine the least count of the instrument. Q3. A Vernier calliper is so constructed that 49 divisions on main scale coincides with 50 divisions on Vernier scale. Find its Vernier constant in cm if each main scale divisions equals 1mm. Q4. The mains scale of a spectrometer is divided into 720 divisions in all. If the vernier scale consist of 30 divisions, the least count of the instrument is: (given that 30 divisions on vernier scale coincide with 29 main scale divisions.)

(a) 0.10 (b) 1’’ (c) 1’ (d) 0.1’’

Q5. The vernier calliper has its main scale of 10cm equally divided into 200 equal parts. Its vernier scale of 25 divisions coincides with 12mm on the main sacle. The least count of the instrument is

(a) 0.020 cm (b) 0.002 cm (c) 0.010 cm (d) 0.001 cm

Q6. The vernier of a circular scale is divided into 30 divisions which coincides against 29 main scale divisions. Each main scale divisions is 1/20. The least count of the instrument is

(a) 30’ (b) 10’ (c) 1’ (d) 0.1’

Q7. In a vernier calliper, 10 smallest divisions of the vernier scale are equal to nine smallest divisions on main scale. If the smallest divisions on main scale are half millimeter, then the vernier constant is

(a) 0.5mm (b) 0.1mm (c) 0.05mm (d) 0.005mm

Q8. One centimeter on the main scale of vernier scale of vernier calliper is divided into ten equal parts. If 10 divisions of vernier scale coincides with 8 small divisions of the main scale, the least count of the calliper is

(a) 0.005 cm (b) 0.05 cm (c) 0.02 cm (d) 0.01 cm


Q9. A vernier calliper has 20 divisions on the vernier scale, which coincide with 19 on the main scale. The least count of the instrument is 0.1 mm. The main scale divisions are of

(a) 0.5 mm (b) 1 mm (c) 2 mm (d) 0.25 mm

Q10. N divisions on vernier scale coincides with (N – 1) main scale divisions on main scale. What is the least count of the instrument if the length of one main scale division is 1mm?

(a) 10N cm (b) N cm (c) 1/10N cm (d) 1/100N cm

Q11. The smallest division on main scale of a Vernier calliper is 1mm and 10 Vernier divisions coincides with 9 scale divisions. While measuring the length of line, the zero mark of the Vernier scale lies between 10.2 cm and 10.3 cm and the third division of Vernier scale coincides with a main scale division.

(a) Determine the least count of the Vernier calliper (b) Find the length of the line.

Q12. Least count of a Vernier calliper is 0.01cm. when the two jaws of the instrument touch each other the 5th division of the Vernier scale coincides with a main scale division and the zero the main scale lies to the left of the zero of main scale. Furthermore while measuring the diameter of the sphere, the zero mark of the Vernier scale lies between 2.4 cm and 2.5 cm and the 6th Vernier division coincides with main scale division. Calculate the diameter of the sphere. Q13. In the diagram shown in the figure, find the magnitude and nature of zero error.

Q14. A Vernier calliper is so constructed that 49 divisions on main scale coincides with 50 divisions on Vernier scale. Find its Vernier constant in cm if each main scale divisions equals 1mm.

Micrometer Screw or Screw Gauge ♦The Screw gauge or micrometer screw is based on the principle of screw. According to which, the translational motion produced in a screw is directly proportional to the rotational motion provided to the screw. ♦A screw gauge essentially consists of

(a) Main scale or linear scale. (b) Circular scale

♦A circular scale is on a circular cap, fitted at one end of the screw. Its circumference is equally divided into 100 or 50 equal divisions.


♦The least count (L.C.) is defined as the pitch of the screw per unit number of circular divisions on the circular scale. Thus;

. .

PitchLCNumber of circular scale division

=

♦ Zero error Zero correction in Screw Gauge: There are two kind of zero error in the device as shown.

(c) Positive Zero error: If zero of circular scale lies below the main scale reference line, when the two jaws are in touch, with no object in between.

(d) Negative zero error: If zero of circular scale lies below the main scale reference line, when the two jaws are in touch, with no object in between.

♦Back Lash Error: When the sense of rotation of the screw is suddenly changed, the screw head may rotate, but the screw itself may not move forward or backwards. This is known as back lash error. This error is due to

Ø Loose fitting of the screw. Ø Due to wear and tear of threading due to prolonged use of screw.

To reduce this error; the screw should always be rotated in the same direction for the particular set of observations.


Assignment # 7 Q1. The pitch of screw gauge is 0.5mm. Its head scale contains 50 divisions. The least count of the screw gauge is

(a) 0.001 mm (b) 0.01 mm (c) 0.02 mm (d) 0.025 mm

Q2. The pitch of the screw gauge is 1mm and there are 100 divisions on the circular scale. In measuring the diameter of the sphere there are six divisions on the linear scale and forty divisions on the circular scale coincide with the reference line. Find the diameter of the sphere. Q3. The pitch of a screw gauge is 1mm and there are 100 divisions on the circular side. When nothing is put between the jaws, the zero of the circular scale lies 6 divisions below the reference line. When a wire is placed between the jaws, 2 linear scale divisions are clearly visible while 62 divisions on circular scale coincide with the reference line. Determine the diameter of the wire. Q4. The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale divisions is 2. First figure is without wire in the jaws.

(a) 1.2 (b) 1.25 (c) 2.20 (d) 2.25

Q5. Which of the following is the most precise device for measuring length? (a) A Vernier calliper with 20 divisions on the sliding scale. (b) An optical instrument that can measure length to within a wavelength of

light. (c) A screw gauge of pitch 1mm and 100 divisions on the circular scale. (d) All of the above are equally precise.


Spherometer ♦A Spherometer is an instrument for measuring the curvature of a surface. It also works on the concept of screw. ♦The Spherometer consists of

Ø A micrometer screw threaded into a small tripod with a vertical scale fastened to it.

Ø A vertical scale is used to measure the height or depth of the curvature of the surface.

Ø A circular scale with 100 or 50 graduations marked in it.

♦The least count (L.C.) is defined as the pitch of the screw per unit number of circular divisions on the circular scale. Thus;

. .

PitchLCNumber of circular scale division

=

♦The formula for calculating radius of curvature with a spherometer is

R = L2

6h+ h2

Where; L = Average distance between two legs of spherometer h = saggitta

Assignment # 8 Q1. A spherometer has 250 equal divisions marked along the periphery of its disc and one full rotation of the disc advances on the main scale by 0.0625 cm. The least count of system is

(a) 2.5 x 10 – 4 cm. (b) 2.5 x 10 – 3 cm. (c) 2.5 x 10 – 2 cm. (d) None of these

Q2. The spherometer has a least count of 0.005 mm and its head scale is divided into 200 equal divisions. The distance between consecutive threads on the spherometer screw is

(a) 0.005 mm (b) 1.0 mm (c) 1.0 cm (d) 0.0025 mm

Q3. To measure radius of curvature with a spherometer, we use the formula

(a) R = L2

h+ h2

(b) R = L2

6h+ h2

(c) R = L2

6+ h2

(d) R = 2L2

h+ 6L


Accuracy and Precision ♦Accuracy refers to the closeness of a measured value to a standard or known value. For example, if in lab you obtain a weight measurement of 3.2 kg for a given substance, but the actual or known weight is 10 kg, then your measurement is not accurate. In this case, your measurement is not close to the known value. ♦Precision refers to the closeness of two or more measurements to each other. Using the example above, if you weigh a given substance five times, and get 3.2 kg each time, then your measurement is very precise. ♦Precision is independent of accuracy. You can be very precise but inaccurate, as described above. You can also be accurate but imprecise. ♦For example, if on average, your measurements for a given substance are close to the known value, but the measurements are far from each other, then you have accuracy without precision. ♦A good analogy for understanding accuracy and precision is to imagine a basketball player shooting baskets. If the player shoots with accuracy, his aim will always take the ball close to or into the basket. If the player shoots with precision, his aim will always take the ball to the same location which may or may not be close to the basket. A good player will be both accurate and precise by shooting the ball the same way each time and each time making it in the basket. ♦Both the concepts can be clearly understood from the following diagram:


Assignment # 9 Additional Questions Including from NCERT

Q1. Two clocks are being tested against a standard clock located in a national laboratory. At 12:00:00 noon by the standard clock, the readings of the two clocks are : Monday Tuesday Wednesday Thursday Friday Saturday Sunday Clock1 12:00:05 12:01:15 11:59:08 12:01:50 11:59:15 12:01:30 12:01:19 Clock2 10:15:06 10:14:59 10:15:18 10:15:07 10:14:53 10:15:24 10:15:11 If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer ? Ans.: The range of variation over the seven days of observations is 162 s for clock 1, and 31 s for clock 2. The average reading of clock 1 is much closer to the standard time than the average reading of clock 2. The important point is that a clock’s zero error is not as significant for precision work as its variation, because a ‘zero-error’ can always be easily corrected. Hence clock 2 is preferred to clock 1. Q2. The resistance R = V/I where V = (100 ± 5)V and I = (10 ± 0.2)A. Find the percentage error in R. Q3. Find the relative error in Z, if Z = A4B1/3CD3/2.

Q4. The period of oscillation of a simple pendulum is T = 2π Lg.Measured value of

L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ? Q5. Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appropriate significant figures? Q6. 5.74 g of a substance occupies 1.2 cm3. Express its density by keeping the significant figures in view. Q7. State the number of significant figures in the following :

(a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm– 3. (d) 6.320 J (e) 6.032 N m – 2. (f) 0.0006032 m2.

Q8. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. Q9. The unit of percentage error is

(a) Same as that of physical quantity (b) Different from that of physical quantity (c) Percentage error is unit less (d) Errors have got their own units which are different from that of physical

quantity measured Q10. The decimal equivalent of 1/20 upto three significant figures is

(a) 0.0500 (b) 0.05000 (c) 0.0050 (d) 5.0 x 10 – 2 .


Q11. If 97.52 is divided by 2.54, the correct result in terms of significant figures is (a) 38.4 (b) 38.3937 (c) 38.394 (d) 38.39

Q12. Accuracy of measurement is determined by (a) Absolute error (b) Percentage error (c) Both (d) None of these

Q13. A formula is given as

P = ba1+ kθt

3

ma

where P = pressure; k = Boltzmann’s constant; θ = temperature; t = time; ‘a’ and ‘b’ are constants. Dimensional formula of ‘b’ is same as

(a) Forcr (b) Linear momentum (c) Angular momentum (d) Torque

Q14. Backlash error may occur in which of the following instrument? (a) Slide callipers and screw gauge (b) Spherometer and screw gauge (c) Slide callipers and spherometer (d) Slide callipers, screw gauge and spherometer

Q15. The ability of an instrument gives consistent reading, when repeated readings are taken is called as

(a) Accuracy (b) Precision (c) Sensitivity (d) Error

Q16. Which of the following should be the small value, so that the precision becomes high?

(a) Actual value (b) Mean value (c) Relative error (d) Relative deviation

Q17. The ability of an instrument to detect a slight change that occurs in the measured quantity is called as

(a) Precision (b) Accuracy (c) Sensitivity (d) Error

Q18. The accuracy of an instrument increases if (a) The number of significant figures increases (b) The relative deviation increases (c) The relative error increases (d) None of these

Q19. Which of the following statements about errors is correct? (a) Zero error is random error


(b) Random errors can be reduced by taking repeated readings. (c) Systematic errors can be due to instruments which are not sensitive. (d) Systematic errors cause the readings scattered on both sides of the actual

value. Q20. Which of the following experiment techniques can reduce systematic error of the quantity being measured?

(a) Measuring the diameter of a wire at different points along the wire. (b) Adjusting an ammeter to read zero before measuring a current. (c) Timing a large number of oscillations to find the period of a pendulum. (d) Measuring the thickness of a large number of pieces of paper to find the

thickness of one piece. Q21. The side of a cube.is measured by vernier calipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). The main scale reads 10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g. Find the density of the cube in appropriate significant figures. [2.66 g/cm3] Q22. A screw gauge having 100 equal divisions and a pitch of length 1 mm is used to measure the diameter of a wire of length 5.6 cm. The main scale reading is 1 mm and 47th circular division coincides with the main scale. Find the curved surfuce area of wire in cm2 to appropriate signiflcant figure. [2.6 cm2] Q23. A new system of units is proposed in which unit of mass is expressed as α kg, β m and γ sec. How much will 5J measure in this new system? Q24. According to Joule's law of heating, heat produced H = I2Rt, where I is current, R is resistance and t is time. If the errors in the measurement of t, R and I are 3%, 4% and 6% respectively find error in the measurement of H.

End of unit # 1

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