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Q1. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is (1) 6% (2) zero (3) 1% (4) 3%

What This Question Tests

This question tests the understanding of error propagation in division, where relative errors add up. Given percentage errors, the total percentage error is their sum.

Concepts Tested

Ohm's LawPropagation of errors

Formulas Used

R = V/I

ΔR/R = ΔV/V + ΔI/I

📚 NCERT Sections This Tests

3.4Ohm’S Law

Physics Class 11 · Chapter 3

78% match

3.4 OHM’S LAW A basic law regarding flow of currents was discovered by G.S. Ohm in 1828, long before the physical mechanism responsible for flow of currents was discovered. Imagine a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor. Then Ohm’s law states that V µ I or, V = R I (3.3) where the constant of proportionality R is called the resistance of the conductor. The SI units of resistance is ohm, and is denoted by the symbol W. The resistance R not only depends on the material of the conductor but also on the dimensions of the conductor. The dependence of R on the dimensions of the conductor can easily be determined as follows. FIGURE 3.2 Consider a conductor satisfying Eq. (3.3) to be in the form of a slab of Illustrating the length l and cross sectional area A [Fig. 3.2(a)]. Imagine placing two such relation R = rl/A for a rectangular slabidentical slabs side by side [Fig. 3.2(b)], so that the length of the of length l and areacombination is 2l. The current flowing through the combination is the of cross-section A. same as that flowing through either of the slabs. If V is the potential difference across the ends of the first slab, then V is also the potential 83difference across the ends of the second slab since the second slab is Reprint 2025-26 Physics identical to the first and the same current I flows through both. The potential difference across the ends of the combination is clearly sum of the potential difference across the two individual slabs and hence equals 2V. The current through the combination is I and the resistance of the combination RC is [from Eq. (3.3)], 2V R C = = 2 R (3.4) I since V/I = R, the resistance of either of the slabs. Thus, doubling the length of a conductor doubles the(1787–1854) resistance. In general, then resistance is proportional to length, R ∝ l (3.5)OHM Georg Simon Ohm (1787– Next, imagine dividing the slab into two by cutting it 1854) German physicist, lengthwise so that the slab can be considered as a professor at Munich. Ohm combination of two identical slabs of length l, but each was led to his law by anSIMON having a cross sectional area of A/2 [Fig. 3.2(c)]. analogy between the For a given voltage V across the slab, if I is the current conduction of heat: the through the entire slab, then clearly the current flowing electric field is analogous to the temperature gradient, through each of the two half-slabs is I/2. Since theGEORG and the electric current is potential difference across the ends of the half-slabs is V, analogous to the heat flow. i.e., the same as across the full slab, the resistance of each of the half-slabs R1 is V V R1 = = 2 = 2 R. (3.6) ( I /2) I Thus, halving the area of the cross-section of a conductor doubles the resistance. In general, then the resistance R is inversely proportional to the cross-sectional area, 1 R ∝ (3.7) A Combining Eqs. (3.5) and (3.7), we have l R ∝ (3.8) A and hence for a given conductor l R = ρ (3.9) A where the constant of proportionality r depends on the material of the conductor but not on its dimensions. r is called resistivity. Using the last equation, Ohm’s law reads I ρl V = I × R = (3.10) A Current per unit area (taken normal to the current), I/A, is called current density and is denoted by j. The SI units of the current density are A/m2. Further, if E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its 84 ends is El. Using these, the last equation reads Reprint 2025-26 Current Electricity E l = j r l or, E = j r (3.11) The above relation for magnitudes E and j can indeed be cast in a vector form. The current density, (which we have defined as the current through unit area normal to the current) is also directed along E, and is also a vector j (º j E/E). Thus, the last equation can be written as, E = jr (3.12) or, j = s E (3.13) where s º1/r is called the conductivity. Ohm’s law is often stated in an equivalent form, Eq. (3.13) in addition to Eq.(3.3). In the next section, we will try to understand the origin of the Ohm’s law as arising from the characteristics of the drift of electrons.

3.2A Battery Of Emf 10 V And Internal Resistance 3 Ω Is Connected To A

Physics Class 11 · Chapter 3

78% match

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

1.3Significant Figures Figures In A Measurement. This Important

Physics Class 11 · Chapter 1

76% match

1.3 SIGNIFICANT FIGURES figures in a measurement. This important remark makes most of the followingAs discussed above, every measurement observations clear:involves errors. Thus, the result of (1) For example, the length 2.308 cm has fourmeasurement should be reported in a way that significant figures. But in different units, theindicates the precision of measurement. same value can be written as 0.02308 m or 23.08Normally, the reported result of measurement is a number that includes all digits in the mm or 23080 µm. number that are known reliably plus the first All these numbers have the same number of digit that is u Reprint 2025-26 4 PHYSICS This shows that the location of decimal point is negative exponent (or power) of 10. In order to of no consequence in determining the number get an approximate idea of the number, we may of significant figures. round off the number a to 1 (for a ≤5) and to 10 The example gives the following rules : (for 5<a ≤10). Then the number can be • All the non-zero digits are significant. expressed approximately as 10b in which the • All the zeros between two non-zero digits exponent (or power) b of 10 is called order of magnitude of the physical quantity. When only are significant, no matter where the an estimate is required, the quantity is of the decimal point is, if at all. order of 10b. For example, the diameter of the • If the number is less than 1, the zero(s) earth (1.28×107m) is of the order of 107m with on the right of decimal point but to the the order of magnitude 7. The diameter of left of the first non-zero digit are not hydrogen atom (1.06 ×10–10m) is of the order of significant. [In 0.00 2308, the underlined 10–10m, with the order of magnitude zeroes are not significant]. –10. Thus, the diameter of the earth is 17 orders • The terminal or trailing zero(s) in a of magnitude larger than the hydrogen atom. It is often customary to write the decimal after number without a decimal point are not the first digit. Now the confusion mentioned in significant. (a) above disappears : [Thus 123 m = 12300 cm = 123000 mm has 4.700 m = 4.700 × 102 cm three significant figures, the trailing zero(s) = 4.700 × 103 mm = 4.700 × 10–3 km being not significant.] However, you can also The power of 10 is irrelevant to the see the next observation. determination of significant figures. However, all • The trailing zero(s) in a number with a zeroes appearing in the base number in the decimal point are significant. scientific notation are significant. Each number [The numbers 3.500 or 0.06900 have four in this case has four significant figures. significant figures each.] Thus, in the scientific notation, no confusion (2) There can be some confusion regarding the arises about the trailing zero(s) in the base trailing zero(s). Suppose a length is reported to number a. They are always significant. be 4.700 m. It is evident that the zeroes here (4) The scientific notation is ideal for reporting are meant to convey the precision of measurement. But if this is not adopted, we use measurement and are, therefore, significant. [If the rules adopted in the preceding example : these were not, it would be superfluous to write • For a number greater than 1, without anythem explicitly, the reported measurement decimal, the trailing zero(s) are notwould have been simply 4.7 m]. Now suppose we change units, then significant. • For a number with a decimal, the trailing4.700 m = 470.0 cm = 4700 mm = 0.004700 km zero(s) are significant. Since the last number has trailing zero(s) in a number with no decimal, we would conclude (5) The digit 0 conventionally put on the left of a erroneously from observation (1) above that the decimal for a number less than 1 (like 0.1250) number has two significant figures, while in is never significant. However, the zeroes at the fact, it has four significant figures and a mere end of such number are significant in a change of units cannot change the number of measurement. significant figures. (6) The multiplying or dividing factors which are (3) To remove such ambiguities in neither rounded numbers nor numbers determining the number of significant representing measured values are exact and figures, the best way is to report every have infinite number of significant digits. For measurement in scientific notation (in the dpower of 10). In this notation, every number is example in r = or s = 2πr, the factor 2 is an expressed as a × 10b, where a is a number 2 between 1 and 10, and b is any positive or exact number and it can be written as 2.0, 2.00 Reprint 2025-26 UNITS AND MEASUREMENT 5 decimal place. The final result should, therefore, t or 2.0000 as required. Similarly, in T = , n is be rounded off to 663.8 g. n Similarly, the difference in length can be an exact number. expressed as : 1.3.1 Rules for Arithmetic Operations with 0.307 m – 0.304 m = 0.003 m = 3 ×10–3 m. Significant Figures Note that we should not use the rule (1) applicable The result of a calculation involving approximate for multiplication and division and write 664 g as measured values of quantities (i.e. values with the result in the example of addition and limited number of significant figures) must 3.00 × 10–3 m in the example of subtraction. They reflect the umeasured values. It cannot be more accurate properly. For addition and subtraction, the rule than the original measured values themselves is in terms of decimal places. on which the result is based. In general, the final result should not have more significant 1.3.2 Rounding off the Ufigures than the original data from which it was The result of computation with approximate obtained. Thus, if mass of an object is measured numbers, which contain more than one to be, say, 4.237 g (four significant figures) and uits volume is measured to be 2.51 cm3, then its for rounding off numbers to the appropriate density, by mere arithmetic division, is significant figures are obvious in most cases. A 1.68804780876 g/cm3 upto 11 decimal places. number 2.746 rounded off to three significant It would be clearly absurd and irrelevant to figures is 2.75, while the number 1.743 would record the calculated value of density to such a be 1.74. The rule by convention is that the precision when the measurements on which the preceding digit is raised by 1 if the value is based, have much less precision. The insignificant digit to be dropped (the following rules for arithmetic operations with underlined digit in this case) is more than significant figures ensure that the final result 5, and is left unchanged if the latter is less of a calculation is shown with the precision that than 5. But what if the number is 2.745 in is consistent with the precision of the input which the insignificant digit is 5. Here, themeasured values : convention is that if the preceding digit is(1) In multiplication or division, the final even, the insignificant digit is simplyresult should retain as many significant dropped and, if it is odd, the preceding digitfigures as are there in the original number with the least significant figures. is raised by 1. Then, the number 2.745 rounded Thus, in the example above, density should off to three significant figures becomes 1.74. On be reported to three significant figures. the other hand, the number 2.735 rounded off to three significant figures becomes 1.74 since 4.237g -3 Density = = 1.69 g cm the preceding digit is odd. 3 2.51 cm In any involved or complex multi-step Similarly, if the speed of light is given as calculation, you should retain, in intermediate 3.00 × 108 m s-1 (three significant figure) and steps, one digit more than the significant digits one year (1y = 365.25 d) has 3.1557 × 107 s (five and round off to proper significant figures at the significant figures), the light year is 9.47 × 1015 m end of the calculation. Similarly, a number (three significant figures). known to be within many significant figures, such as in 2.99792458 × 108 m/s for the speed (2) In addition or subtraction, the final result of light in vacuum, is rounded off to anshould retain as many decimal places as are approximate value 3 × 108 m/s , which is oftenthere in the number with the least employed in computations. Finally, rememberdecimal places. that exact numbers that appear in formulae like For example, the sum of the numbers 436.32 g, 227.2 g and 0.301 g by mere arithmetic L addition, is 663.821 g. But the least precise 2 π in T = 2π , have a large (infinite) number measurement (227.2 g) is correct to only one g Reprint 2025-26 6 PHYSICS of significant figures. The value of π = = 16.2 cm ± 0.6 %. 3.1415926.... is known to a large number of significant figures. You may take the value as Similarly, the breadth b may be written as 3.142 or 3.14 for π, with limited number of b = 10.1 ± 0.1 cm significant figures as required in specific = 10.1 cm ± 1 % cases. Then, the error of the product of two (or more)⊳ Example 1.1 Each side of a cube is experimental values, using the combination of measured to be 7.203 m. What are the errors rule, will be total surface area and the volume of the l b = 163.62 cm2 + 1.6% cube to appropriate significant figures? = 163.62 + 2.6 cm2 Answer The number of significant figures in the measured length is 4. The calculated area This leads us to quote the final result as and the volume should therefore be rounded off l b = 164 + 3 cm2 to 4 significant figures. Here 3 cm2 is the uSurface area of the cube = 6(7.203)2 m2 estimation of area of rectangular sheet. = 311.299254 m2 (2) If a set of experimental data is specified = 311.3 m2 to n significant figures, a result obtained by combining the data will also be valid to n Volume of the cube = (7.203)3 m3 significant figures. = 373.714754 m3 However, if data are subtracted, the number of = 373.7 m3 ⊳ significant figures can be reduced. ⊳ Example 1.2 5.74 g of a substance For example, 12.9 g – 7.06 g, both specified to occupies 1.2 cm3. Express its density by three significant figures, cannot properly be keeping the significant figures in view. evaluated as 5.84 g but only as 5.8 g, as u in a different fashion (smallest number ofmeasured mass whereas there are only 2 decimal places rather than the number of significant figures in the measured volume. significant figures in any of the number added Hence the density should be expressed to only or subtracted). 2 significant figures. (3) The relative error of a value of number 5.74 −3 specified to significant figures depends not Density = g cm 1.2 only on n but also on the number itself. = 4.8 g cm--3 . ⊳ For example, the accuracy in measurement of mass 1.02 g is ± 0.01 g whereas another 1.3.3 Rules for Determining the U in the Results of Arithmetic The relative error in 1.02 g is Calculations = (± 0.01/1.02) × 100 % = ± 1%The rules for determining the u Similarly, the relative error in 9.89 g iserror in the number/measured quantity in = (± 0.01/9.89) × 100 %arithmetic operations can be understood from = ± 0.1 %the following examples. Finally, remember that intermediate results in(1) If the length and breadth of a thin rectangular sheet are measured, using a metre a multi-step computation should be scale as 16.2 cm and, 10.1 cm respectively, there calculated to one more significant figure in are three significant figures in each every measurement than the number of measurement. It means that the length l may digits in the least precise measurement. be written as These should be justified by the data and then l = 16.2 ± 0.1 cm the arithmetic operations may be carried out; Reprint 2025-26 UNITS AND MEASUREMENT 7 otherwise rounding errors can build up. For mass, one dimension in length, and –2 example, the reciprocal of 9.58, calculated (after dimensions in time. The dimensions in all other rounding off) to the same number of significant base quantities are zero. figures (three) is 0.104, but the reciprocal of Note that in this type of representation, the magnitudes are not considered. It is the quality0.104 calculated to three significant figures is of the type of the physical quantity that enters. 9.62. However, if we had written 1/9.58 = 0.1044 Thus, a change in velocity, initial velocity, and then taken the reciprocal to three significant average velocity, final velocity, and speed are figures, we would have retrieved the original all equivalent in this context. Since all these value of 9.58. quantities can be expressed as length/time, This example justifies the idea to retain one their dimensions are [L]/[T] or [L T–1]. more extra digit (than the number of digits in the least precise measurement) in intermediate 1.5 DIMENSIONAL FORMULAE AND steps of the complex multi-step calculations in DIMENSIONAL EQUATIONS order to avoid additional errors in the process The expression which shows how and which of of rounding off the numbers. the base quantities represent the dimensions of a physical quantity is called the dimensional