## Units and Measurements

### Class-11-Science-Physics-Chapter-1-Maharashtra Board

### Solutions

**Question 1. Choose the correct option.**

**(i) [L ^{1}M^{1}T^{-2}] is the dimensional formula for **

**(A) Velocity **

**(B) Acceleration**

**(C) Force **

**(D) Work**

(C) Force

**(ii) The error in the measurement of the sides of a rectangle is 1%. The error in the measurement of its area is**

**(A) 1% **

**(B) 1/2% **

**(C) 2% **

**(D) None of the above.**

(C) 2%

**(iii) Light year is a unit of**

**(A) Time **

**(B) Mass**

**(C) Distance **

**(D) Luminosity**

(C) Distance

**(iv) Dimensions of kinetic energy are the same as that of**

**(A) Force **

**(B) Acceleration**

**(C) Work **

**(D) Pressure**

(C) Work

**(v) Which of the following is not a fundamental unit?**

**(A) cm **

**(B) kg**

**(C) centigrade **

**(D) volt**

(B) kg

**Question 2. Answer the following questions.**

**(i) Star A is farther than star B. Which star will have a large parallax angle?**

Stat B, since parallax angle is inversely proportional to the distance of the star from the Earth.

**(ii) What are the dimensions of the quantity \(l\sqrt{l/g}\)*** ***being the length and g the acceleration due to gravity?**

**Given **: [*l*] = [L], [g] = [LT^{—2}]

∴ [\(l\sqrt{l/g}\)] = [*l*][* l*]^{1/2}[g]^{— ½}

= [L][L]^{1/2}[LT ^{—}^{ 2}]^{— ½}

= [L][L^{1/2}][L^{— ½}^{ }T^{1}]

= [L^{1+ ½-½} T^{1}]

= [L^{1}T^{1}]

Therefore, the dimensions of \(l\sqrt{l/g}\) are 1 in length and 1 in time.

**(iii) Define absolute error, mean absolute error, relative error and percentage error.**

**Absolute error**: For a set of measurements of the same quantity, the positive difference between each individual value and the most probable value gives the absolute error in that value.**Mean absolute error**: For a set of measurements of the same quantity, the arithmetic mean of all the absolute errors is called the mean absolute error in the measurement of that quantity.

Relative error = D_{a }*/*

**Relative error**: The ratio of the mean absolute error in the measurement of a physical quantity to its most probable value is called the relative error in the measurement of that quantity.**Mean percentage error**: The relative error in a measurement multiplied by 100, gives the mean percentage error in the measurement of a quantity.

Mean percentage error = relative error x 100% = \(\frac{Δa}{a}\)×100%

**(iv) Describe what is meant by significant figures and order of magnitude.**

**(a) Significant figures** : The value of a measured or calculated quantity should be reported in a way that indicates the precision with which the quantity is known.

The number of figures or digits used to write this value should include only the digits that are known reliably plus the first uncertain digit. The reliable digits plus the first uncertain digit are known as significant figures.

Example ; A length quoted as 2.4 cm has two significant figures. This implies that the actual length could be as small as 2.3 cm or as large as 2.5 cm. That is, the digit 2 in 2.4 is certain, but there is uncertainty in the last digit.

In general, in a measured value, the last digit to the right is taken to be inexact or uncertain but all digits farther to the left are assumed to be exact. However, it is necessary to record the last digit to indicate the precision of the measurement : a value 2.20 cm has three significant figures and indicates that the length was measured to the nearest 0.01 cm; the last zero is significant, though inexact. Here, the length lies between 2.19 cm and 2.21 cm.

**(b) Order of magnitude** : The order of magnitude of a physical quantity is its magnitude expressed to the nearest integral power of ten.

To find the order of magnitude of a physical quantity, its magnitude is expressed as a number that lies between 0.5 and 5 multiplied by an appropriate integral power of 10. The power of 10 along with the unit then gives the order of magnitude of the quantity.

The magnitude of any physical quantity can be expressed as A×10^{n} where ‘A’ is a number such that 0.5 ≤ A<5 and ‘n’ is an integer called the **order of magnitude**.

**Example **: (i) radius of Earth = 6400 km = 0.64×10^{7}m

The order of magnitude is 7 and the number of significant figures are 2.

(ii) Magnitude of the charge on electron *e *= 1.6×10^{-19} C

Here the order of magnitude is -19 and the number of significant digits are 2.

**(v) If the measured values of two quantities are A ****±**** Δ****A and B ****±**** Δ****B, ****Δ****A and ****Δ****B being the mean absolute errors. What is the maximum possible error in A ****±** **B? Show that if Z = \(\frac{A}{B}\), \(\frac{ΔZ}{Z}=\frac{ΔA}{A}+\frac{ΔB}{B}\) **

Consider two quantities whose measured values are A ± ΔA and B ± ΔB, where ΔA and ΔB are their respective mean absolute errors.

(i) If a quantity Z is equal to the sum of A and B, and ΔZ is the maximum absolute error in Z, then

Z ± ΔZ = (A ± ΔA)+( B ± ΔB)

= (a + b) ± (ΔA + ΔB)

∴ ΔZ = ±ΔA ± ΔB

The four possible values of ΔZ are (+ΔA + ΔB), (+ Aa — Ab), (—ΔA + ΔB) and (—ΔA — ΔB).

∴ The maximum possible absolute error in Z,

ΔZ = ΔA + ΔB

(ii) If Z = A/B, then

Z ± ΔZ = (A ± ΔA)/(B ± ΔB)

= (A ± ΔA)(B ± ΔB)^{—1}

= \(A(1±\frac{ΔA}{A})B^{-1}(1±\frac{ΔB}{B})^{-1}\)

= \(\frac{A}{B}(1±\frac{ΔA}{A})(1±\frac{ΔB}{B}+\text{terms containing higher powers of}\frac{ΔB}{B})\)

\((\frac{ΔB}{B})\) being small, its higher powers can be ignored in comparison to it.

∴ Z ± ΔZ = \(\frac{A}{B}(1±\frac{ΔA}{A})(1±\frac{ΔB}{B})\)

= \(Z(1±\frac{ΔA}{A}±\frac{ΔB}{B}-\frac{ΔAΔB}{AB})\)

Again ignoring the considerably smaller term \(\frac{ΔAΔB}{AB}\), we get

∴ Z ± ΔZ = \(Z(1±\frac{ΔA}{A}±\frac{ΔB}{B})\)

∴ 1 ± \(\frac{ΔZ}{Z}\) = \((1±\frac{ΔA}{A}±\frac{ΔB}{B})\)

∴ \(\frac{ΔZ}{Z}\) = \(±\frac{ΔA}{A}±\frac{ΔB}{B}\)

The four possible values of \(\frac{ΔZ}{Z}\) are \((\frac{ΔA}{A}-\frac{ΔB}{B})\), \((\frac{ΔA}{A}+\frac{ΔB}{B})\), \((-\frac{ΔA}{A}-\frac{ΔB}{B})\), and \((-\frac{ΔA}{A}+\frac{ΔB}{B})\)

∴ The maximum possible relative error in z,

\(\frac{ΔZ}{Z}=\frac{ΔA}{A}+\frac{ΔB}{B}\)

∴ The maximum relative error in the quotient of two measured quantities is the sum of their individual relative errors.

**(vi) Derive the formula for kinetic energy of a particle having mass ***m ***and velocity v using dimensional analysis**

The kinetic energy of a body depends upon the mass m and its speed v. To determine the exact dependence of kinetic energy on m and v,

we write,

E = km^{x} v^{y}

where k is dimensionless constant. The values of x and y can be determined using dimensional analysis as follows :

[E] = [ML^{2}T^{—}^{2}],

[m] = [M]

[v] = [LT^{—}^{1}]

∴ [E] = [M^{x}].[LT^{—1}]^{y}

∴ [ML^{2}T^{— 2}] = [M^{x}L^{y}T^{—}^{y}]

Comparing the powers of respective quantities on both the sides,

x = 1 and y = 2

∴ E = km v^{2}

**Question 3. Solve numerical examples.**

**(i) The masses of two bodies are measured to be 15.7 ****± ****0.2 kg and 27.3 ****± ****0.3 kg. What is the total mass of the two and the error in it?**

**Given** : m_{1} = (15.7 ± 0.2) kg, m_{2} = (27.3 ± 0.3) kg

Total mass, m = 15.7 kg + 27.3 kg = 43.0 kg

Absolute error in m,

Δm = Δm_{1} + Δm_{2} = 0.2 + 0.3 = 0.5 kg

**(ii) The distance travelled by an object in time (100 ****± ****1) s is (5.2 ****± ****0.1) m. What is the speed and it's relative error?**

**Given **: t = (100 ± 1) s, d = (5.2 ± 0.1) m

Average speed, v= d/t = 5.2/100 = 0.052 m/s

Relative error in v, \(\frac{Δv}{v}=\frac{Δd}{d}+\frac{Δt}{t}\) = \(\frac{0.1}{5.2}+\frac{1}{100}\) = 0.01923 + 0.01 = 0.02923

∴ The absolute error in v,

Δv = (0.02923)(0.052) = 1.520 x 10^{—}^{3} m/s

= 0.001520 m/s

= 0.002 m/s, rounded to the precision of v

∴ v = (0.052 ± 0.002) m/s

**(iii) An electron with charge ***e ***enters a uniform. magnetic field ** ** ****with a velocity ** **. ****The velocity is perpendicular to the magnetic field. The force on the charge ***e ***is given by ****| ** **| =**** Bev , ****Obtain the dimensions of **

F = | \(\vec{F}\) | = Bev

∴ B = \(\frac{F}{ev}\),

where [F] = [force] = [MLT^{—}^{2}], [e] = [charge] = [TI], [v] = [speed] = [LT^{—1}]

∴ [B] = \(\frac{[F]}{[e][v]}=\frac{[MLT^{-2}]}{[TI][v]}\) = **[MT**^{—}^{2}**I**^{—}^{1}**]**

**(iv) A large ball 2 m in radius is made up of a rope of square cross section with edge length 4 mm. Neglecting the air gaps in the ball, what is the total length of the rope to the nearest order of magnitude?**

**Given **: R = 2m, d = 4mm = 4 x 10^{—}^{3}m

Let L be the length of the rope in the ball. The total volume of the rope is roughly the volume V of the ball.

Total volume of the string, V = (cross-sectional area)(length)

∴ V = d^{2}L = \(\frac{4}{3}\)πR^{3} ≈ 4R^{3} …('.'π ≈ 3)

∴ L = \(\frac{4R^3}{d^2}=\frac{4(2)^3}{(4×10^{-3})^2}\)\) = 2 x 10^{6} m

∴ O(L) = 10^{6} m = 10^{3} km

**(v) Nuclear radius ***R ***has ***a ***dependence on the mass number (***A***) as ***R ***= 1.3 × 10 ^{-16} **

*A*

^{1/3}**m. For a nucleus of mass number**

*A***= 125, obtain the order of magnitude of**

*R***expressed in metre.**

**Given** : A = 125 = 5^{3}

The formula for nuclear radius is R ≈ R_{0} A^{1/3} where R_{0} ≈ 1.3 x 10^{-15} m.

∴ R = (1.3 x 10^{-15} )(5^{3})^{ 1/3} = (1.3 x 10^{-15})(5)

= 6.5 x 10^{-15} m = 0.65 x 10^{-14} m

O(R) = 10^{-14} m

**(vi) In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the length yielded the following values: 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error and the percentage error in the measured value of the length.**

Given: n = 4, *l*_{1} = 3.11 cm, *l*_{2} = 3.13 cm, *l*_{3 }=3.14 cm, *l*_{4 }= 3.14 cm

**(i)** Mean (or most probable) length,

\(\bar{l}\) = \(\frac{1}{n}\)(*l*_{1 }+ *l*_{2 }+ *l*_{3 }+ *l*_{4})

= (3.11 + 3.13 + 3.14 + 3.14)/4 = 12.52/4

= 3.13 cm, to the precision of 0.01 cm

**(ii)** Mean absolute error,

Δ* l* = \(\frac{1}{n}\sum _{i=1}^{4}|l_i-\bar{l}|\)

= (0.02 + 0.00 + 0.01 + 0.01)/4 = 0.04/4 cm

= **0.01 cm**

**(iii)** Percentage error

\(\frac{Δl}{l}\)×100% = \(\frac{0.1}{3.13}\)×100%

= 0.003195 X 100% = 0.3195%

≈ 0.32%

**(vii) Find the percentage error in kinetic energy of a body having mass 60.0 ****± ****0.3 g moving with a velocity 25.0 ****± ****0.1 cm/s.**

**Given** : m = (60.0 ± 0.3) g, v = (25.0 ± 0.1) cm/s

Kinetic energy, K.E. = \(\frac{1}{2}\)mv^{2} = \(\frac{1}{2}\)(60.0)(25.0)^{2}

= (3.00 x 10)(6.25 x 10^{2})

= 1.8750 x 10^{4} J ≈ 1.88 x 10^{4} J, to 3 significant figures

Relative error in K,

= \(\frac{ΔK}{K} =\frac{Δm}{m}+2(\frac{Δv}{v})\) = \(\frac{0.2}{60.0}+2(\frac{0.1}{25.0})\)

= 0.005 + 2(0.004) = 0.005 + 0.008

= 0.013

∴ Percentage error in K,

\(\frac{ΔK}{K}\) x 100% = 0.013 x 100% = **1.3%**

**(viii) In Ohm's experiments, the values of the unknown resistances were found to be 6.12 ****Ω ****, 6.09 ****Ω****, 6.22 ****Ω****, 6.15 ****Ω****. Calculate the mean absolute error, relative error and percentage error in these measurements.**

**Given **: R_{1} = 6.12 Ω , R_{2} = 6.09 Ω, R_{3} = 6.22 Ω, R_{4} = 6.15 Ω

Most probable value of resistance,

= (6.12 + 6.09 + 6.22 + 6.15)/4

= 6.145 Ω ≈ 6.14 Ω

(i) Mean absolute error,

ΔR = \(\frac{1}{n}\sum _{i=1}^{4}|R_i-\bar{R}|\) = (0.02 + 0.05 + 0.08 + 0.01)/4 = 0.16/4 = 0.04 Ω

(ii) Relative error in R,

\(\frac{ΔR}{R}=\frac{0.04}{6.14}\) = 0.006515 ≈ 0.0065

(iii) Percentage error in R = \(\frac{ΔR}{R}\)x 100% = 0.0065 x 100% = 0.65%

**(ix) An object is falling freely under the gravitational force. Its velocity after travelling a distance h is v. If v depends upon gravitational acceleration g and distance, prove with dimensional analysis that v =** \(k\sqrt{gh}\) **where k is a ****constant.**

We assume a power relation where v is the x^{th} power of g and y^{th} power of h.

Then, v ∝ g^{x}h^{y} = kg^{x}h^{y}

where k is a dimensionless constant.

[v] = [LT^{—1}], [g] = [LT^{—2}], [h] = [L], [k] = 1

∴ [v] = 1[LT^{—2}]^{x} [L]^{y}

∴ [LT^{—1}] = [L^{x+y} T^{—2x}]

Comparing the powers of the respective quantities on both the sides,

x + y = 1 and —2x = —1

∴ x = 1/2 and y = 1—1/2 = 1/2

∴ v = k g^{1/2} h^{1/2} = \(k\sqrt{gh}\)** ** as required.

**(x) v = \(at+\frac{b}{t+c}+v_0\)**** ****is a dimensionally valid ****equation. Obtain the dimensional ****formula for a, b and c where v is velocity, t is time and v _{0} is initial velocity.**

[v] = [v_{o}] = [velocity] = [LT^{—1}], [t] = [T]

Given that the equation, v = \(at+\frac{b}{t+c}+v_0\) is dimensionally valid,

[at] = \(\frac{b}{t+c}\) = [v] = [LT^{—1}]

∴ [a] = \(\frac{[at]}{[t]} = [LT^{—2}]

In an addition, the terms must have the same dimensions.

∴ [c] = [t] = [T]

\([\frac{b}{t+c}]=\frac{[b]}{[t+c]}\) = [LT^{—1}]

∴ [b] = [t + c][LT^{—1}] = [T][LT^{—1}] = **[L]**

**(xi) 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.**

**Given** : *l* = 4.234 m, b = 1.005 m,

d = 2.01 cm = 2.01 x 10^{-2} m

The area of the largest face of the sheet,

A = *l* x b = (4.234)(1.005) = 4.255 m^{2}, correct to 4 significant figures.

The volume of the sheet,

V = *l* bd = (4.234)(1.005)(2.01 x 10 ^{-2})

= 8.553 x 10^{-2} m^{2} ≈ 8.55 x 10^{-2} m^{2}, rounded to 3 significant figures.

**(xii) If the length of a cylinder is ***l***= (4.00****±****0.001) cm, radius r = (0.0250 ****±****0.001) cm and mass m = (6.25****±****0.01) gm. Calculate the percentage error in the determination of density.**

**Given** : *l* = (4.000 ± 0.001) cm, r = (0.025 ± 0.001) cm, m = (6.25 ± 0.01) g

Density, ρ = mass/volume = \(\frac{m}{πr^2l}\)

= \(\frac{6.25}{(3.142)(2.5×10^{-2})^2(4.000)}\)=\(\frac{6.25×10^4}{(3.142)(6.25)(4.000)}\)

= \(\frac{2500}{3.142}\) = 795.7 g/cm^{3}

The relative error in ρ,

\(\frac{Δρ}{ρ}=\frac{Δm}{m}+2\frac{Δr}{r}+\frac{Δl}{l}\) = \(\frac{0.01}{6.25}+2(\frac{0.001}{0.025})+\frac{0.001}{4.000}\)

= 0.00016 + 0.08 + 0.0025 = 0.08185

∴ The percentage in ρ, = \(\frac{Δρ}{ρ}\) x 100% = 0.08185 x 100% = 8.185% ≈ 8%

**(xiii) When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of the Jupiter.**

**Given **:

D = 824.7 x 10^{6} km = 8.247 x 10^{8} km,

α = 35.72 as = 35.72 x 4.847 x 10^{—}^{6 }rad ……(1 as = 4.847 x 10^{—}^{6} rad)

∴ α = 173.14 x 10^{—6 }rad

Diameter of Jupiter,

d = α D = (173.14 x 10^{—}^{6})(8.247 x 10^{8} km)

= 1428 x 10^{2} km = 1.428 x 10^{5} km

**(xiv) If the formula for a physical quantity is **X = \(\frac{a^4b^3}{c^{1/3}d^{1/2}}\) **and if the percentage error in the measurements of a, b, c and d are 2%, 3%, 3% and 4% respectively. Calculate percentage error in X.**

*X = *\(\frac{a^4b^3}{c^{1/3}d^{1/2}}\)

∴ Relative error in X,

\(\frac{ΔX}{X}=4\frac{Δa}{a}+3\frac{Δb}{b}+\frac{1}{3}\frac{Δc}{c}+\frac{1}{2}\frac{Δd}{d}\)

∴ \(\frac{ΔX}{X}\)x 100% = \(4(\frac{Δa}{a}×\text{100%})+3(\frac{Δb}{b}×\text{100%})+\frac{1}{3}(\frac{Δc}{c}×\text{100%})+\frac{1}{2}(\frac{Δd}{d}×\text{100%})\)

= 4(2%) +3(3%) + \(\frac{1}{3}\)(3%) + \(\frac{1}{2}\)(4%)

= 8% + 9% + 1% + 2%

= **20%**

**(xv) Write down the number of significant figures in the following: 0.003 m ^{2}, 0.1250 gm cm^{-2}, 6.4 x 10^{6} m, 1.6 x 10^{-19} C, 9.1 x 10^{-31} kg.**

Number |
Number of significant figures |

0.003 | 1 |

0.1250 | 4 |

6.4 x 10^{6} |
2 |

1.6 x 10^{-19} |
2 |

9.1 x 10^{-31} |
2 |

**(xvi) The diameter of a sphere is 2.14 cm. Calculate the volume of the sphere to the correct number of significant figures.**

**Given** : d = 2.14 cm

Volume, V = \(\frac{4}{3}πr^3\)= \(\frac{1}{6}πd^3\)= (3.142)(2.14)^{3} = 5.132 cm^{3}

≈ 5.13 cm^{3}, correct to 3 signiﬁcant figures.

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