Solution-Class-11-Science-Physics-Chapter-1-Units and Measurements-Maharashtra Board

(C) Force

(C) 2%

(C) Distance

(C) Work

(B) kg

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

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} T¹]

= [L^{1+ ½-½} T¹]

= [L¹T¹]

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

• 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%

(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×10ⁿ 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 = 0.64×10⁷m

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.

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.

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] = [ML²T^—2],

[m] = [M]

[v] = [LT^—1]

∴ [E] = [M^x].[LT^—1]^y

∴  [ML²T^{— 2}] = [M^xL^yT^—y]

Comparing the powers of respective quantities on both the sides,

x = 1 and y = 2

∴ E = km v²

Given : m₁ = (15.7 ± 0.2) kg, m₂ = (27.3 ±  0.3) kg

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

Absolute error in m,

Δm = Δm₁ + Δm₂ = 0.2 + 0.3  = 0.5 kg

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

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^—2I^—1]

Given : R = 2m, d = 4mm = 4 x 10^—3m

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 = d²L = \(\frac{4}{3}\)πR³ ≈ 4R³   …('.'π  ≈ 3)

∴ L = \(\frac{4R^3}{d^2}=\frac{4(2)^3}{(4×10^{-3})^2}\)\) = 2 x 10⁶ m

∴ O(L) = 10⁶ m = 10³ km

Given : A = 125 = 5³

The formula for nuclear radius is R ≈ R₀ A^1/3 where R₀ ≈ 1.3 x 10^-15 m.

∴ R = (1.3 x 10^-15 )(5³) ^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

Given: n = 4, l₁ = 3.11 cm, l₂ = 3.13 cm, l₃ =3.14 cm, l₄ = 3.14 cm

(i) Mean (or most probable) length,

\(\bar{l}\) = \(\frac{1}{n}\)(l₁ + l₂ + l₃ + l₄)

= (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%

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

Kinetic energy, K.E. = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)(60.0)(25.0)²

= (3.00 x 10)(6.25 x 10²)

= 1.8750 x 10⁴ J  ≈ 1.88 x 10⁴ 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%

Given : R₁ = 6.12 Ω , R₂ = 6.09 Ω, R₃ = 6.22 Ω, R₄ = 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%

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

Then, v ∝ g^xh^y = kg^xh^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.

[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]

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 m², 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 m² ≈ 8.55 x 10^-2 m², rounded to 3 significant figures.

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/cm³

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%

Given :

D = 824.7 x 10⁶ km = 8.247 x 10⁸ 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 10⁸ km)

= 1428 x 10² km = 1.428 x 10⁵ km

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%

Given : d = 2.14 cm

Volume, V = \(\frac{4}{3}πr^3\)= \(\frac{1}{6}πd^3\)=  (3.142)(2.14)³ = 5.132 cm³

≈ 5.13 cm³, correct to 3 significant figures.