Chapter 1 Units and Measurements Solutions

Maharashtra State Board 11th Physics Solutions Chapter 1 Units and Measurements

1. Choose the correct option.

2. Answer the following questions.

ii) As star A is farther i.e., D_A> D_B
⇒ θ_A< θ_B
Hence, star B will have larger parallax angle than star A.

Mean absolute error:
For a given set of measurements of a same quantity the arithmetic mean of all the absolute errors is called mean absolute error in the measurement of that physical quantity.
∆a_mean=

Relative error:
The ratio of the mean absolute error in the measurement of a physical quantity to its arithmetic mean value is called relative error.
Relative error =

Percentage error:
The relative error represented by percentage (i.e., multiplied by 100) is called the percentage error.
Percentage error = × 100%
[Note: Considering conceptual conventions question is modified to define percentage error and not mean percentage error.]

Rules for determining significant figures:

Order of magnitude:
The magnitude of any physical quantity can be expressed as A × 10ⁿwhere ‘A’ is a number such that 0.5 ≤ A < 5 then, ‘n’ is an integer called the order of magnitude.
Examples:

ii) Thus, when two quantities are divided, the maximum relative error in the result is the sum of relative errors in each quantity.

3. Solve numerical examples.

ii) ±∆Z = ±∆A ± ∆B
Calculation: From formula (i),
Z = 15.7 + 27.3 = 43 kg
From formula (ii),
± ∆Z (± 0.2) + (± 0.3)
=±(0.2 + 0.3)
= ± 0.5 kg
Total mass is 43 kg and total error is ± 0.5 kg.

Formulae: i. v =
ii.

Calculation: From formula (i),
v = = 0.052 m/s
From formula (ii),

=
= ± 0.029 rn/s
The speed is 0.052 m/s and its maximum relative error is ± 0.029 m/s.
[Note: Framing of numerical is modified to make it specific and meaningful.]

Formulae: i. a_mean=
ii. ∆a_n= |a_mean– a_n|
iii. ∆a_mean=
iv. Percentage error = × 100

Calculation: From formula (i),
a_mean=
= 3.13 cm
From formula (ii),
∆a₁= |3.13 – 3.11| = 0.02 cm
∆a₂= |3.13 – 3.13| = 0
∆a₃= |3.13 – 3.14| = 0.01 cm
∆a₄= |3.13 – 3.14| = 0.01 cm
From formula (iii),
∆a_mean= = 0.01 cm
From formula (iii).
% error = × 100
=
= 0.3196
……..(using reciprocal table)
= 0.32%

i. Mean length is 3.13 cm.
ii. Mean absolute error is 0.01 cm.
iii. Percentage error is 0.32 %.
[Note: As per given data of numerical, percentage error calculation upon rounding off yields percentage error as 0.32%]

Calculation: From formula,
Percentage error in E
= × 100%
= 1.3%
The percentage error in energy is 1.3%.

To find:

i) Absolute error (∆a_mean)
ii) Relative error
iii) Percentage error

Formulae:

i) a_mean=
ii) ∆a_n= |a_mean– a_n|
iii) ∆a_mean=
iv) Percentage error = × 100

From formula (ii),
∆a₁= |6.145 – 6.12|= 0.025
∆a₂= |6.145 – 6.09| = 0.055
∆a₃= |6.145 – 6.22| = 0.075
∆a₄= |6.l45 – 6.15| = 0.005
From formula (iii),
∆a_mean=
= 0.04 Ω
From formula (iv),
Relative error = = 0.0065 Ω
From formula (v).
Percentage error = 0.0065 100 = 0.65%

i. The mean absolute error is 0.04 Ω.
ii. The relative error is 0.0065 Ω.
iii. The percentage error is 0.65%.
[Note: Framing of numerical is modified to reach the answer given to the numerical.]

As only dimensionally identical quantities can be added together or subtracted from each other, each term on R.H.S. has dimensions of L.H.S. i.e., dimensions of velocity.

∴ [LH.S.] = [v] = [L¹T^-1]
This means, [at] = [v] = [L¹T^-1]
Given, t = time has dimension [T^-1]
∴ [a] = = [L¹T^-2] = L¹M⁰T^-2]
Similarly, [c] = [t] = [T¹] = [L⁰M⁰T¹]
∴ = [v] = [L¹T^-1]
∴ [b] = [L¹T^-1] × [T¹] = [L¹] = [L¹M⁰T⁰]

To find:
i) Area of sheet to correct significant figures (A)
ii) Volume of sheet to correct significant figures (V)
Formulae: i. A = 2(lb + bt + tl)
iii) V = l × b × t

Calculation: From formula (i),
A = 2(4.234 × 1.005 + 1.005 × 0.0201 +0.0201 × 4.234)
= 2 |[antilog(log 4.234 + logl.005) + antiiog(log 1.005 + log0.0201) + antilog(log 0.0201 + log 4.234)]}
= 2{[antilog(0.6267 + 0.0021) + antilog(0.0021 + .3010) + antilog (.3010 + 0.6267)]}
= 2 {[antilog(0.6288) + antilog (.9277)]}
= 2 [4.254 + 0.02009 + 0.08467]
= 2 [4.35876]
= 8.71752m²

In correct significant figure,
A = 8.72 m2 From formula (ii),
V =4.234 × 1.005 × 0.0201
= antilog [log (4.234) + log (1.005) + log (0.0201)]
= antilog [0.6269 – 0.0021 – .3032]
= antilog [0.6288 – .3032]
= antilog [ 2.9320]
= 8.551 × 10^-2
= 0.08551 m³
In correct significant figure (rounding off),
V = 0.086 m³

i.) Area of sheet to correct significant figures is 8.72 m².
ii) Volume of sheet to correct significant figures is 0.086 m³.
[Note: The given solution is arrived to by considering a rectangular sheet.]

Formulae:
i) Relative error in volume,
….(∵ Volume of cylinder, V = πr²l)

ii) Releative error

iii) Percentage error= Relative error × 100%

Calculation.
From formulae (i) and (ii),
∴
=
= 0.00 16 + 0.08 + 0.00025
= 0.08 185
From formula (iii).
% error in density = × 100
= 0.08185 × 100
= 8.185%
Percentage error in density is 8.185%.

Distance from Earth (D)
= 824.7 million km
= 824.7 × 10⁶km
= 824.7 × 10⁹m.

To find: Diameter of Jupiter (d)
Formula: d = α D
Calculation: From formula,
d = 1.73 × 10^-4× 824.7 × 10⁹
= 1.428 × 10⁸m
= 1.428 × 10⁵km
Diameter of Jupiter is 1.428 × 10⁵km.

11th Physics Digest Chapter1 Units and MeasurementsIntext Questions and Answers

Can you recall (Textbook Page No. 1)

Can you tell? (Textbook Page No. 8)

But least count of metre scale is 1 mm. As a result, even smallest uncertainty in reading would vary reading significantly. Also, skill of students doing measurement may also introduce uncertainty in observation.
Hence, their answers are likely to be different.

Activity (Textbook Page No. 10)

Perform an experiment using a Vernier callipers of least count 0.01cm to measure the external diameter of a hollow cylinder. Take 3 readings at different positions on the cylinder and find (i) the mean diameter (ii) the absolute mean error and (iii) the percentage error in the measurement of diameter.
Answer:
Given: L.C. = 0.01 cm
To measure external diameter of hollow cylinder readings are taken as follows:

[Note: The above table is made assuming zero error in Vernier calipers. If caliper has positive or negative zero error, the zero error correction needs to be introduced into observed reading.]

Internet my friend (Textbook Page No. 12)

i. ideoiectures.net/mit801f99_lewin_lec0l/
ii. hyperphysicsphy-astr.gsu.ed u/libase/hfra me. html

[Students can use links given above as a reference and collect information about units and measurements]