Maharashtra Board Class 11 Physics Chapter 1 Units and Measurements Solutions

Exercise 1. Choose The Correct Option.

Question 1. [L¹M¹T-2] is the dimensional formula for
(A) Velocity
(B) Acceleration
(C) Force
(D) Work
Answer: (C) Force
In simple words: The given dimensional formula `[L^1M^1T^{-2}]` represents Force, which is derived from mass times acceleration.

🎯 Exam Tip: Remember common dimensional formulas like force, work, and energy. Pay attention to the exponents of L, M, and T.

Question 2. 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) \( \frac{1}{2} \) %
(C) 2%
(D) None of the above.
Answer: (C) 2%
In simple words: For a quantity that depends on a product of measurements (like area = length × width), the percentage error in the final quantity is the sum of the percentage errors in the individual measurements. If the error in sides is 1%, the error in area will be 1% + 1% = 2%.

🎯 Exam Tip: For products or quotients of physical quantities, percentage errors add up. For powers, the error is multiplied by the power.

Question 3. Light year is a unit of
(A) Time
(B) Mass
(C) Distance
(D) Luminosity
Answer: (C) Distance
In simple words: A light-year is the distance light travels in one Earth year, making it a unit of distance, not time.

🎯 Exam Tip: Do not confuse "light-year" with a unit of time due to the word "year". It's a fundamental unit for astronomical distances.

Question 4. Dimensions of kinetic energy are the same as that of
(A) Force
(B) Acceleration
(C) Work
(D) Pressure
Answer: (C) Work
In simple words: Kinetic energy and work are both forms of energy, and all forms of energy have the same dimensional formula, which is `[M^1L^2T^{-2}]`.

🎯 Exam Tip: Quantities with the same dimensions can be added or subtracted, and they represent similar physical entities (e.g., all forms of energy share the same dimensions).

Question 5. Which of the following is not a fundamental unit?
(A) cm
(B) kg
(C) centigrade
(D) volt
Answer: (D) volt
In simple words: Fundamental units are basic units like meter (length), kilogram (mass), second (time), Kelvin (temperature), Ampere (current), candela (luminous intensity), and mole (amount of substance). Volt is a derived unit (for electric potential), while cm, kg, and centigrade (related to temperature) can be considered fundamental or derived from fundamental units.

🎯 Exam Tip: Understand the difference between fundamental and derived units in the SI system. Derived units are combinations of fundamental units.

Exercise 2. Answer The Following Questions.

Question 1. Star A is farther than star B. Which star will have a large parallax angle? Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो तारों A और B को पृथ्वी से देखा जाने पर उनके लंबन कोण को दर्शाता है। तारा B पृथ्वी के करीब है, और तारा A दूर है। दो तारों के बीच की दूरी 'b' है, और पृथ्वी से तारों की दूरी क्रमशः \( D_A \) और \( D_B \) है, जिससे लंबन कोण \( \theta_A \) और \( \theta_B \) बनते हैं।
i). 'b' is constant for the two stars
\( \implies \theta \propto \frac{1}{D} \)
ii) As star A is farther i.e., \( D_A > D_B \)
\( \implies \theta_A < \theta_B \)
Hence, star B will have larger parallax angle than star A.
In simple words: Parallax angle is inversely proportional to the distance of an object. Since Star B is closer to Earth than Star A, Star B will have a larger parallax angle.

🎯 Exam Tip: The concept of parallax is crucial for measuring astronomical distances; a larger parallax angle indicates a closer object.

Question 2. What are the dimensions of the quantity \( I \sqrt{1/g} \), l being the length and g the acceleration due to gravity? Answer:
Quantity = \( I \times \sqrt{\frac{L}{g}} \) ....(i)
gravitational acceleration, \( g = \frac{\text{velocity}}{\text{time}} = \frac{\text{distance}}{\text{time} \times \text{time}} \)
Substituting in equation (i),
Quantity = \( I \times \sqrt{\frac{I \times \text{time}^2}{\text{distance}}} \)
Dimensional formula of quantity
= \( [L] \times \frac{[L^{1/2} T^{2 \times 1/2}]}{[L^{1/2}]} = [L] \times [T^1] = [L^1T^1] \)
[Note: When power of symbol expressing fundamental quantity appearing in the dimensional formula is not given, ills taken as 1.]
In simple words: To find the dimensions of `\( I \sqrt{1/g} \)`, substitute the dimensions of length `[L]` and acceleration due to gravity `[LT^{-2}]`. This simplifies to `[L^{1/2}T]`. The given solution's intermediate steps seem to be derived from a slightly different initial formula or simplification. However, if the expression is interpreted as `I sqrt(L/g)`, the dimensions become `[L] * sqrt([L] / [LT^-2]) = [L] * sqrt([T^2]) = [L] * [T] = [LT]`.

🎯 Exam Tip: Always break down complex quantities into their fundamental dimensions (L, M, T, etc.) and simplify the exponents. Be careful with square roots and powers.

Question 3. Define absolute error, mean absolute error, relative error and percentage error. Answer:
Absolute error:
a. For a given set of measurements of a quantity, the magnitude of the difference between mean value (Most probable value) and each individual value is called absolute error (\( \Delta a \)) in the measurement of that quantity.
b. absolute error = |mean value - measured value|
\( \Delta a_1 = |a_{\text{mean}} - a_1| \)
Similarly, \( \Delta a_2 = |a_{\text{mean}} - a_2| \)
....
....
\( \Delta a_n = |a_{\text{mean}} - a_n| \)
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.
\( \Delta a_{\text{mean}} = \frac{\Delta a_1+\Delta a_2+.......+\Delta a_n}{n} \)
\( = \frac{1}{n} \sum_{i=1}^{n} \Delta a_i \)
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 = \( \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \)
Percentage error:
The relative error represented by percentage (i.e., multiplied by 100) is called the percentage error.
Percentage error = \( \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100\% \)
[Note: Considering conceptual conventions question is modified to define percentage error and not mean percentage error.]
In simple words: Absolute error is the difference between a measured value and the true value. Mean absolute error is the average of all absolute errors. Relative error is the ratio of mean absolute error to the mean value. Percentage error is the relative error expressed as a percentage.

🎯 Exam Tip: Clear and concise definitions are key. Understand the formula for each type of error and when to apply them in calculations.

Question 4. Describe what is meant by significant figures and order of magnitude. Answer:
Significant figures:
1. Significant figures in the measured value of a physical quantity is the sum of reliable digits and the first uncertain digit.
OR
The number of digits in a measurement about which we are certain, plus one additional digit, the first one about which we are not certain is known as significant figures or significant digits.
2. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.
3. If one uses the instrument of smaller least count, the number of significant digits increases.
Rules for determining significant figures:
1. All the non-zero digits are significant, for example if the volume of an object is \( 178.43 \text{ cm}^3 \), there are five significant digits which are 1,7,8,4 and 3.
2. All the zeros between two nonzero digits are significant, eg., \( m = 165.02 \text{ g} \) has 5 significant digits.
3. If the number is less than 1, the zero/zeroes on the right of the decimal point and to the left of the first nonzero digit are not significant e.g. in 0.001405, significant. Thus the above number has four significant digits.
4. The zeroes on the right hand side of the last nonzero number are significant (but for this, the number must be written with a decimal point), e.g. 1.500 or 0.01500 both have 4 significant figures each.
On the contrary, if a measurement yields length L given as L = 125 m = 12500 cm = 125000 mm, it has only three significant digits.
Order of magnitude:
The magnitude of any physical quantity can be expressed as \( A \times 10^n \) where 'A' is a number such that \( 0.5 \le A < 5 \) then, 'n' is an integer called the order of magnitude.
Examples:
1. Speed of light in air = \( 3 \times 10^8 \text{ m/s} \)
\( \implies \) order of magnitude = 8
2. Mass of an electron = \( 9.1 \times 10^{-31} \text{ kg} \)
= \( 0.91 \times 10^{-30} \text{ kg} \)
\( \implies \) order of magnitude = -30
In simple words: Significant figures are the reliable digits in a measurement, indicating its precision. The order of magnitude represents the power of 10 that best approximates a quantity, giving a sense of its scale.

🎯 Exam Tip: Practice the rules for significant figures, especially for leading zeros, trailing zeros, and zeros between non-zero digits. For order of magnitude, ensure the 'A' value is between 0.5 and 5.

Question 5. If the measured values of two quantities are \( A \pm \Delta A \) and \( B \pm \Delta B \), \( \Delta A \) and \( \Delta B \) being the mean absolute errors. What is the maximum possible error in \( A \pm B \)? Show that if \( Z = \frac{A}{B} \), \( \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \)
Answer:
Maximum possible error in \( (A + B) \) is \( (\Delta A + \Delta B) \).
Errors in divisions:
i) A
Suppose, \( Z = \frac{A}{B} \) and measured values of A and B are \( (A \pm \Delta A) \) and \( (B \pm \Delta B) \) then,
\( Z + \Delta Z = \frac{A \pm \Delta A}{B \pm \Delta B} \)
\( Z \left(1 \pm \frac{\Delta Z}{Z}\right) = \frac{A \left(1 \pm \frac{\Delta A}{A}\right)}{B \left(1 \pm \frac{\Delta B}{B}\right)} \)
\( = \frac{A}{B} \frac{1 \pm \frac{\Delta A}{A}}{1 \pm \frac{\Delta B}{B}} \)
As, \( \frac{\Delta B}{B} << 1 \), expanding using Binomial theorem,
\( Z \left(1 \pm \frac{\Delta Z}{Z}\right) = Z \times \left(1 \pm \frac{\Delta A}{A}\right) \times \left(1 \pm \frac{\Delta B}{B}\right) \)
\( \implies 1 \pm \frac{\Delta Z}{Z} = 1 \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \times \frac{\Delta B}{B} \)
Ignoring term \( \frac{\Delta A}{A} \times \frac{\Delta B}{B} \)
This gives four possible values of \( \frac{\Delta Z}{Z} \) as
\( \left(\pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B}\right) \) and \( \left(\pm \frac{\Delta A}{A} \mp \frac{\Delta B}{B}\right) \)
\( \implies \) Maximum relative error of \( \frac{\Delta Z}{Z} = \pm \left(\frac{\Delta A}{A} + \frac{\Delta B}{B}\right) \)
ii) Thus, when two quantities are divided, the maximum relative error in the result is the sum of relative errors in each quantity.
In simple words: When adding or subtracting quantities, the absolute errors add. When dividing two quantities, the maximum relative error in the result is the sum of their individual relative errors.

🎯 Exam Tip: Remember that absolute errors add for sums/differences, while relative errors add for products/quotients. This is a common and important error propagation rule.

Question 6. Derive the formula for kinetic energy of a particle having mass m and velocity v using dimensional analysis
Answer:
Kinetic energy of a body depends upon mass (m) and velocity (v) of the body.
Let K.E. \( \propto m^x v^y \)
\( \implies \) K.E. = \( km^x v^y \) ...... (1)
where,
k = dimensionless constant of proportionality. Taking dimensions on both sides of equation (1),
\( [L^2M^1T^{-2}] = [L^0M^1T^0]^x [L^1M^0T^{-1}]^y \)
\( = [L^0M^xT^0] [L^yM^0T^{-y}] \)
\( = [L^{0+y}M^{x+0}T^{0-y}] \)
\( [L^2M^1T^{-2}] = [L^yM^xT^{-y}] \) ........ (2)
Equating dimensions of L, M, T on both sides of equation (2),
x = 1 and y = 2,
Substituting x, y in equation (1), we have
K.E. = \( kmv^2 \)
In simple words: By assuming kinetic energy depends on mass (m) and velocity (v) raised to unknown powers (x and y) and equating the dimensions of both sides of the equation, we can solve for x and y to find the formula `K.E. = kmv^2`.

🎯 Exam Tip: Dimensional analysis is a powerful tool to verify formulas or derive relations between physical quantities, especially when the constant of proportionality is dimensionless.

Exercise 3. Solve Numerical Examples.

Question 1. The masses of two bodies are measured to be \( 15.7 \pm 0.2 \text{ kg} \) and \( 27.3 \pm 0.3 \text{ kg} \). What is the total mass of the two and the error in it? Answer:
Given: \( A \pm \Delta A = 15.7 \pm 0.2 \text{kg} \) and
\( B \pm \Delta B = 27.3 \pm 0.3 \text{ kg} \).
To find: Total mass (Z), and total error (\( \Delta Z \))
Formulae: i. Z = A + B
ii) \( \pm \Delta Z = \pm \Delta A + \Delta B \)
Calculation: From formula (i),
Z = 15.7 + 27.3 = 43 kg
From formula (ii),
\( \pm \Delta Z = (\pm 0.2) + (\pm 0.3) \)
= \( \pm (0.2 + 0.3) \)
= \( \pm 0.5 \text{ kg} \)
Total mass is 43 kg and total error is \( \pm 0.5 \text{ kg} \).
In simple words: To find the total mass and its error, simply add the individual masses and their respective absolute errors.

🎯 Exam Tip: For addition or subtraction, absolute errors always add up in the worst-case scenario.

Question 2. The distance travelled by an object in time \( (100 \pm 1) \text{ s} \) is \( (5.2 \pm 0.1) \text{ m} \). What is the speed and it's relative error? Answer:
Given: Distance \( (D \pm \Delta D) = (5.2 \pm 0.1) \text{ m} \),
time \( (t \pm \Delta t) = (100 \pm 1) \text{s} \).
To find: Speed (v), maximum relative error \( \left(\frac{\Delta v}{v}\right) \)
Formulae: i. \( v = \frac{D}{t} \)
ii. \( \frac{\Delta v}{v} = \pm \left(\frac{\Delta D}{D} + \frac{\Delta t}{t}\right) \)
Calculation: From formula (i),
\( v = \frac{5.2}{100} = 0.052 \text{ m/s} \)
From formula (ii),
\( \frac{\Delta v}{v} = \pm \left(\frac{0.1}{5.2} + \frac{1}{100}\right) \)
\( = \pm \left(\frac{19}{650}\right) \)
= \( \pm 0.029 \text{ rn/s} \)
The speed is 0.052 m/s and its maximum relative error is \( \pm 0.029 \text{ m/s} \).
[Note: Framing of numerical is modified to make it specific and meaningful.]
In simple words: To calculate speed, divide distance by time. To find the relative error in speed, add the relative errors of distance and time because speed is a quotient of these quantities.

🎯 Exam Tip: When dealing with division, use the formula for relative error addition. Always ensure units are consistent and explicitly stated in the final answer.

Question 3. An electron with charge e enters a uniform. magnetic field \( \vec{B} \) with a velocity \( \vec{v} \). The velocity is perpendicular to the magnetic field. The force on the charge e is given by
\( |F| = Bev \) Obtain the dimensions of B.
Answer:
Given: \( |F| = B e v \)
Considering only magnitude, given equation is simplified to,
F = Bev
\( \implies B = \frac{F}{ev} \)
but, F = ma = \( m \times \frac{\text{distance}}{\text{time}^2} \)
\( \implies [F] = [M] \times \left[\frac{L}{T^2}\right] = [L^1M^1T^{-2}] \)
Electric charge, e = current \( \times \) time
\( \implies [e] = [I^1T^1] \)
Velocity \( v = \frac{\text{distance}}{\text{time}} \)
\( \implies [v] = \left[\frac{L}{T}\right] = [L^1T^{-1}] \)
Now, \( [B] = \left[\frac{F}{ev}\right] \)
\( = \frac{[L^1M^1T^{-2}]}{[I^1T^1][L^1T^{-1}]} \)
\( \implies B = [L^0M^1T^{-2}I^{-1}] \)
In simple words: The dimensions of magnetic field B are found by rearranging the force formula `F = Bev` to `B = F/(ev)`. Then, substitute the fundamental dimensions for force, charge (current × time), and velocity to get the final dimensional formula for B.

🎯 Exam Tip: Dimensional analysis is an effective method to determine the dimensions of a physical quantity, especially when given a formula involving other known quantities.

Question 4. 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? Answer:
Volume of ball = Volume enclosed by rope.
\( \frac{4}{3} \pi (\text{radius})^3 \) = Area of cross-section of rope \( \times \) length of rope.
\( \implies \text{length of rope } l = \frac{\frac{4}{3} \pi r^3}{A} \)
Given:
r = 2 m and
Area = A = \( 4 \times 4 = 16 \text{ mm}^2 \)
= \( 16 \times 10^{-6} \text{ m}^2 \)
\( \implies l = \frac{4 \times 3.142 \times 2^3}{3 \times 16 \times 10^{-6}} \)
\( = \frac{3.142 \times 2^3}{3 \times 10^{-6}} \text{ m} \)
\( \approx 2 \times 10^6 \text{ m} \).
Total length of rope to the nearest order of magnitude = \( 10^6 \text{ m} = 10^3 \text{ km} \)
In simple words: To find the length of the rope, equate the total volume of the rope (its cross-sectional area multiplied by its length) to the volume of the ball. Then, calculate and express the result as an order of magnitude.

🎯 Exam Tip: Ensure consistent units throughout the calculation (e.g., convert mm to m) and clearly identify the physical principle (volume conservation) being applied.

Question 5. Nuclear radius R has a dependence on the mass number (A) as \( R = 1.3 \times 10^{-16} A^{1/3} \text{ m} \). For a nucleus of mass number A=125, obtain the order of magnitude of R expressed in metre. Answer:
R= \( 1.3 \times 10^{-16} \times A^{1/3} \text{ m} \)
For A = 125
R= \( 1.3 \times 10^{-16} \times (125)^{1/3} \)
= \( 1.3 \times 10^{-16} \times 5 \)
= \( 6.5 \times 10^{-16} \)
= \( 0.65 \times 10^{-15} \text{ m} \)
\( \implies \) Order of magnitude = -15
[Note: Taking the standard value of nuclear radius \( R = 1.3 \times 10^{-15} \text{ m} \), the order of magnitude comes to be \( 10^{-14} \text{ m} \).]
In simple words: Substitute the given mass number (A=125) into the formula for nuclear radius, calculate the value, and then determine its order of magnitude by expressing it as a number between 0.5 and 5 multiplied by a power of 10.

🎯 Exam Tip: Pay close attention to powers and scientific notation. When determining the order of magnitude, ensure the pre-factor 'A' is within the \( 0.5 \le A < 5 \) range for the exponent 'n'.

Question 6. 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. Answer:
Given: \( a_1 = 3.11 \text{ cm}, a_2 = 3.13 \text{ cm}, \)
\( a_3 = 3.14 \text{ cm}. a_4 = 3.14 \text{ cm} \)
Least count L.C. = 0.01 cm.
To find.
i. Mean length (\( a_{\text{mean}} \))
ii. Mean absolute error (\( \Delta a_{\text{mean}} \))
iii. Percentage error.
Formulae: i. \( a_{\text{mean}} = \frac{a_1+a_2+a_3+a_4}{4} \)
ii. \( \Delta a_n = |a_{\text{mean}} - a_n| \)
iii. \( \Delta a_{\text{mean}} = \frac{\Delta a_1+\Delta a_2+\Delta a_3+\Delta a_4}{4} \)
iv. Percentage error = \( \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100 \)
Calculation: From formula (i),
\( a_{\text{mean}} = \frac{3.11+3.13+3.14+3.14}{4} \)
= 3.13 cm
From formula (ii),
\( \Delta a_1 = |3.13 - 3.11| = 0.02 \text{ cm} \)
\( \Delta a_2 = |3.13 - 3.13| = 0 \)
\( \Delta a_3 = |3.13 - 3.14| = 0.01 \text{ cm} \)
\( \Delta a_4 = |3.13 - 3.14| = 0.01 \text{ cm} \)
From formula (iii),
\( \Delta a_{\text{mean}} = \frac{0.02+0+0.01+0.01}{4} \)
= 0.01 cm
From formula (iii).
% error = \( \frac{0.01}{3.13} \times 100 \)
\( = \frac{1}{3.13} \)
= 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 %.
In simple words: First, calculate the mean of all measurements. Then, find the absolute error for each measurement by subtracting it from the mean. Average these absolute errors to get the mean absolute error. Finally, divide the mean absolute error by the mean length and multiply by 100 to get the percentage error.

🎯 Exam Tip: Systematically apply the formulas for mean, absolute error, mean absolute error, and percentage error. Ensure correct rounding off at each significant step for accuracy.

Question 7. Find the percentage error in kinetic energy of a body having mass \( 60.0 \pm 0.3 \text{ g} \) moving with a velocity \( 25.0 \pm 0.1 \text{ cm/s} \).
Answer:
Given: m = 60.0 g, v = 25.0 cm/s.
\( \Delta m = 0.3 \text{ g}, \Delta v = 0.1 \text{ cm/s} \)
To find: Percentage error in E
Formula: Percentage error in E
\( = \left(\frac{\Delta m}{m} + 2\frac{\Delta v}{v}\right) \times 100\% \)
Calculation: From formula,
Percentage error in E
\( = \left(\frac{0.3}{60.0} + 2 \times \frac{0.1}{25.0}\right) \times 100\% \)
= 1.3%
The percentage error in energy is 1.3%.
In simple words: Since kinetic energy is proportional to `\( mv^2 \)`, its percentage error is found by adding the percentage error in mass to twice the percentage error in velocity.

🎯 Exam Tip: When a quantity has a power (like `v^2` in kinetic energy), its percentage error is multiplied by that power before adding to other percentage errors.

Question 8. In Ohm's experiments, the values of the unknown resistances were found to be \( 6.12 \text{ } \Omega, 6.09 \text{ } \Omega, 6.22 \text{ } \Omega, 6.15 \text{ } \Omega \). Calculate the mean absolute error, relative error and percentage error in these measurements.
Answer:
Given: \( a_1 = 6.12 \text{ } \Omega, a_2 = 6.09 \text{ } \Omega, a_3 = 6.22 \text{ } \Omega, a_4 = 6.15 \text{ } \Omega \),
To find:
i) Absolute error (\( \Delta a_{\text{mean}} \))
ii) Relative error
iii) Percentage error
Formulae:
i) \( a_{\text{mean}} = \frac{a_1+a_2+a_3+a_4}{4} \)
ii) \( \Delta a_n = |a_{\text{mean}} - a_n| \)
iii) \( \Delta a_{\text{mean}} = \frac{\Delta a_1+\Delta a_2+\Delta a_3+\Delta a_4}{4} \)
iv) Percentage error = \( \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100 \)
From formula (ii),
\( \Delta a_1 = |6.145 - 6.12| = 0.025 \)
\( \Delta a_2 = |6.145 - 6.09| = 0.055 \)
\( \Delta a_3 = |6.145 - 6.22| = 0.075 \)
\( \Delta a_4 = |6.145 - 6.15| = 0.005 \)
From formula (iii),
\( \Delta a_{\text{mean}} = \frac{0.025+0.055+0.075+0.005}{4} = \frac{0.160}{4} \)
= \( 0.04 \text{ } \Omega \)
From formula (iv),
Relative error = \( \frac{0.04}{6.145} = 0.0065 \text{ } \Omega \)
From formula (v).
Percentage error = \( 0.0065 \frac{0.04}{6.145} 100 = 0.65\% \)
i. The mean absolute error is \( 0.04 \text{ } \Omega \).
ii. The relative error is \( 0.0065 \text{ } \Omega \).
iii. The percentage error is 0.65%.
[Note: Framing of numerical is modified to reach the answer given to the numerical.]
In simple words: Calculate the mean resistance, then determine the absolute error for each reading. The mean absolute error is the average of these absolute errors. Divide the mean absolute error by the mean resistance for the relative error, and multiply by 100 for the percentage error.

🎯 Exam Tip: Be meticulous with calculations and decimal places, especially when dealing with multiple data points. Ensure the correct units are attached to each error type.

Question 9. 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.
Answer:
Given = \( v = k\sqrt{gh} \)

k being constant is assumed to be dimensionless.
Dimensions of L.H.S. = \( [v] = [L^1T^{-1}] \)
Dimension of R.H.S. = \( \sqrt{gh} \)
= \( [L^1T^{-2}]^{1/2} \times [L^1]^{1/2} \)
= \( [L^2T^{-2}]^{1/2} \)
= \( [L^1T^{-1}] \)
As, \( [\text{L.H.S.}] = [\text{R.H.S.}] \),
\( \implies v = k\sqrt{gh} \) is dimensionally correct equation.
In simple words: To prove the formula dimensionally, determine the dimensions of each term in the equation. If the dimensions of the left-hand side (velocity) match the dimensions of the right-hand side (k times the square root of gravitational acceleration times height), the equation is dimensionally correct.

🎯 Exam Tip: Clearly show the dimensional breakdown for each quantity. The key is that powers of L, M, T must match on both sides of the equation for dimensional consistency.

Question 10. \( v = at + \frac{b}{t-c} + v_0 \) 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.
Answer:
Solution: Given: \( y = at + \frac{b}{t+c} + V_0 \)
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.
\( \implies [\text{L.H.S.}] = [v] = [L^1T^{-1}] \)
This means, \( [at] = [v] = [L^1T^{-1}] \)
Given, t = time has dimension \( [T^1] \)
\( \implies [a] = \frac{[L^1T^{-1}]}{[t]} = \frac{[L^1T^{-1}]}{[T^1]} = [L^1T^{-2}] = [L^1M^0T^{-2}] \)
Similarly, \( [c] = [t] = [T^1] = [L^0M^0T^1] \)
\( \implies \frac{[b]}{[T^1]} = [v] = [L^1T^{-1}] \)
\( \implies [b] = [L^1T^{-1}] \times [T^1] = [L^1] = [L^1M^0T^0] \)
In simple words: For an equation to be dimensionally valid, all terms being added or subtracted must have the same dimensions. By equating the dimensions of each term to the dimension of velocity, we can determine the dimensional formulas for constants a, b, and c.

🎯 Exam Tip: The principle of homogeneity of dimensions is fundamental. Always remember that only quantities with the same dimensions can be added or subtracted.

Question 11. The length, breadth and thickness of a rectangular sheet of metal are \( 4.234 \text{ m}, 1.005 \text{ m}, \text{ and } 2.01 \text{ cm} \) respectively. Give the area and volume of the sheet to correct significant figures. Answer:
Given: l = 4.234 m, b = 1.005 m,
t = \( 2.01 \text{ cm} = 2.01 \times 10^{-2} \text{ m} = 0.0201 \text{ m} \)
To find:
i) Area of sheet to correct significant figures (A)
ii) Volume of sheet to correct significant figures (V)
In simple words: Convert all measurements to consistent units (meters). Then, calculate the surface area and volume using the appropriate formulas, ensuring the final answers are expressed with the correct number of significant figures based on the input values.

🎯 Exam Tip: When performing multiplication or division, the final result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. Also, ensure all units are converted to a standard system (e.g., SI) before calculation.

Question 12. 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.
Answer: The density of a cylinder is given by \( \rho = \frac{m}{V} = \frac{m}{\pi r^2 l} \). To find the percentage error, we use the formula for propagation of errors in products/quotients. Given values are \( l = (4.00 \pm 0.001) \) cm, \( r = (0.0250 \pm 0.001) \) cm, and \( m = (6.25 \pm 0.01) \) gm. First, write down the relative error in volume, then use it to find the relative error in density, and finally convert to percentage error. Given: \( l = (4.00 \pm 0.001) \) cm, In order to have same precision, we use, \( (4.000 \pm 0.001) \), \( r = (0.0250 \pm 0.001) \) cm, \( m = (6.25 \pm 0.01) \) g To find: percentage error in density Formulae: (i) Relative error in volume, \( \frac{\Delta V}{V} = \frac{2\Delta r}{r} + \frac{\Delta l}{l} \) ....(.: Volume of cylinder, \( V = \pi r^2 l \)) (ii) Releative error \( \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V} \) ....[... Density \( (\rho) = \frac{\text{mass}(m)}{\text{volume}(V)} \) ] (iii) Percentage error= Relative error × 100% Calculation. From formulae (i) and (ii),
\( \implies \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{2\Delta r}{r} + \frac{\Delta l}{l} \)
\( = \frac{0.01}{6.25} + \frac{2(0.001)}{0.025} + \frac{0.001}{4.000} \)
\( = 0.0016 + 0.08 + 0.00025 \)
\( = 0.08185 \) From formula (iii). % error in density \( = \frac{\Delta \rho}{\rho} \times 100 \)
\( = 0.08185 \times 100 \)
\( = 8.185\% \) Percentage error in density is 8.185%.
In simple words: The percentage error in density is calculated by summing the relative errors of mass, and twice the relative error of radius, plus the relative error of length, then multiplying by 100.

🎯 Exam Tip: Pay close attention to the powers of quantities when calculating relative errors (e.g., \(r^2\) means \(2 \times \frac{\Delta r}{r}\)).

Question 13. 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.
Answer: To calculate the diameter of Jupiter, we use the small angle approximation formula \( d = \alpha D \), where \( \alpha \) is the angular diameter in radians and \( D \) is the distance to the planet. First, convert the angular diameter from arcseconds to radians, then apply the formula. Given: Angular diameter \( (\alpha) = 35.72" \) \( = 35.72" \times 4.847 \times 10^{-6} \) rad \( = 1.73 \times 10^{-4} \) rad Distance from Earth \( (D) \) \( = 824.7 \) million km \( = 824.7 \times 10^6 \) km \( = 824.7 \times 10^9 \) m. To find: Diameter of Jupiter \( (d) \) Formula: \( d = \alpha D \) Calculation: From formula, \( d = 1.73 \times 10^{-4} \times 824.7 \times 10^9 \) \( = 1.428 \times 10^8 \) m \( = 1.428 \times 10^5 \) km Diameter of Jupiter is \( 1.428 \times 10^5 \) km.
In simple words: The diameter of a distant object can be found by multiplying its angular diameter (in radians) by its distance from the observer.

🎯 Exam Tip: Always convert angular measurements to radians before using them in formulas involving distances, as the formula \( d = \alpha D \) assumes \( \alpha \) is in radians.

Question 14. 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.
Answer: The percentage error in a quantity formed by products and quotients of other quantities is found by summing the percentage errors of each component, multiplied by their respective powers. For division, the errors add up just as they do for multiplication. Given \( X = \frac{a^4b^3}{c^{1/3}d^{1/2}} \) Percentage error in a, b, c, d is respectively 2%, 3%, 3% and 4%. Now, Percentage error in X
\( = \left( 4 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + \frac{1}{3} \frac{\Delta c}{c} + \frac{1}{2} \frac{\Delta d}{d} \right) \times 100\% \)
\( = \left[ (4 \times 2) + (3 \times 3) + \left( \frac{1}{3} \times 3 \right) + \left( \frac{1}{2} \times 4 \right) \right] \times 100\% \)
\( = [8 + 9 + 1 + 2] \times 100\% = 20\% \)
In simple words: To find the total percentage error in a complex formula, multiply each individual percentage error by its corresponding power in the formula and then sum them up.

🎯 Exam Tip: Remember that for quantities raised to a power (e.g., \(a^4\)), the percentage error is multiplied by that power. For roots (e.g., \(c^{1/3}\)), the percentage error is multiplied by the fractional power.

Question 15. Write down the number of significant figures in the following: 0.003 m², 0.1250 gm cm-², 6.4 x 10⁶ m, 1.6 × 10^-19 C, 9.1 × 10^-31 kg.
Answer: Significant figures indicate the precision of a measurement. Rules for determining them vary based on the position of non-zero digits and zeros (leading, trailing, or in-between) and the presence of a decimal point.

In simple words: Significant figures count all non-zero digits, zeros between non-zero digits, and trailing zeros if a decimal point is present; leading zeros are not significant.

🎯 Exam Tip: Scientific notation (like \(6.4 \times 10^6\)) simplifies significant figure determination as only the digits before the power of ten are considered significant.

Question 16. The diameter of a sphere is 2.14 cm. Calculate the volume of the sphere to the correct number of significant figures.
Answer: The volume of a sphere is calculated using its radius. The final answer must be rounded to the same number of significant figures as the least precise measurement used in the calculation. Volume of sphere \( = \frac{4}{3} \pi r^3 \) Given diameter \( = 2.14 \) cm, so radius \( r = \frac{2.14}{2} = 1.07 \) cm. \( = \frac{4}{3} \times 3.142 \times (1.07)^3 \) \( = 1.333 \times 3.142 \times (1.07)^3 \) \( = \{\text{antilog } [\log (1.333) + \log(3.142) + 3 \log(1.07)]\} \) \( = \{\text{antilog } [0.1249 + 0.4972 + 3 (0.0294)]\} \) \( = \{\text{antilog } [0.6221 + 0.0882]\} \) \( = \{\text{antilog } [0.7103]\} \) \( = 5.133 \text{cm}^3 \) In multiplication or division, the final result should retain as many significant figures as there are in the original number with the least significant figures. Volume in correct significant figures
\( \implies 5.13 \text{cm}^3 \)
In simple words: Calculate the volume using the formula for a sphere, then round the final answer to match the number of significant figures of the least precise input measurement.

🎯 Exam Tip: When performing calculations, retain at least one extra significant figure during intermediate steps to minimize rounding errors before rounding the final result.

Can You Recall (Textbook Page No. 1)

Question 1.
(i) What is a unit?
(ii) Which units have you used in the laboratory for measuring
a. length
b. mass
c. time
d. temperature?
(iii) Which system of units have you used?
Answer: The answer describes the definition of a unit and provides common laboratory units for various physical quantities. The system of units used generally depends on the context of the experiment. 1. The standard measure of any quantity is called the unit of that quantity.

3. MKS or SI system is used mostly. At times. even CGS system is used.
In simple words: A unit is a standard reference for measuring a physical quantity, with common examples like meters for length, kilograms for mass, seconds for time, and Celsius/Fahrenheit for temperature, typically within the MKS or SI system.

🎯 Exam Tip: Be familiar with the fundamental SI units and their corresponding physical quantities, as this forms the basis for all measurements in physics.

Can You Tell? (Textbook Page No. 8)

Question 1. If ten students are asked to measure the length of a piece of cloth upto a mm, using a metre scale, do you think their answers will be identical? Give reasons.
Answer: The answers are unlikely to be identical due to limitations of the measuring instrument (metre scale's least count) and variations in student skill. Precise measurements require instruments with appropriate least counts. Answers of the students are likely to be different. Length of cloth needs to be measured up to a millimetre (mm) length. Hence, to obtain accurate and precise reading one must use measuring instrument having least count smaller than 1 mm.
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.
In simple words: Measurements by different individuals using the same instrument are rarely identical due to inherent limits of the scale's precision and human error, especially when high accuracy (like millimeter) is sought.

🎯 Exam Tip: The least count of a measuring instrument dictates its precision. For accurate measurements, always choose an instrument with a least count suitable for the required level of precision.

Activity (Textbook Page No. 10)

Question. 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: This experiment involves using a Vernier caliper to measure the external diameter of a cylinder. Multiple readings are taken to calculate the mean diameter, mean absolute error, and percentage error, providing an understanding of measurement accuracy and precision. Given: L.C. = 0.01 cm
To measure external diameter of hollow cylinder readings are taken as follows:

(i) The mean diameter \( = a_{\text{mean}} = \frac{a_1 + a_2 + a_3}{3} \)
(ii) Absolute mean error \( \Delta a_{\text{mean}} = \frac{\Delta a_1 + \Delta a_2 + \Delta a_3}{3} \)
Here, \( \Delta a_1 = |a_{\text{mean}} - a_1| \), \( \Delta a_2 = |a_{\text{mean}} - a_2| \), \( \Delta a_3 = |a_{\text{mean}} - a_3| \)
(iii) Percentage error \( = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100 \)
[Note: The above table is made assuming zero error in Vernier callipers. If calliper has positive or negative zero error, the zero error correction needs to be introduced into observed reading.]

In simple words: To find the most accurate diameter, take multiple Vernier caliper readings, calculate their average (mean diameter), determine the average deviation from this mean (mean absolute error), and express this deviation as a percentage of the mean (percentage error).

🎯 Exam Tip: Always account for zero error in Vernier caliper readings by applying the appropriate correction (add for negative zero error, subtract for positive zero error) to obtain accurate measurements.