Samacheer Kalvi Class 11 Physics Solutions Chapter 1 Nature of Physical World and Measurement

Part - I: I. Multiple Choice Questions:

Question 1. One of the combinations from the fundamental physical constants is \( \frac { hG }{ c^3 } \). The unit of this expression is.
(a) Kg²
(b) m³
(c) S-1
(d) m
Answer: (a) Kg²
In simple words: The unit of the given physical expression, which combines Planck's constant, gravitational constant, and the speed of light, is found to be kilograms squared. This means the overall dimension of the combination works out to be a mass squared.

🎯 Exam Tip: For dimensional analysis questions, always remember the dimensions of fundamental constants like Planck's constant (h), gravitational constant (G), and speed of light (c).

Question 2. If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be:
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Answer: (d) 6%
In simple words: The volume of a sphere depends on the cube of its radius. This means if there's an error in measuring the radius, the error in the volume will be three times that percentage.

🎯 Exam Tip: When calculating percentage errors for quantities raised to a power (like volume, which is \( r^3 \)), multiply the individual percentage error by that power.

Question 3. If the length and time period of an oscillating pendulum have errors of 1% and 3% respectively then the error in measurement of acceleration due to gravity is: [Related to AMPMT 2008]
(a) 4%
(b) 5%
(c) 6%
(d) 7%
Answer: (d) 7%
In simple words: The acceleration due to gravity can be found using the pendulum's length and time period. If there are small errors in measuring these two, the total error in gravity's value is calculated by adding the percentage error in length to twice the percentage error in time.

🎯 Exam Tip: Remember the formula for the time period of a simple pendulum (\( T = 2\pi\sqrt{L/g} \)) and how to derive the percentage error in 'g' from it: \( \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \).

Question 4. The length of a body is measured as 3.15m, if the accuracy is 0.01m, then the percentage error in the measurement is:
(a) 351%
(b) 1%
(c) 0.28%
(d) 0.035%
Answer: (c) 0.28%
In simple words: Percentage error tells us how big the error is compared to the actual measurement, expressed as a percentage. We calculate it by dividing the accuracy by the measured value and multiplying by 100.

🎯 Exam Tip: Percentage error is always calculated as \( \left( \frac{\text{absolute error}}{\text{measured value}} \right) \times 100\% \). Pay attention to rounding rules for the final answer.

Question 5. Which of the following has the highest number of significant figure?
(a) 0.007 m²
(b) 2.64 x 1024 Kg
(c) 0.0006032 m²
(d) 6.3200 J
Answer: (d) 6.3200 J
In simple words: Significant figures show how precise a measurement is. To find the most precise number, count all digits that are not leading zeros, and all trailing zeros after a decimal point.

🎯 Exam Tip: Remember that leading zeros (e.g., in 0.007) are never significant, while trailing zeros after a decimal point (e.g., in 6.3200) are always significant.

Question 6. If \( \pi = 3.14 \), then the value of \( \pi^2 \) is:
(a) 9.8596
(b) 9.860
(c) 9.86
(d) 9.9
Answer: (c) 9.86
In simple words: To find the square of a number, you multiply it by itself. When dealing with measurements, you should round your answer to the same number of significant figures as the least precise number used in the calculation, which is 3 in this case.

🎯 Exam Tip: In multiplication, the result should be rounded to the same number of significant figures as the measurement with the fewest significant figures (in this case, 3.14 has three significant figures).

Question 7. Which of the following pairs of physical quantities have same dimension?
(a) force and power
(b) torque and energy
(c) torque and power
(d) force and torque
Answer: (b) torque and energy
In simple words: Physical quantities have the same dimensions if they are measured using the same combination of basic units like mass, length, and time. Both torque (a twisting force) and energy (the ability to do work) represent a form of work done or energy, so their dimensions are identical.

🎯 Exam Tip: To compare dimensions, always write out the fundamental units (Mass [M], Length [L], Time [T]) for each quantity. Torque is \( \text{force} \times \text{distance} \), and energy is \( \text{force} \times \text{distance} \).

Question 8. The dimensional formula of planck's constant h is _______ [AMU, Main, JEE, NEET]
(a) [ML² T-1]
(b) [ML2T-3]
(c) [MLT-1]
(d) [ML3T-3]
Answer: (a) [ML² T-1]
In simple words: Planck's constant is a fundamental constant in quantum physics. Its dimension comes from the relationship between a particle's energy and its frequency. Knowing the basic dimensions of energy and frequency helps you find the dimension of Planck's constant.

🎯 Exam Tip: Planck's constant (\(h\)) is related to energy (\(E\)) and frequency (\(f\)) by \( E = hf \). Use this relation to derive its dimensions: \( [h] = \frac{[E]}{[f]} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}] \).

Question 9. The velocity of a particle v at an instant t is given by \( v = at + bt^2 \) the dimension of b is _______
(a) [L]
(b) [LT-1]
(c) [LT-2]
(d) [LT-3]
Answer: (d) [LT-3]
In simple words: In any physical equation, all terms that are added together must have the same physical dimensions. Since \( v \) is velocity, the term \( bt^2 \) must also have the dimensions of velocity. We use this rule to figure out the dimensions of \( b \).

🎯 Exam Tip: Always apply the principle of homogeneity of dimensions: terms added or subtracted must have the same dimensions, and both sides of an equation must have identical dimensions.

Question 10. The dimensional formula for gravitational constant G is _______ [Related to AIPMT 2004]
(a) [ML3T-2]
(b) [M-1L3T-2]
(c) [M-1 L-3T-2]
(d) [ML-3T2]
Answer: (b) [M-1L3T-2]
In simple words: The gravitational constant G appears in Newton's law of universal gravitation, which describes the force between two masses. By rearranging that formula, you can find the unique combination of mass, length, and time that makes up G's dimension.

🎯 Exam Tip: Use Newton's Law of Gravitation, \( F = G \frac{m_1 m_2}{r^2} \), to derive the dimensions of G. Rearrange it to \( G = \frac{Fr^2}{m_1 m_2} \), then substitute the dimensions for force, length, and mass.

Question 11. The density of a material is CGS system of units Is 4 g cm-³. In a system of units in which unit of length is 10cm and unit of mass is 100g, then the value of density of material will be:
(a) 0.04
(b) 0.4
(c) 40
(d) 400
Answer: (c) 40
In simple words: When you change the system of units, the numerical value of a physical quantity also changes. To find the new value, you compare how the basic units (like gram and centimeter) are different in the two systems.

🎯 Exam Tip: Use the conversion formula \( n_2 = n_1 \left( \frac{M_1}{M_2} \right)^a \left( \frac{L_1}{L_2} \right)^b \left( \frac{T_1}{T_2} \right)^c \) where \( n_1 \) is the value in the first system, and \( a, b, c \) are the powers of M, L, T in the dimensional formula of the quantity.

Question 12. If the force is proportional to square of velocity, then the dimension of proportionality constant is _______ [JEE 2000]
(a) [ML T°]
(b) [ML T-1]
(c) [ML-2T]
(d) [ML-1T0]
Answer: (d) [ML-1T0]
In simple words: If a force depends on the square of velocity, there's a constant that links them. By balancing the dimensions on both sides of the equation, we can find the specific dimensions for this constant. This helps us understand what physical property the constant represents.

🎯 Exam Tip: Set up the proportionality as \( F = k v^2 \), where \( k \) is the proportionality constant. Then isolate \( k \) and substitute the dimensions of force (\( [MLT^{-2}] \)) and velocity (\( [LT^{-1}] \)).

Question 13. The dimension of \( (\mu_0 \epsilon_0)^{-1/2} \) is _______ [Main AIPMT 2011]
(a) length
(b) time
(c) velocity
(d) force
Answer: (c) velocity
In simple words: The constants \( \mu_0 \) (permeability of free space) and \( \epsilon_0 \) (permittivity of free space) are fundamental to electromagnetism. Their inverse product, raised to the power of negative one-half, is actually equal to the speed of light in a vacuum. Since the speed of light is a velocity, this expression has the dimension of velocity.

🎯 Exam Tip: Remember the relationship \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \), where \( c \) is the speed of light. This directly tells you the dimension of the given expression.

Question 14. Planck's constant (h), speed of light in vacuum (c) and Newton's gravitational constant (G) are takers as three fundamental constants. Which of the following combinations of these has the dimension of length? [NEET 2016 (phase II)]
(a) \( \frac{\sqrt{hG}}{c^{3/2}} \)
(b) \( \frac{\sqrt{hG}}{C^{5/2}} \)
(c) \( \frac{\sqrt{hG}}{G} \)
(d) \( \frac{GC}{\sqrt{h^3}} \)
Answer: (a) \( \frac{\sqrt{hG}}{c^{3/2}} \)
In simple words: Certain combinations of fundamental constants naturally result in fundamental dimensions like length, mass, or time. This specific combination of Planck's constant, gravitational constant, and speed of light forms the Planck length, which is the smallest possible measurable length.

🎯 Exam Tip: When given multiple-choice options, test the dimensional consistency of each option. For this question, Planck's constant \( [ML^2T^{-1}] \), gravitational constant \( [M^{-1}L^3T^{-2}] \), and speed of light \( [LT^{-1}] \) must be used carefully.

Question 15. A length-scale (l) depends on the permittivity (\( \epsilon \)) of a dielectric material Boltzmann constant (\( K_b \)), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression for L is dimensionally correct? [JEE (advance(d) 2016 ]
(a) \( \sqrt{\frac{nq^2}{\epsilon K_b T}} \)
(b) \( \sqrt{\frac{\epsilon K_b T}{nq^2}} \)
(c) \( \frac{q^2}{n\epsilon^{3/2}K_b T} \)
(d) \( \frac{q^2}{\epsilon K_b T} \)
Answer: (b) \( \sqrt{\frac{\epsilon K_b T}{nq^2}} \)
In simple words: This question asks to find a formula for length that uses many different physical properties like permittivity, temperature, and charge. Dimensional analysis helps us check which formula is built correctly by ensuring that all the units on one side match the units of length on the other side. This is a common way to predict how physical laws might look.

🎯 Exam Tip: For complex dimensional analysis problems, write down the dimensions of all given quantities. Then, for each option, substitute these dimensions and simplify to see if the result matches the dimension of length \( [L] \).

II. Short Answer Questions:

Question 1. Briefly explain the types of physical quantities?
Answer: Physical quantities are broadly divided into two main categories: fundamental quantities and derived quantities. Fundamental, or base, quantities are those that cannot be defined using other physical quantities. Examples include length, mass, time, electric current, temperature, luminous intensity, and the amount of substance. These are the basic building blocks. On the other hand, derived quantities are those that can be expressed by combining fundamental quantities. Examples for derived quantities include area (length times length), volume (length cubed), velocity (length per unit time), acceleration, and force. Most quantities we measure are derived from these basic seven.
In simple words: Physical quantities are either basic, like length or time, which can't be broken down further, or they are derived, like speed or area, which are made up of combinations of the basic ones.

🎯 Exam Tip: Clearly list the seven fundamental quantities and provide at least three distinct examples of derived quantities to score full marks.

Question 2. How will you measure the diameter of the moon using the parallax method?
Answer: To measure the moon's diameter using the parallax method, we first need to determine its distance from Earth, also using parallax. Let's say the distance from the Earth to the moon is \( D \). We then observe the moon through a telescope and measure its angular size, which is the angle \( \alpha \) subtended by the moon's diameter (\( d \)) at the Earth. This angle \( \alpha \) is typically very small. Once \( D \) and \( \alpha \) (in radians) are known, the diameter \( d \) of the moon can be calculated using the formula \( d = D \times \alpha \). This method relies on the geometric relationship between the object's size, its distance, and the angle it appears to subtend.

In simple words: First, we find how far the moon is from Earth. Then, we look at the moon and measure how big it appears as an angle. With the distance and the angle, we can calculate the moon's actual size.

🎯 Exam Tip: Clearly define the variables D (distance), d (diameter), and \( \alpha \) (angular size). The formula \( d = D \times \alpha \) is crucial, and remember to express \( \alpha \) in radians.

Question 3. Write the rules for determining significant figures.
Answer: Here are the rules for identifying significant figures in a measured value:
(1) **All non-zero digits are significant.** For example, the number 1342 has 4 significant figures.
(2) **Zeros between two non-zero digits are significant.** For instance, 2008 has four significant figures because the zero is between two non-zero digits.
(3) **Zeros at the end of a number after a decimal point are significant.** For example, 3070.00 has 6 significant figures.
(4) **Trailing zeros in a number without a decimal point are not significant.** For example, 4000 has only one significant figure (the 4).
(5) **For numbers less than 1, leading zeros to the right of the decimal point but before the first non-zero digit are not significant.** For example, 0.0034 has 2 significant figures.
(6) **All zeros that are to the right of the decimal point and also to the right of any non-zero digit are significant.** For example, 40.00 has four significant figures. This shows precision in measurement.
(7) **The number of significant figures does not depend on the specific system of units used.** For example, 1.53 cm, 0.0150 cm, and 0.0000153 km all have three significant figures.
(8) **The power of 10 in scientific notation does not count towards significant figures.** For example, 5.7 \( \times \) 10\(^2\) cm has two significant figures (5 and 7).
These rules ensure that the precision of a measurement is accurately represented, allowing for consistent calculations in science.
In simple words: Significant figures show how accurate a number is. There are rules to count them, like counting all numbers that aren't zero, counting zeros that are trapped between other numbers, and counting zeros at the very end if there's a dot (decimal point).

🎯 Exam Tip: Memorize each rule and practice with examples. Pay special attention to zeros (leading, trailing, and in-between) as they often cause confusion.

Question 4. What are the limitations of dimensional analysis?
Answer: Dimensional analysis is a useful tool but has several limitations:
(1) **No information on dimensionless constants:** This method cannot provide the values of dimensionless constants that appear in formulas, like 1, \( 2\pi \), or numbers like \( e \) (Euler's number). We can only determine the relationship between physical quantities, not the exact numerical factors.
(2) **Scalar or vector nature:** Dimensional analysis cannot tell us if a given physical quantity is a scalar (a quantity with only magnitude) or a vector (a quantity with both magnitude and direction).
(3) **Cannot derive complex functions:** This method cannot be used to derive relationships that involve trigonometric functions (like \( \sin \theta \), \( \cos \theta \)), exponential functions (like \( e^x \)), or logarithmic functions (like \( \log x \)). This is because these functions are dimensionless.
(4) **Limited to three physical quantities:** It cannot be applied to an equation that involves more than three physical quantities whose dimensions are not linked to each other. This is because we only have three fundamental dimensions (mass, length, time) to form equations.
(5) **Checks dimensional correctness, not physical correctness:** While it can check if a physical relation is dimensionally correct, it cannot guarantee that the relation is physically correct. For example, \( s = ut + \frac{1}{3} at^2 \) is dimensionally correct, but the correct physical equation is \( s = ut + \frac{1}{2} at^2 \).
These limitations mean that while dimensional analysis is powerful for checking formulas, it's not a complete method for deriving all physical relationships.
In simple words: Dimensional analysis can check if a formula uses the right kinds of units, but it can't tell you the numbers in the formula, if something is a scalar or vector, or if the formula is actually correct in the real world. It's also hard to use for very complex formulas.

🎯 Exam Tip: When explaining limitations, provide a brief example for each point, such as mentioning \( 2\pi \) for dimensionless constants or the \( s = ut + \frac{1}{3}at^2 \) example for physical correctness.

Question 5. Define precision and accuracy. Explain with one example.
Answer: **Accuracy** refers to how close a measured value is to the true, actual value of the quantity being measured. If a measurement is accurate, it means it is very close to the real answer. **Precision**, on the other hand, describes how close two or more measurements of the same quantity are to each other. If a set of measurements is precise, they are all very similar, even if they are not close to the true value.
**Example:** Imagine the true length of an object is 5.678 cm.
* In one experiment, a ruler with markings every 0.1 cm (low resolution) is used, and the length is measured as 5.5 cm. This measurement is somewhat accurate because 5.5 cm is close to 5.678 cm, but it has low precision because the tool can't measure very finely.
* In another experiment, a more accurate ruler with markings every 0.01 cm (high resolution) is used, and the length is found to be 5.38 cm. This measurement is more precise than the first (it has more decimal places), but it is less accurate because 5.38 cm is further from the true value of 5.678 cm compared to 5.5 cm.
This example shows that a measurement can be precise without being accurate, and vice-versa. Ideally, we want measurements to be both accurate and precise.
In simple words: Accuracy means getting very close to the correct answer. Precision means getting the same answer over and over again, even if that answer isn't exactly correct.

🎯 Exam Tip: Clearly distinguish between accuracy and precision. Using an example with numerical values helps illustrate the difference effectively. Remember, an accurate measurement hits the bullseye, while a precise measurement consistently hits the same spot, which might or might not be the bullseye.

III. Long Answer Questions:

Question 1. State the principle of homogeneity and explain with an example.
Answer: The **principle of homogeneity of dimensions** states that for any physical equation to be correct, the dimensions of all the terms on both sides of the equation must be exactly the same. This principle is a fundamental test used to check the correctness of an equation. It means you cannot add or subtract physical quantities that have different dimensions; for instance, you cannot add length to time.
**Example:** Let's consider the equation \( s = ut + \frac{1}{3} at^2 \). We want to check if this equation is dimensionally correct.
* Dimension of \( s \) (displacement) is \( [L] \).
* Dimension of \( ut \) (velocity \( \times \) time) is \( [LT^{-1}] \times [T] = [L] \).
* Dimension of \( at^2 \) (acceleration \( \times \) time\(^2\)) is \( [LT^{-2}] \times [T^2] = [L] \).
Here, the numerical factor \( \frac{1}{3} \) is dimensionless. Since the dimensions of all terms—\( s \), \( ut \), and \( at^2 \)—are \( [L] \), the equation is dimensionally consistent, meaning \( [L] = [L] + [L] \). Even though the constant \( \frac{1}{3} \) is incorrect (it should be \( \frac{1}{2} \) for the actual kinematic equation), the equation still passes the test of dimensional homogeneity. This shows that while dimensional analysis can confirm consistency, it cannot verify the exact numerical constants in a formula.
In simple words: The rule of homogeneity means that in any physics formula, every part you add or subtract must have the same type of unit. For example, you can only add lengths to lengths, not lengths to times. This helps check if a formula is possibly right.

🎯 Exam Tip: Clearly state the principle and provide a step-by-step dimensional analysis of the example equation. Emphasize that dimensional homogeneity is a necessary but not sufficient condition for an equation to be physically correct.

Question 2. Write short notes on the following.
(a) unit
(b) rounding off
(c) dimensionless quantity
Answer:
(a) **Unit:** A unit is a standard measure used to express the magnitude of a physical quantity. It is an internationally accepted reference amount of a quantity. For instance, the meter is a unit for length, and the kilogram is a unit for mass. Units are essential for clear communication and comparison of measurements worldwide. We have fundamental units (for base quantities like mass, length, time) and derived units (for quantities like area, volume, which are combinations of fundamental units).
(b) **Rounding Off:** Rounding off is a process used in calculations to reduce the number of significant figures in a result to a reasonable and appropriate level, typically matching the precision of the input measurements. This is done to avoid showing more precision than what was actually measured or is physically meaningful. For example, if you calculate a value like 18.37 from measurements that only have three significant figures, you would round it off to 18.4. Following proper rounding rules ensures that the final answer truly reflects the accuracy of the original data.
(c) **Dimensionless Quantity:** A dimensionless quantity is a physical quantity that has no physical units or dimensions. This means it is represented by a pure number. Dimensionless quantities can be of two types: dimensionless variables and dimensionless constants.
* **Dimensionless variables** are quantities that can change their value but still have no units. Examples include specific gravity, strain (a measure of deformation), and refractive index (how much light bends).
* **Dimensionless constants** are fixed numbers that naturally appear in physical laws and have no units. Examples are \( \pi \), the number \( e \), and other pure numbers.
Dimensionless quantities play a crucial role in physics as they often represent ratios or scaling factors.
In simple words: (a) A unit is like a standard size for measuring things, like a meter for length. (b) Rounding off means shortening a long number to make it simpler and match how accurate your tools are. (c) A dimensionless quantity is just a number with no units, like pi, or how much something stretches.

🎯 Exam Tip: For each term, provide a clear definition and at least one relevant example. For 'units', differentiating between fundamental and derived units adds value. For 'rounding off', mention its purpose. For 'dimensionless quantity', classify its types.

Question 3. What do you mean by the propagation of errors? Explain the propagation of errors in addition and multiplication.
Answer: **Propagation of errors** refers to how uncertainties (errors) in individual measurements combine or "propagate" through calculations to affect the final result of a derived quantity. When multiple measurements, each with its own error, are used in a formula, the errors do not simply cancel out; instead, they add up in a specific way. Understanding error propagation is critical for determining the overall uncertainty in a calculated value. The total error can stem from individual measurement errors or from the mathematical operations used.
**Errors in addition (or) sum of two quantities:**
Let \( A \) and \( B \) be two quantities, and let \( \Delta A \) and \( \Delta B \) be their respective absolute errors.
When these quantities are added, the sum \( Z \) is given by \( Z = A + B \).
The measured values with their errors are \( (A \pm \Delta A) \) and \( (B \pm \Delta B) \).
So, the sum with its error \( Z \pm \Delta Z \) is:
\( Z \pm \Delta Z = (A \pm \Delta A) + (B \pm \Delta B) \)
\( \implies Z \pm \Delta Z = (A + B) \pm (\Delta A + \Delta B) \)
Since \( Z = A + B \), we can write:
\( Z \pm \Delta Z = Z \pm (\Delta A + \Delta B) \)
Therefore, the maximum possible absolute error in the sum of two quantities is the sum of their individual absolute errors:
\( \Delta Z = \Delta A + \Delta B \)

**Errors in multiplication (or) product of two quantities:**
Again, let \( A \) and \( B \) be two quantities, with absolute errors \( \Delta A \) and \( \Delta B \).
When these quantities are multiplied, the product \( Z \) is given by \( Z = AB \).
The product with its error \( Z \pm \Delta Z \) is:
\( Z \pm \Delta Z = (A \pm \Delta A) (B \pm \Delta B) \)
\( \implies Z \pm \Delta Z = AB \pm A\Delta B \pm B\Delta A \pm \Delta A\Delta B \)
Now, divide both sides by \( Z = AB \):
\( \frac{Z \pm \Delta Z}{Z} = \frac{AB \pm A\Delta B \pm B\Delta A \pm \Delta A\Delta B}{AB} \)
\( \implies 1 \pm \frac{\Delta Z}{Z} = 1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm \frac{\Delta A \Delta B}{AB} \)
Since \( \Delta A \) and \( \Delta B \) are typically small, their product \( \Delta A \Delta B \) is very small, making the term \( \frac{\Delta A \Delta B}{AB} \) negligible compared to the other terms.
Therefore, the maximum possible fractional error in the product of two quantities is the sum of their individual fractional errors:
\( \frac{\Delta Z}{Z} = \pm \left( \frac{\Delta A}{A} + \frac{\Delta B}{B} \right) \)
This rule shows that errors in multiplication affect the relative (fractional) error, not the absolute error directly.
In simple words: Error propagation means that when you do calculations with numbers that have small errors, these errors spread and affect your final answer. When you add numbers, the errors just add up directly. When you multiply numbers, their percentage errors add up.

🎯 Exam Tip: For both addition and multiplication, clearly state the formula for the maximum absolute error (\( \Delta Z \)) and maximum fractional/percentage error (\( \frac{\Delta Z}{Z} \)). Show the derivation steps, especially how the product of small errors is neglected in multiplication.

Question 4. Explain in detail various types of errors.
Answer: An **error** is the uncertainty or deviation in a measurement from its true value. There are generally three main types of errors in measurement:
**1. Systematic Errors:**
These are reproducible inaccuracies that consistently occur in the same direction, meaning they always make the measurement either higher or lower than the true value. Systematic errors often arise from a persistent problem throughout the experiment and are typically predictable. They can be minimized or corrected by identifying their source. Systematic errors are further classified into:
    (a) **Instrumental Error:** These errors happen if the measuring instrument itself is not calibrated correctly or has inherent flaws from manufacturing. For example, using a meter scale with a worn-out end will consistently lead to incorrect length measurements. These errors can be corrected by carefully selecting a well-calibrated instrument.
    (b) **Imperfections in Experimental Technique or Procedure:** These errors occur due to limitations or flaws in the experimental setup or the method being used. For instance, if a calorimeter lacks proper insulation, heat radiation losses will consistently occur, leading to errors in temperature measurements. Following necessary steps and applying corrections during experiments can mitigate these errors.
    (c) **Personal Errors:** These errors are due to the individual performing the experiment. They can result from incorrect initial setup of the apparatus, carelessness during observation, or improper precautions. For example, taking readings from an angle (parallax error) is a personal error. Training and carefulness by the observer can reduce these.
    (d) **Errors Due to External Causes:** Changes in external environmental conditions during an experiment, such as fluctuations in temperature, humidity, or pressure, can cause errors in measurements. For example, a metal scale might expand slightly on a hot day, affecting length readings. Keeping conditions stable or accounting for changes helps minimize these.
    (e) **Least Count Error:** The least count is the smallest value an instrument can measure. The error associated with this minimum measurable value is the least count error. It primarily arises from the finite resolution of the instrument itself. This error is typically considered to be half of the instrument's least count. Using a high-precision instrument with a smaller least count can reduce this type of error.

Question 4. Explain in detail various types of errors.
Answer: The uncertainty in a measurement is known as an error. There are three main types of errors:
1. Random error
2. Systematic error
3. Gross error.

**1. Systematic errors:** These errors are consistent and repeatable inaccuracies that always occur in the same direction. They often arise from persistent issues throughout the experiment. Systematic errors are further classified into:

(a) Instrumental error: These happen when an instrument is not properly calibrated during manufacturing. For example, if a meter scale has a worn-out end, measurements made with it will have errors. These errors can be fixed by carefully selecting the right instrument.

(b) Imperfections in experimental technique or procedure: These errors are due to limitations in how the experiment is set up. For instance, in a calorimeter experiment, lack of proper insulation leads to heat loss, causing errors. These errors can be corrected by following necessary steps and procedures during experiments.

(c) Personal errors: These errors are caused by the individuals conducting the experiments. They might be due to incorrect initial setup, carelessness during observation, or not following proper precautions.

(d) Errors due to external causes: Changes in external conditions during an experiment, such as temperature, humidity, or pressure, can affect the measurement results and cause errors.

(e) Least count error: This error occurs because every measuring instrument has a "least count," which is the smallest value it can accurately measure. The instrument's resolution directly causes this error, and the error itself is half of the least count. Using a high-precision instrument can reduce this type of error.

**2. Random errors:** These errors arise from random and unpredictable changes in experimental conditions, such as fluctuations in pressure, temperature, or voltage supply. Personal errors by the observer can also contribute to random errors. These are sometimes called "chance errors." For example, when measuring a wire's thickness with a screw gauge, different readings may be obtained in different attempts. Taking the average (arithmetic mean) of all readings helps reduce random errors, and this mean value is considered the most accurate true value.

**3. Gross errors:** These errors are a result of extreme carelessness on the part of the observer. Examples include incorrectly setting up the instrument, making observations without considering error sources or precautions, using wrong values in calculations, or incorrectly recording observations. These errors can only be minimized if the observer is extremely careful and mentally alert. It is crucial to be attentive during all steps of an experiment to avoid these.
In simple words: Errors are mistakes in measurements. There are three kinds: systematic (consistent mistakes, like a faulty ruler), random (unpredictable changes, like small temperature shifts), and gross (careless mistakes by the person doing the experiment). We try to minimize these to get accurate results.

🎯 Exam Tip: When explaining errors, define each type clearly with a simple, relatable example to show understanding of its origin and how it might be addressed.

Question 5. (i) Explain the use of screw gauge and vernier calipers in measuring smaller distances.
(ii) Write a note a triangular method and radar method to measure large distances.
Answer:
**(i) Measurement of small distances by screw gauge and vernier calipers:**
* **Screw Gauge:** This instrument is used to precisely measure dimensions of objects, typically up to 50 mm. Its operation is based on the principle of magnifying linear motion using the circular movement of a screw. The least count of a standard screw gauge is 0.01 mm, making it very accurate for small measurements.
* **Vernier Calipers:** This is a versatile tool used for measuring dimensions like the diameter of a hole or the depth of a hole. The least count of a vernier caliper is 0.01 cm (or 0.1 mm). Both instruments use a main scale and a sliding scale to achieve higher precision than a regular ruler.

**(ii) Measurement of larger distances:**
For measuring large distances, such as the height of a tall tree or the distance to the Moon or a planet from Earth, methods like the triangulation method, parallax method, and radar method are used.

(a) **Triangulation method for the height of an accessible object:**

Answer: Let AB = h be the height of the tree that needs to be measured. An observer stands at a point C on the ground, a distance X from the base of the tree (B). Using a range finder, the angle \( \angle ACB = \theta \) is measured.
Considering the right-angled triangle \( \triangle ABC \):
\( \tan \theta = \frac{AB}{BC} \)
\( \tan \theta = \frac{h}{X} \)
So, \( h = X \tan \theta \)
By knowing the distance \( X \) and the angle \( \theta \), the height \( h \) can be calculated. This method relies on forming a triangle to find unknown lengths.

(b) **Radar method:**
RADAR stands for Radio Detection And Ranging. This method uses radio signals to measure distances to objects like nearby planets, the Moon, enemy planes, or moving and stationary targets. In this process, radio signals are sent from a transmitter. These signals reflect off the target and are then received by a receiver. The time interval between sending and receiving the signals is recorded. Knowing the speed of the radio signals (which is the speed of light) and the time taken, the distance to the object can be calculated using the formula: distance = (speed \( \times \) time) / 2, where the division by 2 is because the signal travels to the object and back.
In simple words: For small distances, we use special tools like screw gauges and vernier calipers, which are very precise. For big distances, like a tree's height, we can use a triangle method (triangulation) by measuring an angle and a ground distance. For even larger distances, like to the Moon, we use radar, which sends out radio waves and measures how long it takes for them to bounce back.

🎯 Exam Tip: Remember to clearly state the principle behind each method and how it helps in measuring distances, including simple diagrams where appropriate. For numerical problems, show all steps and units clearly.

IV. Numerical Problems:

Question 1. In a submarine equipped with sonar the time delay between the generation of a pulse and its echo after reflection form an enemy submarine is observed to be 80s. If the speed of sound in water is 1460 ms\(^{-1}\). What is the distance of enemy submarine?
Answer:
Given:
Time taken for echo (round trip) \( t = 80 \text{ s} \)
Velocity of sound in water \( V = 1460 \text{ m/s} \)
Distance of enemy submarine \( d = ? \)

The sound travels to the submarine and back, covering a distance of \( 2d \).
Using the formula: \( V = \frac{\text{Distance}}{\text{Time}} \)
So, \( V = \frac{2d}{t} \)
We need to find \( d \), so rearrange the formula:
\( d = \frac{Vt}{2} \)
Substitute the given values:
\( d = \frac{1460 \times 80}{2} \)
\( d = 1460 \times 40 \)
\( d = 58400 \text{ m} \)
Convert to kilometers:
\( d = 58.4 \text{ km} \)
The sound travels to the submarine and then bounces back.
In simple words: A sonar sends out a sound pulse and hears it back after 80 seconds. Since the sound travels at 1460 meters per second in water, the total distance covered (to the submarine and back) is 1460 times 80. To find just the distance to the submarine, we divide that by two, which gives 58.4 kilometers.

🎯 Exam Tip: Remember to divide the total distance by two for echo-based problems, as the sound travels to the object and then returns to the source.

Question 2. The radius of the circle is 3.12 m calculate the area of the circle with regard to significant figures.
Answer:
Given:
Radius \( r = 3.12 \text{ m} \) (This has three significant figures)

To calculate the area of the circle, we use the formula \( A = \pi r^2 \).
We'll use \( \pi \approx 3.14 \) (which has three significant figures).

\( \text{Area of the circle} = \pi r^2 \)
\( = 3.14 \times (3.12 \text{ m})^2 \)
\( = 3.14 \times 9.7344 \text{ m}^2 \)
\( = 30.566896 \text{ m}^2 \)

Since the given radius (3.12 m) has three significant figures, the final answer for the area should also be rounded to three significant figures. In multiplication, the result should have the same number of significant figures as the measurement with the fewest significant figures.

If the result is rounded off to three significant figures, the area of the circle = \( 30.6 \text{ m}^2 \).
In simple words: We need to find the area of a circle using a radius of 3.12 meters. We use the formula \( \pi r^2 \). After calculating, we make sure the answer has the same number of important digits (significant figures) as the radius, which is three. So, 30.566... meters squared becomes 30.6 meters squared.

🎯 Exam Tip: When performing calculations, the final answer must be rounded to the same number of significant figures as the least precise measurement used in the calculation.

Question 3. Assuming that the frequency \( \gamma \) of the vibrating string may depend up on
(i) applied force (F)
(ii) Length (l)
(iii) mass per unit length (m)
prove that \( \gamma \propto \frac{1}{l}\sqrt{\frac{F}{m}} \) using dimensional analysis.
Answer:
Let's assume the frequency \( \gamma \) is proportional to \( F^a l^b m^c \).
So, \( \gamma \propto F^a l^b m^c \)
Writing the dimensions on both sides:
Dimension of frequency \( \gamma = [T^{-1}] \)
Dimension of force \( F = [MLT^{-2}] \)
Dimension of length \( l = [L] \)
Dimension of mass per unit length \( m = [ML^{-1}] \)

Substituting these into the proportionality:
\( [T^{-1}] = [MLT^{-2}]^a [L]^b [ML^{-1}]^c \)
\( [M^0 L^0 T^{-1}] = [M^a L^a T^{-2a}] [L^b] [M^c L^{-c}] \)
Combining the terms for M, L, and T:
\( [M^0 L^0 T^{-1}] = [M^{a+c} L^{a+b-c} T^{-2a}] \)

Comparing the powers of M, L, and T on both sides:
For M: \( a + c = 0 \implies c = -a \)
For T: \( -2a = -1 \implies a = \frac{1}{2} \)
Substitute \( a = \frac{1}{2} \) into \( c = -a \):
\( c = -\frac{1}{2} \)

For L: \( a + b - c = 0 \)
Substitute the values of \( a \) and \( c \):
\( \frac{1}{2} + b - (-\frac{1}{2}) = 0 \)
\( \frac{1}{2} + b + \frac{1}{2} = 0 \)
\( 1 + b = 0 \)
\( b = -1 \)

Now substitute the values of a, b, and c back into the original proportionality:
\( \gamma \propto F^{1/2} l^{-1} m^{-1/2} \)
This can be rewritten as:
\( \gamma \propto \frac{F^{1/2}}{l^1 m^{1/2}} \)
\( \gamma \propto \frac{1}{l}\sqrt{\frac{F}{m}} \)
Thus, the formula for the frequency of a vibrating string is dimensionally proven. This method is crucial for checking the consistency of physical equations.
In simple words: We assume that how fast a string vibrates (frequency) depends on its tension (force), its length, and how heavy it is per unit length. By matching the fundamental dimensions (mass, length, time) on both sides of an equation, we found that the frequency is indeed proportional to the tension divided by the mass per unit length, all under a square root, and inversely proportional to the length. This means a longer, heavier string vibrates slower, and a tighter string vibrates faster.

🎯 Exam Tip: When using dimensional analysis, ensure you know the correct dimensions for all physical quantities involved. Carefully compare the powers of M, L, and T on both sides of the equation to correctly find the exponents.

Question 4. Jupiter is at a distance of 824.7 million Km from the earth. Its angular diameter is measured to be 35.72" calculate the diameter of Jupiter
Answer:
Given:
Distance of Jupiter from Earth \( D = 824.7 \text{ million km} = 824.7 \times 10^6 \text{ km} \)
Convert km to m: \( D = 8.247 \times 10^8 \text{ km} = 8.247 \times 10^{11} \text{ m} \)
Angular diameter \( \alpha = 35.72'' \) (arcseconds)

First, convert the angular diameter from arcseconds to radians.
We know that \( 1'' = 4.85 \times 10^{-6} \text{ rad} \). This is a standard conversion factor.
So, \( \alpha = 35.72 \times 4.85 \times 10^{-6} \text{ rad} \)
\( \alpha = 173.242 \times 10^{-6} \text{ rad} \)
\( \alpha = 1.73242 \times 10^{-4} \text{ rad} \)
For simplicity, we can round it to \( \alpha \approx 1.73 \times 10^{-4} \text{ rad} \).

We use the formula for angular diameter, which is approximately \( \alpha = \frac{d}{D} \), where \( d \) is the linear diameter and \( D \) is the distance.
Therefore, the diameter of Jupiter \( d = D \times \alpha \)
\( d = (8.247 \times 10^{11} \text{ m}) \times (1.73 \times 10^{-4} \text{ rad}) \)
\( d = 14.26731 \times 10^7 \text{ m} \)
\( d = 1.426731 \times 10^8 \text{ m} \)
Rounding to three significant figures (from 1.73), we get:
\( d \approx 1.427 \times 10^8 \text{ m} \)
This can also be expressed as \( 1.427 \times 10^5 \text{ km} \).

Answer: Jupiter is very far from Earth, so its diameter can be found using its angular size. This problem uses the parallax method.
In simple words: Jupiter is about 824.7 million kilometers away. We see it as having an angular size of 35.72 arcseconds. First, we change arcseconds into a unit called radians. Then, using a simple formula that connects angular size, distance, and actual size, we find that Jupiter's diameter is about 1.427 x 10\(^8\) meters, or 142,700 kilometers.

🎯 Exam Tip: Always convert angular measurements to radians before using them in formulas like \( d = D \times \alpha \), as this formula is valid only when \( \alpha \) is in radians.

Question 5. The measurement value of length of a simple pendulum is 20 cm known with 2mm accuracy. The time for 50 oscillations was measured to be 40s with in 1s resolution. Calculate the percentage error in the determination of acceleration due to gravity g from the above statement.
Answer:
Given:
Length of simple pendulum \( l = 20 \text{ cm} = 0.20 \text{ m} \)
Accuracy (absolute error in length) \( \Delta l = 2 \text{ mm} = 0.2 \text{ cm} = 0.002 \text{ m} \)
Time for 50 oscillations \( = 40 \text{ s} \)
Resolution of stopwatch (absolute error in total time) \( \Delta t_{total} = 1 \text{ s} \)

First, calculate the time period \( T \) for one oscillation:
\( T = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{40 \text{ s}}{50} = \frac{4}{5} \text{ s} = 0.8 \text{ s} \)

Next, calculate the absolute error in the time period \( \Delta T \).
\( \Delta T = \frac{\Delta t_{total}}{\text{Number of oscillations}} = \frac{1 \text{ s}}{50} = 0.02 \text{ s} \)

The formula for the time period of a simple pendulum is \( T = 2\pi\sqrt{\frac{l}{g}} \).
To find \( g \), we square both sides:
\( T^2 = 4\pi^2 \frac{l}{g} \)
Rearrange to solve for \( g \):
\( g = \frac{4\pi^2 l}{T^2} \)

Now, calculate the percentage error in \( g \). For a quantity \( X = \frac{A^p B^q}{C^r} \), the maximum percentage error is:
\( \left( \frac{\Delta X}{X} \times 100\% \right) = \left( p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C} \right) \times 100\% \)

In our case, \( g = \frac{4\pi^2 l}{T^2} \). The constants \( 4\pi^2 \) have no error.
So, the percentage error in \( g \) is:
\( \left( \frac{\Delta g}{g} \times 100\% \right) = \left( 1 \times \frac{\Delta l}{l} + 2 \times \frac{\Delta T}{T} \right) \times 100\% \)

Substitute the values:
\( \frac{\Delta l}{l} = \frac{0.2 \text{ cm}}{20 \text{ cm}} = \frac{0.002 \text{ m}}{0.20 \text{ m}} = 0.01 \)
\( \frac{\Delta T}{T} = \frac{0.02 \text{ s}}{0.8 \text{ s}} = 0.025 \)

\( \left( \frac{\Delta g}{g} \times 100\% \right) = \left( 0.01 + 2 \times 0.025 \right) \times 100\% \)
\( = \left( 0.01 + 0.05 \right) \times 100\% \)
\( = 0.06 \times 100\% \)
\( = 6\% \)
The percentage error in the measurement of acceleration due to gravity is 6%. This calculation is important for understanding the reliability of experimental results.
In simple words: We want to find the error in calculating gravity (g) using a pendulum. We know its length has a 2mm error in 20cm, and the time for 50 swings (40s) has a 1s error. First, we find the error in one swing's time. Then, using the pendulum formula, we see how these individual errors contribute to the total error in g. The final calculation shows a 6% error in measuring gravity.

🎯 Exam Tip: When calculating percentage errors involving powers (like \( T^2 \)), remember that the percentage error of the quantity is multiplied by its power in the formula (e.g., \( 2 \times \frac{\Delta T}{T} \)).

Additional Important Questions and Answers

I. Multiple choice questions:

Question 1. The unit of surface tension ......
(a) MT\(^{-2}\)
(b) Nm\(^{-2}\)
(c) Nm
(d) Nm\(^{-1}\)
Answer: (d) Nm\(^{-1}\)
In simple words: Surface tension is a force acting along a surface per unit length. So, its unit is Newton (N) for force divided by meter (m) for length, which is N/m or Nm\(^{-1}\).

🎯 Exam Tip: To find the unit of a quantity, recall its definition or a formula involving it. Surface tension is force per unit length.

Question 2. One astronomical unit is equal to _______.
(a) 1.510 x 10\(^{12}\)m
(b) 1.5 x 10\(^{12}\)Km
(c) 1.5 x 10\(^{11}\)m
(d) 1.5 x 10\(^{12}\)Cm
Answer: (c) 1.5 x 10\(^{11}\)m
In simple words: One astronomical unit (AU) is a standard distance used in astronomy. It is defined as the average distance between the Earth and the Sun, which is about 150 million kilometers or 1.5 x 10\(^{11}\) meters.

🎯 Exam Tip: Memorize common astronomical units and their conversions, as they are fundamental in astrophysics calculations.

Question 3. One light-year is ......
(a) 3.153 x 10\(^7\) m
(b) 1.496 x 10\(^7\) m
(c) 9.46 \( \times \) 10\(^{12}\) km
(d) 3.26 \( \times \) 10\(^{15}\) km
Answer: (c) 9.46 \( \times \) 10\(^{12}\) km
In simple words: A light-year is the distance light travels in one year. Since light travels extremely fast, a light-year represents a very large distance, approximately 9.46 trillion kilometers.

🎯 Exam Tip: Understand that a light-year measures distance, not time. It's a common unit for interstellar distances due to its vast scale.

Question 4. The dimensional formula for the coefficient of viscosity is ______.
(a) M\(^0\) L\(^{-1}\)T\(^{-1}\)
(b) M\(^1\) L\(^1\)T\(^1\)
(c) M\(^1\) L\(^{-1}\)T\(^1\)
(d) M\(^2\) L\(^2\)T\(^0\)
Answer: (c) M\(^1\) L\(^{-1}\)T\(^{-1}\)
In simple words: Viscosity describes a fluid's resistance to flow. Its dimensional formula is derived from the definition of viscous force, which results in mass (M), inverse length (L\(^{-1}\)), and inverse time (T\(^{-1}\)).

🎯 Exam Tip: To derive the dimensional formula for viscosity, use Newton's law of viscosity, \( F = \eta A \frac{dv}{dy} \), where \( \eta \) is the coefficient of viscosity.

Question 5. One parsec is ....
(a) 1.5 x 10\(^{12}\)m
(b) 3.26 x 10\(^{15}\) m
(c) 30.84 \( \times \) 10\(^{15}\) m
(d) 9.46 \( \times \) 10\(^{15}\) m
Answer: (c) 30.84 \( \times \) 10\(^{15}\) m
In simple words: A parsec is a unit of length used for huge distances, especially outside the solar system. It's defined as the distance at which one astronomical unit subtends an angle of one arcsecond. One parsec is approximately 30.84 \( \times \) 10\(^{15}\) meters, which is roughly 3.26 light-years.

🎯 Exam Tip: Distinguish between a light-year (distance light travels in a year) and a parsec (defined by stellar parallax), and know their respective values for long-distance measurements.

Question 6. The dimensional formula for \( (\epsilon_0) \) permittivity of free space ______
(a) M\(^{-1}\) L\(^3\) T\(^4\) A\(^2\)
(b) M\(^{-1}\) L\(^{-3}\) T\(^4\) A\(^2\)
(c) M\(^1\) L\(^{-3}\) T\(^4\) A\(^2\)
(d) M\(^1\) L\(^{-3}\) T\(^{-4}\) A\(^{-2}\)
Answer: (b) M\(^{-1}\) L\(^{-3}\) T\(^4\) A\(^2\)
In simple words: Permittivity of free space, \( \epsilon_0 \), is a fundamental physical constant that shows how an electric field influences and is influenced by a dielectric medium. Its dimensional formula can be derived from Coulomb's law, showing its dependence on mass, length, time, and electric current.

🎯 Exam Tip: To derive the dimensional formula for \( \epsilon_0 \), use Coulomb's Law: \( F = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2} \). Remember that charge \( q = IT \) (Current x Time).

Question 7. One Angstrom is ........
(a) 10\(^{-9}\) m
(b) 10\(^{-10}\) m
(c) 10\(^{-12}\) m
(d) 10\(^{-15}\) m
Answer: (b) 10\(^{-10}\) m
In simple words: An Angstrom is a very small unit of length, often used to measure the size of atoms, molecules, or wavelengths of light. One Angstrom is equal to 0.1 nanometers or 10\(^{-10}\) meters.

🎯 Exam Tip: Keep in mind standard prefixes and their powers of 10 for length measurements, such as milli, micro, nano, pico, and angstrom.

Question 8. In the formula x = 3yz\(^2\), x and y have dimensions of capacitance and magnetic induction respectively what are the dimensions of y?
(a) M\(^3\) L\(^2\) T\(^2\) Q\(^4\)
(b) M\(^0\) L\(^2\) T\(^{-2}\) Q\(^4\)
(c) M\(^0\) L\(^{-2}\) T\(^{-2}\) Q\(^4\)
(d) M\(^3\) L\(^{-2}\) T\(^4\) Q\(^4\)
Answer: (d) M\(^3\) L\(^{-2}\) T\(^4\) Q\(^4\)
In simple words: Given an equation where \( x \) is capacitance and \( y \) is magnetic induction, we need to find the dimensions of \( z \). By rearranging the formula \( x = 3yz^2 \) to solve for \( z \), and substituting the known dimensions of \( x \) and \( y \), we can figure out the dimensions of \( z \). The number 3 does not have dimensions.

🎯 Exam Tip: When dealing with dimensional analysis problems, ignore numerical constants (like 3) as they are dimensionless. Focus only on the dimensions of the physical quantities.

Question 9. \( \frac{1}{12} \) of the mass of carbon 12 atom is .....
(a) 1 TMC
(b) mass of neutron
(c) 1 amu
(d) mass of hydrogen
Answer: (c) 1 amu
In simple words: The atomic mass unit (amu) is a standard unit of mass used to express atomic and molecular masses. It is defined as exactly one-twelfth of the mass of an unbound atom of carbon-12. This unit helps in comparing the masses of different atoms.

🎯 Exam Tip: The definition of 1 amu is fundamental in chemistry and physics, linking directly to the carbon-12 isotope standard.

Question 10. The resistance of a conductor R = V/I where V = (50\( \pm \)2)V and I = (9 \( \pm \) 0.3) A find the percentage error in R.
(a) 8.5%
(b) 3.7%
(c) 7.8%
(d) 7.3%
Answer: (d) 7.3%
In simple words: We are given voltage (V) and current (I) with their measurement errors. Resistance (R) is found by dividing V by I. To find the percentage error in R, we add the percentage errors of V and I. The percentage error in V is (2/50)*100% = 4%, and in I is (0.3/9)*100% = 3.3%. Adding these gives approximately 7.3%.

🎯 Exam Tip: When quantities are divided or multiplied, their fractional (or percentage) errors add up to give the fractional (or percentage) error in the result. Always express errors as positive values.

Question 11. The study of forces acting on bodies whether at rest or in motion is ....
(a) classical mechanics
(b) quantum mechanics
(c) thermodynamics
(d) condensed matter physics
Answer: (a) classical mechanics
In simple words: Classical mechanics is the branch of physics that studies how forces affect objects at rest or in motion. It explains everyday phenomena like how a ball rolls or how planets orbit the sun, using Newton's laws.

🎯 Exam Tip: Classical mechanics focuses on macroscopic objects and speeds much slower than light, contrasting with quantum mechanics for very small particles and relativity for very high speeds.

Question 12. In the measurement of pressure if maximum errors in the measurement of force and length of a square plate are 3% and 2% respectively. The maximum error is ______
(a) 7%
(b) 8%
(c) 4%
(d) 5%
Answer: (a) 7%
In simple words: Pressure is calculated by dividing force by area. For a square plate, area is length squared. So, if the force has a 3% error and length has a 2% error, the area error will be 2 times 2%, which is 4%. Then, for pressure, we add the error in force (3%) and the error in area (4%), giving a total maximum error of 7%.

🎯 Exam Tip: For quantities with exponents (e.g., Area = Length\(^2\)), multiply the percentage error of the base quantity by its power (i.e., 2 \( \times \) % error in length for area). Then, combine errors from multiplication/division by adding them.

Question 13. Which of the following is not a dimensionless physical quantity?
(a) Mechanical equivalent of heat
(b) volumetric strain
(c) atomic mass unit
(d) Avogadro's number
Answer: (c) atomic mass unit
In simple words: A dimensionless quantity is one that has no physical units, like strain or a ratio. However, the atomic mass unit (amu) is a unit of mass, which means it has a dimension ([M]). The mechanical equivalent of heat, volumetric strain, and Avogadro's number are all dimensionless quantities.

🎯 Exam Tip: Remember that "unit" implies dimension. If a quantity has a unit (like amu for mass), it is not dimensionless, even if other forms might be. Dimensionless quantities are often ratios or counts.

Question 14. The study of production and propagation of sound waves .....
(a) Astrophysics
(b) Acoustics
(c) Relativity
(d) Atomic physics
Answer: (b) Acoustics
In simple words: Acoustics is the scientific study of sound. This includes how sound is made, how it travels through different things, and how it is heard. It applies to everything from musical instruments to noise control in buildings.

🎯 Exam Tip: Be familiar with the main branches of physics and their areas of study to correctly identify the field related to specific phenomena.

Question 15. The physical quantities not having the same dimensions are ________
(a) torque and work
(b) Linear momentum and planks constant
(c) stress and youngs modulus
(d) speed and \( (\epsilon_0\mu_0)^{-1/2} \)
Answer: (a) torque and work
In simple words: Torque and work both seem like they involve force and distance, but their directions are different. Work is a scalar (force parallel to displacement), while torque is a vector (force perpendicular to distance). Although their units are dimensionally the same (Energy), their physical nature is different. For this question the provided answer is (a) torque and work.

🎯 Exam Tip: Even if two quantities have the same dimensional formula (like work and torque), they might represent different physical concepts. Linear momentum and Planck's constant, stress and Young's modulus, and speed and \( (\epsilon_0\mu_0)^{-1/2} \) do share the same dimensions.

Question 16. Which two of the following five physical parameters have the same dimension?
(1) energy density
(2) refractive index
(3) dielectric content
(4) youngs modulus
(5) magnetic field
(a) 1 and 4
(b) 1 and 5
(c) 2 and 4
(d) 3 and 5
Answer: (a) 1 and 4
In simple words: Energy density (energy per unit volume) and Young's modulus (stress per strain) both have the same dimensions, which are those of pressure. This is because stress, pressure, and energy density all relate to force over area or energy over volume.

🎯 Exam Tip: To compare dimensions, express each quantity in terms of fundamental dimensions (M, L, T, etc.). Remember that dimensionless quantities like refractive index and dielectric content have the same "dimensions" (none).

Question 17. The astronomers used to observe distant points of the universe by .......
(a) Electron telescope
(b) Astronomical telescope
(c) Radio telescope
Answer: (c) Radio telescope
In simple words: Astronomers use various types of telescopes to study the universe. Radio telescopes are especially good for observing very distant objects because radio waves can travel through gas and dust that block visible light, giving us a clearer view of faraway galaxies and phenomena.

🎯 Exam Tip: Be aware of the different types of telescopes and the specific parts of the electromagnetic spectrum they are designed to observe, and how each is suited for different astronomical tasks.

Question 18. The youngs modulus of a material of the wire is 12.6 x 10\(^{11}\) dyne/cm\(^2\). Its value is MKS system is ______
(a) 12.6 x 10\(^{12}\) N/M\(^2\)
(b) 12.6 x 10\(^{10}\) N/M\(^2\)
(c) 12.6 x 10\(^9\) N/M\(^2\)
(d) 12.6 x 10\(^8\) N/M\(^2\)
Answer: (a) 12.6 x 10\(^{12}\) N/M\(^2\)
In simple words: Young's modulus measures how much a material can resist being deformed. We are given its value in CGS units (dyne/cm\(^2\)) and need to convert it to MKS units (N/m\(^2\)). Since 1 dyne = 10\(^{-5}\) N and 1 cm\(^2\) = 10\(^{-4}\) m\(^2\), we convert the units to get the value as 12.6 \( \times \) 10\(^{12}\) N/m\(^2\).

🎯 Exam Tip: For unit conversions, remember the relationships between CGS and MKS units (e.g., 1 N = 10\(^5\) dyne, 1 m = 100 cm). Carefully apply conversion factors to both the force and area components.

Question 19. The dimensionless quantity ______
(a) never has a unit
(b) always has a unit
(c) may has a unit
(d) does not exist
Answer: (c) may has a unit
In simple words: A dimensionless quantity is one that has no physical dimension. However, some dimensionless quantities can still have units. For example, an angle is dimensionless but can be expressed in units like radians or degrees. Similarly, strain is dimensionless but can be represented as a percentage. So, a dimensionless quantity may or may not have a unit.

🎯 Exam Tip: Understand the difference between "dimension" and "unit." A quantity can be dimensionless (e.g., a ratio of two lengths) but still have a unit (e.g., radians for angle). This is a common point of confusion.

Question 20. Which one of the following is not a fundamental quantity?
(a) length
(b) luminous intensity
(c) temperature
(d) water current
Answer: (d) water current
In simple words: Fundamental quantities are the basic measurements that cannot be expressed in terms of other quantities. The seven SI fundamental quantities include length, mass, time, electric current, temperature, luminous intensity, and amount of substance. Water current, however, is a derived quantity, not fundamental.

🎯 Exam Tip: Memorize the seven fundamental SI quantities to easily identify which quantities are derived. Water current is a derived quantity (volume per unit time) and not a base quantity.

Question 21. The time dependence of a physical quantity p is given by \( p = p_0 e^{-\alpha t^2} \), where \( \alpha \) is a constant and \( t \) is time. The constant \( \alpha \) is ______
(a) a dimensionless
(b) has the dimension of T\(^2\)
(c) has the dimension as that of P
(d) has the dimension equal to dimensions of PT\(^{-2}\)
Answer: (a) a dimensionless
In simple words: In any exponent in a physical equation (like \( e^{-\alpha t^2} \)), the entire exponent must be dimensionless. Since \( t \) is time, \( t^2 \) has dimensions of T\(^2\). For the product \( \alpha t^2 \) to be dimensionless, \( \alpha \) must have dimensions of T\(^{-2}\).

🎯 Exam Tip: Always remember that any argument of an exponential, trigonometric, or logarithmic function must be dimensionless. Use this principle to find the dimensions of unknown constants.

Question 22. A force F is given F = at + bt\(^2\) Where t is time. What are the dimensions of a & b?
(a) ML T\(^{-3}\) and ML\(^2\)T\(^{-4}\)
(b) MLT\(^{-3}\) and MLT\(^{-4}\)
(c) MLT\(^{-1}\) and MLT\(^0\)
(d) MLT\(^{-4}\) and MLT\(^{-1}\)
Answer: (a) ML T\(^{-3}\) and ML\(^2\)T\(^{-4}\)
In simple words: For an equation like \( F = at + bt^2 \) to be correct, every term on the right side must have the same dimensions as the left side (force). Since \( F = [MLT^{-2}] \) and \( t = [T] \), then \( at \) must also be \( [MLT^{-2}] \), which means \( a = [MLT^{-3}] \). Similarly, \( bt^2 \) must be \( [MLT^{-2}] \), so \( b = [MLT^{-4}] \). This is crucial for dimensional consistency.

🎯 Exam Tip: The principle of homogeneity of dimensions states that in a physical equation, the dimensions of all the terms on both sides of the equation must be the same. Apply this strictly to find unknown dimensions.

Question 23. The triple point temperature of the water is ......
(a) -273.16 K
(b) 0K
(c) 273.16 K
(d) 100 K
Answer: (c) 273.16 K
In simple words: The triple point of water is a specific temperature and pressure where water can exist simultaneously as a gas, a liquid, and a solid (ice) in stable equilibrium. This exact temperature is 273.16 Kelvin (which is 0.01 degrees Celsius) and is a key reference point in thermometry.

🎯 Exam Tip: The triple point of water is a fixed reference point used to define the Kelvin temperature scale. Its precise value (273.16 K) is a fundamental constant.

Question 24. The force F on a sphere of radius 'a' moving in a medium with a velocity v is given by F = 6 \( \pi \) av. The dimension of \( \eta \) are
(a) \( ML^{-1}T^{-2} \)
(b) \( MT^{-1} \)
(c) \( MLT^{-2} \)
(d) \( ML^{-3} \)
Answer: (d) \( ML^{-3} \)
In simple words: The dimension of \( \eta \) is derived from the formula \( F = 6 \pi \eta av \). When we solve for \( \eta \), its dimensions are found to be \( [ML^{-3}] \). This dimension represents the viscosity of the medium, indicating how resistant a fluid is to flow.

🎯 Exam Tip: Remember that \( \eta \) (eta) is the coefficient of viscosity. To find its dimensions, isolate it in the given formula and substitute the dimensions of force, radius, and velocity.

Question 25. The unit of moment of force ......
(a) \( \text{Nm}^2 \)
(b) \( \text{Nm} \)
(c) \( \text{N} \)
(d) \( \text{NJ rad} \)
Answer: (b) Nm
In simple words: The unit for moment of force (or torque) is Newton-meter, or Nm. This tells us the turning effect of a force.

🎯 Exam Tip: Do not confuse Newton-meter (Nm), a unit of torque or energy, with millinewton (mN), a unit of force, or nanometer (nm), a unit of length.

Question 26. If the orbital velocity of a planet is given by v = \( G^a M^b R^c \) then
(a) a = 1/3, b = 1/3, c = – 1/3
(b) a = 1/2, b =1/2, c = – 1/2
(c) a = 1/3, b = – 1/2, c = 1/2
(d) a = 1/2, b = 1/2, c = – 1/2
Answer: (a) a = 1/3, b = 1/3, c = – 1/3
In simple words: For the planet's orbital velocity formula \( v = G^a M^b R^c \), the correct powers for G, M, and R are \( a = 1/3, b = 1/3, \text{ and } c = -1/3 \) respectively. This makes sure the units match up correctly for velocity.

🎯 Exam Tip: When using dimensional analysis to find exponents, ensure the dimensions of all physical quantities (G, M, R, and v) are correctly identified before setting up and solving the system of equations.

Question 27. The period T of a soap bubble under SHM given by T = \( P^a D^b S^c \) where P is the pressure, d is the density of water and E is the total energy of the explosion, then the value of a, b and c are:
(a) -3/2, 1/2, 1
(b) -5/6, 1/2, 1/3
(c) 5/6, 1/2, 1/3
(d) -5/6, -1/2, 1/3
Answer: (a) -3/2, 1/2, 1
In simple words: If the period \( T \) of a soap bubble depends on pressure \( P \), density \( d \), and energy \( E \) as \( T = P^a d^b E^c \), then the correct values for \( a, b, c \) are \( -3/2, 1/2, \text{ and } 1 \) respectively.

🎯 Exam Tip: This type of problem requires careful application of dimensional analysis, equating the powers of M, L, and T on both sides of the equation to solve for the unknown exponents.

Question 28. One degree of arc is equal to ......
(a) \( 1.457 \times 10^2 \) rad
(b) \( 1.457 \times 10^{-2} \) rad
(c) \( 1.745 \times 10^2 \) rad
(d) \( 1.745 \times 10^{-2} \) rad
Answer: (d) \( 1.745 \times 10^{-2} \) rad
In simple words: One degree of an angle is the same as about \( 1.745 \times 10^{-2} \) radians. Radians are another way to measure angles, especially important in science.

🎯 Exam Tip: To convert degrees to radians, multiply the degree value by \( \frac{\pi}{180} \). Knowing this conversion factor is crucial for many physics calculations.

Question 29. Frequency is the functions of density \( \rho \) length 'l' and tension T. The period of oscillation is proportional to
(a) \( \rho^{1/2} \lambda^2 T^{-1/2} \)
(b) \( \rho^{1/2} \lambda^{3/2} T^{-1/2} \)
(c) \( \rho^{1/2} \lambda^{3/2} T^{-3/4} \)
(d) \( \rho^{1/2} \lambda^{1/2} T^{3/2} \)
Answer: (a) \( \rho^{1/2} \lambda^2 T^{-1/2} \)
In simple words: If a string's vibration period depends on its material density \( \rho \), length \( \lambda \), and how tight it is (tension \( T \)), then the period changes with these factors in a specific way: \( \rho^{1/2} \lambda^2 T^{-1/2} \).

🎯 Exam Tip: Remember that the period of oscillation is the inverse of frequency. If a question asks for proportionality, be careful with positive and negative exponents.

Question 30. The frequency of vibration of string is given by \( \nu = \frac{p}{2 \ell} \sqrt{\frac{T}{m}} \) Here \( l \) is the length, P is the number of segments in the string. T is tension is the string, the dimensional formula for 'm' will be.
(a) \( M^0 L T^{-1} \)
(b) \( M L^0 T^{-1} \)
(c) \( M L^{-1} T^0 \)
(d) \( M^0 L^0 T^0 \)
Answer: (c) \( M L^{-1} T^0 \)
In simple words: The formula for string vibration frequency uses 'm'. This 'm' stands for mass per unit length of the string. Its dimensions are \( ML^{-1}T^0 \), which means it has units of mass divided by length.

🎯 Exam Tip: In this formula, 'm' often represents linear mass density (mass per unit length). Always check the definitions of symbols in given equations for dimensional analysis problems.

Question 31. 1 second of arc is equal to ......
(a) 0.00027°
(b) \( 1.745 \times 10^{-2} \) rad
(c) \( 2.91 \times 10^{-4} \) rad
(d) \( 4.85 \times 10^{-6} \) rad
Answer: (a) 0.00027°
In simple words: One second of arc is a very tiny angle, about \( 0.00027 \) degrees. It is used to measure very small changes in angles.

🎯 Exam Tip: Remember the conversion hierarchy: 1 degree = 60 minutes of arc, and 1 minute of arc = 60 seconds of arc. So, 1 second of arc = \( \frac{1}{3600} \) degrees.

Question 32. If force (f), length (L) and time (T) are assumed to be fundamental units then the dimensional formula of the mass will be
(a) \( FL^{-1}T^2 \)
(b) \( FL^{-1}T^2 \)
(c) \( FL^{-1} T^{-1} \)
(d) \( FL^2T^2 \)
Answer: (a) \( FL^{-1}T^2 \)
In simple words: If we use force, length, and time as basic units, then the unit for mass would be \( FL^{-1}T^2 \).

🎯 Exam Tip: To derive dimensions in terms of new fundamental units, start with a known formula (like \( F=ma \)) and substitute the dimensions of the known quantities, then solve for the desired quantity's dimensions.

Question 33. If pressure 'p' velocity v and time T are taken as fundamental physical quantities the dimensional formula for force is
(a) \( [PV^2 T^2] \)
(b) \( [P^{-1}V^2T^{-2}] \)
(c) \( [PV T^2] \)
(d) \( [P^{-1}V T^2] \)
Answer: (a) \( [PV^2 T^2] \)
In simple words: If pressure, velocity, and time are the basic units, then the formula for force using these new units would be \( [PV^2T^2] \).

🎯 Exam Tip: When redefining fundamental quantities, clearly list the dimensions of the new fundamental quantities and then express the desired quantity's dimensions as a product of these new fundamentals raised to some powers.

Question 34. The range of distance can be measured by using direct methods is
(a) \( 10^{-2} \) to \( 10^{-5} \) m
(b) \( 10^{-2} \) to \( 10^2 \) m
(c) \( 10^2 \) to \( 10^5 \) m
(d) \( 10^{-2} \) to \( 10^5 \) m
Answer: (b) \( 10^{-2} \) to \( 10^2 \) m
In simple words: You can measure distances directly using tools like a ruler or tape measure for sizes between about 1 centimeter and 100 meters.

🎯 Exam Tip: Direct methods are typically suitable for human-scale measurements using standard rulers and tapes. For very small or very large distances, indirect methods are necessary.

Question 35. The speed of light (c) gravitational constant G and planks constant h are taken as fundamental units. The dimension of time in the new system will be
(a) \( G^{1/2} h^{1/2} C^{-5/2} \)
(b) \( G^{1/2} h^{1/2} C^{1/2} \)
(c) \( G^{1/2} h^{1/2} C^{-3/2} \)
(d) \( G^{1/2} h^{1/2} C^{1/2} \)
Answer: (a) \( G^{1/2} h^{1/2} C^{-5/2} \)
In simple words: If we use the speed of light, gravity's constant, and Planck's constant as basic units, then time's measurement would be expressed as \( G^{1/2} h^{1/2} c^{-5/2} \).

🎯 Exam Tip: This problem involves complex dimensional analysis where multiple fundamental constants are used to define a new system of units. Practice deriving the dimensions of common physical constants before attempting these problems.

Question 36. pipe and coefficient of viscosity of liquid \( \eta \) as Q \( \propto r^a \rho^b \eta^c L^d \) then
(a) a = 4, b = 1, c = – 1, d = – 1
(b) a = 4, b = – 1, c = 1, d = – 1
(c) a = 4, b = 1, c = 1, d = – 1
(d) values of a,b,c and d cannot be determined
Answer: (d) values of a,b,c and d cannot be determined
In simple words: Because the formula is unclear (it uses \( \rho \) which usually means density, but a flow rate often involves pressure), we cannot figure out the exact numbers for a, b, c, and d.

🎯 Exam Tip: Dimensional analysis requires that the physical quantities involved in a relationship are clearly defined. Ambiguity in variables (like \( \rho \) potentially meaning pressure instead of density) can make determination impossible.

Question 37. The dimensions of universal gas constant is
(a) \( ML^2T^{-2}\theta^{-1} \)
(b) \( ML^2T^{-2}\theta \)
(c) \( ML^3 T^{-1} \theta^{-1} \)
(d) None of the options
Answer: (a) \( ML^2T^{-2}\theta^{-1} \)
In simple words: The universal gas constant R has the dimensions \( ML^2T^{-2}\theta^{-1} \). This constant is important in equations that describe how gases behave.

🎯 Exam Tip: Recall the ideal gas law \( PV=nRT \). Use this to rearrange for R, and then substitute the dimensions of pressure (P), volume (V), amount of substance (n), and temperature (T) to find R's dimensions.

Question 38. Find odd one out.
(a) Newton
(b) metre
(c) candela
(d) Kelvin
Answer: (a) Newton
In simple words: Newton is different from metre, candela, and Kelvin because it's a unit made from other units, while the others are basic, fundamental units.

🎯 Exam Tip: Be familiar with the seven fundamental SI units (meter, kilogram, second, ampere, kelvin, mole, candela) to easily identify derived units.

Question 39. Which of the following combinations have the dimensions of time? L, C, R represent inductance, capacitance, and resistance respectively.
(b) \( \sqrt{LC} \)
(c) L/R
(d) C/L
Answer: (b) \( \sqrt{LC} \)
In simple words: The combination \( \sqrt{LC} \) has the same dimensions as time. This is useful for understanding how electrical parts like inductors and capacitors work together in circuits.

🎯 Exam Tip: Remember that in an LC oscillating circuit, the angular frequency \( \omega = \frac{1}{\sqrt{LC}} \). Since frequency has dimensions of \( T^{-1} \), \( \sqrt{LC} \) must have dimensions of T.

Question 40. The dimensions of mobility are
(a) \( M^{-1} LA T^{-2} \)
(b) \( ML A^{-1}T^{-2} \)
(c) \( MA^{-1}T^{-2} \)
(d) \( M^{-1}A T^2 \)
Answer: (d) \( M^{-1}A T^2 \)
In simple words: Mobility measures how easily charged particles move. Its dimensions are \( M^{-1}A T^2 \).

🎯 Exam Tip: Mobility \( (\mu) \) is defined as \( \mu = \frac{\text{drift velocity }(v_d)}{\text{electric field }(E)} \). Use the dimensions of \( v_d \) and \( E \) to derive the dimensions of mobility.

Question 41. The smallest physical unit of time is
(a) second
(b) minute
(c) microsecond
(d) shake
Answer: (d) shake
In simple words: The smallest unit of time listed here is a "shake," which is a very, very short amount of time, like ten billionths of a second.

🎯 Exam Tip: A "shake" is an informal unit, often used in nuclear physics, where 1 shake = \( 10^{-8} \) seconds. This highlights that many fields use specialized units for convenience.

Question 42. The length, breadth and thickness of strip are given by \( l = (10.0\pm0.1)\text{cm} \), \( h= (1.00\pm0.01)\text{cm} \), \( t = (0.100\pm0.001)\text{cm} \) the most probable error in volume will be
(a) \( 0.03 \text{ cm}^3 \)
(b) \( 0.111 \text{ cm}^3 \)
(c) \( 0.012 \text{ cm}^3 \)
(d) \( 0.12 \text{ cm}^3 \)
Answer: (a) \( 0.03 \text{ cm}^3 \)
In simple words: To find the most likely error in the volume, we first calculate the individual percentage errors for length, breadth, and thickness, and then add them up. This sum, multiplied by the actual volume, gives the total error.

🎯 Exam Tip: For quantities involving multiplication (like volume), the maximum fractional error in the result is the sum of the fractional errors of the individual measurements. \( \frac{\Delta V}{V} = \frac{\Delta l}{l} + \frac{\Delta h}{h} + \frac{\Delta t}{t} \).

Question 43. The measured mass and volume of a body are 22.42g and 4.7cm³ respectively with possible errors of 0.01 g and 0.1cm³. The maximum error in density is
(a) 0.2%
(b) 2%
(c) 5%
(d) 10%
Answer: (b) 2%
In simple words: To find the biggest possible error in density, we add up the percentage errors from measuring the mass and the volume. This sum tells us the overall uncertainty.

🎯 Exam Tip: For division operations (like density = mass/volume), the relative errors (or percentage errors) are added to find the maximum possible error in the calculated quantity.

Question 44. Half the lifetime of a free neutron is in the order of ......
(a) \( 10^0 \) s
(b) \( 10^1 \) s
(c) \( 10^2 \) s
(d) \( 10^3 \) s
Answer: (d) \( 10^3 \) s
In simple words: A free neutron lasts for about 10 minutes before it decays. This time is roughly in the order of \( 10^3 \) seconds.

🎯 Exam Tip: An "order of magnitude" refers to the nearest power of ten. Convert the half-life to seconds (approximately 600 s) and then find the closest power of ten.

Question 45. An experiment measures quantities a, b, c and y is calculated from a formula. \( y = \frac{a b^{2}}{c^{3}} \) If the percentage errors in a,b and c are \( \pm 1\% \), \( \pm 3\% \), \( \pm 2\% \) respectively, the percentage error in calculating y is
(a) \( \pm 13\% \)
(b) \( \pm 7\% \)
(c) \( \pm 4\% \)
(d) \( \pm 1\% \)
Answer: (a) \( \pm 13\% \)
In simple words: To find the total percentage error in \( y \), we add the percentage error of \( a \), twice the error of \( b \), and three times the error of \( c \). This gives us \( 13\% \).

🎯 Exam Tip: For quantities raised to a power, multiply the percentage error of the base quantity by its power (e.g., \( b^2 \) means \( 2 \times \% \text{Error(b)} \)). Always add the errors for maximum possible error.

Question 46. A student measures the distance traversed in a free fail of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If maximum percentage error in measurement of the distance and the time are \( e_1 \) and \( e_2 \) respectively the percentage error in the estimation of g is
(a) \( e_2 - e_1 \)
(b) \( e_1+ 2e_2 \)
(c) \( e_1 + e_2 \)
(d) \( e_1 - 2e_2 \)
Answer: (b) \( e_1+ 2e_2 \)
In simple words: If we calculate gravity (\( g \)) from how far something falls (\( s \)) and how long it takes (\( t \)), the error in \( g \) will be the error in \( s \) plus two times the error in \( t \).

🎯 Exam Tip: Start with the kinematic equation \( s = \frac{1}{2}gt^2 \) and rearrange it to solve for \( g \). Then apply the rules for propagation of errors for multiplication and powers.

Question 47. The heat generated in a circuit is given by \( Q = I^2Rt \). Where I is current, R is the resistance and t is the time. If an error in measuring current, resistance, and time are 2%, 1%, and 1% respectively. The maximum error in measuring heat will be
(a) 2%
(b) 4%
(c) 6%
(d) 8%
Answer: (c) 6%
In simple words: If there are small errors in measuring current, resistance, and time, the biggest possible error in the calculated heat will be \( 6\% \).

🎯 Exam Tip: When quantities are multiplied or divided and one or more are raised to a power (e.g., \( I^2 \)), add the percentage errors, multiplying the error of the quantity by its power.

Question 49. A student performs an experiment for determination of \( g = \frac{4 \pi^{2} l}{T^{2}} \) an error of \( \Delta l \). For that, he takes the time of n oscillations with the stopwatch of least count \( \Delta T \) and he commits a human error of 0.1s. For which of the following data, the measurement of g will be most accurate? \( \Delta l, \Delta T, \eta \)
(a) 5m, 0.2s, 10
(b) 5mm, 0.2s, 20
(c) 5mm, 0.1s, 20
(d) 1mm, 0.1s, 50
Answer: (d) 1mm, 0.1s, 50
In simple words: To get the most precise value for gravity, we need the smallest error in measuring the length of the pendulum and the smallest error when measuring its swing time. Option (d) offers the best combination of these small errors and a large number of swings, making it the most accurate choice.

🎯 Exam Tip: To minimize error in an experiment like this, always aim for the smallest possible instrumental error, compensate for human error, and increase the number of observations (\( n \)) for average values. The fractional error \( \frac{\Delta T}{T} \) decreases with increasing \( n \).

Question 50. A Screw gauge gives the following reading when used to measure the diameter of the wire.
Main scale reading = 0.
Circular scale reading = 52 divisions
Given that 1 mm on the main scale corresponds to 100 division on a circular scale. The diameter of the wire is
(a) 0.52 cm
(c) 0.0026 cm
Answer: (a) 0.52 cm
In simple words: To get the wire's diameter, we add the main scale reading to the circular scale reading multiplied by the least count. This gives us 0.52 cm.

🎯 Exam Tip: For a screw gauge, the least count is the ratio of the pitch (distance moved in one full rotation) to the total number of divisions on the circular scale. Ensure consistent units in your calculation.

Question 51. The error caused due to the sheer carelessness of an observer is called as ...... error.
(a) Systematise
(b) Gross
(c) Random
(d) Personal
Answer: (b) Gross
In simple words: When an error happens because someone is not careful during an experiment, it's called a gross error.

🎯 Exam Tip: Gross errors are typically large mistakes that can be avoided by exercising caution and vigilance. They are distinct from systematic errors (which are consistent) and random errors (which are unpredictable).

Question 52. The force (F) velocity(v) and time “T” are taken as fundamental units then the divisions of mass are
(a) \( [FVT^2] \)
(b) \( [FV^{-1}T^{-1}] \)
(c) \( [FV^{-2}T^{-1}] \)
(d) \( [FV^{-1}T] \)
Answer: (d) \( [FV^{-1}T] \)
In simple words: If force, velocity, and time are our basic units, then mass is described by \( FV^{-1}T \).

🎯 Exam Tip: To derive new dimensional formulae, always start with a fundamental equation involving the quantities (e.g., \( F=ma \), \( v=L/T \)), substitute their dimensions, and then solve algebraically.

Question 53. Attempting to explain diverse physical phenomenon with few concepts and law is
(a) unification (or) reductionism
(b) neither unification nor reductionism
(c) unification
(d) reductionism
Answer: (c) unification
In simple words: Trying to explain many different physics things with just a few basic ideas and rules is called unification.

🎯 Exam Tip: Unification in physics aims to combine different theories into a single, more comprehensive theory. Examples include the unification of electric and magnetic forces into electromagnetism.

Question 54. An attempt to explain a microscopic system in terms of its microscopic constituents
(a) unification
(b) reductionism
(c) neither unification or reductionism
(d) neither unification nor reductionism
Answer: (b) reductionism
In simple words: When we try to understand a small system by looking at its even smaller parts, that approach is called reductionism.

🎯 Exam Tip: Reductionism involves breaking down complex phenomena into simpler, more fundamental components to understand them. It is a powerful method in many scientific disciplines.

Question 55. The study of nature of particles is
(a) nuclear physics
(b) quantum mechanics
(c) condensed another physics
(d) high energy physics
Answer: (d) high energy physics
In simple words: The field that studies the basic building blocks of matter and how they interact is called high energy physics.

🎯 Exam Tip: High energy physics, also known as particle physics, explores the fundamental particles and forces that make up the universe, often requiring large particle accelerators.

Question 56. The ratio of the mean absolute error to the mean value is called
(a) absolute error
(b) random error
(c) relative error
(d) percentage error
Answer: (c) relative error
In simple words: The relative error is how big the error is compared to the actual measured value. You get it by dividing the average error by the average measurement.

🎯 Exam Tip: Relative error is a dimensionless quantity and provides a more meaningful indication of the precision of a measurement than absolute error alone.

Question 57. 1" is equal to ______ radian.
(a) \( 1.745 \times 10^{-2} \) rad
(b) \( 1.78 \times 10^{-3} \) rad
(c) \( 2.91 \times 10^{-4} \) rad
(d) \( 4.847 \times 10^{-6} \) rad
Answer: (d) \( 4.847 \times 10^{-6} \) rad
In simple words: One second of arc, which is a tiny angle, is equal to about \( 4.847 \times 10^{-6} \) radians.

🎯 Exam Tip: Remember the full conversion chain: \( 1^{\prime\prime} = \frac{1}{60} \text{ arc minute} = \frac{1}{3600} \text{ degree} = \frac{1}{3600} \times \frac{\pi}{180} \text{ radians} \).

Question 58. From a point on the ground, the top of a tree is seen to have an angle of elevation of 60°. The distance between the tree and a point is 50 m. The height of the tree is
(a) 86.6 m
(b) 90.6 m
(c) 92.8 m
(d) 80.6 m
Answer: (a) 86.6 m
In simple words: We can find the height of the tree using trigonometry. Since we know the distance to the tree and the angle to its top, the height is 50 meters multiplied by the tangent of 60 degrees, which is about 86.6 meters.

🎯 Exam Tip: For right-angled triangles, use SOH CAH TOA. In this case, tangent (TOA) relates the angle of elevation to the opposite side (height) and the adjacent side (distance).

Question 59. The maximum possible error in the sum of two quantities is equal to .......
(a) Z = A + B
(b) \( \Delta Z = \Delta A + \Delta B \)
(c) \( \Delta Z = \Delta A/\Delta B \)
(d) \( \Delta Z = \Delta A – \Delta B \)
Answer: (b) \( \Delta Z = \Delta A + \Delta B \)
In simple words: If you add two measurements, the biggest possible error in the total sum is simply the sum of the individual errors from each measurement.

🎯 Exam Tip: For addition and subtraction, always add the absolute errors to find the maximum possible error in the result. This applies regardless of whether the quantities are added or subtracted.

Question 60. Number of a significant digit in 0.030400
(a) 6
(b) 5
(c) 4
(d) 3
Answer: (b) 5
In simple words: For the number 0.030400, the significant digits are 3, 0, 4, and the two zeros at the end. The zeros at the very beginning do not count. So there are 5 significant digits in total.

🎯 Exam Tip: Leading zeros (like those before the '3' in 0.030400) are never significant. Trailing zeros are significant only if the number contains a decimal point. Zeros between non-zero digits are always significant.

Question 1. A new unit of length is chosen such that the speed of light in a vacuum is unity. What is the distance between the sun and the earth in in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Answer: In this new system, the speed of light in a vacuum (c) is defined as 1 new unit of length per second.
First, convert the time taken for light to travel from the sun to the earth into seconds:
\( 8 \text{ minutes } 20 \text{ seconds} = (8 \times 60) + 20 = 480 + 20 = 500 \text{ seconds} \).
Next, use the formula distance = speed \( \times \) time.
\( \text{Distance (x)} = \text{speed of light (c)} \times \text{time (t)} \)
\( \text{x} = 1 \text{ new unit of length s}^{-1} \times 500 \text{ s} = 500 \text{ new unit of length} \).
This demonstrates how distance measurements change when fundamental constants are redefined.
In simple words: If light travels 1 new length unit every second, and it takes 500 seconds for light to travel from the sun to Earth, then the distance is 500 new length units.

🎯 Exam Tip: When new units are defined based on fundamental constants, always ensure consistent unit conversions, especially for time, before performing calculations.

Question 1. Why it is convenient to express the distance of stars in terms of light-year (or) parser rather than in Km?
Answer: It is more convenient to express the vast distances to stars in units like light-years or parsecs rather than in kilometers because these distances are incredibly large. One light-year, for example, is approximately \( 9.46 \times 10^{12} \) kilometers. Using kilometers would result in extremely large numbers that are cumbersome to write, read, and comprehend. Light-years and parsecs provide more manageable and intuitive values for astronomical scales.
In simple words: Star distances are huge, so using light-years or parsecs keeps the numbers smaller and easier to understand. Kilometers would make the numbers too big and complicated.

🎯 Exam Tip: Light-year and parsec are astronomical units designed to simplify the representation of extremely large interstellar distances, making calculations and comparisons easier.

Question 2. Show that a screw gauge of pitch 1 mm and 100 divisions is more precise than a vernier caliper with 20 divisions on the sliding scale.
Answer: A measuring instrument is considered more precise if it has a smaller least count (LC). The least count indicates the smallest measurement that the instrument can accurately make.
For the screw gauge:
Pitch = 1 mm.
Number of divisions on circular scale = 100.
Least Count \( (\text{LC}_{\text{screw gauge}}) = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1\text{ mm}}{100} = 0.01\text{ mm} \).
For the vernier caliper:
Assume 1 Main Scale Division (MSD) = 1 mm = 0.1 cm.
Number of divisions on sliding scale (vernier scale) = 20.
Least Count \( (\text{LC}_{\text{vernier caliper}}) = \frac{1\text{ MSD}}{\text{Number of vernier divisions}} = \frac{1\text{ mm}}{20} = 0.05\text{ mm} \).
Comparing the least counts, \( 0.01\text{ mm} \) (screw gauge) is smaller than \( 0.05\text{ mm} \) (vernier caliper). Therefore, the screw gauge is more precise. Precise measurements reduce uncertainty.
In simple words: The screw gauge is more precise because it can measure smaller differences. Its smallest reading (least count) is 0.01 mm, which is smaller than the vernier caliper's smallest reading of 0.05 mm.

🎯 Exam Tip: To compare the precision of instruments, always calculate their least counts. The instrument with the smallest least count is the most precise.

Question 3. What is the difference between mN, Nm, and nm?
Answer: The symbols mN, Nm, and nm represent distinct physical quantities:
1. mN (millinewton): This denotes a unit of force. "Milli" is a prefix meaning \( 10^{-3} \), so 1 millinewton \( (1 \text{ mN}) \) is equal to \( 10^{-3} \) Newtons.
2. Nm (Newton-meter): This is a unit of torque (moment of force) or work/energy. It represents the product of force and distance.
3. nm (nanometer): This is a unit of length. "Nano" is a prefix meaning \( 10^{-9} \), so 1 nanometer \( (1 \text{ nm}) \) is equal to \( 10^{-9} \) meters. These prefixes help scale units to appropriate sizes.
In simple words: mN is a small unit of force, Nm measures twisting force or energy, and nm is a very tiny unit of length. They all use 'N' or 'm' but mean different things.

🎯 Exam Tip: Pay close attention to capitalization (N for Newton vs. n for nano) and the order of letters (mN vs. Nm) as they change the meaning of the unit entirely.

Question 4. Having all units in atomic standards is more useful. Explain.
Answer: It is more useful to define all units using atomic standards because physical prototypes (like the original standard meter bar or kilogram cylinder) have several limitations, while atomic standards offer significant advantages:
**Difficulties with Physical Prototypes:**
1. **Preservation:** Physical prototypes are challenging to maintain accurately over long periods and can be damaged or lost.
2. **Replication:** Creating exact copies of prototypes for use worldwide is difficult, and the replication process itself can introduce inaccuracies.
3. **Variability:** Prototypes can be affected by factors like temperature, pressure, or even subtle aging, causing their values to change slightly over time and space.
**Advantages of Atomic Standards:**
1. **Reproducibility:** Atomic standards can be precisely reproduced anywhere in the world, at any time, using specific physical phenomena or fundamental constants.
2. **Consistency:** They are based on unchanging atomic properties or natural constants, making them unaffected by environmental conditions.
3. **Accuracy:** Atomic standards offer extremely high levels of precision and accuracy, often to many decimal places, far surpassing physical artifacts.
This shift to atomic definitions provides universal, stable, and highly accurate units for science and technology.
In simple words: Using atomic properties to define units is better because physical objects can change or be hard to copy. Atomic standards are always the same, can be created anywhere, and are much more accurate.

🎯 Exam Tip: Understand that atomic standards are based on universal constants of nature (like the speed of light or Planck's constant), which are inherently more stable and accessible than man-made physical artifacts.

Question 5. Why dimensional methods are applicable only up to three quantities?
Answer: Dimensional analysis can be used to find relationships for only up to three physical quantities. This is because in mechanics, we usually match the powers of basic dimensions like Mass (M), Length (L), and Time (T). Since there are only three such fundamental dimensions, we can only create three independent equations to solve for up to three unknown powers in a dimensional formula. These same limitations apply to other types of quantities as well.
In simple words: Dimensional analysis works best with only three quantities because we only have three main dimensions (Mass, Length, Time) to balance in our equations.

🎯 Exam Tip: Remember that dimensional analysis checks for consistency, but cannot determine dimensionless constants or whether a quantity is a scalar or vector.