RBSE Solutions Class 11 Physics Chapter 1 Physical World and Measurement

 

RBSE Class 11 Physics Chapter 1 Textbook Exercises with Solutions

RBSE Class 11 Physics Chapter 1 Very Short Answer Type Questions

 

Question 1. What is the unit of luminous intensity in S.I. system?
Answer: The S.I. unit for luminous intensity is candela (cd). This unit measures how much light a source emits in a specific direction.
In simple words: The standard unit for measuring light brightness in a certain direction is called candela.

🎯 Exam Tip: Remember both the full name "candela" and its symbol "cd" for S.I. units of luminous intensity.

 

Question 2. What is the unit of amount of a substance?
Answer: The unit for the amount of a substance is mole. A mole is a standard way to count a very large number of atoms or molecules.
In simple words: The unit used to measure how much of a substance there is, like counting atoms, is called a mole.

🎯 Exam Tip: The mole (mol) is a fundamental S.I. unit, essential for calculations in chemistry and physics.

 

Question 3. 1 Processing math: 8% gs?
Answer: The question appears incomplete in the source. Assuming it relates to the concept of processing math. The processing math value of 8% is provided.
In simple words: The question seems unfinished, but it refers to a math processing percentage which is 8%.

🎯 Exam Tip: Incomplete questions should be noted, and any available numerical information should be included in the answer.

 

Question 5. How many significant figures are there in 0.005?
Answer: There is one (1) significant figure in 0.005. The leading zeros before a non-zero digit are not counted as significant.
In simple words: Only the digit '5' counts, so there is one significant figure.

🎯 Exam Tip: Remember that leading zeros (zeros before the first non-zero digit) are never significant when determining significant figures.

 

Question 6. How many base and supplementary units are there in S.I. system?
Answer: In the S.I. system, there are 7 base or fundamental units and 2 supplementary units. These units form the foundation for all other measurements.
In simple words: The S.I. system has seven basic units and two extra units that help describe measurements.

🎯 Exam Tip: Know the exact number of base and supplementary units in the S.I. system, as this is a common factual question.

 

Question 7. Write the dimensional formula for Boltzmann constant?
Answer: The dimensional formula for the Boltzmann constant is \( \left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2} \mathrm{K}^{-1}\right] \). This formula shows how mass, length, time, and temperature relate to the constant.
In simple words: The Boltzmann constant's formula uses mass, length, time, and temperature in a specific way.

🎯 Exam Tip: Accurately writing dimensional formulas requires understanding the fundamental units involved in the constant's definition.

 

Question 8. How many types of errors are there?
Answer: There are three types of errors in measurement. These typically include instrumental errors, natural errors, and observational errors.
In simple words: There are three main kinds of mistakes that can happen when you measure something.

🎯 Exam Tip: Be able to name and briefly describe the main categories of errors encountered in scientific measurements.

 

Question 9. What is the dimensional formula of surface tension?
Answer: The dimensional formula of surface tension is \( \left[\mathrm{M}^{1} \mathrm{L}^{0} \mathrm{T}^{-2}\right] \). This shows that surface tension depends on mass and time, but not length.
In simple words: Surface tension's formula only uses mass and time, not length.

🎯 Exam Tip: For surface tension, remember that the length dimension cancels out, resulting in \( \mathrm{L}^{0} \).

 

Question 10. What is the wavelength of Kr86 in 1 m?
Answer: The wavelength of Kr86 (Krypton-86) in 1 meter is 1650763.73. This specific wavelength was once used to define the meter.
In simple words: If you measure how many Kr86 wavelengths fit into one meter, it's about 1.65 million.

🎯 Exam Tip: Note this historical value for the wavelength of Kr86, which was used in the past to define the standard meter.

 

Question 11. N_Processing math: 8% Jantity it's dimensional formula is same as that of impules.
Answer: The question appears incomplete, but it is asking for a physical quantity whose dimensional formula is the same as that of impulse. Impulse has the dimensional formula \( \left[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-1}\right] \). Linear momentum has the same dimensional formula.
In simple words: The question asks for a measurement with the same formula as impulse; linear momentum is the answer.

🎯 Exam Tip: Always recognize quantities that share the same dimensional formula, such as impulse and linear momentum.

 

Question 13. How much is the percentage error in constant K?
Answer: The percentage error in a constant K is zero. By definition, a constant value does not have any variation or uncertainty.
In simple words: A constant value never changes, so there is no error in it.

🎯 Exam Tip: Understand that by definition, true constants have no error or uncertainty associated with them.

 

Question 14. If Ax is the error in measurement of a physical quantity x, then how much is the percentage error in xm?
Answer: If \( \Delta x \) is the error in the measurement of a physical quantity \( x \), then the percentage error in \( x^m \) is given by \( m \frac{\Delta x}{x} \times 100 \% \). This means the power multiplies the relative error.
In simple words: When you raise a measurement with an error to a power, the error percentage for the new value is the original error percentage multiplied by that power.

🎯 Exam Tip: Remember the rule for propagation of errors in powers: the relative error is multiplied by the power.

 

Question 15. If the units of each force and length is doubled, then how much times would be the unit of energy?
Answer: If the units of both force and length are doubled, the unit of energy would become 4 times larger. Energy is calculated as force multiplied by length (\( E = F \times L \)). So, if \( F \) becomes \( 2F \) and \( L \) becomes \( 2L \), then \( E \) becomes \( (2F) \times (2L) = 4FL \).
In simple words: Since energy is force times length, doubling both force and length units makes the energy unit four times bigger.

🎯 Exam Tip: Always recall the relationship between energy, force, and length (\( E = F \times L \)) when dealing with unit changes.

 

RBSE Class 11 Physics Chapter 1 Short Answer Type Questions

 

Question 1. What is Science?
Answer: The word "Science" comes from the Latin word "Scientia," which means "to know." Science is knowledge gained by observing things and doing experiments, then organizing this information in a systematic way. It helps us understand the natural world.
In simple words: Science is organized knowledge that we get by watching things and doing tests.

🎯 Exam Tip: When defining science, highlight keywords like "knowledge," "observations," "experiments," and "systematically organized."

 

Question 2. What is a unit? How many systems are there for units?
Answer: A unit is a standard amount of a particular type of quantity that we use to compare and measure other physical quantities. When we measure something, we say it is a certain number of units. There are three main systems for units: the C.G.S. system (centimeter, gram, second), the M.K.S. system (meter, kilogram, second), and the F.P.S. system (foot, pound, second).
In simple words: A unit is a standard way to measure things. There are three main ways to organize these units, like CGS, MKS, and FPS.

🎯 Exam Tip: Clearly define "unit" as a standard for comparison and list the primary unit systems (C.G.S., M.K.S., F.P.S.) as historical examples.

 

Question 3. Define a radian.
Answer: A radian is a unit used to measure plane angles. One radian is defined as the angle made at the center of a circle by an arc that has the same length as the radius of that circle. For a small angle \( d\theta \), the plane angle is given by \( d\theta = \frac{ds}{r} \), where \( ds \) is the arc length and \( r \) is the radius. If \( ds = r \), then \( d\theta = 1 \) radian.
In simple words: A radian is an angle in a circle where the curved part is as long as the straight line from the center to the edge.

🎯 Exam Tip: The key to defining a radian is the relationship between the arc length and the radius being equal.

 

Question 4. Write the quantities of a S.I. system of units.
Answer: The S.I. system of units, officially named "Le Système International d'Unités" (The International System of Units), is the modern metric system. It has seven fundamental (or basic) units and two supplementary units.
Fundamental quantities: These include Length, Mass, Time, Electric Current, Temperature, Luminous Intensity, and Amount of Substance.
Supplementary quantities: These are Plane Angle and Solid Angle.
In simple words: The S.I. system has seven basic units like length and mass, plus two extra units for angles.

🎯 Exam Tip: Make sure to list all seven fundamental quantities and the two supplementary quantities correctly to score full marks.

 

Question 5. Write the advantages of dimensional analysis.
Answer: Dimensional analysis is a useful tool in physics. Its advantages include: (1) It helps convert units of a physical quantity from one system to another. (2) It can check if a physical equation is correct dimensionally. (3) It helps to derive relationships between different physical quantities. The "Processing math: 8% ses or advantages of dimensional analysis." highlights its usefulness.
In simple words: Dimensional analysis helps change units, check if formulas are right, and find connections between different physical measurements.

🎯 Exam Tip: When listing advantages, focus on conversion, checking equation correctness, and deriving relationships between quantities.

 

Question 6. Is it possible to have dimensionless and unitless physical quantities?
Answer: Yes, it is possible to have physical quantities that are both dimensionless and unitless. This means they do not have any physical dimensions (like mass, length, or time) and therefore no specific unit. For example:
1. Strain: Strain is a measure of deformation, calculated as a ratio of change in length to original length, making it dimensionless and unitless.
2. Angle and solid angle: Although they have units (radian and steradian, respectively), they are sometimes considered dimensionless as they are ratios of lengths or areas.
In simple words: Yes, some measurements have no dimensions or units, like strain, which is a ratio.

🎯 Exam Tip: Strain is the best example for a quantity that is both dimensionless and unitless because it is a ratio of two similar quantities.

 

RBSE Class 11 Physics Chapter 1 Long Answer Type Questions

 

Question 1. Explain the International System of Units and define the different fundamental units.
Answer: The International System of Units, abbreviated as S.I. (from the French "Système International d'Unités"), is the modern version of the metric system and is the most widely used system for measurements globally. The S.I. system is built upon seven fundamental units and two supplementary units.

These seven basic units, along with two supplementary units, are enough to create units for all physical quantities across different fields of physics.

Supplementary quantities and their units

The above seven basic and two supplementary units are sufficient to derive units for physical quantities from all branches of physics.

(ii) In addition to the fundamental units and the supplementary units, there are many derived units in this system. Some of these derived units are listed in tables.

Supplementary Units: Definitions

1. Radian: One radian is the angle formed at the center of a circle by an arc whose length is equal to the radius of the circle.

The plane angle \( d\theta = \frac{dS}{r} \) rad.
If \( dS = r \), then \( d\theta = 1 \) rad.

2. Steradian: One steradian is the solid angle made at the center of a sphere by a surface area on the sphere that is equal to the square of its radius.

Fig. 1.3

Solid angle \( (d\Omega) = \frac{\text{Normal surface area}}{(\text{Radius})^2} \)
\( \implies \) Solid angle \( (d\Omega) = \frac{dA}{r^2} \) steradian
If \( dA = r^2 \), then \( d\Omega = 1 \) steradian. In the case of a complete spherical surface area, the total solid angle subtended by it is \( \Omega = \frac{4\pi r^2}{r^2} = 4\pi \) steradian.
In simple words: The International System of Units (SI) is the main way we measure things, using seven basic units like meters for length and kilograms for mass, and two extra units for angles. Radian is an angle where the arc length equals the radius. Steradian is a 3D angle where the surface area equals the radius squared.

🎯 Exam Tip: Clearly list all seven fundamental units with their symbols and also the two supplementary units. Provide precise definitions for radian and steradian, including their mathematical relations.

 

Question 2. Differentiate fundamental units and derived units by giving examples. Write popular units.
Answer: Physical quantities are classified into two main types: fundamental quantities and derived quantities.
(i) Fundamental quantities: Fundamental units are basic units that cannot be obtained from or broken down into other units. They are independent. For example, mass, length, and time are fundamental physical quantities, and their units (kilogram, meter, second) are fundamental units because:
1. Mass, length, and time cannot be derived from each other.
2. All other physical quantities in mechanics can be expressed using these three.
(ii) Derived quantities: Derived units are units for physical quantities that can be expressed using combinations of fundamental units (like mass, length, and time). Examples include force, work, power, energy, pressure, momentum, acceleration, and density. These are obtained by writing their defining equations in terms of fundamental quantities. For instance, speed is defined as:
Speed \( = \frac{\text{Distance covered}}{\text{Time taken}} \)
So, the unit of speed \( = \frac{\text{Unit of distance i.e. length}}{\text{Unit of time}} = \frac{\text{meter}}{\text{second}} = \text{m/s} \).
In simple words: Fundamental units are basic measurements that can't be split up, like mass and time. Derived units are made by combining fundamental units, like speed (length divided by time).

🎯 Exam Tip: When differentiating, give clear definitions and provide distinct examples for both fundamental and derived units, explaining how derived units are formed from fundamental ones.

 

Question 3. Derive the relation for the conversion of one unit system to another using the dimensions.
Answer: Dimensional analysis can be used to convert a physical quantity's value from one system of units to another. The measurement of a physical quantity (Q) is always expressed as the product of a numerical value (n) and its unit (u), so \( Q = nu \).
If the unit of a physical quantity in the first system is \( u_1 \) and its numerical value is \( n_1 \), then \( Q = n_1 u_1 \) ...... (1)
Similarly, in another system, if the unit is \( u_2 \) and its numerical value is \( n_2 \), then \( Q = n_2 u_2 \) ...... (2)
From equations (1) and (2), we get \( n_1 u_1 = n_2 u_2 \) ...... (3)
If \( a, b, c \) are the dimensions of a physical quantity in terms of mass (M), length (L), and time (T), then the equation becomes:
\( n_1 \left[\mathrm{M}_{1}^{a} \mathrm{L}_{1}^{b} \mathrm{T}_{1}^{c}\right] = n_2 \left[\mathrm{M}_{2}^{a} \mathrm{L}_{2}^{b} \mathrm{T}_{2}^{c}\right] \)
Here, \( M_1, L_1, T_1 \) and \( M_2, L_2, T_2 \) are the units of mass, length, and time in the two different systems. We can then find \( n_2 \) as:
\( n_2 = n_1 \left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]^{a} \left[\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}\right]^{b} \left[\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}\right]^{c} \)
This equation allows us to calculate the value of a physical quantity in a new system if its value in the first system is known. For example, common unit systems include the C.G.S. System (centimeters, grams, seconds), M.K.S. System (meters, kilograms, seconds), and F.P.S. System (foot, pound, seconds).
In simple words: To change a measurement from one unit system to another, we use a formula that compares the old and new units for mass, length, and time. This helps us find the new numerical value.

🎯 Exam Tip: Clearly state the principle \( n_1 u_1 = n_2 u_2 \) and the derived formula for \( n_2 \). Ensure you explain each variable in the formula.

 

Question 5. What is meant by errors of measurement? How do errors vary with a combination of errors? Explain.
Answer: Measurement errors refer to the difference between the measured value of a physical quantity and its true or actual value. All measurements involve some degree of error because perfect measurement is impossible. Even with our best efforts, the measured value will always be slightly different from the real value. Error is quantity = (True value – measurement value) of the quantity. These errors can be classified into different types, such as instrumental error, natural error, and observational error.

Absolute Error, Relative Error and Percentage Error.

1. Absolute error: This is the size of the difference between the true value and the measured value of a quantity. If a quantity is measured n times as \( a_1, a_2, a_3 \ldots a_n \), the arithmetic mean \( a_m = \frac{a_1+a_2+a_3+\dots+a_n}{n} \) is usually taken as the true value. The absolute errors for each measurement are \( \Delta a_1 = a_m - a_1 \), \( \Delta a_2 = a_m - a_2 \), and so on, up to \( \Delta a_n = a_m - a_n \). Absolute errors can be positive or negative.
2. Mean absolute error: This is the average of the magnitudes (absolute values) of the absolute errors from all measurements. It is represented by \( \Delta \overline{a} \). Thus, \( \Delta \overline{a} = \frac{|\Delta a_1| + |\Delta a_2| + \dots + |\Delta a_n|}{n} \). The final measurement result is often written as \( a = a_m \pm \Delta \overline{a} \), meaning the true value lies between \( (a_m - \Delta \overline{a}) \) and \( (a_m + \Delta \overline{a}) \).
4. Percentage error: When the relative error (fractional error) is expressed as a percentage, it is called percentage error. It is calculated as \( \text{percentage error} = \frac{\Delta \overline{a}}{a_m} \times 100 \% \).
In simple words: Measurement errors are how much your measured value is different from the true value. Errors can be absolute (the direct difference), mean absolute (average of these differences), or percentage (the error as a part of 100).

🎯 Exam Tip: Define measurement error clearly. Then, explain absolute error, mean absolute error, and percentage error, including their formulas and how they are calculated from multiple measurements.

 

Question 6. What are the rules to determine the significant figures? Explain each rule by giving examples.
Answer: Significant figures tell us how accurate a physical measurement is. More significant figures mean a more accurate measurement. Here are the rules for determining significant figures:
1. All non-zero digits are significant. For example, 243.48 has five significant figures.
2. Any zeros that appear between two non-zero digits are significant. For example, 46.0009 has six significant figures.
3. Zeros to the right of a decimal point and to the left of a non-zero digit are not significant. For example, 0.00678 has three significant figures (the '6', '7', '8'). The single zero usually placed before the decimal point (e.g., 0.00678) is also not significant.
4. (a) Zeros to the right of a decimal point are significant if they are not followed by a non-zero digit. For example, 30.00 has four significant figures.
(b) Zeros to the right of the last non-zero digit after the decimal point are significant. For example, 0.054300 has five significant figures.
5. (a) Zeros to the right of the last (rightmost) non-zero digit are not significant unless they come from a measurement. For example, 3030 typically has three significant figures (3, 0, 3).
(b) However, if the distance between two objects is measured as 3030 m, then 3030 m contains four significant figures (2, 3, 0, and 8 from the example text, which seems to imply the last zero is significant if measured).
In simple words: Significant figures show how precise a number is. Rules determine which digits count, like all non-zero digits, zeros between non-zeros, and zeros after a decimal point in some cases.

🎯 Exam Tip: Memorize each rule and its corresponding example to accurately apply significant figure determination in calculations and analysis.

 

Question 7. What are the rules for rounding off the numbers? Explain each rule with suitable examples.
Answer: Rounding off a number helps to express its value with a specific number of significant figures. Here are the rules:
1. If the digit to be dropped is less than 5, the preceding digit remains unchanged. For example, 1.24 is rounded off to 1.2.
2. If the digit to be dropped is greater than 5, the preceding digit is increased by 1. For example, 19.48 is rounded off to 19.5.
3. If the digit to be dropped is 5 and the preceding digit is even, it remains unchanged. For example, 1.25 is rounded off to 1.2.
4. If the digit to be dropped is 5 and the preceding digit is odd, it is increased by 1. For example, 3.35 is rounded off to 3.4.

Example 18. Let's round off 92.810576 to:

Example 19. Let's round off 0.0046753 to:

In simple words: To round numbers, if the digit to drop is less than 5, keep the last digit as is. If it's more than 5, increase the last digit by one. If it's exactly 5, keep it even if the last digit is even, or make it even if it's odd.

🎯 Exam Tip: Practice each rounding rule with various numbers, especially those ending in 5, to master how the preceding digit changes based on whether it's odd or even.

 

Question 8. How is the correctness of a formula or a relation is checked using the dimensional equations? Explain with examples.
Answer: Dimensional analysis can be used to check the correctness of a physical formula or relation. The principle of homogeneity of dimensions states that a physical equation is dimensionally correct if the dimensions of all the terms on both sides of the equation are the same. If the dimensions do not match, the equation is incorrect. However, a dimensionally correct equation is not necessarily physically correct. The method works as follows: you find the dimensional formula for each term in the equation and verify if they are consistent.
For example, to check the formula for force: \( n_1 \left[\mathrm{M}_{1}^{1} \mathrm{L}_{1}^{1} \mathrm{T}_{1}^{-2}\right] = n_2 \left[\mathrm{M}_{2}^{1} \mathrm{L}_{2}^{1} \mathrm{T}_{2}^{-2}\right] \).
The dimensional formula for force is \( \left[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}\right] \). If both sides of an equation have this same dimensional formula, then the equation is dimensionally correct. For instance, both newton and dyne are units of force.
\( n_2 = n_1 \left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]^{a} \left[\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}\right]^{b} \left[\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}\right]^{c} \)
If \( a=1, b=1, c=-2 \) for force, and converting from N (MKS) to dyne (CGS):
\( n_2 = 1 \left[\frac{1 \text{ kg}}{1 \text{ g}}\right]^{1} \left[\frac{1 \text{ m}}{1 \text{ cm}}\right]^{1} \left[\frac{1 \text{ s}}{1 \text{ s}}\right]^{-2} \)
\( \implies n_2 = 1 \left[\frac{1000 \text{ g}}{1 \text{ g}}\right] \left[\frac{100 \text{ cm}}{1 \text{ cm}}\right] \left[1\right]^{-2} \)
\( \implies n_2 = 1 \times 1000 \times 100 \times 1 = 10^5 \)
So, \( 1 \text{ N} = 10^5 \text{ Dyne} \). This shows how dimensional analysis helps convert units and check relationships.
In simple words: Dimensional analysis checks if a formula is correct by making sure all parts of the equation have the same basic measurements (like mass, length, time). If they don't match, the formula is wrong.

🎯 Exam Tip: State the principle of homogeneity clearly. Use an example like converting Newton to Dyne to demonstrate how dimensional analysis checks the relation and converts units. Be sure to show the dimensional formula for force correctly.

 

Question 9. Write an essay on the importance of Physics.
Answer: Physics is a fundamental science that studies matter, energy, and their interactions. It is all around us, influencing everyday life in ways we might not even realize. The word "Physics" comes from the Greek word "Fusis," meaning "knowledge of nature." This field aims to analyze and understand all natural phenomena in the universe.
Physics covers a vast range of topics, from the smallest particles like atoms to enormous structures like galaxies. It helps us understand why things fall, how light travels, why the sky is blue, and how technologies like computers and phones work. Its scope includes classical physics (like laws of motion and gravity) and modern physics (like quantum mechanics and relativity).

For your information

1. Macroscopic physics: This branch deals with large-scale events, such as astronomical phenomena and Earth-related incidents. Classical mechanics principles are key here.
2. Microscopic physics: This branch focuses on very small things, like molecular, atomic, and nuclear events. Quantum mechanics is studied to understand microscopic physics.

Branches of Physics

Physics is broadly classified into two parts: Classical physics and Modern physics. Classical physics was well-developed before the 20th century and mainly deals with the laws of motion and gravitation (Isaac Newton), kinetic theory, and thermodynamics (James Clerk Maxwell). In classical physics, matter and energy are seen as separate things.
Classical mechanics, Optics, Acoustics, and Electromagnetics are traditional branches of classical physics.

Mechanics: This is the study of how objects move and how forces affect them. It helps us understand the world by explaining why things move. Mechanics can be divided into Relativistic Mechanics, Quantum Mechanics, Rigid body Mechanics, Deformable body Mechanics, and Fluid Mechanics.

Thermodynamics: This branch studies heat and its connection to energy and work. It also looks at how heat is transferred, how efficient heat engines and refrigerators are, and how temperature, internal energy, and entropy change.

Electromagnetism: This involves studying electricity, magnetism, and electromagnetic waves. It deals with the electromagnetic force, which is a type of interaction between charged particles, and how it creates electric and magnetic fields and light.

Acoustics: This comes from the Greek word 'akouen' (to hear). It is the branch of physics that studies how sound is made, travels, is heard, and can be controlled. It also covers the effects of sound in different materials like gases, liquids, and solids.

Optics: This branch studies different phenomena related to light and optical tools like microscopes and telescopes. It includes topics like how images are formed by lenses and mirrors.

In simple words: Physics is super important because it helps us understand everything around us, from tiny atoms to huge galaxies, and how energy works. It's behind all technology and splits into old (classical) and new (modern) ideas, with branches like mechanics (how things move) and optics (how light works).

🎯 Exam Tip: Structure your essay with an introduction, discuss the broad scope of physics (macro and micro), and then detail several key branches of classical and modern physics, giving a brief explanation for each.

 

Question 10. Explain the parallax method for measuring large distances. With the help of it, explain the method to determine the shape of a heavenly body.
Answer: The parallax method is used to measure very large distances, like the distance to a planet or a star. When you look at an object first with one eye and then with the other, you see a slight shift in its position against the background. This shift is called parallax.
(i) **Parallax Method:** To measure large distances, we observe the object from two different points on Earth. These two points act like your two eyes. The angle formed at the object by these two observation points is called the parallax angle.
(ii) **Measuring the Diameter of the Moon:** We can use a similar idea to find the diameter of the Moon. We observe the Moon through a telescope from a point (say, F) on Earth. We measure the angle \( \theta \) that the Moon's diameter (PQ) makes at our observation point F. This angle \( \theta \) is called the angular diameter of the Moon. If 'd' is the distance from Earth to the Moon, then the angular diameter \( \theta \) can be given by the formula:
\[ \theta = \frac{\text{Length of the arc}}{\text{radius}} \]
So, for the Moon, if 'D' is its actual diameter, then:
\( \theta = \frac{D}{d} \)
From this, we can find the diameter of the Moon:
\( D = \theta d \)
This way, by knowing the distance 'd' and measuring the angular diameter \( \theta \), we can calculate the actual diameter 'D' of the Moon.
In simple words: The parallax method helps us measure how far away big things like stars or planets are by looking at them from two different places. We can also use this idea to find out how big a faraway object like the Moon actually is, if we know how far it is from us.

🎯 Exam Tip: Remember that parallax is the apparent shift in position. For astronomical bodies, the baseline (distance between observation points) is crucial for accurate measurements. Angular diameter is key for finding the actual size of distant objects.

RBSE Class 11 Physics Chapter 1 Numerical Problems

 

Question 1. The velocity of a body is \( v = A + Bt \), then what are the dimensions of A and B ?
Answer: The equation given is \( v = A + Bt \). According to the principle of homogeneity, all terms in an equation must have the same dimensions. Since 'v' is velocity, its dimension is \( [\mathrm{L}^{1} \mathrm{T}^{-1}] \).
Therefore, the dimensions of 'A' must be equal to the dimensions of 'v'.
So, Dimension of \( A = [\mathrm{M}^{0} \mathrm{L}^{1} \mathrm{T}^{-1}] \).
Similarly, the dimensions of the term 'Bt' must also be equal to the dimensions of 'v'.
Dimension of \( Bt = [\mathrm{L}^{1} \mathrm{T}^{-1}] \)
Since 't' is time, its dimension is \( [\mathrm{T}^{1}] \).
So, Dimension of \( B \times [\mathrm{T}^{1}] = [\mathrm{L}^{1} \mathrm{T}^{-1}] \)
This means, Dimension of \( B = \frac{[\mathrm{L}^{1} \mathrm{T}^{-1}]}{[\mathrm{T}^{1}]} = [\mathrm{L}^{1} \mathrm{T}^{-2}] \)
Therefore, Dimension of \( B = [\mathrm{M}^{0} \mathrm{L}^{1} \mathrm{T}^{-2}] \).
In simple words: For a physics equation to be correct, every part added together must measure the same type of thing. So, if velocity is length per time, then A must also be length per time, and B times time must also be length per time.

🎯 Exam Tip: Always apply the principle of homogeneity when finding dimensions of constants in an equation. Dimensions of terms added or subtracted must be identical.

 

Question 2. In relation, \( F = a + bx \), where F is the force, x is the distance. Calculate the dimensions of a and b.
Answer: The given relation is \( F = a + bx \). Here, 'F' is force, and its dimensional formula is \( [\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}] \). 'x' is distance, with dimensional formula \( [\mathrm{L}^{1}] \).
According to the principle of homogeneity, each term in the sum must have the same dimensions as the force 'F'.
First, for 'a':
Dimension of \( a = \) Dimension of \( F \)
So, Dimension of \( a = [\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}] \).
Next, for 'bx':
Dimension of \( bx = \) Dimension of \( F \)
Dimension of \( b \times \) Dimension of \( x = \) Dimension of \( F \)
Dimension of \( b \times [\mathrm{L}^{1}] = [\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}] \)
Therefore, Dimension of \( b = \frac{[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}]}{[\mathrm{L}^{1}]} = [\mathrm{M}^{1} \mathrm{L}^{0} \mathrm{T}^{-2}] \).
In simple words: Since force is made of mass, length, and time in a specific way, 'a' must have the same make-up as force. For 'b', when it's multiplied by distance, the result must also have the same make-up as force.

🎯 Exam Tip: When using dimensional analysis, clearly state the principle of homogeneity and ensure each term's dimensions match the main quantity of the equation.

 

Question 3. In the gas equation, \( \left(P+\frac{a}{V^{2}}\right)(V-b)=RT \), where P is pressure, V is volume, R is gas constant and T is temperature. Calculate the dimensional formula of a and b.
Answer: The given gas equation is \( \left(P+\frac{a}{V^{2}}\right)(V-b)=RT \).
From the principle of homogeneity, terms added or subtracted must have the same dimensions.
First, for 'b':
Since 'b' is subtracted from 'V' (volume), the dimension of 'b' must be the same as the dimension of 'V'.
Dimension of \( V = [\mathrm{L}^{3}] \)
So, Dimension of \( b = [\mathrm{M}^{0} \mathrm{L}^{3} \mathrm{T}^{0}] \).
Next, for 'a':
Since \( \frac{a}{V^{2}} \) is added to 'P' (pressure), the dimension of \( \frac{a}{V^{2}} \) must be the same as the dimension of 'P'.
Dimension of Pressure \( P = \frac{\text{Force}}{\text{Area}} = \frac{[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}]}{[\mathrm{L}^{2}]} = [\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{T}^{-2}] \).
Dimension of \( V^{2} = ([\mathrm{L}^{3}])^{2} = [\mathrm{L}^{6}] \).
So, Dimension of \( \frac{a}{V^{2}} = \) Dimension of \( P \)
Dimension of \( a = \) Dimension of \( P \times \) Dimension of \( V^{2} \)
Dimension of \( a = [\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{T}^{-2}] \times [\mathrm{L}^{6}] \)
Dimension of \( a = [\mathrm{M}^{1} \mathrm{L}^{5} \mathrm{T}^{-2}] \).
In simple words: When things are added or taken away from each other in an equation, they must be of the same type. So, 'b' must be a volume, and 'a' divided by volume squared must be a pressure.

🎯 Exam Tip: Remember to use the dimensional formula for each physical quantity (like pressure and volume) correctly. Pay attention to powers when calculating dimensions of derived terms.

 

Question 4. If the value of universal gravitational constant G is \( 6.67 \times 10^{-11} \mathrm{N} \mathrm{m}^{2} / \mathrm{kg}^{2} \) in M.K.S. system, then determine its value in C.G.S. system using the dimensions.
Answer: The universal gravitational constant G in M.K.S. system is \( n_1 = 6.67 \times 10^{-11} \mathrm{N} \mathrm{m}^{2} / \mathrm{kg}^{2} \). We want to find its value \( n_2 \) in the C.G.S. system (dyne \( \mathrm{cm}^{2} \mathrm{g}^{-2} \)).
First, let's find the dimensional formula of G.
From Newton's law of gravitation, \( F = \frac{G M_1 M_2}{r^2} \).
So, \( G = \frac{F r^2}{M_1 M_2} \).
Dimensions of \( F = [\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}] \).
Dimensions of \( r^2 = [\mathrm{L}^{2}] \).
Dimensions of \( M_1 M_2 = [\mathrm{M}^{2}] \).
So, Dimensions of \( G = \frac{[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}] [\mathrm{L}^{2}]}{[\mathrm{M}^{2}]} = [\mathrm{M}^{-1} \mathrm{L}^{3} \mathrm{T}^{-2}] \).
Now, we use the conversion formula:
\( n_2 = n_1 \left[ \frac{\mathrm{M}_1}{\mathrm{M}_2} \right]^{a} \left[ \frac{\mathrm{L}_1}{\mathrm{L}_2} \right]^{b} \left[ \frac{\mathrm{T}_1}{\mathrm{T}_2} \right]^{c} \)
Here, System 1 is M.K.S. (N, m, kg, s) and System 2 is C.G.S. (dyne, cm, g, s).
\( n_1 = 6.67 \times 10^{-11} \).
From the dimensional formula \( [\mathrm{M}^{-1} \mathrm{L}^{3} \mathrm{T}^{-2}] \), we have \( a = -1, b = 3, c = -2 \).
\( n_2 = 6.67 \times 10^{-11} \left[ \frac{\mathrm{kg}}{\mathrm{g}} \right]^{-1} \left[ \frac{\mathrm{m}}{\mathrm{cm}} \right]^{3} \left[ \frac{\mathrm{s}}{\mathrm{s}} \right]^{-2} \)
We know that \( 1 \mathrm{kg} = 1000 \mathrm{g} \) and \( 1 \mathrm{m} = 100 \mathrm{cm} \).
\( n_2 = 6.67 \times 10^{-11} \left[ \frac{1000 \mathrm{g}}{\mathrm{g}} \right]^{-1} \left[ \frac{100 \mathrm{cm}}{\mathrm{cm}} \right]^{3} \left[ 1 \right]^{-2} \)
\( n_2 = 6.67 \times 10^{-11} [1000]^{-1} \times [100]^{3} \times 1 \)
\( n_2 = 6.67 \times 10^{-11} \times \frac{1}{1000} \times (100 \times 100 \times 100) \)
\( n_2 = 6.67 \times 10^{-11} \times \frac{1}{10^3} \times (10^2)^3 \)
\( n_2 = 6.67 \times 10^{-11} \times 10^{-3} \times 10^{6} \)
\( n_2 = 6.67 \times 10^{-11 - 3 + 6} \)
\( n_2 = 6.67 \times 10^{-8} \)
So, in C.G.S. system, \( G = 6.67 \times 10^{-8} \text{ dyne } \mathrm{cm}^{2} \mathrm{g}^{-2} \).
In simple words: To change a value from one unit system (like M.K.S.) to another (like C.G.S.), we use its dimensional formula. We convert kilograms to grams and meters to centimeters while keeping the powers from the formula. This calculation gives us the new value in the C.G.S. system.

🎯 Exam Tip: Clearly write down the dimensional formula, identify \( n_1, a, b, c \), and meticulously convert each unit using the correct factors and powers. Common errors include sign mistakes in powers or incorrect conversion factors.

 

Question 5. The density of Mercury is \( 13.6 \text{ g.cm}^{-3} \). Calculate its value in M.K.S. system using the dimensions.
Answer: The density of Mercury in C.G.S. system is \( n_1 = 13.6 \text{ g.cm}^{-3} \). We want to find its value \( n_2 \) in the M.K.S. system (kg.\( \mathrm{m}^{-3} \)).
The dimensional formula of density is \( [\mathrm{M}^{1} \mathrm{L}^{-3} \mathrm{T}^{0}] \).
So, \( a = 1, b = -3, c = 0 \).
We use the conversion formula:
\( n_2 = n_1 \left[ \frac{\mathrm{M}_1}{\mathrm{M}_2} \right]^{a} \left[ \frac{\mathrm{L}_1}{\mathrm{L}_2} \right]^{b} \left[ \frac{\mathrm{T}_1}{\mathrm{T}_2} \right]^{c} \)
Here, System 1 is C.G.S. (g, cm) and System 2 is M.K.S. (kg, m).
\( n_1 = 13.6 \).
\( n_2 = 13.6 \left[ \frac{\mathrm{g}}{\mathrm{kg}} \right]^{1} \left[ \frac{\mathrm{cm}}{\mathrm{m}} \right]^{-3} \left[ \frac{\mathrm{s}}{\mathrm{s}} \right]^{0} \)
We know that \( 1 \mathrm{g} = \frac{1}{1000} \mathrm{kg} \) and \( 1 \mathrm{cm} = \frac{1}{100} \mathrm{m} \).
\( n_2 = 13.6 \left[ \frac{1 \mathrm{g}}{1000 \mathrm{g}} \right]^{1} \left[ \frac{1 \mathrm{cm}}{100 \mathrm{cm}} \right]^{-3} \times 1 \)
\( n_2 = 13.6 \times \left( \frac{1}{1000} \right)^{1} \times \left( \frac{1}{100} \right)^{-3} \)
\( n_2 = 13.6 \times \frac{1}{1000} \times (100)^{3} \)
\( n_2 = 13.6 \times \frac{1}{1000} \times (100 \times 100 \times 100) \)
\( n_2 = 13.6 \times \frac{1}{10^3} \times (10^2)^3 \)
\( n_2 = 13.6 \times 10^{-3} \times 10^{6} \)
\( n_2 = 13.6 \times 10^{3} \)
So, the density of Mercury in M.K.S. system is \( 13.6 \times 10^{3} \text{ kg.m}^{-3} \).
In simple words: To change density from grams per cubic centimeter to kilograms per cubic meter, we use a formula that helps convert the units by considering their powers. This gives us a larger number because kilograms are much heavier than grams and cubic meters are much larger than cubic centimeters.

🎯 Exam Tip: Be careful with the powers of 10 when converting between units. A negative power in dimensions means the unit is in the denominator, and its conversion factor needs to be inverted.

 

Question 6. Check the correctness of the relation: \( Y=\frac{MgL}{\pi r^2 l} \) using the dimensional analysis.
Answer: The given relation is \( Y=\frac{M g L}{\pi r^{2} l} \). We need to check its correctness using dimensional analysis. Here, 'Y' is Young's Modulus, 'M' is mass, 'g' is acceleration due to gravity, 'L' and 'l' are lengths, and 'r' is radius.
The dimensional formula of Young's Modulus (Y) is \( [\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{T}^{-2}] \). (This is the dimension of L.H.S.)
Now, let's find the dimensional formula of the R.H.S.:
Dimensions of \( M = [\mathrm{M}^{1}] \)
Dimensions of \( g = [\mathrm{L}^{1} \mathrm{T}^{-2}] \)
Dimensions of \( L = [\mathrm{L}^{1}] \)
Dimensions of \( \pi \) (a constant) are \( [\mathrm{M}^{0} \mathrm{L}^{0} \mathrm{T}^{0}] \)
Dimensions of \( r^2 = [\mathrm{L}^{2}] \)
Dimensions of \( l = [\mathrm{L}^{1}] \)
So, Dimensions of R.H.S. \( = \frac{[\mathrm{M}^{1}] [\mathrm{L}^{1} \mathrm{T}^{-2}] [\mathrm{L}^{1}]}{[\mathrm{L}^{2}] [\mathrm{L}^{1}]} \)
\( = \frac{[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}]}{[\mathrm{L}^{3}]} \)
\( = [\mathrm{M}^{1} \mathrm{L}^{2-3} \mathrm{T}^{-2}] \)
\( = [\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{T}^{-2}] \).
Since the dimensions of both the Left Hand Side (L.H.S.) and the Right Hand Side (R.H.S.) are the same \( [\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{T}^{-2}] \), the given relation is dimensionally correct.
In simple words: To see if a physics formula is correct, we check if both sides have the same fundamental building blocks like mass, length, and time. If they match, the formula is likely correct in terms of units.

🎯 Exam Tip: Dimensional analysis can only check the consistency of units, not the numerical constants. Always list the dimensions of all quantities involved clearly.

 

Question 7. To check the correctness of the relation : \( \frac{1}{2} m v=m g h \) using the dimensional analysis.
Answer: The given relation is \( \frac{1}{2} m v=m g h \). We need to check its correctness using dimensional analysis. Here, 'm' is mass, 'v' is velocity, 'g' is acceleration due to gravity, and 'h' is height.
Let's find the dimensional formula of the Left Hand Side (L.H.S.):
L.H.S. \( = \frac{1}{2} m v \)
Dimensions of \( m = [\mathrm{M}^{1}] \)
Dimensions of \( v = [\mathrm{L}^{1} \mathrm{T}^{-1}] \)
(The constant \( \frac{1}{2} \) has no dimensions.)
So, Dimensions of L.H.S. \( = [\mathrm{M}^{1}] [\mathrm{L}^{1} \mathrm{T}^{-1}] = [\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-1}] \).
Now, let's find the dimensional formula of the Right Hand Side (R.H.S.):
R.H.S. \( = m g h \)
Dimensions of \( m = [\mathrm{M}^{1}] \)
Dimensions of \( g = [\mathrm{L}^{1} \mathrm{T}^{-2}] \)
Dimensions of \( h = [\mathrm{L}^{1}] \)
So, Dimensions of R.H.S. \( = [\mathrm{M}^{1}] [\mathrm{L}^{1} \mathrm{T}^{-2}] [\mathrm{L}^{1}] \)
\( = [\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}] \).
Since the dimensions of the L.H.S. \( [\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-1}] \) are not equal to the dimensions of the R.H.S. \( [\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}] \), the given relation \( \frac{1}{2} m v=m g h \) is dimensionally incorrect.
In simple words: The equation tries to say that half of mass times velocity equals mass times gravity times height. But when we check the basic units, the left side is like momentum (mass times speed) and the right side is like energy (mass times gravity times height). Since these are different types of physical quantities, the equation cannot be correct.

🎯 Exam Tip: Remember that \( \frac{1}{2}mv \) (momentum) is distinct from \( \frac{1}{2}mv^2 \) (kinetic energy). Dimensional analysis is a powerful tool to spot errors in physical equations quickly.

 

Question 7. The radius of the test-tube (r), height of the liquid column in tube (h), density of the liquid filled (d) and the acceleration due to gravity (g). Establish a relation for surface tension using the method of dimensions.
Answer: We need to establish a relation for surface tension (T) based on the radius (r), height (h), density (d), and acceleration due to gravity (g). We will use the method of dimensions.
Let's assume that surface tension T depends on these quantities as:
\( T \propto r^x h^a d^b g^c \)
So, \( T = K r^x h^a d^b g^c \), where K is a dimensionless constant. For simplicity, we can consider r as part of the constant or a general length term. The problem statement itself mentions 'r' separately, so let's keep it in the power 'x'. The source derivation used `r^1` specifically, let's follow that and assume `x=1`. The source then used `h^a d^b g^c`, essentially making `r` a separate proportionality. Let's adapt to match the source's subsequent algebra.
Let's assume \( T \propto h^a d^b g^c \), with 'r' being a general length scale implicitly handled by constant K. For the purpose of finding powers, the method remains similar.
Let's follow the source's algebraic setup: \( T = K r^1 h^a d^b g^c \). (The source states `T \propto r^1`, `T \propto h^a`, `T \propto d^b`, `T \propto g^c`, and then combines them as `T = K r^1 h^a d^b g^c`).
The dimensional formula for each quantity is:
Surface Tension \( T = \frac{\text{Force}}{\text{Length}} = \frac{[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}]}{[\mathrm{L}^{1}]} = [\mathrm{M}^{1} \mathrm{L}^{0} \mathrm{T}^{-2}] \). (This is the L.H.S.)
Radius \( r = [\mathrm{L}^{1}] \).
Height \( h = [\mathrm{L}^{1}] \).
Density \( d = \frac{\text{Mass}}{\text{Volume}} = \frac{[\mathrm{M}^{1}]}{[\mathrm{L}^{3}]} = [\mathrm{M}^{1} \mathrm{L}^{-3} \mathrm{T}^{0}] \).
Acceleration due to gravity \( g = [\mathrm{L}^{1} \mathrm{T}^{-2}] \).
Now, let's write the dimensional formula for the R.H.S. of \( T = K r^1 h^a d^b g^c \):
Dimensions of R.H.S. \( = [\mathrm{L}^{1}] [\mathrm{L}^{1}]^a [\mathrm{M}^{1} \mathrm{L}^{-3} \mathrm{T}^{0}]^b [\mathrm{L}^{1} \mathrm{T}^{-2}]^c \)
\( = [\mathrm{L}^{1}] [\mathrm{L}^a] [\mathrm{M}^b \mathrm{L}^{-3b}] [\mathrm{L}^c \mathrm{T}^{-2c}] \)
Combining the powers for M, L, and T:
\( = [\mathrm{M}^b \mathrm{L}^{1+a-3b+c} \mathrm{T}^{-2c}] \).
For the relation to be dimensionally correct, the dimensions of L.H.S. and R.H.S. must be equal.
Comparing powers of M, L, and T:
For Mass (M): \( b = 1 \).
For Time (T): \( -2c = -2 \implies c = 1 \).
For Length (L): \( 1+a-3b+c = 0 \).
Substitute the values of \( b=1 \) and \( c=1 \) into the length equation:
\( 1 + a - 3(1) + 1 = 0 \)
\( 1 + a - 3 + 1 = 0 \)
\( a - 1 = 0 \)
\( a = 1 \).
So, the powers are \( r^1, h^1, d^1, g^1 \).
Therefore, the relation for surface tension becomes:
\( T = K r h d g \)
Based on practical results, the constant \( K = \frac{1}{2} \).
So, the final relation for surface tension is \( T = \frac{\text{rhdg}}{2} \).
In simple words: We find out how surface tension depends on radius, height, density, and gravity by matching the fundamental units of mass, length, and time on both sides of the equation. This helps us find the correct powers for each physical quantity.

🎯 Exam Tip: When using dimensional analysis to derive a relation, ensure all quantities are included with unknown powers. Carefully solve the system of equations derived from comparing dimensions of M, L, and T.