Match the LIST-I with LIST-II LIST-I LIST-II A. Gravitational constant I. [LT-2] B. Gravitational potential energy II. [L2T-2] C. Gravitational potential III. [ML2T-2] D. Acceleration due to gravity IV. [M-1L3T-2] Choose the correct answer from the options given below:

Newton's Law of Gravitation: Potential Energy: Potential: Acceleration: Let's perform dimensional analysis for each item: 1. **A. Gravitational constant (G)**: Matches with **IV**. 2. **B. Gravitational potential energy (U)**: Matches with **III**. 3. **C. Gravitational potential (V)**: Matches with **II**. 4. **D. Acceleration due to gravity (g)**: Matches with **I**. Let's align our matches: - A arrow IV - B arrow III - C arrow II - D arrow I This sequence matches option (1). To save precious exam time on match-the-column questions, start with the easiest dimensional terms first. You know Acceleration due to gravity is g arrow [LT-2] (D-I) and energy is [ML2T-2] (B-III). Looking at the options, only Option 1 matches this sequence immediately! Class 11 Physics: Units and Measurements Class 11 Physics: Gravitation

Match the LIST-I with LIST-II <th style="border: 1px

Newton's Law of Gravitation: Potential Energy: Potential: Acceleration: Let's perform dimensional analysis for each item: 1. **A. Gravitational constant (G)**: Matches with **IV**. 2. **B. Gravitational potential energy (U)**: Matches with **III**. 3. **C. Gravitational potential (V)**: Matches with **II**. 4. **D. Acceleration due to gravity (g)**: Matches with **I**. Let's align our matches: - A arrow IV - B arrow III - C arrow II - D arrow I This sequence matches option (1). To save precious

Match the LIST-I with LIST-II LIST-I LIST-II A. Gravitational constant I. [LT-2] B. Gravitational potential energy II. [L2T-2] C. Gravitational potential III. [ML2T-2] D. Acceleration due to gravity IV. [M-1L3T-2] Choose the correct answer from the options given below:

Newton's Law of Gravitation: Potential Energy: Potential: Acceleration: Let's perform dimensional analysis for each item: 1. **A. Gravitational constant (G)**: Matches with **IV**. 2. **B. Gravitational potential energy (U)**: Matches with **III**. 3. **C. Gravitational potential (V)**: Matches with **II**. 4. **D. Acceleration due to gravity (g)**: Matches with **I**. Let's align our matches: - A arrow IV - B arrow III - C arrow II - D arrow I This sequence matches option (1). To save precious exam time on match-the-column questions, start with the easiest dimensional terms first. You know Acceleration due to gravity is g arrow [LT-2] (D-I) and energy is [ML2T-2] (B-III). Looking at the options, only Option 1 matches this sequence immediately! Class 11 Physics: Units and Measurements Class 11 Physics: Gravitation

Match the LIST-I with LIST-II <th style="border: 1px

Newton's Law of Gravitation: Potential Energy: Potential: Acceleration: Let's perform dimensional analysis for each item: 1. **A. Gravitational constant (G)**: Matches with **IV**. 2. **B. Gravitational potential energy (U)**: Matches with **III**. 3. **C. Gravitational potential (V)**: Matches with **II**. 4. **D. Acceleration due to gravity (g)**: Matches with **I**. Let's align our matches: - A arrow IV - B arrow III - C arrow II - D arrow I This sequence matches option (1). To save precious

Match the LIST-I with LIST-II
LIST-I LIST-II
A. Gravitational constant I. $[LT^-2]$ $[LT^{-2}]$
B. Gravitational potential energy II. $[L^2T^-2]$ $[L^2T^{-2}]$
C. Gravitational potential III. $[ML^2T^-2]$ $[ML^2T^{-2}]$
D. Acceleration due to gravity IV. $[M^-1L^3T^-2]$ $[M^{-1}L^3T^{-2}]$
Choose the correct answer from the options given below:

LIST-I	LIST-II
A. Gravitational constant	I. $[LT^-2]$ $[LT^{-2}]$
B. Gravitational potential energy	II. $[L^2T^-2]$ $[L^2T^{-2}]$
C. Gravitational potential	III. $[ML^2T^-2]$ $[ML^2T^{-2}]$
D. Acceleration due to gravity	IV. $[M^-1L^3T^-2]$ $[M^{-1}L^3T^{-2}]$

A. A-IV, B-III, C-II, D-I
B. A-III, B-II, C-I, D-IV
C. A-II, B-IV, C-III, D-I
D. A-I, B-III, C-IV, D-II

Solution & Explanation

### Related Formula Newton's Law of Gravitation:

F = Gfracm_1 m_2r^2 implies G = fracFr^2m_1 m_2

Potential Energy:

U = mgh quad [textWork]

Potential:

V = fracWm

Acceleration:

g = fractextVelocitytextTime

### Core Logic Let's perform dimensional analysis for each item: 1. **A. Gravitational constant (

G

)**:

[G] = frac[F][r^2][M^2] = frac[MLT^-2][L^2][M^2] = [M^-1L^3T^-2]

Matches with **IV**. 2. **B. Gravitational potential energy (

U

)**:

[U] = textDimensions of Work = [ML^2T^-2]

Matches with **III**. 3. **C. Gravitational potential (

V

)**:

[V] = frac[textEnergy][M] = frac[ML^2T^-2][M] = [L^2T^-2]

Matches with **II**. 4. **D. Acceleration due to gravity (

g

)**:

[g] = [textAcceleration] = [LT^-2]

Matches with **I**. ### Step 1: Match and Selection Let's align our matches: - A

rightarrow

IV - B

rightarrow

III - C

rightarrow

II - D

rightarrow

I This sequence matches option (1). ### Pattern Recognition To save precious exam time on match-the-column questions, start with the easiest dimensional terms first. You know Acceleration due to gravity is

g rightarrow [LT^-2]

(D-I) and energy is

[ML^2T^-2]

(B-III). Looking at the options, only Option 1 matches this sequence immediately! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Gravitation

Reference Study Guides

Units And Measurements

More Units and Measurements Previous-Year Questions

Q7 2025 Dimensional Analysis

Given a charge

q

, current

I

and permeability of vacuum

mu_0

. Which of the following quantity has the dimension of momentum?

A. $q I / mu_0$ $q I / \mu_0$
B. $q mu_0 I$ $q \mu_0 I$
C. $q^2 mu_0 I$ $q^2 \mu_0 I$
D. $q mu_0 / I$ $q \mu_0 / I$

Solution

### Related Formula Let momentum be

P

(

[P] = [M L T^-1]

). We assume:

P = q^x mu_0^y I^z

### Core Logic Let's find the dimensional formulas of the individual variables: 1. **Charge (

q

):**

[q] = [A T]

2. **Current (

I

):**

[I] = [A]

3. **Permeability of vacuum (

mu_0

):** From Biot-Savart law or force between parallel wires:

F = fracmu_0 I^2 L2pi d

[mu_0] = frac[F][I]^2 = frac[M L T^-2][A]^2 = [M L T^-2 A^-2]

### Step 1: Apply Dimensional Homogeneity Substitute these into our assumed dimensional equation:

[M L T^-1] = [A T]^x [M L T^-2 A^-2]^y [A]^z

[M L T^-1] = [M]^y [L]^y [T]^x - 2y [A]^x - 2y + z

Comparing exponents on both sides: - For

[M]

y = 1

- For

[L]

y = 1 quad text(consistent)

- For

[T]

x - 2y = -1 implies x - 2(1) = -1 implies x = 1

- For

[A]

x - 2y + z = 0 implies 1 - 2(1) + z = 0 implies z = 1

Thus,

x = 1, y = 1, z = 1

. Therefore, the required quantity is:

q^1 mu_0^1 I^1 = q mu_0 I

### Pattern Recognition Sees: Permeability, charge, and current linked to momentum. Trap: Deriving the dimensions of

mu_0

using complex magnetic formulas. Remember

[mu_0] = [textForce]/[textCurrent]^2

is the quickest way to get its dimensions. Shortcut: Since

[q] = AT

and

[I] = A

[q mu_0 I] = [A T] [M L T^-2 A^-2] [A] = [M L T^-1]

, which is exactly the dimensions of momentum. Hence, option (2) is correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Moving Charges and Magnetism

Q 2025 Dimensional Analysis

Match List-I with List-II.

beginarray|l|l| hline textbfList-I & textbfList-II \\ hline text(A) Coefficient of viscosity & text(I) [mathrmML^0mathrmT^-3] \\ hline text(B) Intensity of wave & text(II) [mathrmML^-2mathrmT^-2] \\ hline text(C) Pressure gradient & text(III) [mathrmM^-1mathrmLT^2] \\ hline text(D) Compressibility & text(IV) [mathrmML^-1mathrmT^-1] \\ hline endarray

Choose the correct answer from the options given below:

A. (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
B. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
C. (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
D. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Solution

### Related Formula Definitions of physical quantities: 1. Viscosity:

F = -eta A fracdvdx

2. Intensity:

I = fractextPowertextArea

3. Pressure gradient:

fracdPdx

4. Compressibility:

K = frac1B = fractextStraintextStress

### Core Logic Let's calculate each dimensional formula: 1. **Coefficient of viscosity (

eta

):**

[eta] = frac[F][A]left[fracdvdxright] = fractextM L T^-2textL^2 left(fractextL T^-1textLright) = textM L^-1textT^-1

Matches **(IV)**. 2. **Intensity of wave (

I

):**

[I] = frac[textPower][textArea] = fractextM L^2textT^-3textL^2 = textM T^-3 = textM L^0textT^-3

Matches **(I)**. 3. **Pressure gradient (

fracdPdx

):**

left[fracdPdxright] = frac[textPressure][textLength] = fractextM L^-1textT^-2textL = textM L^-2textT^-2

Matches **(II)**. 4. **Compressibility (

K

):** Compressibility is the inverse of Bulk Modulus:

[K] = frac1[textPressure] = frac1textM L^-1textT^-2 = textM^-1textLtextT^2

Matches **(III)**. Thus: (A)-(IV), (B)-(I), (C)-(II), (D)-(III). ### Step 1: Final Conclusion The correct option is (2). ### Pattern Recognition Target basic matching terms first: Compressibility is the reciprocal of pressure, giving

[M^-1LT^2]

. Wave intensity has units of power per unit area, giving

[MT^-3]

. This immediately isolates option (2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids Class 11 Physics: Mechanical Properties of Solids

Q2 2025 Dimensional Analysis

The equation for real gas is given by

left(P + fracaV^2right)(V - b) = RT

, where

P, V, T

and

R

are the pressure, volume, temperature and gas constant, respectively. The dimension of

ab^-2

is equivalent to that of:

A. Planck's constant
B. Compressibility
C. Strain
D. Energy density

Solution

### Related Formula By the principle of dimensional homogeneity, terms added or subtracted must have the same dimensions:

[P] = left[fracaV^2right] implies [a] = [P][V]^2

[V] = [b]

### Core Logic Let's find the dimensional formula of the quantities: - Pressure

P

[P] = textM L^-1textT^-2

- Volume

V

[V] = textL^3

Substituting these to find

[a]

and

[b]

[a] = (textM L^-1textT^-2)(textL^6) = textM L^5textT^-2

[b] = textL^3 implies [b^-2] = textL^-6

Now, compute the dimensions of

ab^-2

[ab^-2] = (textM L^5textT^-2)(textL^-6) = textM L^-1textT^-2

This matches the dimensions of pressure. Let's evaluate the options: 1. Planck's constant:

[h] = textM L^2textT^-1

2. Compressibility:

[beta] = textM^-1textLtextT^2

3. Strain: dimensionless 4. Energy density (energy per unit volume):

left[fracEVright] = fractextM L^2textT^-2textL^3 = textM L^-1textT^-2

### Step 1: Final Conclusion Therefore, the dimension of

ab^-2

is equivalent to that of Energy density. ### Pattern Recognition By writing the relation directly as

[ab^-2] = frac[a][b]^2

, and noting

[a] = [P][V]^2

and

[b] = [V]

, we get

[ab^-2] = frac[P][V]^2[V]^2 = [P]

(Pressure). Since pressure and energy density have identical dimensions, the answer is immediately Energy density. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Kinetic Theory of Gases

Q 2025 Dimensional Analysis and Constants

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. Boltzmann constant	I. $ML^2T^-1$ $ML^2T^{-1}$
B. Coefficient of viscosity	II. $MLT^-3K^-1$ $MLT^{-3}K^{-1}$
C. Planck's constant	III. $ML^2T^-2K^-1$ $ML^2T^{-2}K^{-1}$
D. Thermal conductivity	IV. $ML^-1T^-1$ $ML^{-1}T^{-1}$

Choose the correct answer from the options given below :

A. A-III, B-IV, C-I, D-II
B. A-II, B-III, C-IV, D-I
C. A-III, B-II, C-I, D-IV
D. A-III, B-IV, C-II, D-I

Solution

### Related Formula Formulas to find dimensional formulas: - Boltzmann constant:

k_B = fractextEnergytextTemperature

- Coefficient of viscosity:

eta = fracFA fracdvdx

- Planck's constant:

h = fracEnu

- Thermal conductivity:

fracdQdt = K A fracdTdx Rightarrow K = fractextHeat flow cdot textthicknesstextArea cdot textTemperature difference

### Core Logic Evaluate each constant individually: ### Step 1: Dimensions of Boltzmann constant (

k_B

)

[k_B] = frac[ML^2T^-2][K] = [ML^2T^-2K^-1] quad Rightarrow textMatches III

### Step 2: Dimensions of Coefficient of viscosity (

eta

)

[eta] = frac[MLT^-2][L^2] [T^-1] = [ML^-1T^-1] quad Rightarrow textMatches IV

### Step 3: Dimensions of Planck's constant (

h

)

[h] = frac[ML^2T^-2][T^-1] = [ML^2T^-1] quad Rightarrow textMatches I

### Step 4: Dimensions of Thermal conductivity (

K

)

[K] = frac[ML^2T^-3] [L][L^2] [K] = [MLT^-3K^-1] quad Rightarrow textMatches II

This sequence yields A-III, B-IV, C-I, D-II, matching Option (1). ### Pattern Recognition To solve matching sets efficiently, search for the most recognizable dimensions first. Planck's constant

h

(

ML^2T^-1

) and viscosity coefficient

eta

(

ML^-1T^-1

) are highly unique and usually resolve the options instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

Q25 2025 Error Analysis

A physical quantity C is related to four other quantities p, q, r and s as follows

C = fracpq^2r^3sqrts

The percentage errors in the measurement of p, q, r and s are

1\% , 2\% , 3\%

and

2\%

respectively. The percentage error in the measurement of C will be ________ \%.

Numerical Answer. Answer: 15 to 15

Solution

### Related Formula For a physical quantity defined by algebraic powers

C = fracp^a q^br^c s^d

, the maximum fractional error is calculated by summing absolute scaled fractional errors:

fracDelta CC = a fracDelta pp + b fracDelta qq + c fracDelta rr + d fracDelta ss

Expressed as percentages:

\% text error in C = a(\% text error in p) + b(\% text error in q) + c(\% text error in r) + d(\% text error in s)

### Core Logic Given expression:

C = p^1 q^2 r^-3 s^-1/2

Max fractional error equation:

fracDelta CC = 1 left(fracDelta ppright) + 2 left(fracDelta qqright) + 3 left(fracDelta rrright) + frac12 left(fracDelta ssright)

### Step 1: Calculate the total percentage error Substitute the individual percentage errors: - Error in

p = 1\%

- Error in

q = 2\%

- Error in

r = 3\%

- Error in

s = 2\%

\% text error in C = 1(1\%) + 2(2\%) + 3(3\%) + frac12(2\%)

\% text error in C = 1\% + 4\% + 9\% + 1\% = 15\%

The total percentage error in

C

15\%

. ### Pattern Recognition In error propagation, individual errors always combine constructively to produce the maximum possible uncertainty limit. Hence, negative powers (like division by

r^3

s^1/2

) are integrated using positive coefficients during maximum absolute error summation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

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