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 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 I

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 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 I

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…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Year

2026

2025

2024

Total

Questions

LIST-I	LIST-II
A. Gravitational constant	I. $[LT⁻²]$ $[LT^{-2}]$
B. Gravitational potential energy	II. $[L²T⁻²]$ $[L^2T^{-2}]$
C. Gravitational potential	III. $[ML²T⁻²]$ $[ML^2T^{-2}]$
D. Acceleration due to gravity	IV. $[M⁻¹L³T⁻²]$ $[M^{-1}L^3T^{-2}]$

LIST-I

LIST-II

A. Gravitational constant

[LT⁻²]

B. Gravitational potential energy

II.

[L²T⁻²]

C. Gravitational potential

III.

[ML²T⁻²]

D. Acceleration due to gravity

IV.

[M⁻¹L³T⁻²]

Solution & Explanation

Related Formula

Newton's Law of Gravitation:

F = G(m₁ m₂)/(r²) G = (Fr²)/(m₁ m₂)

Potential Energy:

U = mgh [Work]

Potential:

V = (W)/(m)

Acceleration:

g = VelocityTime

Core Logic

Let's perform dimensional analysis for each item:

A. Gravitational constant ( $G$ $G$):

[G] = ([F][r²])/([M²]) = [MLT⁻²][L²][M²] = [M⁻¹L³T⁻²]

Matches with IV.

B. Gravitational potential energy ( $U$ $U$):

[U] = Dimensions of Work = [ML²T⁻²]

Matches with III.

C. Gravitational potential ( $V$ $V$):

[V] = [Energy][M] = [ML²T⁻²][M] = [L²T⁻²]

Matches with II.

D. Acceleration due to gravity ( $g$ $g$):

[g] = [Acceleration] = [LT⁻²]

Matches with I.

Step 1: Match and Selection

Let's align our matches:

A $arrow$ $\rightarrow$ IV
B $arrow$ $\rightarrow$ III
C $arrow$ $\rightarrow$ II
D $arrow$ $\rightarrow$ 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 $g arrow [LT⁻²]$ $g \rightarrow [LT^{-2}]$ (D-I) and energy is $[ML²T⁻²]$ $[ML^2T^{-2}]$ (B-III). Looking at the options, only Option 1 matches this sequence immediately!

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Gravitation

More Units and Measurements Previous-Year Questions

Q30 jee_main_2026_21_jan_morning Errors in Measurement

In an experiment the values of two spring constants were measured as

k₁=(10\pm0.2) N/m

and

k₂=(20\pm0.3) N/m

. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

A. 2.67%
B. 2.33%
C. 1.33%
D. 1.67%

Solution

Related Formula

Keq = K₁ + K₂ (Parallel Combination)

Δ Keq = Δ K₁ + Δ K₂

Percentage Error = Δ KeqKeq \times 100

Core Logic

For a parallel combination of springs, the equivalent spring constant is simply the sum of individual constants.

Keq = K₁ + K₂ = 10 + 20 = 30 N/m

When quantities are added, their absolute errors are also added:

Δ Keq = Δ K₁ + Δ K₂ = 0.2 + 0.3 = 0.5 N/m

Step 1: Calculating Percentage Error

The percentage error in $K$ $K$ is:

% Error in K = (0.5)/(30) \times 100 = (5)/(3)% \approx 1.67%

Pattern Recognition

Addition operation = add absolute errors. For parallel springs, it's just plain addition. Then to get percentage, divide the absolute error sum by the nominal value sum and multiply by 100.

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

Q32 jee_main_2026_21_jan_morning Dimensional Analysis

Consider a modified Bernoulli equation.

(P+ ABt²)+ (h+Bt)+(1)/(2) ²=constant

If t has the dimension of time then the dimensions of A and B are ____, ____ respectively.

A. $[ML⁰T⁻¹] and [M⁰LT]$ $[ML^{0}T^{-1}]\text{ and }[M^{0}LT]$
B. $[ML⁰T⁻¹] and [M⁰LT⁻¹]$
C. $[ML⁰T⁻²] and [M⁰LT⁻²]$
D. $[ML⁰T⁻²] and [M⁰LT⁻¹]$

Solution

Related Formula

Principle of Homogeneity: Only quantities with same dimensions can be added/subtracted.

Core Logic

From the term $(h + Bt)$ $(h + Bt)$, $h$ $h$ (height/length) and $Bt$ $Bt$ must have the same dimension. $[h] = [Bt]$ $[h] = [Bt]$

[B] = ([h])/([t]) = ([L])/([T]) = [L T⁻¹]

So, $B = [M⁰ L T⁻¹]$ $B = [M^0 L T^{-1}]$.

Step 1: Finding Dimension of A

From the term $(P + (A)/(Bt²))$ $\left(P + \frac{A}{Bt^2}\right)$, pressure $P$ $P$ and $(A)/(Bt²)$ $\frac{A}{Bt^2}$ must have the same dimension. Dimension of Pressure $P = ForceArea = [M L⁻¹ T⁻²]$ .

[P] = [(A)/(Bt²)]

[M L⁻¹ T⁻²] = [A][L T⁻¹] \times [T²]

[M L⁻¹ T⁻²] = ([A])/([L T])

[A] = [M L⁻¹ T⁻²] \times [L T] = [M L⁰ T⁻¹]

So, $A = [M L⁰ T⁻¹]$ $A = [M L^0 T^{-1}]$.

Pattern Recognition

Apply the Principle of Homogeneity across any $(X + Y)$ $(X + Y)$ terms inside brackets. Extract one unknown, plug it into the next additive bracket group.

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids

Q36 jee_main_2026_21_jan_evening Significant Figures

Keeping the significant figures in view, the sum of the physical quantities

52.01 m

153.2 m

and

0.123 m

is :

A. $205 m$ $205 \text{ m}$
B. $205.333 m$ $205.333 \text{ m}$
C. $205.33 m$ $205.33 \text{ m}$
D. $205.3 m$ $205.3 \text{ m}$

Solution

Related Formula

Addition Rule for Significant Figures: The result cannot have more digits to the right of the decimal point than any of the original numbers.

Core Logic

Let's align the numbers and perform exact addition: $52.01$ $52.01$ (2 decimal places) $153.2$ $153.2$ (1 decimal place - least precise) $+ 0.123$ $+ 0.123$ (3 decimal places) -------- $205.333$ $205.333$

Step 1: Final Conclusion

According to the rules of significant figures for addition, the sum must be rounded to the least number of decimal places present in the given measurements. Here, $153.2$ $153.2$ has only $1$ $1$ decimal place. Rounding $205.333$ $205.333$ to one decimal place yields $205.3 m$ $205.3 \text{ m}$.

Pattern Recognition

Addition/Subtraction is bounded by decimal places. Multiplication/Division is bounded by total significant figures. In this case, 1 decimal place rules.

Chapter Mix

Class 11 Physics: Units and Measurements

Q37 jee_main_2026_21_jan_evening Errors in Measurement

A spherical body of radius r and density

σ

falls freely through a viscous liquid having density

ρ

and viscosity

η

and attains a terminal velocity

v₀

. Estimated maximum error in the quantity

η

is : (Ignore errors associated with

σ

ρ

and g, gravitational acceleration)

A. $2(Δ r)/(r)-(Δ v₀)/(v₀)$
B. $(2Δ r)/(r)+(Δ v₀)/(v₀)$
C. $2[(Δ r)/(r)+(Δ v₀)/(v₀)]$
D. $2[(Δ r)/(r)-(Δ v₀)/(v₀)]$

Solution

Related Formula

v₀ = (2)/(9)(r² g)/(η) (σ - ρ)

Core Logic

Rearranging the formula for viscosity $η$ $\eta$:

η = (2)/(9) (r² g)/(v₀) (σ - ρ)

Since $(2)/(9)$ $\frac{2}{9}$, $g$ $g$, $σ$ $\sigma$, and $ρ$ $\rho$ are constants or have negligible error, the relative error depends only on $r$ $r$ and $v₀$ $v_0$.

Step 1: Error Propagation

Taking natural logarithm on both sides:

ln(η) = ln(constant) + 2ln(r) - ln(v₀)

Differentiating for maximum possible error (adding absolute errors):

(Δ η)/(η) = 2(Δ r)/(r) + (Δ v₀)/(v₀)

Step 2: Final Conclusion

The maximum fractional error is $(2Δ r)/(r) + (Δ v₀)/(v₀)$ .

Pattern Recognition

For maximum fractional error of a derived quantity $Z = X^a Y^b$ $Z = X^a Y^b$, the formula is $Δ Z/Z = |a|Δ X/X + |b|Δ Y/Y$ . Always sum the absolute relative errors.

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids

Q44 jee_main_2026_22_january_morning Dimensional Analysis

Match the LIST-I with LIST-II

List-I	List-II
(A) Spring constant	(I) $ML²T⁻²K⁻¹$ $ML^{2}T^{-2}K^{-1}$
(B) Thermal conductivity	(II) $ML⁰T⁻²$ $ML^{0}T^{-2}$
(C) Boltzmann constant	(III) $ML²T⁻³A⁻²$ $ML^{2}T^{-3}A^{-2}$
(D) Inductive reactance	(IV) $MLT⁻³K⁻¹$ $MLT^{-3}K^{-1}$

Choose the correct answer from the options given below:

A. \text{A-II, B-I, C-IV, D-III}
B. \text{A-I, B-IV, C-II, D-III}
C. \text{A-III, B-II, C-IV, D-I}
D. \text{A-II, B-IV, C-I, D-III}

Solution

Related Formula

F = Kx, (dQ)/(dt) = -KA(dT)/(dx), E = kB T, XL = ω L

Core Logic

(A) Spring constant $K$ $K$: $[ML⁰ T⁻²]$ $[ML^0 T^{-2}]$ -> (II) (B) Thermal conductivity $K$ $K$: $[MLT⁻³ K⁻¹]$ $[MLT^{-3} K^{-1}]$ -> (IV) (C) Boltzmann constant $kB$ $k_B$: $[ML² T⁻² K⁻¹]$ $[ML^2 T^{-2} K^{-1}]$ -> (I) (D) Inductive reactance $XL$ $X_L$: $[ML² T⁻³ A⁻²]$ $[ML^2 T^{-3} A^{-2}]$ -> (III)

Therefore, matching is A-II, B-IV, C-I, D-III.

Pattern Recognition

Sees: Match List dimensional formulas. Shortcut: Derive or recall key physical constants dimensions (Spring constant, Thermal conductivity, Boltzmann constant, Inductive reactance). Check: Matches option (4). ✓

Chapter Mix

Class 11 Physics: Units and Measurements

Solution & Explanation

Related Formula

Core Logic

Step 1: Match and Selection

Pattern Recognition

Chapter Mix

Reference Study Guides

More Units and Measurements Previous-Year Questions

Solution

Related Formula

Core Logic

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Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Finding Dimension of A

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

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Core Logic

Step 1: Error Propagation

Step 2: Final Conclusion

Pattern Recognition

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Solution

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