In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Q31 jee_main_2024_30_jan_morning Dimensional Analysis

Match List-I with List-II.

List-I	List-II
A. Coefficient of viscosity	I. $[M L²T⁻²]$ $[M L^2T^{-2}]$
B. Surface Tension	II. $[M L²T⁻¹]$ $[M L^2T^{-1}]$
C. Angular momentum	III. $[M L⁻¹T⁻¹]$ $[M L^{-1}T^{-1}]$
D. Rotational kinetic energy	IV. $[M L⁰T⁻²]$ $[M L^0T^{-2}]$

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

Solution

Related Formula

F = η A (dv)/(dy)

Surface Tension = (F)/(l)

$L = mvr$ $L = mvr$

K.E = (1)/(2) I ω²

Core Logic

Let us determine the dimensional formula for each quantity sequentially:

A. Coefficient of viscosity ( $η$ $\eta$): Using $F = η A (dv)/(dy)$ $F = \eta A \frac{dv}{dy}$, we have:

[M L T⁻²] = η [L²] [T⁻¹]

η = [M L⁻¹ T⁻¹] \Rightarrow (III)

B. Surface Tension (S.T.):

S.T = (F)/( ) = [M L T⁻²][L] = [M L⁰ T⁻²] \Rightarrow (IV)

C. Angular momentum (L):

L = mvr = [M] [L T⁻¹] [L] = [M L² T⁻¹] \Rightarrow (II)

D. Rotational kinetic energy (K.E.):

K.E = (1)/(2) I ω² = [M L² T⁻²] \Rightarrow (I)

Step 1: Final Matching

Matching the derived dimensional formulas: A $arrow$ $\rightarrow$ III B $arrow$ $\rightarrow$ IV C $arrow$ $\rightarrow$ II D $arrow$ $\rightarrow$ I

Pattern Recognition

Kinetic energy (whether translational or rotational) always carries the dimension of Work: $[M L² T⁻²]$ $[M L^2 T^{-2}]$. Surface tension is force per unit length, dropping the $L$ $L$ term. Viscosity commonly includes $L⁻¹$ $L^{-1}$.

Chapter Mix

Class 11 Physics: Units and Measurements

Q33 jee_main_2024_31_jan_evening Errors in Measurement

The measured value of the length of a simple pendulum is

20 cm

with

2 mm

accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is

N%

. The value of

N

is:

A. $4$ $4$
B. $8$ $8$
C. $6$ $6$
D. $5$ $5$

Solution

Related Formula

T = 2π \sqrt(( )/(g)) g = (4π² )/(T²)

Core Logic

By taking logarithms and differentiating to find relative error (accuracy):

(Δ g)/(g) = (Δ )/( ) + 2(Δ T)/(T)

Step 1: Extrapolating Errors

Given values: $= 20 cm = 200 mm$ $\ell = 20 \text{ cm} = 200 \text{ mm}$ $Δ = 2 mm$ $\Delta \ell = 2 \text{ mm}$ $Ttotal = 40 s$ $T_{\text{total}} = 40 \text{ s}$ for 50 oscillations $Δ Ttotal = 1 s$ $\Delta T_{\text{total}} = 1 \text{ s}$ Note: The relative error in time period $T$ $T$ is equal to the relative error in total time $t$ $t$: $(Δ T)/(T) = (Δ t)/(t)$ .

Step 2: Substitution

(Δ g)/(g) = 0.2 cm20 cm + 2 ( 1 s40 s)

(Δ g)/(g) = (2)/(200) + (2)/(40)

(Δ g)/(g) = (1)/(100) + (5)/(100) = (6)/(100)

Step 3: Percentage Conversion

Percentage change $= (Δ g)/(g) \times 100% = (6)/(100) \times 100% = 6%$ . Thus, $N = 6$ $N = 6$.

Pattern Recognition

For pendulum gravity error, always use $%g = % + 2(%T)$ $\%g = \%\ell + 2(\%T)$. Remember that measuring 50 oscillations reduces absolute error on a single swing, but the relative error $Δ t / t$ $\Delta t / t$ remains unchanged whether you use total time or single period.

Chapter Mix

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

Q50 jee_main_2024_31_jan_evening Dimensional Analysis

Consider two physical quantities A and B related to each other as

E = (B - x²)/(At)

where E, x and t have dimensions of energy, length and time respectively. The dimension of AB is

A. $L⁻²M¹T⁰$ $\text{L}^{-2}\text{M}^{1}\text{T}^{0}$
B. $L²M⁻¹T¹$ $\text{L}^2\text{M}^{-1}\text{T}^1$
C. $L⁻²M⁻¹T¹$
D. $L⁰M⁻¹T¹$ $\text{L}^0\text{M}^{-1}\text{T}^1$

Solution

Related Formula

By the Principle of Homogeneity, terms added or subtracted must have the same dimensions: $[B] = [x²]$ $[B] = [x^2]$

Core Logic

Known dimensional formulas: Length $x \to [L]$ $x \to [L]$ Energy $E \to [ML²T⁻²]$ $E \to [ML^2T^{-2}]$ Time $t \to [T]$ $t \to [T]$

Step 1: Dimension of B

Since $x²$ $x^2$ is subtracted from $B$ $B$:

[B] = [x²] = [L²]

Step 2: Dimension of A

From the equation $E = (B - x²)/(At)$ $E = \frac{B - x^2}{At}$:

[A] = ([B - x²])/([E][t])

[A] = [L²][ML²T⁻²][T] = [L²][ML²T⁻¹]

[A] = [M⁻¹T¹]

Step 3: Dimension of AB

[AB] = [A] \times [B]

[AB] = [M⁻¹T¹] \times [L²]

[AB] = [L² M⁻¹ T¹]

Pattern Recognition

Identify sums/differences first to instantly isolate $B$ $B$. Once $[B]$ $[B]$ is fixed, the entire numerator is just $L²$ $L^2$. Swap out variables to isolate $[A]$ $[A]$. Combining is just standard exponent addition.

Chapter Mix

Class 11 Physics: Units and Measurements

Q jee_main_2024_31_jan_morning Errors In Measurement

If the percentage errors in measuring the length and the diameter of a wire are

0.1%

each. The percentage error in measuring its resistance will be:

A. $0.2%$ $0.2\%$
B. $0.3%$ $0.3\%$
C. $0.1%$ $0.1\%$
D. $0.144%$ $0.144\%$

Solution

Related Formula

R = (ρ L)/(A) = (ρ L)/(π ((d)/(2))²) = (4ρ L)/(π d²)

Core Logic

To find the maximum percentage error in resistance, apply logarithmic differentiation:

(Δ R)/(R) = (Δ L)/(L) + 2(Δ d)/(d)

Given percentage errors:

$(Δ L)/(L) \times 100% = 0.1%$
$(Δ d)/(d) \times 100% = 0.1%$

Step 2: Substitution

Substituting the values:

(Δ R)/(R) \times 100% = 0.1% + 2(0.1%) = 0.1% + 0.2% = 0.3%

Pattern Recognition

Resistance scales inversely with the square of the diameter. The error multiplier for diameter is 2. Sum the linear components directly: $Error = Lₑᵣᵣₒᵣ + 2 \times dₑᵣᵣₒᵣ$ .

Chapter Mix

Class 12 Physics: Current Electricity

Q41 jee_main_2024_31_jan_morning Dimensional Analysis

A force is represented by

F = ax² + bt1/2

Where

x =

distance and

t =

time. The dimensions of

b² / a

are:

A. $[ML³T⁻³]$ $[ML^3T^{-3}]$
B. $[MLT⁻²]$ $[MLT^{-2}]$
C. $[ML⁻¹T⁻¹]$ $[ML^{-1}T^{-1}]$
D. $[ML²T⁻³]$ $[ML^2T^{-3}]$

Solution

Related Formula

Principle of Homogeneity: [F] = [ax²] = [bt1/2]

Core Logic

By the principle of dimensional homogeneity, each additive term must have the same dimension as the left hand side. Dimension of force $F = [M L T⁻²]$ $F = [M L T^{-2}]$.

For the term $ax²$ $ax^2$:

[a] = ([F])/([x²]) = [M L T⁻²][L²] = [M L⁻¹ T⁻²]

For the term $bt1/2$ $bt^{1/2}$:

[b] = [F][t1/2] = [M L T⁻²][T1/2] = [M L T-5/2]

Step 2: Computing Required Ratio

We need the dimension of $(b²)/(a)$ $\frac{b^2}{a}$:

[ (b²)/(a) ] = [M L T-5/2]²[M L⁻¹ T⁻²]

[ (b²)/(a) ] = [M² L² T⁻⁵][M L⁻¹ T⁻²]

[ (b²)/(a) ] = [M²⁻¹ L2 - (-1) T-5 - (-2)]

[ (b²)/(a) ] = [M L³ T⁻³]

Chapter Mix

Class 11 Physics: Units And Measurements

In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of [MaLbTc]$\left[\mathrm{M}^{\mathrm{a}}\mathrm{L}^{\mathrm{b}}\mathrm{T}^{\mathrm{c}}\right]$ . If b = 3$b = 3$ , the value of c$c$ is

Solution & Explanation

Core Logic

Step 1: Exponent Matching

Pattern Recognition

Chapter Mix

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Final Matching

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Extrapolating Errors

Step 2: Substitution

Step 3: Percentage Conversion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Dimension of B

Step 2: Dimension of A

Step 3: Dimension of AB

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Substitution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Computing Required Ratio

Chapter Mix

More Units and Measurements Questions — jee_main_2025_28_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of $[MaLbTc]$ . If $b = 3$ $b = 3$ , the value of $c$ $c$ is