A tiny metallic rectangular sheet has length and breadth of 5 ~mm and 2.5 ~mm , respectively. Using a specially…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

First, find the least count of the measurement tool:

Least Count = PitchNumber of circular scale divisions = 0.75 ~mm15 = 0.05 ~mm

Least count calculation tracking diagram for Q21

The area of the rectangular metallic sheet is calculated as:

A = L \cdot W

Expressing the absolute error via fractional configuration parts:

dAA = dLL + dWW

Substituting the instrument limits ( $dL = dW = 0.05 ~mm$ ):

dAA = (0.05)/(5) + (0.05)/(2.5) = (1)/(100) + (2)/(100) = (3)/(100)

Step 1: Final Value Match

Comparing this to the target format $x100$ $\frac{\mathrm{x}}{100}$ gives:

$x = 3$ $\mathrm{x} = 3$

Pattern Recognition

The absolute measurement uncertainty matches the instrument's least count value directly. Sum up individual fractional errors to compute the total area uncertainty parameter.

Chapter Mix

Class 11 Physics: Units and Measurements

More Units and Measurements Previous-Year Questions — Page 5

Q16 jee_main_2025_29_jan_evening Dimensional Analysis

Match List-I with List-II.

array|l|l|l|l| List-I & & List-II & (A) & Young's Modulus & (I) & ML⁻¹T⁻¹ (B) & Torque & (II) & ML⁻¹T⁻² (C) & Coefficient of Viscosity & (III) & M⁻¹L³T⁻² (D) & Gravitational Constant & (IV) & ML²T⁻² array

Choose the correct answer from the options given below:

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

Solution

Related Formula

Young's Modulus: Y = (F/A)/(Δ / )

Torque: τ = F \cdot r

Viscosity Force: F = η A (dv)/(dx)

Gravitational Force: F = (G m₁ m₂)/(r²)

Core Logic

Evaluating dimensions component-by-component:

(A) Young's Modulus:

[Y] = ([F])/([A]) = MLT⁻²L² = ML⁻¹T⁻² arrow (II)

(B) Torque:

[τ] = [F][r] = ( MLT⁻²)(L) = ML²T⁻² arrow (IV)

(C) Coefficient of Viscosity:

[η] = ([F])/([A][dv/dx]) = MLT⁻²(L²)( T⁻¹) = ML⁻¹T⁻¹ arrow (I)

(D) Gravitational Constant:

[G] = ([F][r²])/([m₁][m₂]) = ( MLT⁻²)(L²)M² = M⁻¹L³T⁻² arrow (III)

Matching path yields: (A)-(II), (B)-(IV), (C)-(I), (D)-(III).

Pattern Recognition

Torque and energy share the identical dimensional formula $ML²T⁻²$ $\mathrm{ML^2T^{-2}}$. Modulus and pressure share $ML⁻¹T⁻²$ $\mathrm{ML^{-1}T^{-2}}$. Spotting these matching associations cuts solving time significantly.

Chapter Mix

Class 11 Physics: Units and Measurements

Q23 jee_main_2025_29_jan_evening Combination of Errors

A physical quantity

Q

is related to four observables

a, b, c, d

as follows:

Q = (ab⁴)/(cd)

where,

a = (60 \pm 3)~Pa

;

b = (20 \pm 0.1)~m

;

c = (40 \pm 0.2)~Nsm⁻²

and

d = (50 \pm 0.1)~m

, then the percentage error in

Q

(x)/(1000)

, where

x =

______.

Numerical Answer. Answer: 7700 to 7700

Solution

Related Formula

(Δ Q)/(Q) = (Δ a)/(a) + 4(Δ b)/(b) + (Δ c)/(c) + (Δ d)/(d)

Core Logic

Write down fractional errors from the raw text configurations:

$(Δ a)/(a) = (3)/(60) = 0.05$
$(Δ b)/(b) = (0.1)/(20) = 0.005$
$(Δ c)/(c) = (0.2)/(40) = 0.005$
$(Δ d)/(d) = (0.1)/(50) = 0.002$

Compute the total fractional error expression:

(Δ Q)/(Q) = [0.05 + 4(0.005) + 0.005 + 0.002]

(Δ Q)/(Q) = 0.05 + 0.02 + 0.005 + 0.002 = 0.077

Percentage error expression configuration:

% Error = (Δ Q)/(Q) \times 100 = 7.7 %

Given that percentage error equals $(x)/(1000)$ $\frac{x}{1000}$:

(x)/(1000) = 7.7 x = 7700

Pattern Recognition

Powers scale up error contributions via direct multiplication multipliers. The term $b⁴$ $b^4$ contributes exactly 4 times its basic fraction error component to the compilation step.

Chapter Mix

Class 11 Physics: Units and Measurements

Q23 jee_main_2025_28_jan_morning Dimensional Analysis

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

, the value of

c

Numerical Answer. Answer: 0 to 0

Solution

Core Logic

Let's find the dimensional formula for the ratio of Modulus of Elasticity to Torque:

Target Dimensions = [Modulus of Elasticity][Torque]

Target Dimensions = [M L⁻¹ T⁻²][M L² T⁻²] = [M⁰ L⁻³ T⁰]

Step 1: Exponent Matching

Comparing this output to the target layout formula $[Ma Lb Tc]$ :

$c = 0$ $\mathrm{c} = 0$

Pattern Recognition

Both dimensions share identical time dependence factors ( $T⁻²$ $\mathrm{T}^{-2}$), meaning they cancel out completely. This leaves the time exponent value as exactly zero.

Chapter Mix

Class 11 Physics: Units and Measurements

Q9 jee_main_2025_03_april_morning Dimensional Formulae

Match the LIST-I with LIST-II

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. $[M⁻¹L³T⁻²]$ $[M^{-1}L^3T^{-2}]$

Choose the correct answer from the options given below:

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

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

Solution & Explanation

Core Logic

Step 1: Final Value Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

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Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Exponent Matching

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Match and Selection

Pattern Recognition

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More Units and Measurements Previous-Year Questions — Page 5

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Related Formula

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Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Exponent Matching

Pattern Recognition

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Related Formula

Core Logic

Step 1: Match and Selection

Pattern Recognition

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