JEE Main · Physics ↓ Falling

Units and Measurements appeared 50 times across 3 years — 5.8% of Physics. This question is from Errors in Measurement.

Year 2026 2025 2024 Total
Questions 14 22 14 50

A tiny metallic rectangular sheet has length and breadth of 5 ~mm and 2.5 ~mm , respectively. Using a specially designed screw gauge which has pitch of 0.75 ~mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be x100 where x is ________.

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3 +4 marks

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
Least count calculation tracking diagram for Q21

The area of the rectangular metallic sheet is calculated as:

A = L · 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 gives:

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

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 7

Q12 jee_main_2025_07_april_evening Dimensional Formula
Match List-I with List-II.
List-IList-II
(A) Mass density(I) [ML²T⁻³]
(B) Impulse(II) [MLT⁻¹]
(C) Power(III) [ML²T⁰]
(D) Moment of inertia(IV) [ML⁻³T⁰]
Choose the correct answer from the options given below: [cite: 107]
  • A. (A)-(IV), (B)-(II), (C)-(III), (D)-(I) [cite: 108]
  • B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II) [cite: 109]
  • C. (A)-(IV), (B)-(II), (C)-(I), (D)-(III) [cite: 110]
  • D. (A)-(II), (B)-(III), (C)-(IV), (D)-(I) [cite: 111]

Solution

Core Logic

Let's derive the dimensional formulas systematically:

  • (A) Mass density: ρ = MassVolume = (M)/(L³) = [M¹ L⁻³ T⁰] (IV) [cite: 765].
  • (B) Impulse: I = F · Δ t = [M¹ L¹ T⁻²] · [T] = [M¹ L¹ T⁻¹] (II) [cite: 767].
  • (C) Power: P = WorkTime = [M¹ L² T⁻²][T] = [M¹ L² T⁻³] (I) [cite: 770].
  • (D) Moment of inertia: I = M r² = [M¹ L² T⁰] (III) [cite: 772].
  • Matching all four pairings establishes the layout: (A)-(IV), (B)-(II), (C)-(I), (D)-(III)[cite: 110, 758].

Pattern Recognition

Isolating the unique matching option for a straightforward parameter like Mass Density (A-IV) or Moment of Inertia (D-III) easily allows exclusion of multiple invalid answer branches instantly[cite: 765, 772].

Chapter Mix

Class 11 Physics: Units and Measurements

Q jee_main_2025_24_jan_morning Screw Gauge and Least Count
The least count of a screw guage is 0.01 ~mm . If the pitch is increased by 75 % and number of divisions on the circular scale is reduced by 50 % , the new least count will be _ × 10⁻³ ~mm .
Numerical Answer. Answer: 35 to 35

Solution

Related Formula

The Least Count (LC) of a screw gauge tool is defined by:

L.C. = PitchTotal Number of Circular Divisions (N)
Core Logic

The initial least count is given as:

L.C.initial = (P)/(N) = 0.01 mm

Now, calculate the modified parameters from the text details :

  • New Pitch: P' = P(1 + 0.75) = 1.75P
  • New Divisions: N' = N(1 - 0.50) = 0.5N
Step 1: Calculating the New Least Count

Set up the updated least count expression ratio :

L.C.new = (P')/(N') = (1.75P)/(0.5N) = 3.5 × ((P)/(N))

Substitute the initial least count value :

L.C.new = 3.5 × 0.01 mm = 0.035 mm

Converting into the requested scientific prefix units (10⁻³ mm) :

L.C.new = 35 × 10⁻³ mm

Therefore, the requested value is 35.

Pattern Recognition

Least count scales proportionally with pitch increases, and inversely with reductions in circular divisions.

Chapter Mix

Class 11 Physics: Units and Measurements

Q4 jee_main_2025_24_jan_morning Significant Figures
For an experimental expression y=(32.3×1125)/(27.4) , where all the digits are significant. Then to report the value of y we should write :-
  • A. y=1326.2
  • B. y=1326.19
  • C. y=1326.186
  • D. y=1330

Solution

Related Formula

In multiplication and division arithmetic rules, the final product or quotient must be rounded off to retain as many significant figures as are present in the least precise operand.

Core Logic

Let us check the significant digit count of the operands in the expression :

  • 32.3 has 3 significant figures.
  • 1125 has 4 significant figures.
  • 27.4 has 3 significant figures.
  • The minimum number of significant figures among the numbers is 3.

Step 1: Rounding Off

Direct calculation yield :

y = 1326.186...

Rounding this value off to contain exactly 3 significant figures means changing it to 1330, since the digit after 2 is 6 (which is greater than 5), updating the hundreds spot upwards.

Pattern Recognition

Never keep unearned precision from automated calculation. The output is bounded strictly by your least precise entry.

Chapter Mix

Class 11 Physics: Units and Measurements

Q3 jee_main_2025_28_jan_evening Dimensional Analysis
Match List-I with List-II:
List-IList-II
(A) Angular Impulse(I) [M⁰L²T⁻²]
(B) Latent Heat(II) [M L²T⁻³A⁻¹]
(C) Electrical resistivity(III) [M L²T⁻¹]
(D) Electromotive force(IV) [M L³T-3"A⁻²]
Choose the correct answer from the options given below:
  • A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  • B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  • C. (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  • D. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Solution

Related Formula
  • Angular Impulse = Change in Angular Momentum = τ · Δ t = [M L² T⁻²] · [T] = [M L² T⁻¹] [cite: 670, 672]
  • Latent Heat (L) = (Q)/(m) = [M L² T⁻²][M] = [M⁰ L² T⁻²]
  • Electrical resistivity (\rho) = (R · A)/(l) = [M L² T⁻³ A⁻²] · [L²][L] = [M L³ T⁻³ A⁻²]
  • Electromotive force (V) = (W)/(q) = [M L² T⁻²][A T] = [M L² T⁻³ A⁻¹]
Core Logic

By comparing the formulas derived for each physical quantity with the options given in List-II [cite: 670, 673, 674]:

  • (A) matches with (III) [cite: 670, 672]
  • (B) matches with (I)
  • (C) matches with (IV)
  • (D) matches with (II)
  • Hence, the correct matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Pattern Recognition

In match-the-column dimensional analysis questions, identifying even one or two straightforward quantities like Latent Heat (L = Q/m) often immediately eliminates three incorrect options, securing a quick correct answer.

Chapter Mix

Class 11 Physics: Units and Measurements

Q jee_main_2025_29_jan_morning Dimensional Analysis
The pair of physical quantities not having same dimensions is :
  • A. Torque and energy
  • B. Surface tension and impulse
  • C. Angular momentum and Planck's constant
  • D. Pressure and Young's modulus

Solution

Core Logic

Let\'s check the dimensions of each pair :

  • Torque = [Energy] = [ML²T⁻²]
  • Surface Tension = [MT⁻²] vs Impulse = [MLT⁻¹]
  • [Angular Momentum] = [Planck's Constant] = [ML²T⁻¹]
  • [Pressure] = [Young's Modulus] = [ML⁻¹T⁻²]
Step 1: Identify Non-Matching Pair

Surface tension and impulse do not share matching dimension frameworks.

Pattern Recognition

Surface tension is force per unit length ([MT⁻²]), while impulse is force times time ([MLT⁻¹])

Chapter Mix

Class 11 Physics: Units and Measurements

More Units and Measurements Questions — jee_main_2025_28_jan_morning

Practice all Units and Measurements previous-year questions →

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)