JEE Main · Physics ↓ Falling

Units and Measurements appeared 50 times across 3 years — 5.8% of Physics. This question is from Dimensional Analysis.

Year 2026 2025 2024 Total
Questions 14 22 14 50

In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of MPLQTRAS, where value of 'Q' and 'R' are

Solution & Explanation

Related Formula

Ratio formulation:

(φE)/(φM) = (E · A)/(B · A) = (E)/(B)

From Maxwell's electromagnetic wave equations:

E = c · B (E)/(B) = c

where c is the speed of light.

Core Logic

Since the ratio reduces to the dimension of speed (c):

[(φE)/(φM)] = [c] = M⁰ L¹ T⁻¹ A⁰
Step 1: Identify Exponent Values

Matching indices with MPLQTRAS:

  • P = 0
  • Q = 1
  • R = -1
  • S = 0
  • Hence, (Q, R) = (1, -1).

Pattern Recognition

Flux areas cancel out immediately. The ratio (E)/(B) always carries the dimension of velocity (LT⁻¹).

Chapter Mix

Class 11 Physics: Units and Measurements Class 12 Physics: Electromagnetic Waves

More Units and Measurements Previous-Year Questions — Page 9

Q38 jee_main_2024_27_jan_morning Measuring Instruments
Identify the physical quantity that cannot be measured using a spherometer:
  • A. Radius of curvature of concave surface
  • B. Specific rotation of liquids
  • C. Thickness of thin plates
  • D. Radius of curvature of convex surface

Solution

Core Logic

A spherometer is a mechanical instrument designed to measure small vertical displacements to calculate the thickness of thin plates or the radius of curvature of spherical (convex/concave) surfaces.

Specific rotation of liquids is an optical property measured via a polarimeter, completely outside the scope of a spherometer.

Pattern Recognition

Spherometers operate purely on linear micrometer screw scale geometry, hence restricted strictly to spatial dimensions.

Chapter Mix

Class 11 Physics: Units and Measurements

Q47 jee_main_2024_27_jan_morning Dimensional Analysis
Given below are two statements: Statement (I): Planck's constant and angular momentum have same dimensions. Statement (II): Linear momentum and moment of force have same dimensions. In the light of the above statements, choose the correct answer from the options given below:
  • A. Statement I is true but Statement II is false
  • B. Both Statement I and Statement II are false
  • C. Both Statement I and Statement II are true
  • D. Statement I is false but Statement II is true

Solution

Core Logic

Evaluate dimensions step-by-step:

  • Planck's constant (h):
  • E = h u [h] = ([E])/([ u]) = ML²T⁻²T⁻¹ = ML²T⁻¹

  • Angular momentum (L):
L = mvr [L] = M · LT⁻¹ · L = ML²T⁻¹

Since [h] = [L], Statement I is true.

  • Linear momentum (P):
P = mv [P] = MLT⁻¹
  • Moment of force (Torque τ):
τ = F r [τ] = MLT⁻² · L = ML²T⁻²

Since [P] ≠ [τ], Statement II is false.

Pattern Recognition

Planck's constant can always be paired with angular momentum units (Joule-seconds), while moment of force matches work/energy footprints, not translational momentum.

Chapter Mix

Class 11 Physics: Units and Measurements

Q38 jee_main_2024_29_jan_morning Error Analysis
The resistance R = (V)/(I) where V = (200 ± 5) ~V and I = (20 ± 0.2) ~A, the percentage error in the measurement of R is:
  • A. 3.5%
  • B. 7%
  • C. 3%
  • D. 5.5%

Solution

Related Formula

By propagation of maximum relative error in division:

R = (V)/(I) (Δ R)/(R) = (Δ V)/(V) + (Δ I)/(I)

Percentage error in R is given by:

% error in R = ( (Δ V)/(V) + (Δ I)/(I) ) × 100
Core Logic

Given values:

V = 200 ~V, Δ V = 5 ~V I = 20 ~A, Δ I = 0.2 ~A
Step 1: Evaluate Relative Error
(Δ R)/(R) = (5)/(200) + (0.2)/(20) (Δ R)/(R) = (5)/(200) + (2)/(200) (Δ R)/(R) = (7)/(200)
Step 2: Calculate Percentage Error
% error in R = (Δ R)/(R) × 100 = (7)/(200) × 100 = 3.5%

Thus, the percentage error is 3.5%.

Pattern Recognition

Whenever independent physical quantities are multiplied or divided, their fractional/relative errors always add up. Make sure to keep the base denominator values aligned to make mental calculations quick.

Chapter Mix

Class 11 Physics: Units and Measurements

Q31 jee_main_2024_30_january_evening Vernier Callipers
If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 ~mm, the Vernier constant of travelling microscope is:
  • A. 0.1 ~mm
  • B. 0.1 ~cm
  • C. 0.01 ~cm
  • D. 0.01 ~mm

Solution

Related Formula
Vernier Constant (Least Count) = 1 ~MSD - 1 ~VSD
Core Logic

Given that 50 Vernier Scale Divisions (VSD) equal 49 Main Scale Divisions (MSD).

50 ~VSD = 49 ~MSD 1 ~VSD = (49)/(50) ~MSD

Also, the smallest reading of the main scale (1 ~MSD) is 0.5 ~mm.

Step 1: Calculate Vernier Constant
Vernier Constant = 1 ~MSD - 1 ~VSD = 1 ~MSD - (49)/(50) ~MSD = (1)/(50) ~MSD

Substitute the value of 1 ~MSD:

= (1)/(50) × 0.5 ~mm = (0.5)/(50) ~mm = (1)/(100) ~mm = 0.01 ~mm
Pattern Recognition

In Vernier calipers problems where N VSD = (N-1) MSD, the Least Count is always exactly (1)/(N) MSD.

Chapter Mix

Class 11 Physics: Units and Measurements

Q44 jee_main_2024_30_january_evening Dimensional Analysis
If mass is written as m = k c^p G-1/2 h1/2 then the value of P will be: (Constants have their usual meaning with k a dimensionless constant)
  • A. 1/2
  • B. 1 / 3
  • C. 2
  • D. -1 / 3

Solution

Related Formula
[m] = [M]¹ [L]⁰ [T]⁰ [c] = [L T⁻¹] [G] = [M⁻¹ L³ T⁻²] [h] = [M L² T⁻¹]
Core Logic

By applying the principle of dimensional homogeneity, the dimensions on both sides of the equation must be identical.

[M]¹ [L]⁰ [T]⁰ = [L T⁻¹]^p [M⁻¹ L³ T⁻²]-1/2 [M L² T⁻¹]1/2
Step 1: Substitute Dimensions
[M] = L^p T-p · M1/2 L-3/2 T¹ · M1/2 L¹ T-1/2
Step 2: Collect Powers of L

Equating the powers of [L] on both sides:

0 = p - (3)/(2) + 1 0 = p - (1)/(2) p = (1)/(2)
Pattern Recognition

The expression m ∝ √(hc/G) is a known fundamental relation representing the Planck mass. The exponent on c inside the square root gives p = 1/2.

Chapter Mix

Class 11 Physics: Units and Measurements

More Units and Measurements Questions — jee_main_2025_04_april_morning

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