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

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Questions 14 22 14 50

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 is

Numerical Answer Type:
Enter a numerical value Answer: 0 to 0 +4 marks

Solution & Explanation

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

Pattern Recognition

Both dimensions share identical time dependence factors (T⁻²), meaning they cancel out completely. This leaves the time exponent value as exactly zero.

Chapter Mix

Class 11 Physics: Units and Measurements

Reference Study Guides

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±0.2) N/m and k₂=(20±0.3) 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 × 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 is:

% Error in K = (0.5)/(30) × 100 = (5)/(3)% ≈ 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]
  • 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 (height/length) and Bt must have the same dimension. [h] = [Bt]

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

So, B = [M⁰ L T⁻¹].

Step 1: Finding Dimension of A

From the term (P + (A)/(Bt²)), pressure P and (A)/(Bt²) 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] × [T²] [M L⁻¹ T⁻²] = ([A])/([L T]) [A] = [M L⁻¹ T⁻²] × [L T] = [M L⁰ T⁻¹]

So, A = [M L⁰ T⁻¹].

Pattern Recognition

Apply the Principle of Homogeneity across any (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
  • B. 205.333 m
  • C. 205.33 m
  • D. 205.3 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 (2 decimal places) 153.2 (1 decimal place - least precise) + 0.123 (3 decimal places) -------- 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 has only 1 decimal place. Rounding 205.333 to one decimal place yields 205.3 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:

η = (2)/(9) (r² g)/(v₀) (σ - ρ)

Since (2)/(9), g, σ, and ρ are constants or have negligible error, the relative error depends only on r and v₀.

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, 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-IList-II
(A) Spring constant(I) ML²T⁻²K⁻¹
(B) Thermal conductivity(II) ML⁰T⁻²
(C) Boltzmann constant(III) ML²T⁻³A⁻²
(D) Inductive reactance(IV) MLT⁻³K⁻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: [ML⁰ T⁻²] -> (II) (B) Thermal conductivity K: [MLT⁻³ K⁻¹] -> (IV) (C) Boltzmann constant kB: [ML² T⁻² K⁻¹] -> (I) (D) Inductive reactance XL: [ML² 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

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