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$\text{Statement I is true but Statement II is false}$
B.Both Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
C.Both Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
D.Statement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
Solution & Explanation
Core Logic
Evaluate dimensions step-by-step:
Planck's constant (h$h$):
E = h u [h] = ([E])/([ u]) = ML²T⁻²T⁻¹ = ML²T⁻¹$$E = h
u \implies [h] = \frac{[E]}{[
u]} = \frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-1}} = \text{ML}^2\text{T}^{-1}$$
Angular momentum (L$L$):
L = mvr [L] = M · LT⁻¹ · L = ML²T⁻¹$$L = mvr \implies [L] = \text{M} \cdot \text{LT}^{-1} \cdot \text{L} = \text{ML}^2\text{T}^{-1}$$
τ = F r [τ] = MLT⁻² · L = ML²T⁻²$$\tau = F r \implies [\tau] = \text{MLT}^{-2} \cdot \text{L} = \text{ML}^2\text{T}^{-2}$$
Since [P] ≠ [τ]$[P] \neq [\tau]$, 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.
Keywords:#Planck's constant and angular momentum dimensions#JEE Main 2024 Morning Q47#Units and Measurements JEE Main 2024#Dimensional Analysis JEE Main 2024
More Units and Measurements Previous-Year Questions — Page 5
Young's Modulus: Y = (F/A)/(Δ / )$$\text{Young's Modulus: } Y = \frac{F/A}{\Delta \ell / \ell}$$Torque: τ = F · r$$\text{Torque: } \tau = F \cdot r$$Viscosity Force: F = η A (dv)/(dx)$$\text{Viscosity Force: } F = \eta A \frac{dv}{dx}$$Gravitational Force: F = (G m₁ m₂)/(r²)$$\text{Gravitational Force: } F = \frac{G m_1 m_2}{r^2}$$
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
Q23jee_main_2025_29_jan_eveningCombination of Errors
A physical quantity Q$Q$ is related to four observables a, b, c, d$a, b, c, d$ as follows: Q = (ab⁴)/(cd)$Q = \frac{ab^4}{cd}$ where, a = (60 ± 3)~Pa$a = (60 \pm 3)\mathrm{~Pa}$ ; b = (20 ± 0.1)~m$b = (20 \pm 0.1)\mathrm{~m}$ ; c = (40 ± 0.2)~Nsm⁻²$c = (40 \pm 0.2)\mathrm{~Nsm}^{-2}$ and d = (50 ± 0.1)~m$d = (50 \pm 0.1)\mathrm{~m}$ , then the percentage error in Q$Q$ is (x)/(1000)$\frac{x}{1000}$ , where x =$x = $ ______.
Given that percentage error equals (x)/(1000)$\frac{x}{1000}$:
(x)/(1000) = 7.7 x = 7700$$\frac{x}{1000} = 7.7 \implies x = 7700$$
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
Q21jee_main_2025_28_jan_morningErrors in Measurement
A tiny metallic rectangular sheet has length and breadth of 5 ~mm$5 \mathrm{~mm}$ and 2.5 ~mm$2.5 \mathrm{~mm}$ , respectively. Using a specially designed screw gauge which has pitch of 0.75 ~mm$0.75 \mathrm{~mm}$ 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$\frac{\mathrm{x}}{100}$ where x$\mathrm{x}$ is ________.
Numerical Answer.Answer: 3 to 3
Solution
Core Logic
First, find the least count of the measurement tool:
Least Count = PitchNumber of circular scale divisions = 0.75 ~mm15 = 0.05 ~mm$$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.75 \mathrm{~mm}}{15} = 0.05 \mathrm{~mm}$$
Least count calculation tracking diagram for Q21
The area of the rectangular metallic sheet is calculated as:
A = L · W$$\mathrm{A} = \mathrm{L} \cdot \mathrm{W}$$
Expressing the absolute error via fractional configuration parts:
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.
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
Numerical Answer.Answer: 0 to 0
Solution
Core Logic
Let's find the dimensional formula for the ratio of Modulus of Elasticity to Torque:
Comparing this output to the target layout formula [Ma Lb Tc]$[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}]$:
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.
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
More Units and Measurements Questions — jee_main_2024_27_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.