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JEE • PHYSICS • UNITS & MEASUREMENTS

Units & measurements practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 361-390 of 616 questions

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361 Easy
361 of 616
Understanding +4 | -1
Statement-I: Plane angle is a dimensionless quantity.
Statement-II: All supplementary quantíties are dimensionless.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A is correct because plane angle is defined as the ratio of arc length to radius, making it dimensionless, and supplementary quantities are dimensionless.
362 Easy
362 of 616
Application +4 | -1
The force F on the sphere of radius ‘a’ moving in a medium with velocity 'vv' is given by F=6πηaυF = 6\pi \eta a\upsilon. The dimensions of η are
View answer and explanation
Correct option: A
ML−1T−1M{L^{ - 1}}{T^{ - 1}}
Option A (ML−1T−1M{L^{ - 1}}{T^{ - 1}}) is correct because it correctly derives the dimensions of viscosity (η\eta) from the given formula using dimensional analysis.
363 Easy
363 of 616
Application +4 | -1
The dimensional formula of farad is
View answer and explanation
Correct option: B
[M−1 L−2 T2Q2]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2 \mathrm{Q}^2\right]
Option B ([M⁻¹ L⁻² T² Q²]) is correct because it represents the correct dimensional formula for capacitance (Farad) derived from Q = CV and W = QV.
364 Medium
364 of 616
Understanding +4 | -1
Which one of the following pairs of quantities and their unit is properly matched?
View answer and explanation
Correct option: B
Magnetic flux - Weber /m2/ {m}^2
The correct answer is Magnetic flux - Weber /m2/ {m}^2 because Weber /m2/ {m}^2 is the correct unit for magnetic flux density, often called magnetic field.
365 Medium
365 of 616
Application +4 | -1
In a new system of units, the net force applied on a block of mass 10 kg10 \mathrm{~kg} moving with acceleration 10 m/s210 \mathrm{~m} / \mathrm{s}^2 is given as 100 unit of force. When the same block is moving at a speed of 20 m/s20 \mathrm{~m} / \mathrm{s}, its kinetic energy become 20 unit of energy. A liquid has a surface tension of 10 SI units. What is the magnitude of its surface tension in new system? (Dimensional formula of surface tension is ML0T−2M L^0 T^{-2}.)
View answer and explanation
Correct option: A
1000
Option A (1000) is correct because the surface tension in the new system is 1000 times the value in SI units, derived from dimensional analysis and the given force and kinetic energy relationships.
366 Easy
366 of 616
Understanding +4 | -1
Statement-I : If u1u_1 and u2u_2 are units and n1,n2n_1, n_2 are their numerical values in two different systems then n1>n2⇒u1<u2n_1 > n_2 \Rightarrow u_1 < u_2.
Statement-II: The numerical value of physical quantity is inversely proportional to unit.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A (Statement-I is true and statement-II is true) is correct because the numerical value of a physical quantity is inversely proportional to the unit used to measure it.
367 Medium
367 of 616
Remembering +4 | -1
Wave number has the dimensions of
View answer and explanation
Correct option: B
Length−1^{-1}
Option B (Length−1^{-1}) is correct because wave number is the reciprocal of wavelength, which has the dimension of length.
368 Medium
368 of 616
Understanding +4 | -1
Two quantities A and B have different dimensions. Which mathematical operation given below is physically meaningful
View answer and explanation
Correct option: A
A/B
Option A (A/B) is correct because quantities with different dimensions can be divided or multiplied, but not added or subtracted.
369 Medium
369 of 616
Remembering +4 | -1
Which of the following unit is not of length?
View answer and explanation
Correct option: D
Becquerel
Option D (Becquerel) is correct because Becquerel is the SI unit of radioactivity, not length.
370 Medium
370 of 616
Understanding +4 | -1
The physical quantity that has no dimensions is
View answer and explanation
Correct option: D
Strain
Option D is correct because strain is a dimensionless quantity, being the ratio of two lengths.
371 Medium
371 of 616
Understanding +4 | -1
Assertion: Linear mass density has the dimensions of [M1L−1T0]\left[\mathrm{M}^1 \mathrm{L}^{-1} \mathrm{T}^0\right].
Reason: This is so because density is mass per unit volume.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not the correct explanation of assertion
Option B (If both assertion and reason are true but reason is not the correct explanation of assertion) is correct because both the assertion about linear mass density's dimensions and the reason about density are true, but the reason doesn't explain the assertion.
372 Medium
372 of 616
Understanding +4 | -1
Assertion : Force cannot be added to pressure.
Reason : The dimensions of force and pressure are
different.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because force and pressure have different dimensions, so they cannot be added.
373 Medium
373 of 616
Understanding +4 | -1
Which two of the following quantities are
dimensionally equivalent?
(i) Force (ii) Pressure (iii) Young’s Modulus (iv) Energy
View answer and explanation
Correct option: C
(ii) and (iii)
Option C ((ii) and (iii)) is correct because both pressure and Young's modulus have the same dimensional formula, which is [ML⁻¹T⁻²].
374 Medium
374 of 616
Understanding +4 | -1
The number of significant figures in 0.06900 is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the trailing zeros after the decimal point are significant.
375 Medium
375 of 616
Analyzing +4 | -1
If (P−aV2)(V−b)=nRT\left(P-\frac{a}{V^2}\right)(V-b)=n R T, where P,V,RP, V, R and TT are pressure, volume, universal gas constant and temperature, then ab2\frac{a}{b^2} has same dimensional formula as that of
View answer and explanation
Correct option: D
PP
Option D is correct because dimensional analysis shows that a/b² has the same dimensions as pressure.
376 Medium
376 of 616
Application +4 | -1
The surface tension is T=Fl,T = \frac{F}{l}, then the dimensions of surface tension are
View answer and explanation
Correct option: B
[MT−2]\left[ {\mathrm{M}{\mathrm{T}^{ - 2}}} \right]
Option B (MT−2{\mathrm{M}{\mathrm{T}^{ - 2}}}) is correct because surface tension is force per unit length, and its dimensions are mass times time to the power of -2.
377 Medium
377 of 616
Analyzing +4 | -1
If L\mathrm{L} is the inductance, C\mathrm{C} capacitance and R\mathrm{R} resistance the ratios L/R\mathrm{L} / \mathrm{R} and R−C\mathrm{R}-\mathrm{C} have the same dimensions as those of
View answer and explanation
Correct option: B
time
Option B (time) is correct because the ratio L/R has dimensions of time, and the product RC also has dimensions of time.
378 Medium
378 of 616
Remembering +4 | -1
The dimensions of stress are equal to
View answer and explanation
Correct option: B
Pressure
The correct answer is B (Pressure) because stress is defined as force per unit area, which is the same as pressure.
379 Medium
379 of 616
Understanding +4 | -1
The unit of poynting vector is
View answer and explanation
Correct option: B
Jm2 s\frac{\mathrm{J}}{\mathrm{m}^2 \mathrm{~s}}
Option B (Jm2 s\frac{\mathrm{J}}{\mathrm{m}^2 \mathrm{~s}}) is correct because the Poynting vector represents power per unit area, which is measured in Watts per square meter (W/m²), equivalent to Joules per second per square meter (J/m²s).
380 Medium
380 of 616
Analyzing +4 | -1
A student when discussing the properties of a medium (except vacuum) writes

Velocity of light in vacuum = Velocity of light in medium. This formula is
View answer and explanation
Correct option: D
Both (1) and (3)
Option D (Both (1) and (3)) is correct because the equation is dimensionally correct but numerically incorrect as the refractive index is always greater than 1.
381 Medium
381 of 616
Application +4 | -1
A physical quantity P is described by the relation P=a1/2b2c3d−4P = {a^{1/2}}{b^2}{c^3}{d^{ - 4}}. If the relative errors in the measurement of a, b, c and d respectively are 2%, 1%, 3% and 5% then the relative error in P will be
View answer and explanation
Correct option: D
0.32
Option D (32%) is correct because it is the relative error in P calculated using the given relative errors in a, b, c, and d.
382 Medium
382 of 616
Understanding +4 | -1
A physical quantity depends upon five factors, all of which have dimensions; then method of dimensional analysis
View answer and explanation
Correct option: B
Cannot be applied
Option B is correct because dimensional analysis can only determine relationships with up to three independent variables.
383 Medium
383 of 616
Application +4 | -1
In a vernier callipers, (N+1)(N+1) divisions of vernier scale coincide with NN divisions of main scale. If 1 MSD represents 0.1 mm,0.1 \mathrm{~mm}, the vernier constant (in cm) is:
View answer and explanation
Correct option: A
1100(N+1)\frac{1}{100(N+1)}
Option A (1100(N+1)\frac{1}{100(N+1)}) is correct because the vernier constant is the difference between one main scale division and one vernier scale division.
384 Easy
384 of 616
Understanding +4 | -1
Dimensions of frequency are
View answer and explanation
Correct option: B
M0L0T−1{M^0}{L^0}{T^{ - 1}}
Option B (M0L0T−1{M^0}{L^0}{T^{ - 1}}) is correct because frequency is the inverse of time, and time has dimensions of T.
385 Hard
385 of 616
Application +4 | -1
Velocity (v)\left( v \right) and acceleration (a)\left( a \right) in two systems of units 1 and 2 are related as v2=nm2v1{v_2} = \frac{n}{{{m^2}}}{v_1} and a2=a1mn{a_2} = \frac{{{a_1}}}{{mn}} respectively. Here mm and nn are constants. The relations for distance and time in two systems respectively are:
View answer and explanation
Correct option: A
n3m3L1=L2  and  n2mT1=T2\frac{{{n^3}}}{{{m^3}}}{L_1} = {L_2}\,\,{\text{and}}\,\,\frac{{{n^2}}}{m}{T_1} = {T_2}
Option A (n3m3L1=L2  and  n2mT1=T2\frac{{{n^3}}}{{{m^3}}}{L_1} = {L_2}\,\,{\text{and}}\,\,\frac{{{n^2}}}{m}{T_1} = {T_2}) is correct because it correctly relates the distance and time units between the two systems based on the given velocity and acceleration relationships.
386 Medium
386 of 616
Understanding +4 | -1
Which of the technique is not used for measuring small time intervals?
View answer and explanation
Correct option: C
Spring oscillator
The correct answer is Spring oscillator because it has the lowest frequency among the given options, making it unsuitable for measuring small time intervals.
387 Medium
387 of 616
Understanding +4 | -1
Accidental error can be minimised by
View answer and explanation
Correct option: D
Selecting instrument with small least count
Option D (Selecting instrument with small least count) is correct because a smaller least count allows for more precise measurements, reducing accidental errors.
388 Hard
388 of 616
Analyzing +4 | -1
The mass of the liquid flowing per second per unit area of cross-section of the tube is proportional to (pressure difference across the ends) n^n and (average velocity m^m of the liquid. Which one of the following relation is correct?
View answer and explanation
Correct option: B
m=−nm=-n
Option B (m=−nm=-n) is correct because dimensional analysis shows that the exponents must be equal in magnitude but opposite in sign.
389 Medium
389 of 616
Analyzing +4 | -1
The dimensions of e2/4πε0hc,{e^2}/4\mathrm{\pi }{\mathrm{\varepsilon }_0}hc,, where e,  ε0,  h  and  ce,\;{\mathrm{\varepsilon }_0},\mathrm{\;}h\;\mathrm{and\;}c are electronic charge, electric permittivity, Planck’s constant and velocity of light in vacuum respectively, are
View answer and explanation
Correct option: A
[M0L0T0]\left[ {{\mathrm{M}^0}{\mathrm{L}^0}{\mathrm{T}^0}} \right]
Option A is correct because the expression is dimensionless, representing the fine-structure constant.
390 Medium
390 of 616
Application +4 | -1
To determine the Young’s modulus of a wire, the formula is Y=FA×LΔLY = \frac{F}{A} \times \frac{L}{{\Delta L}}; where L= length, A= area of cross-section of the wire, ΔL{\Delta L}= change in length of the wire when stretched with a force F. The conversion factor to change it from CGS to MKS system is
View answer and explanation
Correct option: C
0.1
Option C (0.1) is correct because the conversion factor from CGS to MKS for Young's modulus is 0.1, derived from the unit conversion of dyne/cm² to N/m².
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