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JEE • PHYSICS • UNITS & MEASUREMENTS

Units & measurements practice · Medium

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Practice format and chapter quiz

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 451-480 of 502 questions
451 Medium
451 of 502
Understanding +4 | -1
A physicist performs an experiment and takes 200 readings. He repeats the same experiment and now takes 800 readings. By doing so the probable error
View answer and explanation
Correct option: D
Is reduced by a factor of 1/4
Option D (Is reduced by a factor of 1/4) is correct because the error is inversely proportional to the square root of the number of readings. Since the number of readings is multiplied by 4, the error is divided by the square root of 4, which is 2. The variance is reduced by a factor of 4.
452 Medium
452 of 502
Understanding +4 | -1
State the number of significant figures in the
following:
(i) 2.000 m (ii) 5100 kg (iii) 0.050 cm.
View answer and explanation
Correct option: A
(i) - 4; (ii) - 4; (iii) - 2
Option A is correct because it accurately counts the significant figures in each given measurement based on the rules of significant figures.
453 Medium
453 of 502
Remembering +4 | -1
The dimensions of Wien’s constant are
View answer and explanation
Correct option: B
[M0LT0K]\left[ {{M^0}L{T^0}K} \right]
Option B ([M0LT0K{M^0}L{T^0}K]) is correct because Wien's constant has dimensions of length times temperature.
454 Medium
454 of 502
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 (dyne/cm²) to N/m² is 0.1.
455 Medium
455 of 502
Analyzing +4 | -1
The physical quantity that does not have the dimensional formula [ML−1 T−2]\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] is
View answer and explanation
Correct option: A
force
Option A is correct because force has the dimensional formula [MLT⁻²], while pressure, stress, and modulus of elasticity have [ML⁻¹T⁻²].
456 Medium
456 of 502
Understanding +4 | -1
The unit of Young's modulus is

[Y=StressStrain=force/areaΔℓ/L]\left[\mathrm{Y}=\frac{\text {Stress}}{\text {Strain}}=\frac{\text {force} / \text {area}}{\Delta \ell / \mathrm{L}}\right]
View answer and explanation
Correct option: C
Newton/metre2{ }^2
Option C (Newton/metre2{ }^2) is correct because Young's modulus is defined as stress/strain, where stress has units of force/area (N/m²) and strain is dimensionless.
457 Medium
457 of 502
Understanding +4 | -1
Young’s modulus of a material has the same units as
View answer and explanation
Correct option: A
Pressure
Option A (Pressure) is correct because Young's modulus is defined as stress/strain, and stress has the same units as pressure.
458 Medium
458 of 502
Remembering +4 | -1
Unit of magnetic flux is:
View answer and explanation
Correct option: C
Weber
Option C (Weber) is correct because the Weber is the SI unit of magnetic flux.
459 Medium
459 of 502
Understanding +4 | -1
The dimensional formula of intensity of magnetisation is
View answer and explanation
Correct option: D
[L−1 A]\left[{L}^{-1} {~A}\right]
Option D ([L−1 A{L}^{-1} {~A}]) is correct because the intensity of magnetization is defined as the magnetic moment per unit volume, which has the dimensions of current per unit length.
460 Medium
460 of 502
Application +4 | -1
If the acceleration due to gravity is 10  ms−2\;\mathrm{m}{\mathrm{s}^{ - 2}} and the units of length and time are changed in kilometer and hour respectively, the numerical value of acceleration is
View answer and explanation
Correct option: D
129600
Option D (129600) is correct because the acceleration is transformed using the conversion factors for length and time to the new system of units.
461 Medium
461 of 502
Analyzing +4 | -1
The velocity of a particle as a function of time is given as v(t)=at2+bt+ct2+dv(t)=a t^2+b t+\frac{c}{t^2+d}, where a,b,ca, b, c and dd are constants, The dimensions of [abcd]\left[\frac{a b}{c d}\right] is
View answer and explanation
Correct option: B
[LT−8]\left[L T^{-8}\right]
Option A is correct because by equating dimensions of each term in the given equation, we can find the dimensions of a, b, c, and d, and then calculate the dimensions of ab/cd.
462 Medium
462 of 502
Understanding +4 | -1
The SI unit of universal gas constant (R) is
View answer and explanation
Correct option: C
Joule  K−1mol−1Joule\;{K^{ - 1}}mo{l^{ - 1}}
Option C (Joule;K−1mol−1Joule \\;{K^{ - 1}}mo{l^{ - 1}}) is correct because the universal gas constant (R) is expressed in terms of energy per mole per Kelvin.
463 Medium
463 of 502
Analyzing +4 | -1
Assertion A: If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge in 5 mm and there are 50 total divisions on circular scale, then least count of screw gauge is 0.002 cm.
Reason R: Least count = Pitch  Total divisions on circular scale =\frac{\text { Pitch }}{\text { Total divisions on circular scale }}
View answer and explanation
Correct option: C
Both A and R are true and R is the correct explanation of A
Option C is correct because both the assertion and reason are true, and the reason correctly explains how the least count of a screw gauge is calculated.
464 Medium
464 of 502
Analyzing +4 | -1
If the speed of light (c), acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities, then the dimension of gravitational constant is
View answer and explanation
Correct option: B
c0g2p−1{c^0}{g^2}{p^{ - 1}}
Option B (c0g2p−1{c^0}{g^2}{p^{ - 1}}) is correct because it is derived by equating the dimensions of the gravitational constant with the given fundamental quantities.
465 Medium
465 of 502
Analyzing +4 | -1
The dimension of a×ba \times b in the relation E=b−x2at, E = \frac{{b - {x^2}}}{{at}},\, where EE is the energy, xx is the displacement and tt is time are
View answer and explanation
Correct option: B
M−1L2T{M^{ - 1}}{L^2}T
Option B (M−1L2T{M^{ - 1}}{L^2}T) is correct because by equating dimensions, we find the dimension of 'a' and 'b' and then multiply them to get the dimension of a x b.
466 Medium
466 of 502
Remembering +4 | -1
The circular head of screw gauge is divided into 200 division and move 1 mm a head in one revolution. Find the pitch of screw gauge:
View answer and explanation
Correct option: A
1 mm
Option A (1 mm) is correct because the pitch of a screw gauge is the distance it moves forward in one complete revolution, which is given as 1 mm.
467 Medium
467 of 502
Application +4 | -1
Given, Force =αdensity+  β3= \frac{\alpha }{{\mathrm{density} + \mathrm{\;}{\mathrm{\beta }^3}}} What are the dimensions of α,β?\mathrm{\alpha },\mathrm{\beta }?
View answer and explanation
Correct option: D
[M2L−2T2],  [ML−3]\left[ {{\mathrm{M}^2}{\mathrm{L}^{ - 2}}{\mathrm{T}_2}} \right],\mathrm{\;}\left[ {\mathrm{M}{\mathrm{L}^{ - 3}}} \right]
Option C ([M2 L-2 T-2], [M1/3 L-1]) is correct because it is derived from equating dimensions of force and density with the given expression.
468 Medium
468 of 502
Analyzing +4 | -1
The physical quantity which does not have the same dimensions as the other three is
View answer and explanation
Correct option: D
acceleration due to gravity
Option D is correct because acceleration due to gravity has dimensions [LT⁻²], while spring constant, surface tension, and surface energy have dimensions [MT⁻²].
469 Medium
469 of 502
Understanding +4 | -1
In a slide calliper, nn divisions of vernier scale coincides with (n−1)(n-1) divisions of main scale. The least count of the instrument is
View answer and explanation
Correct option: A
1nMSD\frac{1}{n} \mathrm{MSD}
Option A (1nMSD\frac{1}{n} \mathrm{MSD}) is correct because the least count of a slide caliper is calculated as the difference between one main scale division and one vernier scale division.
470 Medium
470 of 502
Analyzing +4 | -1
The fundamental unit, which has the same power in the dimensional formulae of surface tension and viscosity is
View answer and explanation
Correct option: A
Mass
Option A is correct because both surface tension and viscosity have mass to the power of 1 in their dimensional formulas.
471 Medium
471 of 502
Analyzing +4 | -1
Assertion: The number of significant figures in 0.100 is 1 .

Reason: The zeros at the end of a number are always meaningless.
View answer and explanation
Correct option: D
If both assertion and reason are false.
Option D (If both assertion and reason are false) is correct because 0.100 has three significant figures, and zeros at the end of a number after the decimal point are significant.
472 Medium
472 of 502
Analyzing +4 | -1
Given that the amplitude A of scattered light is :

(i) Directly proportional to the amplitude (Ao)\left(A_o\right) of incident light.

(ii) Directly proportional to the volume (V)(V) of the scattering particle

(iii) Inversely proportional to the distance (r) from the scattered particle

(iv) Depend upon the wavelength (λ)(\lambda) of the scattered light. then:
View answer and explanation
Correct option: B
A∝1λ2A \propto \frac{1}{\lambda^2}
Option B is correct because the amplitude of scattered light is inversely proportional to the square of the wavelength, A∝1λ2A \propto \frac{1}{\lambda^2}.
473 Medium
473 of 502
Understanding +4 | -1
Which of the following is not a unit of energy
View answer and explanation
Correct option: B
kg–m/sec
Option B is correct because kg-m/sec is the unit of momentum, not energy.
474 Medium
474 of 502
Application +4 | -1
Rounding off the value 324.13821 upto four significant digits is
View answer and explanation
Correct option: A
324.1
Option A (324.1) is correct because when rounding to four significant digits, the digit after the fourth significant digit is less than 5, so the fourth digit remains unchanged.
475 Medium
475 of 502
Analyzing +4 | -1
Dimensional formula for h2v2/mc21+hvmc2\frac{h^2 v^2 / m c^2}{1+\frac{h v}{m c^2}} is (where hh is planck's constant, vv is frequency of light, mm is mass and cc is speed of light).
View answer and explanation
Correct option: C
[ML2 T−2]\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
Option C ([ML2 T−2]\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]) is correct because after simplifying the expression using dimensional analysis, the result matches the dimensions of energy.
476 Medium
476 of 502
Application +4 | -1
The physical quantity which has the dimensional formula [M1T−3]\left[ {{\mathrm{M}^1}{\mathrm{T}^{ - 3}}} \right] is
View answer and explanation
Correct option: C
Solar constant
Option C (Solar constant) is correct because its dimensions are energy per unit area per unit time, which matches [M1T-3].
477 Medium
477 of 502
Understanding +4 | -1
Statement-I: Positive zero error is added in measurement.
Statement-II: Defect may occurs during manufacturing of measuring instruments.
View answer and explanation
Correct option: B
Statement-I is wrong while statement-II is correct
Option B (Statement-I is wrong while statement-II is correct) is correct because positive zero error is subtracted, not added, and manufacturing defects can occur.
478 Medium
478 of 502
Application +4 | -1
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is
View answer and explanation
Correct option: A
6%6\%
Option A (6%6\%) is correct because the percentage error in resistance is the sum of the percentage errors in voltage and current.
479 Medium
479 of 502
Understanding +4 | -1
Which of the following measurement is most precise?
View answer and explanation
Correct option: D
3.00 mm
Option D is correct because the measurement with the smallest unit (3.00 mm) provides the highest precision.
480 Medium
480 of 502
Understanding +4 | -1
Assertion: Mass of a body and the velocity, with which it is moving, cannot be multiplied.

Reason: It is because, mass is a scalar and velocity is a vector quantity. In addition to this, the dimensions of mass and velocity are different.
View answer and explanation
Correct option: D
If both assertion and reason are false.
Option D (If both Assertion & Reason are false) is correct because mass and velocity can be multiplied to give momentum, and dimensions can be different for quantities that are multiplied.
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