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JEE • PHYSICS • UNITS & MEASUREMENTS

Units & measurements practice · Medium

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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 301-330 of 502 questions

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301 Medium
301 of 502
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.
302 Medium
302 of 502
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.
303 Medium
303 of 502
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.
304 Medium
304 of 502
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.
305 Medium
305 of 502
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].
306 Medium
306 of 502
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.
307 Medium
307 of 502
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.
308 Medium
308 of 502
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.
309 Medium
309 of 502
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.
310 Medium
310 of 502
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.
311 Medium
311 of 502
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).
312 Medium
312 of 502
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.
313 Medium
313 of 502
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.
314 Medium
314 of 502
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.
315 Medium
315 of 502
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.
316 Medium
316 of 502
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.
317 Medium
317 of 502
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.
318 Medium
318 of 502
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.
319 Medium
319 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 is 0.1, derived from the unit conversion of dyne/cm² to N/m².
320 Medium
320 of 502
Understanding +4 | -1
The maximum static friction on a body is F=μNF = \mathrm{\mu }N. Here, N= normal reaction force on the body μ=\mathrm{\mu } = coefficient of static friction. The dimensions of μ are
View answer and explanation
Correct option: C
Dimensionless
The correct answer is C (Dimensionless) because the coefficient of friction is a ratio of two forces, making it dimensionless.
321 Medium
321 of 502
Analyzing +4 | -1
The physical quantities not having same dimensions are
View answer and explanation
Correct option: B
Momentum and Planck’s constant
Option B (Momentum and Planck’s constant) is correct because momentum has dimensions of [MLT⁻¹] while Planck's constant has dimensions of [ML²T⁻¹], which are different.
322 Medium
322 of 502
Understanding +4 | -1
Assertion : When we change the unit of measurement of a
quantity, its numerical value changes.
Reason : Smaller the unit of measurement smaller is its
numerical value.
View answer and explanation
Correct option: C
Assertion is correct, reason is incorrect
Assertion is correct because changing units changes the numerical value. Reason is incorrect because smaller units have larger numerical values.
323 Medium
323 of 502
Understanding +4 | -1
A suitable unit for gravitational constant is
View answer and explanation
Correct option: C
N  m2  kg−2N\;{m^2}\;k{g^{ - 2}}
Option C is correct because the gravitational constant G has units of Nm2kg−2N m^2 kg^{-2}, derived from the gravitational force equation.
324 Medium
324 of 502
Application +4 | -1
A screw gauge has least count of 0.01mm\mathrm{0}\mathrm{.01 mm} and there are 50\mathrm{50} divisions in its circular scale. The pitch of the screw gauge is :
View answer and explanation
Correct option: C
0.5  mm\mathrm{0}\mathrm{.5}\,\,\mathrm{mm}
Option C (0.5 mm) is correct because the pitch is calculated by multiplying the least count by the number of divisions on the circular scale.
325 Medium
325 of 502
Application +4 | -1
Wavelength of ray of light is 0.00006 m. It is equal to
View answer and explanation
Correct option: B
60 micron
Option B (60 micron) is correct because 0.00006 m is equal to 60 micrometers.
326 Medium
326 of 502
Analyzing +4 | -1
Assertion : Rate of flow of a liquid represents velocity of flow.

Reason : The dimensions of rate of flow are [MLT][M L T].
View answer and explanation
Correct option: D
If the assertion and reason both are false.
Option D (If the assertion and reason both are false) is correct because rate of flow is volume per unit time, not velocity, and its dimensions are [L³T⁻¹], not [MLT].
327 Medium
327 of 502
Analyzing +4 | -1
A thin 1 m1 \mathrm{~m} long rod has a radius of 5 mm5 \mathrm{~mm}. A force of 50π kN50 \pi \, \mathrm{kN} is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm0.01 \mathrm{~mm}, which of the following statements is false?
View answer and explanation
Correct option: A
The maximum value of YY that can be determined is 1014 N/m210^{14} \mathrm{~N} / \mathrm{m}^2
Option A (The maximum value of YY that can be determined is 1014 N/m210^{14} \mathrm{~N} / \mathrm{m}^2) is false because the calculated Young's modulus is significantly lower than this value, given the provided parameters and uncertainties.
328 Medium
328 of 502
Remembering +4 | -1
Which of the following is the smallest unit
View answer and explanation
Correct option: C
Fermi
Option C (Fermi) is correct because 1 Fermi is equal to 10-15 meters, which is the smallest unit among the given options.
329 Medium
329 of 502
Understanding +4 | -1
1kWh  =1kWh\; =
View answer and explanation
Correct option: B
36×105J36 \times {10^5}J
Option B (36×105J36 \times {10^5}J) is correct because 1 kWh is equal to 1000 watts multiplied by 3600 seconds, which results in 36×10536 \times {10^5} Joules.
330 Medium
330 of 502
Understanding +4 | -1
Velocity of sound in a gas is given by v=γpρv=\sqrt{\frac{\gamma p}{\rho}}. Dimensional formula for γ\gamma is
View answer and explanation
Correct option: B
[M0 L0 T0]\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]
Option B ([M⁰ L⁰ T⁰]) is correct because γ is a dimensionless quantity, being the ratio of two specific heats.
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