Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Physics PYQs
  3. /Units and Measurements

Units and Measurements — JEE Previous Year Questions

Every Units and Measurements question asked in JEE Main and JEE Advanced across the last 186 papers — 193 questions, each with its correct answer. Free to read, no account needed.

Questions

193

Papers it appeared in

149/186

Appearance rate

80%

All 193 Units and Measurements questions

Most recent papers first.

Q1·PhysicsNumericalJEE Advanced 2026
In a new system of units, the units of mass, length, time and current are 5 kg, 5 m, 5 s and 5 A, respectively. If μ0\mu_{0}μ0​ and ϵ0\epsilon_{0}ϵ0​ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of μ0/ϵ0\sqrt{\mu_{0}/\epsilon_{0}}μ0​/ϵ0​​, is:

Correct answer: 25

Step-by-step solution →
Q2·PhysicsNumericalJEE Advanced 2026
In a single slit diffraction experiment, a slit of width (0.016±0.002)(0.016 \pm 0.002)(0.016±0.002) mm is used to measure the wavelength of a monochromatic light source. In the diffraction pattern, the angular distance between the central maximum and first minimum is measured to be (2°±40′)(2° \pm 40')(2°±40′). The value of the fractional error in the measurement of wavelength is: [Given: sin⁡(2°)=0.035\sin(2°) = 0.035sin(2°)=0.035]

Correct answer: 0.46

Step-by-step solution →
Q3·PhysicsSingle correctJEE Main 2026
A new unit (α)(\alpha)(α) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of α\alphaα units if light takes 6 min. 40 s to cover this distance?
  1. (A)200 α200\,\alpha200α
  2. (B)400 α400\,\alpha400α
  3. (C)300 α300\,\alpha300α
  4. (D)500 α500\,\alpha500α

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
Consider the equation H=xpϵqErtsH = \dfrac{x^p \epsilon^q E^r}{t^s}H=tsxpϵqEr​, where HHH = magnetic field; EEE = electric field, ϵ\epsilonϵ = permittivity, xxx = distance, ttt = time. The values of p,q,rp, q, rp,q,r and sss respectively are:
  1. (A)1,1,1,11, 1, 1, 11,1,1,1
  2. (B)−1,1,2,1-1, 1, 2, 1−1,1,2,1
  3. (C)1,−1,−2,11, -1, -2, 11,−1,−2,1
  4. (D)−1,−2,−2,1-1, -2, -2, 1−1,−2,−2,1

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
Match List - I with List - II. Choose the correct answer from the options given below :
List - IList - II
A.Boltzmann constantI.[M−1L3T−2][M^{-1}L^{3}T^{-2}][M−1L3T−2]
B.Stefan's constantII.[ML2T−1][ML^{2}T^{-1}][ML2T−1]
C.Planck's constantIII.[ML2T−2K−1][ML^{2}T^{-2}K^{-1}][ML2T−2K−1]
D.Gravitational constantIV.[ML0T−3K−4][ML^{0}T^{-3}K^{-4}][ML0T−3K−4]
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-IV, B-III, C-II, D-I
  3. (C)A-III, B-IV, C-II, D-I
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
The potential energy of a particle changes with distance x from a fixed origin as V=Axx+BV = \frac{A\sqrt{x}}{x + B}V=x+BAx​​, where A and B are constant with appropriate dimensions. The dimensions of AB are ________.
  1. (A)[M1L5/2T−2][M^{1}L^{5/2}T^{-2}][M1L5/2T−2]
  2. (B)[M3/2L5/2T−2][M^{3/2}L^{5/2}T^{-2}][M3/2L5/2T−2]
  3. (C)[M1L2T−2][M^{1}L^{2}T^{-2}][M1L2T−2]
  4. (D)[M1L7/2T−2][M^{1}L^{7/2}T^{-2}][M1L7/2T−2]

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
The percentage error in the calculated volume of a sphere, if there is 2% error in its diameter measurement, is __________.
  1. (A)1
  2. (B)2
  3. (C)6
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
The density ρ of a uniform cylinder is determined by measuring its mass m, length l and diameter d. The measured values of m, l and d are 97.42 ± 0.02 g, 8.35 ± 0.05 mm and 20.20 ± 0.02 mm, respectively. Calculated percentage fractional error in ρ is ________.
  1. (A)0.63%
  2. (B)0.82%
  3. (C)0.72%
  4. (D)0.25%

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correctJEE Main 2026
LLL, CCC and RRR represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula ML2T−4A−2ML^2T^{-4}A^{-2}ML2T−4A−2 corresponds to __________.
  1. (A)RLC\frac{R}{\sqrt{LC}}LC​R​
  2. (B)RLC\frac{R}{LC}LCR​
  3. (C)CLR\frac{C}{\sqrt{LR}}LR​C​
  4. (D)1RLC\frac{1}{R}\sqrt{\frac{L}{C}}R1​CL​​

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
Match List-I with List-II. where h (Planck's constant), G (gravitational constant) and c (speed of light in vacuum) as fundamental units. Choose the correct answer from the options given below :
List-IList-II
A.Meter (L)I.hcG\sqrt{\dfrac{hc}{G}}Ghc​​
B.Second (S)II.Ghc5\sqrt{\dfrac{Gh}{c^{5}}}c5Gh​​
C.Kilogram (M)III.K2L2c3Gh\sqrt{\dfrac{K^{2}L^{2}c^{3}}{Gh}}GhK2L2c3​​
D.Kelvin (K)IV.Ghc3\sqrt{\dfrac{Gh}{c^{3}}}c3Gh​​
  1. (A)A-II, B-IV, C-I, D-III
  2. (B)A-IV, B-II, C-I, D-III
  3. (C)A-IV, B-I, C-II, D-III
  4. (D)A-III, B-I, C-II, D-IV

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A, respectively. The least counts of voltmeter and ammeter are 500 mV and 200 mA, respectively. The estimated error in the resistance measurement is ________ Ω.
  1. (A)0.25
  2. (B)2
  3. (C)2.5
  4. (D)0.18

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
Match the LIST-I with LIST-II. Choose the correct answer from the options given below:
List-IList-II
A.Planck's constantI.ML2T−2ML^{2}T^{-2}ML2T−2
B.Stopping potentialII.T−1T^{-1}T−1
C.Work functionIII.ML2T−1ML^{2}T^{-1}ML2T−1
D.Threshold frequencyIV.ML2T−3A−1ML^{2}T^{-3}A^{-1}ML2T−3A−1
  1. (A)A-III, B-IV, C-I, D-II
  2. (B)A-I, B-II, C-III, D-IV
  3. (C)A-IV, B-III, C-I, D-II
  4. (D)A-I, B-IV, C-III, D-II

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
The dimensional formula of 12ϵ0E2\frac{1}{2}\epsilon_{0}E^{2}21​ϵ0​E2 (ϵ0\epsilon_{0}ϵ0​ = permittivity of vacuum and EEE = electric field) is MaLbTcM^{a}L^{b}T^{c}MaLbTc. The value of 2a−b+c2a - b + c2a−b+c = ______.
  1. (A)0
  2. (B)1
  3. (C)−1
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm. The length measured by a scale of least count 0.1 cm is 150 cm. When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm, measured by a micrometer of least count 0.001 cm. The error in the measured Young's modulus is α×109\alpha \times 10^{9}α×109 N/m2^{2}2. The value of α\alphaα is ______. (Ignore the contribution of the load to Young's modulus error calculation)
  1. (A)1.3
  2. (B)1.65
  3. (C)0.13
  4. (D)0.25

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correctJEE Main 2026
Dimensions of universal gravitational constant (G)(G)(G) in terms of Planck's constant (h)(h)(h), distance (L)(L)(L), mass (M)(M)(M) and time (T)(T)(T) are ______.
  1. (A)[hTLM−2][hTLM^{-2}][hTLM−2]
  2. (B)[hT−1LM−2][hT^{-1}LM^{-2}][hT−1LM−2]
  3. (C)[hTL2M−2][hTL^{2}M^{-2}][hTL2M−2]
  4. (D)[h−1T−1LM−2][h^{-1}T^{-1}LM^{-2}][h−1T−1LM−2]

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correctJEE Main 2026
The time period of a simple harmonic oscillator is T=2πkmT = 2\pi\sqrt{\dfrac{k}{m}}T=2πmk​​ . The measured value of mass (m) of the object is 10 g with an accuracy of 10 mg, and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant(k) is _______%.
  1. (A)3.43
  2. (B)3.35
  3. (C)7.60
  4. (D)6.76

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2026
Match List-I with List-II. Choose the correct answer from the options given below :
List-IList-II
A.Coefficient of viscosityI.[ML–1^{–1}–1T–2^{–2}–2]
B.Surface tensionII.[ML2^{2}2T–2^{–2}–2]
C.PressureIII.[ML0^{0}0T–2^{–2}–2]
D.Surface energyIV.[ML–1^{–1}–1T–1^{–1}–1]
  1. (A)A-I, B-II, C-IV, D-III
  2. (B)A-IV,B-III,C-I,D-II
  3. (C)A-I, B-III, C-II, D-IV
  4. (D)A-IV, B-I, C-II, D-III

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2026
Match the List-I with List-II Choose the correct answer from the options given below:
List-IList-II
A.Magnetic inductionI.M L T−2A−2\mathrm{M\,L\,T^{-2}A^{-2}}MLT−2A−2
B.Magnetic fluxII.M L2 T−2A−2\mathrm{M\,L^{2}\,T^{-2}A^{-2}}ML2T−2A−2
C.Magnetic permeabilityIII.M L0 T−2A−1\mathrm{M\,L^{0}\,T^{-2}A^{-1}}ML0T−2A−1
D.Self inductanceIV.M L2 T−2A−1\mathrm{M\,L^{2}\,T^{-2}A^{-1}}ML2T−2A−1
  1. (A)A-IV,B-III,C-I,D-II
  2. (B)A-III,B-IV,C-II,D-I
  3. (C)A-I,B-III,C-IV,D-II
  4. (D)A-III,B-IV,C-I,D-II

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsNumericalJEE Main 2026
A ball of radius r and density ρ dropped through a viscous liquid of density σ and viscosity η attains its terminal velocity at time t, given by t = A ρa\rho^{a}ρa rbr^{b}rb ηc\eta^{c}ηc σd\sigma^{d}σd, where A is a constant and a, b c and d are integers. The value of b+ca+d\frac{b+c}{a+d}a+db+c​ is ______.

Correct answer: 1

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2026
Four persons measure the length of a rod as 20.00 cm, 19.75 cm, 17.01 cm and 18.25 cm. The relative error in the measurement of average length of the rod is :
  1. (A)0.24
  2. (B)0.18
  3. (C)0.06
  4. (D)0.08

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2026
Match the LIST-I with LIST-II Choose the correct answer from the options given below:
List-IList-II
A.Spring constantI.ML2T−2K−1ML^2T^{-2}K^{-1}ML2T−2K−1
B.Thermal conductivityII.ML0T−2ML^0T^{-2}ML0T−2
C.Boltzmann constantIII.ML2T−3A−2ML^2T^{-3}A^{-2}ML2T−3A−2
D.Inductive reactanceIV.MLT−3K−1MLT^{-3}K^{-1}MLT−3K−1
  1. (A)A-II, B-I, C-IV, D-III
  2. (B)A-I, B-IV, C-II, D-III
  3. (C)A-III, B-II, C-IV, D-I
  4. (D)A-II, B-IV, C-I, D-III

Correct answer: (D)

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2026
Consider a modified Bernoulli equation. (P+ABt2)+ρg(h+Bt)+12ρV2=constant\left(P + \frac{A}{Bt^{2}}\right) + \rho g(h + Bt) + \frac{1}{2}\rho V^{2} = \mathrm{constant}(P+Bt2A​)+ρg(h+Bt)+21​ρV2=constant If t has the dimension of time then the dimensions of A and B are ______, ______respectively.
  1. (A)[ML0T−1][ML^{0}T^{-1}][ML0T−1] and [M0LT][M^{0}LT][M0LT]
  2. (B)[ML0T−1][ML^{0}T^{-1}][ML0T−1] and [M0LT−1][M^{0}LT^{-1}][M0LT−1]
  3. (C)[ML0T−2][ML^{0}T^{-2}][ML0T−2] and [M0LT−2][M^{0}LT^{-2}][M0LT−2]
  4. (D)[ML0T−2][ML^{0}T^{-2}][ML0T−2] and [M0LT−1][M^{0}LT^{-1}][M0LT−1]

Correct answer: (B)

Step-by-step solution →
Q23·PhysicsSingle correctJEE Main 2026
In an experiment the values of two spring constants were measured as k1=(10±0.2)k_{1} = (10 \pm 0.2)k1​=(10±0.2) N/m and k2=(20±0.3)k_{2} = ( 20 \pm 0.3)k2​=(20±0.3) N/m. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
  1. (A)2.67%
  2. (B)2.33%
  3. (C)1.33%
  4. (D)1.67%

Correct answer: (D)

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2026
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 v0v_0v0​. Estimated maximum error in the quantity η is : (Ignore errors associated with σ, ρ and g, gravitational acceleration)
  1. (A)2Δrr−Δv0v02\frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}2rΔr​−v0​Δv0​​
  2. (B)2Δrr+Δv0v0\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}r2Δr​+v0​Δv0​​
  3. (C)2[Δrr+Δv0v0]2\left[\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}\right]2[rΔr​+v0​Δv0​​]
  4. (D)2[Δrr−Δv0v0]2\left[\frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}\right]2[rΔr​−v0​Δv0​​]

Correct answer: (B)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2026
Keeping the significant figures in view, the sum of the physical quantities 52.01m, 153.2 m and 0.123 m is :
  1. (A)205 m
  2. (B)205.333 m
  3. (C)205.33 m
  4. (D)205.3 m

Correct answer: (D)

Step-by-step solution →
Q26·PhysicsMultiple correctJEE Advanced 2025
Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm , 0.05 mm , and 6.0 μm, respectively. Which of the following option(s) give(s) the volume of the strip in cm3\text{cm}^{3}cm3 with correct significant figures:
  1. (A)3.2×10−53.2 \times 10^{-5}3.2×10−5
  2. (B)32.0×10−632.0 \times 10^{-6}32.0×10−6
  3. (C)3.0×10−53.0 \times 10^{-5}3.0×10−5
  4. (D)3×10−53 \times 10^{-5}3×10−5

Correct answer: (D)

Step-by-step solution →
Q27·PhysicsSingle correctJEE Advanced 2025
A temperature difference can generate e.m.f. in some materials. Let S be the e.m.f. produced per unit temperature difference between the ends of a wire, σ the electrical conductivity and κ the thermal conductivity of the material of the wire. Taking M, L, T, I and K as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity Z=S2σκZ = \frac{S^2\sigma}{\kappa}Z=κS2σ​ is:
  1. (A)[M0L0T0I0K0][M^0L^0T^0I^0K^0][M0L0T0I0K0]
  2. (B)[M0L0T0I0K−1][M^0L^0T^0I^0K^{-1}][M0L0T0I0K−1]
  3. (C)[M1L2T−2I−1K−1][M^1L^2T^{-2}I^{-1}K^{-1}][M1L2T−2I−1K−1]
  4. (D)[M1L2T−4I−1K−1][M^1L^2T^{-4}I^{-1}K^{-1}][M1L2T−4I−1K−1]

Correct answer: (B)

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2025
A quantity QQQ is formulated as X−2Y3/2Z−2/5X^{-2}Y^{3/2}Z^{-2/5}X−2Y3/2Z−2/5. XXX, YYY and ZZZ are independent parameters which have fractional errors of 0.1, 0.2 and 0.5, respectively in measurement. The maximum fractional error of QQQ is:
  1. (A)0.10.10.1
  2. (B)0.80.80.8
  3. (C)0.70.70.7
  4. (D)0.60.60.6

Correct answer: (C)

Step-by-step solution →
Q29·PhysicsSingle correctJEE Main 2025
Match the LIST-I (Mechanical quantity) with LIST-II (Dimensional formula). Choose the correct answer from the options given below:
LIST-I (Mechanical quantity)LIST-II (Dimensional formula)
A.Mass densityI.[ML2T−3][ML^2T^{-3}][ML2T−3]
B.ImpulseII.[MLT−1][MLT^{-1}][MLT−1]
C.PowerIII.[ML2T0][ML^2T^0][ML2T0]
D.Moment of inertiaIV.[ML−3T0][ML^{-3}T^0][ML−3T0]
  1. (A)(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  2. (B)(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  3. (C)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  4. (D)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct answer: (C)

Step-by-step solution →
Q30·PhysicsSingle correctJEE Main 2025
In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of MPLQTRASM^P L^Q T^R A^SMPLQTRAS, where value of QQQ and RRR are
  1. (A)(3,−5)(3,-5)(3,−5)
  2. (B)(−2,2)(-2,2)(−2,2)
  3. (C)(−2,1)(-2,1)(−2,1)
  4. (D)(1,−1)(1,-1)(1,−1)

Correct answer: (D)

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2025
In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of [MP LQ TA][M^P\,L^Q\,T^A][MPLQTA]. The value of P and Q are:
  1. (A)1, 0
  2. (B)−1,1-1, 1−1,1
  3. (C)1,−11, -11,−1
  4. (D)0,−10, -10,−1

Correct answer: (D)

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2025
For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm):
  1. (A)0.001
  2. (B)0.002
  3. (C)0.0005
  4. (D)0.0025

Correct answer: (B)

Step-by-step solution →
Q33·PhysicsIntegerJEE Main 2025
A physical quantity C is related to four other quantities p, q, r and s as follows C=pq2r3sC=\dfrac{pq^2}{r^3\sqrt s}C=r3s​pq2​. The percentage errors in the measurement of p, q, r and s are 1%, 2%, 3% and 2% respectively. The percentage error in the measurement of C will be __________ %.

Correct answer: 15

Step-by-step solution →
Q34·PhysicsSingle correctJEE Main 2025
Match the LIST-I (Physical quantities and constants) with LIST-II (Dimensional formula). Choose the correct answer from the options given below:
LIST-I (Physical quantities and constants)LIST-II (Dimensional formula)
A.Boltzmann constantI.ML2T−1ML^2T^{-1}ML2T−1
B.Coefficient of viscosityII.MLT−3K−1MLT^{-3}K^{-1}MLT−3K−1
C.Planck's constantIII.ML2T−2K−1ML^2T^{-2}K^{-1}ML2T−2K−1
D.Thermal conductivityIV.ML−1T−1ML^{-1}T^{-1}ML−1T−1
  1. (A)A-III, B-IV, C-I, D-II
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-III, B-II, C-I, D-IV
  4. (D)A-III, B-IV, C-II, D-I

Correct answer: (A)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2025
A person measures the mass of 3 different particles as 435.42 g, 226.3 g and 0.125 g. According to the rules for arithmetic operations with significant figures, the addition of the masses of the 3 particles will be:
  1. (A)661.845 g
  2. (B)662 g
  3. (C)661.8 g
  4. (D)661.84 g

Correct answer: (C)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2025
Match the LIST-I (Physical quantity) with LIST-II (Dimensional formula). Choose the correct answer from the options given below:
LIST-I (Physical quantity)LIST-II (Dimensional formula)
A.Gravitational constantI.[LT−2][LT^{-2}][LT−2]
B.Gravitational potential energyII.[L2T−2][L^2T^{-2}][L2T−2]
C.Gravitational potentialIII.[ML2T−2][ML^2T^{-2}][ML2T−2]
D.Acceleration due to gravityIV.[M−1L3T−2][M^{-1}L^3T^{-2}][M−1L3T−2]
  1. (A)(A)-IV, (B)-III, (C)-II, (D)-I
  2. (B)(A)-III, (B)-II, (C)-I, (D)-IV
  3. (C)(A)-II, (B)-IV, (C)-III, (D)-I
  4. (D)(A)-I, (B)-III, (C)-IV, (D)-II

Correct answer: (A)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2025
The equation for real gas is given by (P+aV2)(V−b)=RT\left(P+\dfrac{a}{V^2}\right)(V-b)=RT(P+V2a​)(V−b)=RT, where P,V,TP,V,TP,V,T and RRR are the pressure, volume, temperature and gas constant, respectively. The dimension of ab−2ab^{-2}ab−2 is equivalent to that of:
  1. (A)Planck’s constant
  2. (B)Compressibility
  3. (C)Strain
  4. (D)Energy density

Correct answer: (D)

Step-by-step solution →
Q38·PhysicsSingle correctJEE Main 2025
Match List-I (physical quantities) with List-II (dimensional formulae). Choose the correct answer from the options given below:
List-IList-II
A.Coefficient of viscosityI.[ML0T−3][ML^0T^{-3}][ML0T−3]
B.Intensity of waveII.[ML−2T−2][ML^{-2}T^{-2}][ML−2T−2]
C.Pressure gradientIII.[M−1LT2][M^{-1}LT^2][M−1LT2]
D.CompressibilityIV.[ML−1T−1][ML^{-1}T^{-1}][ML−1T−1]
  1. (A)(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  2. (B)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  3. (C)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  4. (D)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct answer: (B)

Step-by-step solution →
Q39·PhysicsSingle correctJEE Main 2025
Given a charge qqq, current III and permeability of vacuum μ0\mu_0μ0​. Which of the following quantity has the dimension of momentum?
  1. (A)qI/μ0qI/\mu_0qI/μ0​
  2. (B)qμ0Iq\mu_0 Iqμ0​I
  3. (C)q2μ0Iq^2\mu_0 Iq2μ0​I
  4. (D)qμ0/Iq\mu_0/Iqμ0​/I

Correct answer: (B)

Step-by-step solution →
Q40·PhysicsSingle correctJEE Main 2025
The pair of physical quantities not having same dimensions is:
  1. (A)Torque and energy
  2. (B)Surface tension and impulse
  3. (C)Angular momentum and Planck’s constant
  4. (D)Pressure and Young’s modulus

Correct answer: (B)

Step-by-step solution →
Q41·PhysicsIntegerJEE Main 2025
A physical quantity Q is related to four observables a, b, c, d as follows: Q=ab4cdQ=\dfrac{ab^4}{cd}Q=cdab4​, where a=(60±3)a=(60\pm 3)a=(60±3) Pa; b=(20±0.1)b=(20\pm 0.1)b=(20±0.1) m; c=(40±0.2)c=(40\pm 0.2)c=(40±0.2) Nsm⁻² and d=(50±0.1)d=(50\pm 0.1)d=(50±0.1) m, then the percentage error in Q is x1000\dfrac{x}{1000}1000x​, where x=x=x= ______.

Correct answer: 7700

Step-by-step solution →
Q42·PhysicsSingle correctJEE Main 2025
Match List-I (Mechanical Quantity) with List-II (Dimensional Formula). Choose the correct answer from the options given below:
List-I (Mechanical Quantity)List-II (Dimensional Formula)
A.Young’s ModulusI.ML−1T−1ML^{-1}T^{-1}ML−1T−1
B.TorqueII.ML−1T−2ML^{-1}T^{-2}ML−1T−2
C.Coefficient of ViscosityIII.M−1L3T−2M^{-1}L^{3}T^{-2}M−1L3T−2
D.Gravitational ConstantIV.ML2T−2ML^{2}T^{-2}ML2T−2
  1. (A)(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  2. (B)(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  3. (C)(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  4. (D)(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Correct answer: (D)

Step-by-step solution →
Q43·PhysicsSingle correctJEE Main 2025
The expression given below shows the variation of velocity (v) with time (t), v=At2+BtC+tv=At^2+\dfrac{Bt}{C+t}v=At2+C+tBt​. The dimension of ABC is:
  1. (A)[M0L2T−3][M^0 L^2 T^{-3}][M0L2T−3]
  2. (B)[M0L1T−3][M^0 L^1 T^{-3}][M0L1T−3]
  3. (C)[M0L1T−2][M^0 L^1 T^{-2}][M0L1T−2]
  4. (D)[M0L2T−2][M^0 L^2 T^{-2}][M0L2T−2]

Correct answer: (A)

Step-by-step solution →
Q44·PhysicsIntegerJEE Main 2025
A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 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{x}{100}100x​ where x is______

Correct answer: 3

Step-by-step solution →
Q45·PhysicsSingle correctJEE Main 2025
Match List-I (Physical Quantity) with List-II (Dimensional Formula). Choose the correct answer from the options given below:
List-I (Physical Quantity)List-II (Dimensional Formula)
A.Angular ImpulseI.[M0L2T−2][M^0L^2T^{-2}][M0L2T−2]
B.Latent HeatII.[ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
C.Electrical resistivityIII.[ML2T−1][ML^2T^{-1}][ML2T−1]
D.Electromotive forceIV.[ML3T−3A−2][ML^3T^{-3}A^{-2}][ML3T−3A−2]
  1. (A)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (B)(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  3. (C)(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  4. (D)(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Correct answer: (A)

Step-by-step solution →
Q46·PhysicsSingle correctJEE Main 2025
For an experimental expression y=32.3×112527.4y=\dfrac{32.3\times1125}{27.4}y=27.432.3×1125​, where all the digits are significant. Then to report the value of y we should write :-
  1. (A)y=1326.2y=1326.2y=1326.2
  2. (B)y=1326.19y=1326.19y=1326.19
  3. (C)y=1326.186y=1326.186y=1326.186
  4. (D)y=1330y=1330y=1330

Correct answer: (D)

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2025
The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation.
  1. (A)E2=pc2+m2c4E^2=pc^2+m^2c^4E2=pc2+m2c4
  2. (B)E2=pc2+m2c2E^2=pc^2+m^2c^2E2=pc2+m2c2
  3. (C)E2=p2c2+m2c2E^2=p^2c^2+m^2c^2E2=p2c2+m2c2
  4. (D)E2=p2c2+m2c4E^2=p^2c^2+m^2c^4E2=p2c2+m2c4

Correct answer: (D)

Step-by-step solution →
Q48·PhysicsSingle correctJEE Main 2025
The energy of a system is given as E(t)=α3e−βtE(t)=\alpha^3 e^{-\beta t}E(t)=α3e−βt, where ttt is the time and β=0.3 s−1\beta=0.3\ \mathrm{s^{-1}}β=0.3 s−1. The errors in the measurement of α\alphaα and ttt are 1.2%1.2\%1.2% and 1.6%1.6\%1.6%, respectively. At t=5t=5t=5 s, maximum percentage error in the energy is :
  1. (A)4%4\%4%
  2. (B)11.6%11.6\%11.6%
  3. (C)6%6\%6%
  4. (D)8.4%8.4\%8.4%

Correct answer: (C)

Step-by-step solution →
Q49·PhysicsSingle correctJEE Main 2025
Match List-I (Electromagnetic Quantity) with List-II (Dimensional Formula). Choose the correct answer from the options given below:
List-I (Electromagnetic Quantity)List-II (Dimensional Formula)
A.Permeability of free spaceI.[M L2 T−2][\mathrm{M\,L^2\,T^{-2}}][ML2T−2]
B.Magnetic fieldII.[M T−2 A−1][\mathrm{M\,T^{-2}\,A^{-1}}][MT−2A−1]
C.Magnetic momentIII.[M L T−2 A−2][\mathrm{M\,L\,T^{-2}\,A^{-2}}][MLT−2A−2]
D.Torsional constantIV.[L2 A][\mathrm{L^2\,A}][L2A]
  1. (A)(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
  2. (B)(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
  3. (C)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  4. (D)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Correct answer: (D)

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2025
The position of a particle moving on x-axis is given by x(t)=Asin⁡t+Bcos⁡2t+Ct2+Dx(t)=A\sin t+B\cos^2 t+Ct^2+Dx(t)=Asint+Bcos2t+Ct2+D, where ttt is time. The dimension of ABCD\dfrac{ABC}{D}DABC​ is-
  1. (A)L
  2. (B)L3T−2L^3T^{-2}L3T−2
  3. (C)L2T−2L^2T^{-2}L2T−2
  4. (D)L2L^2L2

Correct answer: (C)

Step-by-step solution →
Q51·PhysicsSingle correctJEE Main 2025
If B is the magnetic field and μ₀ is the permeability of free space, then the dimensions of (B/μ₀) are:
  1. (A)M T⁻¹ A⁻¹
  2. (B)L⁻¹ A
  3. (C)L T⁻¹ A⁻¹
  4. (D)M L T⁻¹ A⁻¹

Correct answer: (B)

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2025
Which one of the following is the correct dimensional formula for the capacitance in F? M, L, T and C stand for unit of mass, length, time and charge,
  1. (A)[F]=[C2M−2L2T2][F]=[C^{2}M^{-2}L^{2}T^{2}][F]=[C2M−2L2T2]
  2. (B)[F]=[CM−2L−2T−2][F]=[CM^{-2}L^{-2}T^{-2}][F]=[CM−2L−2T−2]
  3. (C)[F]=[CM−1L−2T2][F]=[CM^{-1}L^{-2}T^{2}][F]=[CM−1L−2T2]
  4. (D)[F]=[C2M−1L−2T2][F]=[C^{2}M^{-1}L^{-2}T^{2}][F]=[C2M−1L−2T2]

Correct answer: (D)

Step-by-step solution →
Q53·PhysicsSingle correctJEE Main 2025
The maximum percentage error in the measurement of density of a wire is [Given, mass of wire =(0.60±0.003)=(0.60\pm0.003)=(0.60±0.003) g, radius of wire =(0.50±0.01)=(0.50\pm0.01)=(0.50±0.01) cm, length of wire =(10.00±0.05)=(10.00\pm0.05)=(10.00±0.05) cm]
  1. (A)4
  2. (B)5
  3. (C)8
  4. (D)7

Correct answer: (B)

Step-by-step solution →
Q54·PhysicsNumericalJEE Advanced 2024
The dimensions of a cone are measured using a scale with a least count of 2mm . The diameter of the base and the height are both measured to be 20.0cm . The maximum percentage error in the determination of the volume is -

Correct answer: 3

Step-by-step solution →
Q55·PhysicsSingle correctJEE Advanced 2024
A dimensionless quantity is constructed in terms of electronic charge eee, permittivity of free space ε0\varepsilon_{0}ε0​, Planck's constant hhh, and speed of light ccc. If the dimensionless quantity is written as eαε0βhγcδe^{\alpha}\varepsilon_{0}^{\beta}h^{\gamma}c^{\delta}eαε0β​hγcδ and nnn is a non-zero integer, then (α,β,γ,δ)(\alpha, \beta, \gamma, \delta)(α,β,γ,δ) is given by
  1. (A)(2n,−n,−n,−n)(2n, -n, -n, -n)(2n,−n,−n,−n)
  2. (B)(n,−n,−2n,−n)(n, -n, -2n, -n)(n,−n,−2n,−n)
  3. (C)(n,−n,−n,−2n)(n, -n, -n, -2n)(n,−n,−n,−2n)
  4. (D)(2n,−n,−2n,−2n)(2n, -n, -2n, -2n)(2n,−n,−2n,−2n)

Correct answer: (A)

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2024
The dimensional formula of latent heat is:
  1. (A)[M0LT−2][M^0 L T^{-2}][M0LT−2]
  2. (B)[MLT−2][M L T^{-2}][MLT−2]
  3. (C)[M0L2T−2][M^0 L^2 T^{-2}][M0L2T−2]
  4. (D)[ML2T−2][M L^2 T^{-2}][ML2T−2]

Correct answer: (C)

Step-by-step solution →
Q57·PhysicsSingle correctJEE Main 2024
The de-Broglie wavelength associated with a particle of mass mmm and energy EEE is h/2mEh/\sqrt{2mE}h/2mE​. The dimensional formula for Planck's constant is:
  1. (A)[ML−1T−1][ML^{-1}T^{-1}][ML−1T−1]
  2. (B)[ML2T−1][ML^{2}T^{-1}][ML2T−1]
  3. (C)[MLT−1][MLT^{-1}][MLT−1]
  4. (D)[M4L2T−3][M^{4}L^{2}T^{-3}][M4L2T−3]

Correct answer: (B)

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2024
There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is:
  1. (A)4.65 mm
  2. (B)4.55 mm
  3. (C)4.60 mm
  4. (D)3.35 mm

Correct answer: (B)

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2024
In an expression a×10ba\times10^ba×10b:
  1. (A)aaa is order of magnitude for b≤5b\le5b≤5
  2. (B)bbb is order of magnitude for a≤5a\le5a≤5
  3. (C)bbb is order of magnitude for 5<a≤105<a\le105<a≤10
  4. (D)bbb is order of magnitude for a≥5a\ge5a≥5

Correct answer: (B)

Step-by-step solution →
Q60·PhysicsSingle correctJEE Main 2024
Young's modulus is determined by the equation Y=49000mℓdynecm2Y=49000\dfrac{m}{\ell}\dfrac{\text{dyne}}{cm^2}Y=49000ℓm​cm2dyne​ where MMM is the mass and ℓ\ellℓ is the extension of wire used in the experiment. Now error in Young modulus (Y)(Y)(Y) is estimated by taking data from M-ℓ\ellℓ plot in graph paper. The smallest scale divisions are 5 g5\,g5g and 0.02 cm0.02\,cm0.02cm along load axis and extension axis respectively. If the value of MMM and ℓ\ellℓ are 500 g500\,g500g and 2 cm2\,cm2cm respectively then percentage error of YYY is:
  1. (A)0.2%0.2\%0.2%
  2. (B)0.02%0.02\%0.02%
  3. (C)2%2\%2%
  4. (D)0.5%0.5\%0.5%

Correct answer: (C)

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2024
Least count of a vernier caliper is 120N\dfrac{1}{20N}20N1​ cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is:
  1. (A)(2N−120N)\left(\dfrac{2N-1}{20N}\right)(20N2N−1​)
  2. (B)(2N−12)\left(\dfrac{2N-1}{2}\right)(22N−1​)
  3. (C)(2N−1)(2N-1)(2N−1)
  4. (D)(2N−12N)\left(\dfrac{2N-1}{2N}\right)(2N2N−1​)

Correct answer: (B)

Step-by-step solution →
Q62·PhysicsSingle correctJEE Main 2024
If ε0\varepsilon_0ε0​ is the permittivity of free space and E is the electric field, then ε0E2\varepsilon_0 E^2ε0​E2 has the dimensions:
  1. (A)[M0L−2TA][M^0L^{-2}TA][M0L−2TA]
  2. (B)[ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
  3. (C)[M−1L−3T4A2][M^{-1}L^{-3}T^4A^2][M−1L−3T4A2]
  4. (D)[ML2T−2][ML^2T^{-2}][ML2T−2]

Correct answer: (B)

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2024
Given below are two statements : Statement (I) : Dimensions of specific heat is [L2T−2K−1][L^{2}T^{-2}K^{-1}][L2T−2K−1]. Statement (II) : Dimensions of gas constant is [ML2T−1K−1][M L^{2}T^{-1}K^{-1}][ML2T−1K−1].
  1. (A)Statement (I) is incorrect but statement (II) is correct
  2. (B)Both statement (I) and statement (II) are incorrect
  3. (C)Statement (I) is correct but statement (II) is incorrect
  4. (D)Both statement (I) and statement (II) are correct

Correct answer: (C)

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2024
To find the spring constant (kkk) of a spring experimentally, a student commits 2% positive error in the measurement of time and 1% negative error in the measurement of mass. The percentage error in determining the value of kkk is:
  1. (A)3%3\%3%
  2. (B)1%1\%1%
  3. (C)4%4\%4%
  4. (D)5%5\%5%

Correct answer: (D)

Step-by-step solution →
Q65·PhysicsSingle correctJEE Main 2024
Match List-I (physical quantities) with List-II (their dimensional formulae). Choose the correct answer from the options given below:
List-IList-II
A.TorqueI.[M1L1T−2A−2][M^{1}L^{1}T^{-2}A^{-2}][M1L1T−2A−2]
B.Magnetic fieldII.[L2A1][L^{2}A^{1}][L2A1]
C.Magnetic momentIII.[M1T−2A−1][M^{1}T^{-2}A^{-1}][M1T−2A−1]
D.Permeability of free spaceIV.[M1L2T−2][M^{1}L^{2}T^{-2}][M1L2T−2]
  1. (A)A-I, B-III, C-II, D-IV
  2. (B)A-IV, B-III, C-II, D-I
  3. (C)A-III, B-I, C-II, D-IV
  4. (D)A-IV, B-II, C-III, D-I

Correct answer: (B)

Step-by-step solution →
Q66·PhysicsSingle correctJEE Main 2024
What is the dimensional formula of ab−1ab^{-1}ab−1 in the equation (P+aV2)(V−b)=RT\left(P+\dfrac{a}{V^2}\right)(V-b)=RT(P+V2a​)(V−b)=RT, where the letters have their usual meaning?
  1. (A)[M0L3T−2][M^0L^3T^{-2}][M0L3T−2]
  2. (B)[ML2T−2][ML^2T^{-2}][ML2T−2]
  3. (C)[M−1L5T3][M^{-1}L^5T^3][M−1L5T3]
  4. (D)[M6L7T4][M^6L^7T^4][M6L7T4]

Correct answer: (B)

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2024
If G is the gravitational constant and uuu is the energy density then which of the following quantity has the same dimensions as uG\sqrt{uG}uG​:
  1. (A)Pressure gradient per unit mass
  2. (B)Force per unit mass
  3. (C)Gravitational potential
  4. (D)Energy per unit mass

Correct answer: (B)

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2024
Applying the principle of homogeneity of dimensions, determine which one is correct, where T is time period, G is gravitational constant, M is mass, r is radius of orbit.
  1. (A)T2=4π2rGM2T^2=\tfrac{4\pi^2 r}{GM^2}T2=GM24π2r​
  2. (B)T2=4π2r3T^2=4\pi^2 r^3T2=4π2r3
  3. (C)T2=4π2r3GMT^2=\tfrac{4\pi^2 r^3}{GM}T2=GM4π2r3​
  4. (D)T2=4π2r2GMT^2=\tfrac{4\pi^2 r^2}{GM}T2=GM4π2r2​

Correct answer: (C)

Step-by-step solution →
Q69·PhysicsSingle correctJEE Main 2024
The dimensional formula of angular impulse is:
  1. (A)[ML−2T−1][ML^{-2}T^{-1}][ML−2T−1]
  2. (B)[ML2T−1][ML^2T^{-1}][ML2T−1]
  3. (C)[MLT−1][MLT^{-1}][MLT−1]
  4. (D)[ML2T−2][ML^2T^{-2}][ML2T−2]

Correct answer: (B)

Step-by-step solution →
Q70·PhysicsSingle correctJEE Main 2024
The radius (r)(r)(r), length (l)(l)(l) and resistance (R)(R)(R) of a metal wire was measured in the laboratory as r=(0.35±0.05) cmr=(0.35\pm0.05)\,cmr=(0.35±0.05)cm, R=(100±10)R=(100\pm10)R=(100±10) ohm, l=(15±0.2) cml=(15\pm0.2)\,cml=(15±0.2)cm. The percentage error in resistivity of the material of the wire is:
  1. (A)25.6%25.6\%25.6%
  2. (B)39.9%39.9\%39.9%
  3. (C)37.3%37.3\%37.3%
  4. (D)35.6%35.6\%35.6%

Correct answer: (B)

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2024
Match List-I (Number) with List-II (Significant figures). Choose the correct answer from the options given below:
List-I (Number)List-II (Significant figure)
A.1001I.3
B.010.1II.4
C.100.100III.5
D.0.0010010IV.6
  1. (A)(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  2. (B)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  3. (C)(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  4. (D)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Correct answer: (C)

Step-by-step solution →
Q72·PhysicsSingle correctJEE Main 2024
If the percentage errors in measuring the length and the diameter of a wire are 0.1% each. The percentage error in measuring its resistance will be :
  1. (A)0.2%0.2\%0.2%
  2. (B)0.3%0.3\%0.3%
  3. (C)0.1%0.1\%0.1%
  4. (D)0.144%0.144\%0.144%

Correct answer: (B)

Step-by-step solution →
Q73·PhysicsSingle correctJEE Main 2024
Consider two physical quantities A and B related to each other as E=B−x2AtE=\dfrac{B-x^2}{At}E=AtB−x2​ where E, xxx and ttt have dimensions of energy, length and time respectively. The dimension of AB is
  1. (A)L−2M1T0L^{-2}M^1T^0L−2M1T0
  2. (B)L2M−1T1L^2M^{-1}T^1L2M−1T1
  3. (C)L−2M−1T1L^{-2}M^{-1}T^1L−2M−1T1
  4. (D)L0M1T−1L^0M^1T^{-1}L0M1T−1

Correct answer: (B)

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2024
A force is represented by F=ax2+bt1/2F=ax^2+bt^{1/2}F=ax2+bt1/2 where x = distance and t = time. The dimensions of b2a\dfrac{b^2}{a}ab2​ are :
  1. (A)[ML3T−3][ML^3T^{-3}][ML3T−3]
  2. (B)[MLT−2][MLT^{-2}][MLT−2]
  3. (C)[ML−1T−1][ML^{-1}T^{-1}][ML−1T−1]
  4. (D)[ML2T−3][ML^2T^{-3}][ML2T−3]

Correct answer: (A)

Step-by-step solution →
Q75·PhysicsSingle correctJEE Main 2024
The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:
  1. (A)4
  2. (B)8
  3. (C)6
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q76·PhysicsSingle correctJEE Main 2024
Match List-I (physical quantities) with List-II (their dimensional formulae). Choose the correct answer:
List-I (Physical quantity)List-II (Dimensional formula)
A.Coefficient of viscosityI.[M L2 T−2][M\,L^2\,T^{-2}][ML2T−2]
B.Surface TensionII.[M L2 T−1][M\,L^2\,T^{-1}][ML2T−1]
C.Angular momentumIII.[M L−1 T−1][M\,L^{-1}\,T^{-1}][ML−1T−1]
D.Rotational kinetic energyIV.[M L0 T−2][M\,L^0\,T^{-2}][ML0T−2]
  1. (A)A-II, B-I, C-IV, D-III
  2. (B)A-I, B-II, C-III, D-IV
  3. (C)A-III, B-IV, C-II, D-I
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (C)

Step-by-step solution →
Q77·PhysicsSingle correctJEE Main 2024
If mass is written as m=k cPG−1/2h1/2m=k\,c^P G^{-1/2} h^{1/2}m=kcPG−1/2h1/2 then the value of PPP will be: (Constants have their usual meaning with kkk a dimensionless constant)
  1. (A)12\dfrac{1}{2}21​
  2. (B)13\dfrac{1}{3}31​
  3. (C)2
  4. (D)−13-\dfrac{1}{3}−31​

Correct answer: (A)

Step-by-step solution →
Q78·PhysicsSingle correctJEE Main 2024
A physical quantity QQQ is found to depend on quantities aaa, bbb, ccc by the relation Q=a4b3c2Q=\dfrac{a^{4}b^{3}}{c^{2}}Q=c2a4b3​. The percentage error in aaa, bbb and ccc are 3%, 4% and 5% respectively. Then, the percentage error in QQQ is:
  1. (A)66%
  2. (B)43%
  3. (C)34%
  4. (D)14%

Correct answer: (C)

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2024
The resistance R=VIR=\dfrac{V}{I}R=IV​ where V=(200±5)V=(200\pm5)V=(200±5) V and I=(20±0.2)I=(20\pm0.2)I=(20±0.2) A, the percentage error in the measurement of R is :
  1. (A)3.5%3.5\%3.5%
  2. (B)7%7\%7%
  3. (C)3%3\%3%
  4. (D)5.5%5.5\%5.5%

Correct answer: (A)

Step-by-step solution →
Q80·PhysicsSingle correctJEE Main 2024
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:
  1. (A)Statement I is true but Statement II is false
  2. (B)Both Statement I and Statement II are false
  3. (C)Both Statement I and Statement II are true
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q81·PhysicsSingle correctJEE Advanced 2023
Young's modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational constant G, Planck's constant h and the speed of light c, as Y=cαhβGγY = c^{\alpha} h^{\beta} G^{\gamma}Y=cαhβGγ. Which of the following is the correct option?
  1. (A)α=7,β=−1,γ=−2\alpha = 7, \beta = -1, \gamma = -2α=7,β=−1,γ=−2
  2. (B)α=−7,β=−1,γ=−2\alpha = -7, \beta = -1, \gamma = -2α=−7,β=−1,γ=−2
  3. (C)α=7,β=−1,γ=2\alpha = 7, \beta = -1, \gamma = 2α=7,β=−1,γ=2
  4. (D)α=−7,β=1,γ=−2\alpha = -7, \beta = 1, \gamma = -2α=−7,β=1,γ=−2

Correct answer: (A)

Step-by-step solution →
Q82·PhysicsNumericalJEE Advanced 2023
In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is 10 ±\pm± 0.1 cm and the distance of its real image from the lens is 20 ±\pm± 0.2 cm. The error in the determination of focal length of the lens is n%. The value of n is ________.

Correct answer: 1

Step-by-step solution →
Q83·PhysicsSingle correctJEE Main 2023
The speed of a wave produced in water is given by v=λagbρcv = \lambda^a g^b \rho^cv=λagbρc. Where λ\lambdaλ, ggg and ρ\rhoρ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of a, b and c respectively, are :
  1. (A)12,12,0\frac{1}{2}, \frac{1}{2}, 021​,21​,0
  2. (B)1,1,01, 1, 01,1,0
  3. (C)1,−1,01, -1, 01,−1,0
  4. (D)12,0,12\frac{1}{2}, 0, \frac{1}{2}21​,0,21​

Correct answer: (A)

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2023
In the equation [X+aY2][Y−b]=RT\left[X + \dfrac{a}{Y^2}\right][Y - b] = RT[X+Y2a​][Y−b]=RT, XXX is pressure, YYY is volume, R is universal gas constant and TTT is temperature. The physical quantity equivalent to the ratio ab\dfrac{a}{b}ba​ is:
  1. (A)Energy
  2. (B)Impulse
  3. (C)Pressure gradient
  4. (D)Coefficient of viscosity

Correct answer: (A)

Step-by-step solution →
Q85·PhysicsSingle correctJEE Main 2023
Match List I (physical quantities: spring constant, angular speed, angular momentum, moment of inertia) with List II (dimensional formulae). Choose the correct answer from the options given below:
List IList II
A.Spring constantI.[T−1][T^{-1}][T−1]
B.Angular speedII.[MT−2][MT^{-2}][MT−2]
C.Angular momentumIII.[ML2][ML^2][ML2]
D.Moment of InertiaIV.[ML2T−1][ML^2T^{-1}][ML2T−1]
  1. (A)A-II, B-I, C-IV, D-III
  2. (B)A-IV, B-I, C-III, D-II
  3. (C)A-II, B-III, C-I, D-IV
  4. (D)A-II, B-III, C-II, D-IV

Correct answer: (A)

Step-by-step solution →
Q86·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statements I: Astronomical unit (Au). Parsec (Pc) and Light year (ly) are units for measuring astronomical distances. Statements II: Au<Parsec (Pc)<ly\text{Au} < \text{Parsec (Pc)} < \text{ly}Au<Parsec (Pc)<ly. In the light of the above statements. choose the most appropriate answer from the options given below:
  1. (A)Both Statements I and Statements II are correct.
  2. (B)Statements I is correct but Statements II is incorrect.
  3. (C)Both Statements I and Statements II are incorrect.
  4. (D)Statements I is incorrect but statements II is correct.

Correct answer: (B)

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2023
If force (F), velocity (V) and time (T) are considered as fundamental physical quantities, then the dimensional formula of density will be:
  1. (A)FV−2T2FV^{-2}T^{2}FV−2T2
  2. (B)FV−4T−2FV^{-4}T^{-2}FV−4T−2
  3. (C)FV4T−6FV^{4}T^{-6}FV4T−6
  4. (D)F2V−2T6F^{2}V^{-2}T^{6}F2V−2T6

Correct answer: (B)

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2023
A physical quantity P is given as P=a2b3cdP=\dfrac{a^2 b^3}{c\sqrt d}P=cd​a2b3​. The percentage error in the measurement of a, b, c and d are 1%,2%,3%1\%, 2\%, 3\%1%,2%,3% and 4%4\%4% respectively. The percentage error in the measurement of quantity P will be
  1. (A)13%
  2. (B)14%
  3. (C)15%
  4. (D)16%

Correct answer: (A)

Step-by-step solution →
Q89·PhysicsSingle correctJEE Main 2023
In an experiment with Vernier callipers of least count 0.10.10.1 mm, when the two jaws are joined together the zero of the Vernier scale lies to the right of the zero of the main scale and the 6th6^{th}6th division of the Vernier scale coincides with a main scale division. While measuring the diameter of a spherical bob, the zero of the Vernier scale lies in between 3.23.23.2 cm and 3.33.33.3 cm marks, and the 4th4^{th}4th division of the Vernier scale coincides with a main scale division. The diameter of the bob is measured as:
  1. (A)3.183.183.18 cm
  2. (B)3.253.253.25 cm
  3. (C)3.263.263.26 cm
  4. (D)3.223.223.22 cm

Correct answer: (A)

Step-by-step solution →
Q90·PhysicsSingle correctJEE Main 2023
A cylindrical wire of mass (0.4±0.01) g(0.4\pm0.01)\,\text{g}(0.4±0.01)g has length (8±0.04) cm(8\pm0.04)\,\text{cm}(8±0.04)cm and radius (6±0.03) mm(6\pm0.03)\,\text{mm}(6±0.03)mm. The maximum error in its density will be
  1. (A)1%1\%1%
  2. (B)3.5%3.5\%3.5%
  3. (C)4%4\%4%
  4. (D)5%5\%5%

Correct answer: (C)

Step-by-step solution →
Q91·PhysicsSingle correctJEE Main 2023
Match List-I (physical quantity) with List-II (dimensional formula). Choose the correct answer from the options given below:
List-I (Physical quantity)List-II (Dimensional formula)
A.TorqueI.ML−2T−2ML^{-2}T^{-2}ML−2T−2
B.StressII.ML2T−2ML^{2}T^{-2}ML2T−2
C.Pressure gradientIII.ML−1T−1ML^{-1}T^{-1}ML−1T−1
D.Coefficient of viscosityIV.ML−1T−2ML^{-1}T^{-2}ML−1T−2
  1. (A)A-III, B-IV, C-I, D-II
  2. (B)A-IV, B-II, C-III, D-I
  3. (C)A-II, B-IV, C-I, D-III
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (C)

Step-by-step solution →
Q92·PhysicsSingle correctJEE Main 2023
Dimension of 1μ0 ϵ0\dfrac{1}{\mu_{0}\,\epsilon_{0}}μ0​ϵ0​1​ should be equal to
  1. (A)T2/L2T^{2}/L^{2}T2/L2
  2. (B)L/TL/TL/T
  3. (C)L2/T2L^{2}/T^{2}L2/T2
  4. (D)T/LT/LT/L

Correct answer: (C)

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2023
If the velocity of light c, universal gravitational constant G and Planck's constant h are chosen as fundamental quantities. The dimensions of mass in the new system is:
  1. (A)[h1/2c−1/2G2][h^{1/2}c^{-1/2}G^2][h1/2c−1/2G2]
  2. (B)[h−1/2c1/2G1/2][h^{-1/2}c^{1/2}G^{1/2}][h−1/2c1/2G1/2]
  3. (C)[h1/2c1/2G−1/2][h^{1/2}c^{1/2}G^{-1/2}][h1/2c1/2G−1/2]
  4. (D)[h1c1G−1][h^{1}c^{1}G^{-1}][h1c1G−1]

Correct answer: (C)

Step-by-step solution →
Q94·PhysicsSingle correctJEE Main 2023
(P+aV2)(V−b)=RT\left(P+\tfrac{a}{V^2}\right)(V-b)=RT(P+V2a​)(V−b)=RT represents the equation of state of some gases, where PPP is the pressure, VVV is the volume, TTT is the temperature and a,b,Ra, b, Ra,b,R are the constants. The physical quantity, which has dimensional formula as that of b2a\tfrac{b^2}{a}ab2​, will be:
  1. (A)Compressibility
  2. (B)Energy density
  3. (C)Modulus of rigidity
  4. (D)Bulk modulus

Correct answer: (A)

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2023
Match List I (physical quantities) with List II (dimensional formulae). Choose the correct answer from the options given below:
List IList II
A.Angular momentumI.[ML2T−2][ML^2T^{-2}][ML2T−2]
B.TorqueII.[ML−2T−2][ML^{-2}T^{-2}][ML−2T−2]
C.StressIII.[ML2T−1][ML^2T^{-1}][ML2T−1]
D.Pressure gradientIV.[ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-III, B-II, C-IV, D-I
  3. (C)A-IV, B-II, C-I, D-III
  4. (D)A-I, B-IV, C-III, D-II

Correct answer: (A)

Step-by-step solution →
Q96·PhysicsSingle correctJEE Main 2023
Match List-I (physical quantities) with List-II (their dimensional formulae). Choose the correct answer from the options given below:
List-IList-II
A.TorqueI.kg m−1 s−2kg\,m^{-1}\,s^{-2}kgm−1s−2
B.Energy densityII.kg m s−1kg\,m\,s^{-1}kgms−1
C.Pressure gradientIII.kg m−2 s−2kg\,m^{-2}\,s^{-2}kgm−2s−2
D.ImpulseIV.kg m2 s−2kg\,m^{2}\,s^{-2}kgm2s−2
  1. (A)A-IV, B-I, C-III, D-II
  2. (B)A-IV, B-III, C-I, D-II
  3. (C)A-IV, B-I, C-II, D-III
  4. (D)A-I, B-IV, C-III, D-II

Correct answer: (A)

Step-by-step solution →
Q97·PhysicsSingle correctJEE Main 2023
The equation of a circle is given by x2+y2=a2x^2+y^2=a^2x2+y2=a2, where aaa is the radius. If the equation is modified to change the origin other than (0,0)(0,0)(0,0), then find out the correct dimensions of AAA and BBB in a new equation: (x−At)2+(y−tB)2=a2(x-At)^2+\left(y-\dfrac{t}{B}\right)^2=a^2(x−At)2+(y−Bt​)2=a2. The dimensions of ttt is given as [T−1][T^{-1}][T−1].
  1. (A)A=[LT], B=[L−1T−1]A=[LT],\,B=[L^{-1}T^{-1}]A=[LT],B=[L−1T−1]
  2. (B)A=[L−1T−1], B=[LT]A=[L^{-1}T^{-1}],\,B=[LT]A=[L−1T−1],B=[LT]
  3. (C)A=[L−1T], B=[LT−1]A=[L^{-1}T],\,B=[LT^{-1}]A=[L−1T],B=[LT−1]
  4. (D)A=[L−1T−1], B=[LT−1]A=[L^{-1}T^{-1}],\,B=[LT^{-1}]A=[L−1T−1],B=[LT−1]

Correct answer: (A)

Step-by-step solution →
Q98·PhysicsSingle correctJEE Main 2023
Match List I with List II and choose the correct answer from the options given below:
List I (Physical Quantity)List II (Dimensional Formula)
A.Pressure gradientI.[M0L2T−2][M^0 L^2 T^{-2}][M0L2T−2]
B.Energy densityII.[M1L−1T−2][M^1 L^{-1} T^{-2}][M1L−1T−2]
C.Electric FieldIII.[M1L−2T−2][M^1 L^{-2} T^{-2}][M1L−2T−2]
D.Latent heatIV.[M1L1T−3A−1][M^1 L^1 T^{-3} A^{-1}][M1L1T−3A−1]
  1. (A)A-II, B-III, C-I, D-IV
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-III, B-II, C-IV, D-I
  4. (D)A-III, B-II, C-I, D-IV

Correct answer: (C)

Step-by-step solution →
Q99·PhysicsSingle correctJEE Main 2023
Match List I (physical quantities) with List II (their dimensional formulae) and choose the correct answer.
List-IList-II
A.Young's Modulus (Y)I.[M L−1 T−1][M\,L^{-1}\,T^{-1}][ML−1T−1]
B.Co-efficient of Viscosity (η\etaη)II.[M L2 T−1][M\,L^2\,T^{-1}][ML2T−1]
C.Planck's Constant (h)IV.[M L2 T−2][M\,L^2\,T^{-2}][ML2T−2]
D.Work Function (φ\varphiφ)III.[M L−1 T−2][M\,L^{-1}\,T^{-2}][ML−1T−2]
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-I, B-III, C-IV, D-II
  4. (D)A-III, B-I, C-II, D-IV

Correct answer: (D)

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2023
The graph between two temperature scales PPP and QQQ is shown in the figure. Between upper fixed point and lower fixed point there are 150150150 equal divisions of scale P and 100100100 divisions on scale Q. The relationship for conversion between the two scales is given by:
  1. (A)tP100=tQ−180150\frac{t_P}{100} = \frac{t_Q - 180}{150}100tP​​=150tQ​−180​
  2. (B)tQ150=tP−180100\frac{t_Q}{150} = \frac{t_P - 180}{100}150tQ​​=100tP​−180​
  3. (C)tP180−tQ−40100\frac{t_P}{180} - \frac{t_Q - 40}{100}180tP​​−100tQ​−40​
  4. (D)tQ100=tP−30150\frac{t_Q}{100} = \frac{t_P - 30}{150}100tQ​​=150tP​−30​

Correct answer: (D)

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2023
Match List-I (physical quantity) with List-II (SI base-unit dimensional form). Choose the correct answer from the options given below:
List-IList-II
A.Surface tensionI.kg m−1 s−1kg\,m^{-1}\,s^{-1}kgm−1s−1
B.PressureII.kg m s−1kg\,m\,s^{-1}kgms−1
C.ViscosityIII.kg m−1 s−2kg\,m^{-1}\,s^{-2}kgm−1s−2
D.ImpulseIV.kg s−2kg\,s^{-2}kgs−2
  1. (A)A-II, B-I, C-III, D-IV
  2. (B)A-IV, B-III, C-I, D-II
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (B)

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2023
The frequency (ν)(\nu)(ν) of an oscillating liquid drop may depend upon radius (r)(r)(r) of the drop, density (ρ)(\rho)(ρ) of liquid and the surface tension (s)(s)(s) of the liquid as: ν=raρbsc\nu=r^a\rho^b s^cν=raρbsc. The values of aaa, bbb and ccc respectively are
  1. (A)(−32,12,12)\left(-\dfrac32,\dfrac12,\dfrac12\right)(−23​,21​,21​)
  2. (B)(32,−12,12)\left(\dfrac32,-\dfrac12,\dfrac12\right)(23​,−21​,21​)
  3. (C)(−32,−12,12)\left(-\dfrac32,-\dfrac12,\dfrac12\right)(−23​,−21​,21​)
  4. (D)(32,12,−12)\left(\dfrac32,\dfrac12,-\dfrac12\right)(23​,21​,−21​)

Correct answer: (C)

Step-by-step solution →
Q103·PhysicsSingle correctJEE Main 2023
Match List I (physical quantities) with List II (their dimensional formulae). Choose the correct answer from the options given below:
List IList II
A.Planck's constant (h)I.[M1L2T−2][M^1L^2T^{-2}][M1L2T−2]
B.Stopping potential (VsV_sVs​)II.[M1L1T−1][M^1L^1T^{-1}][M1L1T−1]
C.Work function (ϕ\phiϕ)III.[M1L2T−1][M^1L^2T^{-1}][M1L2T−1]
D.Momentum (p)IV.[M1L2T−3A−1][M^1L^2T^{-3}A^{-1}][M1L2T−3A−1]
  1. (A)A-I, B-III, C-IV, D-II
  2. (B)A-III, B-I, C-II, D-IV
  3. (C)A-II, B-IV, C-III, D-I
  4. (D)A-III, B-IV, C-I, D-II

Correct answer: (D)

Step-by-step solution →
Q104·PhysicsIntegerJEE Advanced 2022
In a particular system of units, a physical quantity can be expressed in terms of the electric charge e , electron mass me_ee​ , Planck's constant h , and Coulomb's constant k=14πϵ0k = \frac{1}{4\pi\epsilon_0}k=4πϵ0​1​ , where ε0_00​ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]α^{\alpha}α[me_ee​]β^{\beta}β[h]γ^{\gamma}γ[k]δ^{\delta}δ. The value of α + β + γ + δ is ______________.

Correct answer: 4

Step-by-step solution →
Q105·PhysicsSingle correctJEE Main 2022
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is p and radius of the drop is r, then T=kpr3S3/2T = k\sqrt{\frac{pr^{3}}{S^{3/2}}}T=kS3/2pr3​​ is dimensionally correct, where K is dimensionless. Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period. In the light of above statements, choose the correct answer from the options given below.
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (C)(A) is true but (R) is false
  4. (D)(A) is false but (R) is true

Correct answer: (D)

Step-by-step solution →
Q106·PhysicsSingle correctJEE Main 2022
Match List I with List II. List I A. Torque B. Stress C. Latent Heat D. Power List II I. Nms−1^{-1}−1 II. J kg−1^{-1}−1 III. Nm IV. Nm−2^{-2}−2 Choose the correct answer from the options given below:
  1. (A)A-III, B-II, C-I, D-IV
  2. (B)A-III, B-IV, C-II, D-I
  3. (C)A-IV, B-I, C-III, D-II
  4. (D)A-II, B-III, C-I, D-IV

Correct answer: (B)

Step-by-step solution →
Q107·PhysicsSingle correctJEE Main 2022
Consider the efficiency of Carnot's engine is given by η=αβsin⁡θlog⁡eβxkT\eta = \dfrac{\alpha\beta}{\sin\theta}\log_e \dfrac{\beta x}{kT}η=sinθαβ​loge​kTβx​, where α\alphaα and β\betaβ are constants. If T is temperature, k is Boltzman constant, θ\thetaθ is angular displacement and x has the dimensions of length. Then, choose the incorrect option.
  1. (A)Dimensions of β\betaβ is same as that of force.
  2. (B)Dimensions of α−1\alpha^{-1}α−1 x is same as that of energy.
  3. (C)Dimensions of η−1sin⁡θ\eta^{-1}\sin\thetaη−1sinθ is same as that of αβ\alpha\betaαβ
  4. (D)Dimensions of α\alphaα is same as that of β\betaβ

Correct answer: (D)

Step-by-step solution →
Q108·PhysicsSingle correctJEE Main 2022
The dimensions of (B2μ0)\left(\frac{B^{2}}{\mu_{0}}\right)(μ0​B2​) will be : (if μ0_{0}0​ : permeability of free space and B : magnetic field)
  1. (A)[M L2^{2}2 T−2^{-2}−2]
  2. (B)[M L T−2^{-2}−2]
  3. (C)[M L−1^{-1}−1 T−2^{-2}−2]
  4. (D)[M L2^{2}2 T−2^{-2}−2 A−1^{-1}−1]

Correct answer: (C)

Step-by-step solution →
Q109·PhysicsNumericalJEE Main 2022
In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1s resolution. If measured value of length is 10 cm known to 1mm accuracy. The accuracy in the determination of g is found to be x %. The value of x is

Correct answer: 5

Step-by-step solution →
Q110·PhysicsSingle correctJEE Main 2022
A torque meter is calibrated to reference standards of mass, length and time each with 5% accuracy. After calibration, the measured torque with this torque meter will have net accuracy of :
  1. (A)15%
  2. (B)25%
  3. (C)75%
  4. (D)5%

Correct answer: (B)

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2022
An expression of energy density is given by u=αβsin⁡(αxkt)u = \frac{\alpha}{\beta}\sin\left(\frac{\alpha x}{kt}\right)u=βα​sin(ktαx​), where α, β are constants, x is displacement, k is Boltzmann constant and t is the temperature. The dimensions of β will be :
  1. (A)[ML2T−2θ−1][ML^{2}T^{-2}\theta^{-1}][ML2T−2θ−1]
  2. (B)[M0L2T−2][M^{0}L^{2}T^{-2}][M0L2T−2]
  3. (C)[M0L0T0][M^{0}L^{0}T^{0}][M0L0T0]
  4. (D)[M0L2T0][M^{0}L^{2}T^{0}][M0L2T0]

Correct answer: (D)

Step-by-step solution →
Q112·PhysicsNumericalJEE Main 2022
In an experiment of determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of ±0.02 mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of ±0.01 mm. The error in the measurement of Young's modulus (ΔY) is found to be x × 1010^{10}10 Nm−2^{-2}−2. The value of x is ______ [Take g = 10m/s2^{2}2]

Correct answer: 2

Step-by-step solution →
Q113·PhysicsSingle correctJEE Main 2022
Which of the following physical quantities have the same dimensions ?
  1. (A)Electric displacement (D⃗)(\vec{D})(D) and surface charge density
  2. (B)Displacement current and electric field
  3. (C)Current density and surface charge density
  4. (D)Electric potential and energy

Correct answer: (A)

Step-by-step solution →
Q114·PhysicsSingle correctJEE Main 2022
If momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :
  1. (A)[P A−1 T0][P\ A^{-1}\ T^{0}][P A−1 T0]
  2. (B)[P A T−1][P\ A\ T^{-1}][P A T−1]
  3. (C)[P A−1 T][P\ A^{-1}\ T][P A−1 T]
  4. (D)[P A−1 T−1][P\ A^{-1}\ T^{-1}][P A−1 T−1]

Correct answer: (A)

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2022
In Vander Waals equation [P+aV2][V−b]=RT\left[P+\frac{a}{V^2}\right][V-b]=RT[P+V2a​][V−b]=RT; P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants ab\frac{a}{b}ba​ is dimensionally equal to :
  1. (A)PV\frac{P}{V}VP​
  2. (B)VP\frac{V}{P}PV​
  3. (C)PV
  4. (D)PV3PV^3PV3

Correct answer: (C)

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2022
Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as v2=nm2v1v_2 = \frac{n}{m^2} v_1v2​=m2n​v1​ and a2=a1mna_2 = \frac{a_1}{mn}a2​=mna1​​ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are:
  1. (A)n3m3L1=L2\frac{n^3}{m^3} L_1 = L_2m3n3​L1​=L2​ and n2mT1=T2\frac{n^2}{m} T_1 = T_2mn2​T1​=T2​
  2. (B)L1=n4m2L2L_1 = \frac{n^4}{m^2} L_2L1​=m2n4​L2​ and T1=n2mT2T_1 = \frac{n^2}{m} T_2T1​=mn2​T2​
  3. (C)L1=n2mL2L_1 = \frac{n^2}{m} L_2L1​=mn2​L2​ and T1=n4m2T2T_1 = \frac{n^4}{m^2} T_2T1​=m2n4​T2​
  4. (D)n2mL1=L2\frac{n^2}{m} L_1 = L_2mn2​L1​=L2​ and n4m2T1=T2\frac{n^4}{m^2} T_1 = T_2m2n4​T1​=T2​

Correct answer: (A)

Step-by-step solution →
Q117·PhysicsNumericalJEE Main 2022
A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is x121\frac{x}{121}121x​% . The value of x is ___

Correct answer: 150

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2022
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity. Reason R: Coefficient of viscosity =ForceVelocity gradient= \frac{\text{Force}}{\text{Velocity gradient}}=Velocity gradientForce​ Question: Choose the correct answer from the options given below :
  1. (A)Both A and R true, and R is correct explanation of A.
  2. (B)Both A and R are true but R is NOT the correct explanation of A.
  3. (C)A is true but R is false.
  4. (D)A is false but R is true.

Correct answer: (C)

Step-by-step solution →
Q119·PhysicsSingle correctJEE Main 2022
The distance of the Sun from earth is 1.5×10111.5\times10^{11}1.5×1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be :
  1. (A)2.45×10102.45\times10^{10}2.45×1010 m
  2. (B)1.45×10101.45\times10^{10}1.45×1010 m
  3. (C)1.45×1091.45\times10^{9}1.45×109 m
  4. (D)0.14×1090.14\times10^{9}0.14×109 m

Correct answer: (C)

Step-by-step solution →
Q120·PhysicsSingle correctJEE Main 2022
A sliver wire has mass (0.6±0.006)(0.6 \pm 0.006)(0.6±0.006) g, radius (0.5±0.005)(0.5 \pm 0.005)(0.5±0.005) mm and length (4±0.04)(4 \pm 0.04)(4±0.04) cm. The maximum percentage error in the measurement of its density will be :
  1. (A)4%
  2. (B)3%
  3. (C)6%
  4. (D)7%

Correct answer: (A)

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2022
The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be
  1. (A)[ML−1T−1][ML^{-1}T^{-1}][ML−1T−1]
  2. (B)[ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
  3. (C)[ML2T−1][ML^{2}T^{-1}][ML2T−1]
  4. (D)[M−1L3T0][M^{-1}L^{3}T^{0}][M−1L3T0]

Correct answer: (A)

Step-by-step solution →
Q122·PhysicsSingle correctJEE Main 2022
An expression for a dimensionless quantity P is given by P=αβlog⁡e(ktβx)P = \frac{\alpha}{\beta}\log_{e}\left(\frac{kt}{\beta x}\right)P=βα​loge​(βxkt​); where α and β are constants, x is distance ; k is Boltzmann constant and t is the temperature. Then the dimensions of α will be :
  1. (A)[M0L−1T0][M^{0}L^{-1}T^{0}][M0L−1T0]
  2. (B)[ML0T−2][ML^{0}T^{-2}][ML0T−2]
  3. (C)[MLT−2][MLT^{-2}][MLT−2]
  4. (D)[ML2T−2][ML^{2}T^{-2}][ML2T−2]

Correct answer: (C)

Step-by-step solution →
Q123·PhysicsSingle correctJEE Main 2022
The dimension of mutual inductance is :
  1. (A)[ML2T−2A−1][ML^2 T^{-2} A^{-1}][ML2T−2A−1]
  2. (B)[ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
  3. (C)[ML2T−2A−2][ML^2T^{-2}A^{-2}][ML2T−2A−2]
  4. (D)[ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2]

Correct answer: (C)

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2022
If Z=A2B3C4Z=\frac{A^2B^3}{C^4}Z=C4A2B3​ , then the relative error in Z will be :
  1. (A)ΔAA+ΔBB+ΔCC\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}AΔA​+BΔB​+CΔC​
  2. (B)2ΔAA+3ΔBB−4ΔCC\frac{2\Delta A}{A}+\frac{3\Delta B}{B}-\frac{4\Delta C}{C}A2ΔA​+B3ΔB​−C4ΔC​
  3. (C)2ΔAA+3ΔBB+4ΔCC\frac{2\Delta A}{A}+\frac{3\Delta B}{B}+\frac{4\Delta C}{C}A2ΔA​+B3ΔB​+C4ΔC​
  4. (D)ΔAA+ΔBB−ΔCC\frac{\Delta A}{A}+\frac{\Delta B}{B}-\frac{\Delta C}{C}AΔA​+BΔB​−CΔC​

Correct answer: (C)

Step-by-step solution →
Q125·PhysicsNumericalJEE Main 2022
For z=a2x3y12z = a^2 x^3 y^{\frac{1}{2}}z=a2x3y21​, where 'a' is a constant. If percentage error in measurement of 'x' and 'y' are 4% and 12%, respectively, then the percentage error for 'z' will be %.

Correct answer: 18

Step-by-step solution →
Q126·PhysicsSingle correctJEE Main 2022
Identify the pair of physical quantities that have same dimensions :
  1. (A)velocity gradient and decay constant
  2. (B)wien's constant and Stefan constant
  3. (C)angular frequency and angular momentum
  4. (D)wave number and Avogadro number

Correct answer: (A)

Step-by-step solution →
Q127·PhysicsSingle correctJEE Main 2022
Identify the pair of physical quantities which have different dimensions :
  1. (A)Wave number and Rydberg's constant
  2. (B)Stress and Coefficient of elasticity
  3. (C)Coercivity and Magnetisation
  4. (D)Specific heat capacity and Latent heat

Correct answer: (D)

Step-by-step solution →
Q128·PhysicsMultiple correctJEE Advanced 2021
A physical quantity S⃗\vec{S}S is defined as S⃗=(E⃗×B⃗)/μ0\vec{S} = (\vec{E} \times \vec{B}) / \mu_0S=(E×B)/μ0​, where E⃗\vec{E}E is electric field, B⃗\vec{B}B is magnetic field and μ0\mu_0μ0​ is the permeability of free space. The dimensions of S⃗\vec{S}S are the same as the dimensions of which of the following quantity (ies) ?
  1. (A)Energycharge×current\frac{\text{Energy}}{\text{charge} \times \text{current}}charge×currentEnergy​
  2. (B)ForceLength×Time\frac{\text{Force}}{\text{Length} \times \text{Time}}Length×TimeForce​
  3. (C)EnergyVolume\frac{\text{Energy}}{\text{Volume}}VolumeEnergy​
  4. (D)PowerArea\frac{\text{Power}}{\text{Area}}AreaPower​

Correct answer: (B), (D)

Step-by-step solution →
Q129·PhysicsSingle correctJEE Main 2021
A student determined Young's Modulus of elasticity using the formula Y=MgL34bd3δY = \frac{MgL^{3}}{4bd^{3}\delta}Y=4bd3δMgL3​. The value of g is taken to be 9.8 m/s2^{2}2, without any significant error, his observation are as following. Then the fractional error in the measurement of Y is :
Physical QuantityLeast count of the Equipment used for measurementObserved value
Mass (M)1 g2 kg
Length of bar (L)1 mm1 m
Breadth of bar (b)0.1 mm4 cm
Thickness of bar (d)0.01 mm0.4 cm
Depression (δ)0.01 mm5 mm
  1. (A)0.0083
  2. (B)0.0155
  3. (C)0.155
  4. (D)0.083

Correct answer: (B)

Step-by-step solution →
Q130·PhysicsSingle correctJEE Main 2021
Two resistors R1_{1}1​ = (4 ± 0.8) Ω and R2_{2}2​ = (4 ± 0.4) Ω are connected in parallel. The equivalent resistance of their parallel combination will be :
  1. (A)(4 ± 0.4) Ω
  2. (B)(2 ± 0.4) Ω
  3. (C)(2 ± 0.3) Ω
  4. (D)(4 ± 0.3) Ω

Correct answer: (C)

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2021
Which of the following equations is dimensionally incorrect ? Where t = time, h = height, s = surface tension, θ\thetaθ = angle, ρ\rhoρ = density, a, r = radius, g = acceleration due to gravity, v = volume , p = pressure, W = work done, Γ\GammaΓ = torque, ∈\in∈ = permittivity, E = electric field, J = current density, L = length.
  1. (A)v=π p a48η Lv=\frac{\pi\,p\,a^{4}}{8\eta\,L}v=8ηLπpa4​
  2. (B)h=2scos⁡θρ r gh=\frac{2s\cos\theta}{\rho\,r\,g}h=ρrg2scosθ​
  3. (C)J=∈∂E∂tJ=\in\frac{\partial E}{\partial t}J=∈∂t∂E​
  4. (D)W=ΓθW=\Gamma\thetaW=Γθ

Correct answer: (A)

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2021
Match List-I with List-II. List-I (a) Torque (b) Impulse (c) Tension (d) Surface Tension List-II (i) MLT−1MLT^{-1}MLT−1 (ii) MT−2MT^{-2}MT−2 (iii) ML2T−2ML^{2}T^{-2}ML2T−2 (iv) MLT−2MLT^{-2}MLT−2 Choose the most appropriate answer from the option given below :
  1. (A)(a)–(iii), (b)–(i), (c)–(iv), (d)–(ii)
  2. (B)(a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)
  3. (C)(a)–(i), (b)–(iii), (c)–(iv), (d)–(ii)
  4. (D)(a)–(iii), (b)–(iv), (c)–(i), (d)–(ii)

Correct answer: (A)

Step-by-step solution →
Q133·PhysicsSingle correctJEE Main 2021
If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be :
  1. (A)[FT−1V−1][FT^{-1} V^{-1}][FT−1V−1]
  2. (B)[FTV−1][FTV^{-1}][FTV−1]
  3. (C)[FT2V][FT^{2} V][FT2V]
  4. (D)[FVT−1][FVT^{-1}][FVT−1]

Correct answer: (B)

Step-by-step solution →
Q134·PhysicsSingle correctJEE Main 2021
A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second ? Given : 1 ly = 9.46×10159.46 \times 10^{15}9.46×1015 m \qquad 1 AU = 1.5×10111.5 \times 10^{11}1.5×1011 m
  1. (A)4.1×1084.1 \times 10^{8}4.1×108 s
  2. (B)4.5×10104.5 \times 10^{10}4.5×1010 s
  3. (C)3.5×1063.5 \times 10^{6}3.5×106 s
  4. (D)7.2×1087.2 \times 10^{8}7.2×108 s

Correct answer: (B)

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2021
Which of the following is not a dimensionless quantity ?
  1. (A)Relative magnetic permeability (μr\mu_rμr​)
  2. (B)Power factor
  3. (C)Permeability of free space (μ0\mu_0μ0​)
  4. (D)Quality factor

Correct answer: (C)

Step-by-step solution →
Q136·PhysicsSingle correctJEE Main 2021
Match List-I (physical quantity) with List-II (SI unit). Choose the most appropriate answer from the options given below :
List-IList-II
a.RHR_{H}RH​ (Rydberg constant)i.kg m−1^{-1}−1 s−1^{-1}−1
b.h (Planck's constant)ii.kg m2^{2}2 s−1^{-1}−1
c.μB\mu_{B}μB​ (Magnetic field energy density)iii.m−1^{-1}−1
d.η (coefficient of viscocity)iv.kg m−1^{-1}−1 s−2^{-2}−2
  1. (A)(a)–(ii), (b)–(iii), (c)–(iv), (d)–(i)
  2. (B)(a)–(iii), (b)–(ii), (c)–(iv), (d)–(i)
  3. (C)(a)–(iv), (b)–(ii), (c)–(i), (d)–(iii)
  4. (D)(a)–(iii), (b)–(ii), (c)–(i), (d)–(iv)

Correct answer: (B)

Step-by-step solution →
Q137·PhysicsSingle correctJEE Main 2021
If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density :
  1. (A)[FL−4^{-4}−4 T2^{2}2]
  2. (B)[FL−3^{-3}−3 T2^{2}2]
  3. (C)[FL−5^{-5}−5 T2^{2}2]
  4. (D)[FL−3^{-3}−3 T3^{3}3]

Correct answer: (A)

Step-by-step solution →
Q138·PhysicsSingle correctJEE Main 2021
If E and H represents the intensity of electric field and magnetising field respectively, then the unit of E/H will be :
  1. (A)ohm
  2. (B)mho
  3. (C)joule
  4. (D)newton

Correct answer: (A)

Step-by-step solution →
Q139·PhysicsSingle correctJEE Main 2021
If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P = EL2M−5G−2EL^{2}M^{-5}G^{-2}EL2M−5G−2 are :-
  1. (A)[M0 L1 T0][M^{0}\,L^{1}\,T^{0}][M0L1T0]
  2. (B)[M−1 L−1 T2][M^{-1}\,L^{-1}\,T^{2}][M−1L−1T2]
  3. (C)[M1 L1 T−2][M^{1}\,L^{1}\,T^{-2}][M1L1T−2]
  4. (D)[M0 L0 T0][M^{0}\,L^{0}\,T^{0}][M0L0T0]

Correct answer: (D)

Step-by-step solution →
Q140·PhysicsSingle correctJEE Main 2021
Match List–I with List–II. Choose the most appropriate answer from the options given below :
List-IList-II
a.Magnetic Inductioni.ML2T−2A−1ML^{2}T^{-2}A^{-1}ML2T−2A−1
b.Magnetic Fluxii.M0L−1AM^{0}L^{-1}AM0L−1A
c.Magnetic Permeabilityiii.MT−2A−1MT^{-2}A^{-1}MT−2A−1
d.Magnetizationiv.MLT−2A−2MLT^{-2}A^{-2}MLT−2A−2
  1. (A)(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
  2. (B)(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
  3. (C)(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
  4. (D)(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

Correct answer: (D)

Step-by-step solution →
Q141·PhysicsNumericalJEE Main 2021
The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ..........%

Correct answer: 14

Step-by-step solution →
Q142·PhysicsSingle correctJEE Main 2021
Match List I with List II. List I (a) Capacitance, C (b) Permittivity of free space, ε0\varepsilon_0ε0​ (c) Permeability of free space, μ0\mu_0μ0​ (d) Electric field, E List II (i) M1^11L1^11T−3^{-3}−3 A−1^{-1}−1 (ii) M−1^{-1}−1L−3^{-3}−3 T4^44 A2^22 (iii) M−1^{-1}−1L−2^{-2}−2 T4^44 A2^22 (iv) M1^11L1^11T−2^{-2}−2A−2^{-2}−2 Choose the correct answer from the options given below :
  1. (A)(a) → (iii), (b) → (ii), (c) → (iv), (d) → (i)
  2. (B)(a) → (iii), (b) → (iv), (c) → (ii), (d) → (i)
  3. (C)(a) → (iv), (b) → (ii), (c) → (iii), (d) → (i)
  4. (D)(a) → (iv), (b) → (iii), (c) → (ii), (d) → (i)

Correct answer: (A)

Step-by-step solution →
Q143·PhysicsSingle correctJEE Main 2021
A physical 'y' is represented by the formula y = m2^22 r−4^{-4}−4 gx^xx ℓ−32\ell^{-\frac{3}{2}}ℓ−23​ if the percentage errors found in y, m, r, ℓ\ellℓ and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.
  1. (A)5 and ± 2
  2. (B)163\frac{16}{3}316​ and ± 32\frac{3}{2}23​
  3. (C)8 and ± 2
  4. (D)4 and ± 3

Correct answer: (B)

Step-by-step solution →
Q144·PhysicsSingle correctJEE Main 2021
The force is given in terms of time t and displacement x by the equation F = A cos Bx + C sin Dt The dimensional formula of ADB\frac{AD}{B}BAD​ is :
  1. (A)[M2L2T−3]\left[\mathrm{M}^2\mathrm{L}^2\mathrm{T}^{-3}\right][M2L2T−3]
  2. (B)[M1L1T−2]\left[\mathrm{M}^1\mathrm{L}^1\mathrm{T}^{-2}\right][M1L1T−2]
  3. (C)[ML2T−3]\left[\mathrm{M}\mathrm{L}^2\mathrm{T}^{-3}\right][ML2T−3]
  4. (D)[M0LT−1]\left[\mathrm{M}^0\mathrm{L}\mathrm{T}^{-1}\right][M0LT−1]

Correct answer: (C)

Step-by-step solution →
Q145·PhysicsSingle correctJEE Main 2021
Two vectors X⃗\vec{X}X and Y⃗\vec{Y}Y have equal magnitude. The magnitude of (X⃗−Y⃗)\left(\vec{X}-\vec{Y}\right)(X−Y) is n times the magnitude of (X⃗+Y⃗)\left(\vec{X}+\vec{Y}\right)(X+Y). The angle between X⃗\vec{X}X and Y⃗\vec{Y}Y is :
  1. (A)cos⁡−1(n2+1n2−1)\cos^{-1}\left(\frac{n^2+1}{n^2-1}\right)cos−1(n2−1n2+1​)
  2. (B)cos⁡−1(−n2−1n2−1)\cos^{-1}\left(\frac{-n^2-1}{n^2-1}\right)cos−1(n2−1−n2−1​)
  3. (C)cos⁡−1(n2−1−n2−1)\cos^{-1}\left(\frac{n^2-1}{-n^2-1}\right)cos−1(−n2−1n2−1​)
  4. (D)cos⁡−1(n2+1−n2−1)\cos^{-1}\left(\frac{n^2+1}{-n^2-1}\right)cos−1(−n2−1n2+1​)

Correct answer: (C)

Step-by-step solution →
Q146·PhysicsSingle correctJEE Main 2021
The entropy of any system is given by S=α2β ℓn[μkRJβ2+3]S = \alpha^2\beta\ \ell n\left[\frac{\mu kR}{J\beta^2}+3\right]S=α2β ℓn[Jβ2μkR​+3] Where α\alphaα and β\betaβ are the constants. μ, J, k and R no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively . [ Take S=dQTS = \frac{dQ}{T}S=TdQ​ ] Choose the incorrect option from the following :
  1. (A)S, β, k and μR have the same dimensions.
  2. (B)α and k have the same dimensions.
  3. (C)S and α have different dimensions.
  4. (D)α and J have the same dimensions.

Correct answer: (B)

Step-by-step solution →
Q147·PhysicsSingle correctJEE Main 2021
If time (t)(t)(t), velocity (υ)(\upsilon)(υ), and angular momentum (ℓ)(\ell)(ℓ) are taken as the fundamental units. Then the dimension of mass (m)(m)(m) in terms of t,υ,t, \upsilon,t,υ, and ℓ\ellℓ is :
  1. (A)[t−2 υ−1 ℓ1]\left[ t^{-2}\,\upsilon^{-1}\,\ell^{1} \right][t−2υ−1ℓ1]
  2. (B)[t−1 υ1 ℓ−2]\left[ t^{-1}\,\upsilon^{1}\,\ell^{-2} \right][t−1υ1ℓ−2]
  3. (C)[t−1 υ−2 ℓ1]\left[ t^{-1}\,\upsilon^{-2}\,\ell^{1} \right][t−1υ−2ℓ1]
  4. (D)[t1 υ2 ℓ−1]\left[ t^{1}\,\upsilon^{2}\,\ell^{-1} \right][t1υ2ℓ−1]

Correct answer: (C)

Step-by-step solution →
Q148·PhysicsSingle correctJEE Main 2021
The time period of a simple pendulum is given by T=2πℓgT = 2\pi\sqrt{\dfrac{\ell}{g}}T=2πgℓ​​ . The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is:-
  1. (A)2%
  2. (B)3%
  3. (C)5%
  4. (D)4%

Correct answer: (B)

Step-by-step solution →
Q149·PhysicsNumericalJEE Main 2021
Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3^{3}3 of oleic acid per cm3^{3}3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2^{2}2 by considering 100 spherical drops of radius (340π)13×10−3\left(\frac{3}{40\pi}\right)^{\frac{1}{3}} \times 10^{-3}(40π3​)31​×10−3 cm. Then the thickness of oleic acid layer will be x × 10−14^{-14}−14 m. Where x is___________.

Correct answer: 25

Step-by-step solution →
Q150·PhysicsSingle correctJEE Main 2021
In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1kg (measured using a scale of least count = 1g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment ?
  1. (A)0.14%
  2. (B)0.9%
  3. (C)9%
  4. (D)1.4%

Correct answer: (D)

Step-by-step solution →
Q151·PhysicsNumericalJEE Main 2021
The resistance R = VI\dfrac{V}{I}IV​, where V = (50 ± 2)V and I = (20 ± 0.2)A. The percentage error in R is 'x' %. The value of 'x' to the nearest integer is _________.

Correct answer: 5

Step-by-step solution →
Q152·PhysicsSingle correctJEE Main 2021
In a typical combustion engine the workdone by a gas molecule is given by W=α2βe−βx2kTW = \alpha^{2}\beta e^{\frac{-\beta x^{2}}{kT}}W=α2βekT−βx2​, where x is the displacement, k is the Boltzmann constant and T is the temperature. If α and β are constants, dimensions of α will be :
  1. (A)[M0 L T0][M^{0}\,L\,T^{0}][M0LT0]
  2. (B)[M2 L T−2][M^{2}\,L\,T^{-2}][M2LT−2]
  3. (C)[M L T−2][M\,L\,T^{-2}][MLT−2]
  4. (D)[M L T−1][M\,L\,T^{-1}][MLT−1]

Correct answer: (A)

Step-by-step solution →
Q153·PhysicsSingle correctJEE Main 2021
If ‘C’ and ‘V’ represent capacity and voltage respectively then what are the dimensions of λ where C/V = λ?
  1. (A)[M−2L−4I3T7]\left[M^{-2}L^{-4}I^{3}T^{7}\right][M−2L−4I3T7]
  2. (B)[M−2L−3I2T6]\left[M^{-2}L^{-3}I^{2}T^{6}\right][M−2L−3I2T6]
  3. (C)[M−1L−3I−2T−7]\left[M^{-1}L^{-3}I^{-2}T^{-7}\right][M−1L−3I−2T−7]
  4. (D)[M−3L−4I3T7]\left[M^{-3}L^{-4}I^{3}T^{7}\right][M−3L−4I3T7]

Correct answer: (A)

Step-by-step solution →
Q154·PhysicsSingle correctJEE Main 2021
If e is the electronic charged, c is the speed of light in free space and h is planck's constant, the 1e2^{2}2 quantity has dimensions of : 4πε0_{0}0​hc
  1. (A)[ LC−1^{-1}−1]
  2. (B)[M0^{0}0 L0^{0}0 T0^{0}0]
  3. (C)[ M L T0^{0}0]
  4. (D)[M L T−1^{-1}−1]

Correct answer: (B)

Step-by-step solution →
Q155·PhysicsSingle correctJEE Main 2021
Match List –I with List- II : List-I (a)h (Planck's constant) (b)E (kinetic energy) (c)V (electric potential) (d)P (linear momentum) List-II (i) [M L T−1][M\,L\,T^{-1}][MLT−1] (ii) [M L2 T−1][M\,L^2\,T^{-1}][ML2T−1] (iii) [M L2 T−2][M\,L^2\,T^{-2}][ML2T−2] (iv) [ML2I−1 T−3][ML^2I^{-1}\,T^{-3}][ML2I−1T−3] Choose the correct answer from the options given below :
  1. (A)(a) → (ii), (b) → (iii), (c) → (iv), (d) → (i)
  2. (B)(a) → (i), (b) → (ii), (c) → (iv), (d) → (iii)
  3. (C)(a) → (iii), (b) → (ii), (c) → (iv), (d) → (i)
  4. (D)(a) → (iii), (b) → (iv), (c) → (ii), (d) → (i)

Correct answer: (A)

Step-by-step solution →
Q156·PhysicsSingle correctJEE Main 2021
The period of oscillation of a simple pendulum is T=2πLgT=2\pi\sqrt{\frac{L}{g}}T=2πgL​​. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :
  1. (A)1.33 %
  2. (B)1.30 %
  3. (C)1.13 %
  4. (D)1.03 %

Correct answer: (C)

Step-by-step solution →
Q157·PhysicsSingle correctJEE Main 2021
The workdone by a gas molecule in an isolated system is given by, W=αβ2e−x2αkTW = \alpha\beta^2 e^{-\frac{x^2}{\alpha kT}}W=αβ2e−αkTx2​ , where x is the displacement, k is the Boltzmann constant and T is the temperature.α\alphaα and β\betaβ are constants. Then the dimensions of β\betaβ will be -
  1. (A)[M0LT0][M^0LT^0][M0LT0]
  2. (B)[M2LT2][M^2LT^2][M2LT2]
  3. (C)[MLT−2][MLT^{-2}][MLT−2]
  4. (D)[ML2T−2][ML^2T^{-2}][ML2T−2]

Correct answer: (C)

Step-by-step solution →
Q158·PhysicsMultiple correctJEE Advanced 2020
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity X as follows: [position] = [Xα][X^{\alpha}][Xα]; [speed] = [Xβ][X^{\beta}][Xβ]; [acceleration] = [Xp][X^{p}][Xp]; [linear momentum] = [Xq][X^{q}][Xq]; [force] = [Xr][X^{r}][Xr]. Then -
  1. (A)α+p=2β\alpha + p = 2\betaα+p=2β
  2. (B)p+q−r=βp + q - r = \betap+q−r=β
  3. (C)p−q+r=αp - q + r = \alphap−q+r=α
  4. (D)p+q+r=βp + q + r = \betap+q+r=β

Correct answer: (A), (B)

Step-by-step solution →
Q159·PhysicsNumericalJEE Advanced 2020
Two capacitors with capacitance values C1=2000±10C_1 = 2000 \pm 10C1​=2000±10 pF and C2=3000±15C_2 = 3000 \pm 15C2​=3000±15 pF are connected in series. The voltage applied across this combination is V=5.00±0.02V = 5.00 \pm 0.02V=5.00±0.02 V. The percentage error in the calculation of the energy stored in this combination of capacitors is ________.

Correct answer: 1.30

Step-by-step solution →
Q160·PhysicsSingle correctJEE Main 2020
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm ; 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:
  1. (A)(5.5375 ± 0.0739) mm
  2. (B)(5.5375 ± 0.0740) mm
  3. (C)(5.538 ± 0.074) mm
  4. (D)(5.54 ± 0.07) mm

Correct answer: (D)

Step-by-step solution →
Q161·PhysicsNumericalJEE Main 2020
The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is (x100)%\left(\dfrac{x}{100}\right)\%(100x​)%. If the relative errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of x is __________.

Correct answer: 1050.00

Step-by-step solution →
Q162·PhysicsSingle correctJEE Main 2020
Assume that the displacement (s) of air is proportional to the pressure difference (Λp)(\Lambda p)(Λp) created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air (ρ) and the frequency (f). If Λp∼10Pa\Lambda p \sim 10\text{Pa}Λp∼10Pa, υ∼300\upsilon \sim 300υ∼300 m/s, ρ∼π1 lg/m3\rho \sim \pi 1 \text{ lg} / \text{m}^3ρ∼π1 lg/m3 and f ∼ 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1)
  1. (A)110\dfrac{1}{10}101​ mm
  2. (B)10 mm
  3. (C)3100\dfrac{3}{100}1003​ mm
  4. (D)1 mm

Correct answer: (C)

Step-by-step solution →
Q163·PhysicsSingle correctJEE Main 2020
A physical quantity z depends on four observables a, c, c and d, as z = a2b23c d3\dfrac{a^{2}b^{\frac{2}{3}}}{\sqrt{c}\,d^{3}}c​d3a2b32​​. The percentages of error in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is:
  1. (A)13.5%
  2. (B)16.5%
  3. (C)12.25%
  4. (D)14.5%

Correct answer: (D)

Step-by-step solution →
Q164·PhysicsSingle correctJEE Main 2020
Dimensional formula for thermal conductivity is (here K denotes the temperature):
  1. (A)MLT−3K−1MLT^{-3}K^{-1}MLT−3K−1
  2. (B)MLT−2K−2MLT^{-2}K^{-2}MLT−2K−2
  3. (C)MLT−2KMLT^{-2}KMLT−2K
  4. (D)MLT−3KMLT^{-3}KMLT−3K

Correct answer: (A)

Step-by-step solution →
Q165·PhysicsSingle correctJEE Main 2020
A quantity X is given by (IFυ2/WL4)(IF\upsilon^{2}/WL^{4})(IFυ2/WL4) in terms of moment of inertia I, force F, velocity υ, work W and Length L. The dimensional formula for x is same as that of:
  1. (A)coefficient of viscosity
  2. (B)planck's constant
  3. (C)energy density
  4. (D)force constant

Correct answer: (C)

Step-by-step solution →
Q166·PhysicsSingle correctJEE Main 2020
Amount of solar energy received on the earth's surface per unit area per unit time is defined a solar constant. Dimension of solar constant is:
  1. (A)M2L0T−1M^{2}L^{0}T^{-1}M2L0T−1
  2. (B)ML2T−2ML^{2}T^{-2}ML2T−2
  3. (C)ML0T−3ML^{0}T^{-3}ML0T−3
  4. (D)MLT−2MLT^{-2}MLT−2

Correct answer: (C)

Step-by-step solution →
Q167·PhysicsSingle correctJEE Main 2020
A quantity f is given by f=hc5Gf=\sqrt{\dfrac{hc^{5}}{G}}f=Ghc5​​ where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of:
  1. (A)energy
  2. (B)momentum
  3. (C)area
  4. (D)volume

Correct answer: (A)

Step-by-step solution →
Q168·PhysicsSingle correctJEE Main 2020
A simple pendulum is being used to determine the vale of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:
  1. (A)4.40%
  2. (B)2.40%
  3. (C)3.40%
  4. (D)5.40%

Correct answer: (A)

Step-by-step solution →
Q169·PhysicsSingle correctJEE Main 2020
The dimension of B22μ0\frac{B^{2}}{2\mu_{0}}2μ0​B2​, where B is magnetic field and μ0\mu_{0}μ0​ is the magnetic permeability of vacuum, is
  1. (A)ML2^{2}2T−1^{-1}−1
  2. (B)ML2^{2}2T−2^{-2}−2
  3. (C)ML−1^{-1}−1T−2^{-2}−2
  4. (D)MLT−2^{-2}−2

Correct answer: (C)

Step-by-step solution →
Q170·PhysicsMultiple correctJEE Advanced 2019
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of L, which of the following statement(s) is/are correct?
  1. (A)The dimension of energy is L−2L^{-2}L−2
  2. (B)The dimension of force is L−3L^{-3}L−3
  3. (C)The dimension of power is L−5L^{-5}L−5
  4. (D)The dimension of linear momentum is L−1L^{-1}L−1

Correct answer: (A), (B), (D)

Step-by-step solution →
Q171·PhysicsSingle correctJEE Main 2019
Which of the following combinations has the dimension of electrical resistance (ϵ0\epsilon_0ϵ0​ is the permittivity of vacuum and μ0\mu_0μ0​ is the permeability of vacuum)?
  1. (A)ϵ0μ0\sqrt{\dfrac{\epsilon_0}{\mu_0}}μ0​ϵ0​​​
  2. (B)μ0ϵ0\dfrac{\mu_0}{\epsilon_0}ϵ0​μ0​​
  3. (C)ϵ0μ0\dfrac{\epsilon_0}{\mu_0}μ0​ϵ0​​
  4. (D)μ0ϵ0\sqrt{\dfrac{\mu_0}{\epsilon_0}}ϵ0​μ0​​​

Correct answer: (D)

Step-by-step solution →
Q172·PhysicsSingle correctJEE Main 2019
The formula X=5YZ2X = 5YZ^{2}X=5YZ2 X and Z have dimensions of capacitance and magnetic field respectively. What are the dimensions of Y in SI units?
  1. (A)[M−2L0T−4A−2][M^{-2} L^{0} T^{-4} A^{-2}][M−2L0T−4A−2]
  2. (B)[M−3L−2T8A−1][M^{-3} L^{-2} T^{8} A^{-1}][M−3L−2T8A−1]
  3. (C)[M−2L−2T6A3][M^{-2} L^{-2} T^{6} A^{3}][M−2L−2T6A3]
  4. (D)[M−1L−2T4A2][M^{-1} L^{-2} T^{4} A^{2}][M−1L−2T4A2]

Correct answer: (B)

Step-by-step solution →
Q173·PhysicsSingle correctJEE Main 2019
In SI units, the dimensions of ϵ0μ0\sqrt{\dfrac{\epsilon_0}{\mu_0}}μ0​ϵ0​​​ is:
  1. (A)AT−3ML3/2AT^{-3}ML^{3/2}AT−3ML3/2
  2. (B)A−1TML3A^{-1}TML^{3}A−1TML3
  3. (C)A2T3M−1L−2A^{2}T^{3}M^{-1}L^{-2}A2T3M−1L−2
  4. (D)AT2M−1L−1AT^{2}M^{-1}L^{-1}AT2M−1L−1

Correct answer: (C)

Step-by-step solution →
Q174·PhysicsSingle correctJEE Main 2019
If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be:
  1. (A)S1/2I1/2h0S^{1/2}I^{1/2}h^{0}S1/2I1/2h0
  2. (B)S1/2I3/2h−1S^{1/2}I^{3/2}h^{-1}S1/2I3/2h−1
  3. (C)S3/2I1/2h0S^{3/2}I^{1/2}h^{0}S3/2I1/2h0
  4. (D)S1/2I1/2h−1S^{1/2}I^{1/2}h^{-1}S1/2I1/2h−1

Correct answer: (A)

Step-by-step solution →
Q175·PhysicsSingle correctJEE Main 2019
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μ\muμm diameter of a wire is:
  1. (A)50
  2. (B)200
  3. (C)100
  4. (D)500

Correct answer: (B)

Step-by-step solution →
Q176·PhysicsSingle correctJEE Main 2019
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be:
  1. (A)V−2A2F−2V^{-2}A^{2}F^{-2}V−2A2F−2
  2. (B)V−2A2F2V^{-2}A^{2}F^{2}V−2A2F2
  3. (C)V−4A−2FV^{-4}A^{-2}FV−4A−2F
  4. (D)V−4A2FV^{-4}A^{2}FV−4A2F

Correct answer: (D)

Step-by-step solution →
Q177·PhysicsSingle correctJEE Main 2019
The force of interaction between two atoms is given by F=αβ exp(−x2αkt)F = \alpha\beta\, exp\left(-\dfrac{x^{2}}{\alpha kt}\right)F=αβexp(−αktx2​); where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimension of β is:
  1. (A)M⁰L²T⁻⁴
  2. (B)M²LT⁻⁴
  3. (C)MLT⁻²
  4. (D)M²L²T⁻²

Correct answer: (B)

Step-by-step solution →
Q178·PhysicsSingle correctJEE Main 2019
The density of a material in SI units is 128 kg m−3^{-3}−3. In certain units in which the unit of length is 25 cm and the unit of mass 50 g, the numerical value of density of the material is:
  1. (A)40
  2. (B)16
  3. (C)640
  4. (D)410

Correct answer: (A)

Step-by-step solution →
Q179·PhysicsSingle correctJEE Main 2019
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to:
  1. (A)hc5G\sqrt{\dfrac{hc^{5}}{G}}Ghc5​​
  2. (B)c3Gh\sqrt{\dfrac{c^{3}}{Gh}}Ghc3​​
  3. (C)Ghc5\sqrt{\dfrac{Gh}{c^{5}}}c5Gh​​
  4. (D)Ghc3\sqrt{\dfrac{Gh}{c^{3}}}c3Gh​​

Correct answer: (C)

Step-by-step solution →
Q180·PhysicsSingle correctJEE Advanced 2018
PARAGRAPH "A" If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z=x/yz = x/yz=x/y. If the errors in x,yx, yx,y and zzz are Δx\Delta xΔx, Δy\Delta yΔy and Δz\Delta zΔz, respectively, then z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)−1.z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1}.z±Δz=y±Δyx±Δx​=yx​(1±xΔx​)(1±yΔy​)−1. The series expansion for (1±Δyy)−1\left(1 \pm \frac{\Delta y}{y}\right)^{-1}(1±yΔy​)−1, to first power in Δy/y\Delta y/yΔy/y, is 1∓(Δy/y)1 \mp (\Delta y/y)1∓(Δy/y). The relative errors in independent variables are always added. So the error in zzz will be Δz=z(Δxx+Δyy).\Delta z = z\left(\frac{\Delta x}{x} + \frac{\Delta y}{y}\right).Δz=z(xΔx​+yΔy​). The above derivation makes the assumption that Δx/x≪1\Delta x/x \ll 1Δx/x≪1, Δy/y≪1\Delta y/y \ll 1Δy/y≪1. Therefore, the higher powers of these quantities are neglected. (There are two questions based on PARAGRAPH "A", the question given below is one of them) Consider the ratio r=(1−a)(1+a)r = \frac{(1 - a)}{(1 + a)}r=(1+a)(1−a)​ to be determined by measuring a dimensionless quantity aaa. If the error in the measurement of aaa is Δa\Delta aΔa (Δa/a≪1)(\Delta a/a \ll 1)(Δa/a≪1), then what is the error Δr\Delta rΔr in determining rrr ?
  1. (A)Δa(1+a)2\frac{\Delta a}{(1 + a)^{2}}(1+a)2Δa​
  2. (B)2Δa(1+a)2\frac{2\Delta a}{(1 + a)^{2}}(1+a)22Δa​
  3. (C)2Δa(1−a2)\frac{2\Delta a}{(1 - a^{2})}(1−a2)2Δa​
  4. (D)2aΔa(1−a2)\frac{2a\Delta a}{(1 - a^{2})}(1−a2)2aΔa​

Correct answer: (B)

Step-by-step solution →
Q181·PhysicsSingle correctJEE Advanced 2018
PARAGRAPH "X" In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E][E][E] and [B][B][B] stand for dimensions of electric and magnetic fields respectively, while [∈0][\in_{0}][∈0​] and [μ0][\mu_{0}][μ0​] stand for dimensions of the permittivity and permeability of free space respectively. [L][L][L] and [T][T][T] are dimensions of length and time respectively. All the quantities are given in SI units. (There are two questions based on PARAGRAPH "X", the question given below is one of them) The relation between [E][E][E] and [B][B][B] is
  1. (A)[E]=[B] [L] [T][E] = [B]\,[L]\,[T][E]=[B][L][T]
  2. (B)[E]=[B] [L]−1 [T][E] = [B]\,[L]^{-1}\,[T][E]=[B][L]−1[T]
  3. (C)[E]=[B] [L] [T]−1[E] = [B]\,[L]\,[T]^{-1}[E]=[B][L][T]−1
  4. (D)[E]=[B] [L]−1 [T]−1[E] = [B]\,[L]^{-1}\,[T]^{-1}[E]=[B][L]−1[T]−1

Correct answer: (C)

Step-by-step solution →
Q182·PhysicsSingle correctJEE Advanced 2018
PARAGRAPH "X" In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E][E][E] and [B][B][B] stand for dimensions of electric and magnetic fields respectively, while [∈0][\in_{0}][∈0​] and [μ0][\mu_{0}][μ0​] stand for dimensions of the permittivity and permeability of free space respectively. [L][L][L] and [T][T][T] are dimensions of length and time respectively. All the quantities are given in SI units. (There are two questions based on PARAGRAPH "X", the question given below is one of them) The relation between [∈0][\in_{0}][∈0​] and [μ0][\mu_{0}][μ0​] is
  1. (A)[μ0]=[∈0] [L]2 [T]−2[\mu_{0}] = [\in_{0}]\,[L]^{2}\,[T]^{-2}[μ0​]=[∈0​][L]2[T]−2
  2. (B)[μ0]=[∈0] [L]−2 [T]2[\mu_{0}] = [\in_{0}]\,[L]^{-2}\,[T]^{2}[μ0​]=[∈0​][L]−2[T]2
  3. (C)[μ0]=[∈0]−1 [L]2 [T]−2[\mu_{0}] = [\in_{0}]^{-1}\,[L]^{2}\,[T]^{-2}[μ0​]=[∈0​]−1[L]2[T]−2
  4. (D)[μ0]=[∈0]−1 [L]−2 [T]2[\mu_{0}] = [\in_{0}]^{-1}\,[L]^{-2}\,[T]^{2}[μ0​]=[∈0​]−1[L]−2[T]2

Correct answer: (D)

Step-by-step solution →
Q183·PhysicsSingle correctJEE Advanced 2018
PARAGRAPH "A" If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z=x/yz = x/yz=x/y. If the errors in x,yx, yx,y and zzz are Δx\Delta xΔx, Δy\Delta yΔy and Δz\Delta zΔz, respectively, then z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)−1.z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1}.z±Δz=y±Δyx±Δx​=yx​(1±xΔx​)(1±yΔy​)−1. The series expansion for (1±Δyy)−1\left(1 \pm \frac{\Delta y}{y}\right)^{-1}(1±yΔy​)−1, to first power in Δy/y\Delta y/yΔy/y, is 1∓(Δy/y)1 \mp (\Delta y/y)1∓(Δy/y). The relative errors in independent variables are always added. So the error in zzz will be Δz=z(Δxx+Δyy).\Delta z = z\left(\frac{\Delta x}{x} + \frac{\Delta y}{y}\right).Δz=z(xΔx​+yΔy​). The above derivation makes the assumption that Δx/x≪1\Delta x/x \ll 1Δx/x≪1, Δy/y≪1\Delta y/y \ll 1Δy/y≪1. Therefore, the higher powers of these quantities are neglected. (There are two questions based on PARAGRAPH "A", the question given below is one of them) In an experiment the initial number of radioactive nuclei is 3000. It is found that 1000±401000 \pm 401000±40 nuclei decayed in the first 1.0s1.0s1.0s. For ∣x∣≪1|x| \ll 1∣x∣≪1, ln⁡(1+x)=x\ln(1 + x) = xln(1+x)=x up to first power in xxx. The error Δλ\Delta\lambdaΔλ, in the determination of the decay constant λ\lambdaλ, in s−1s^{-1}s−1, is
  1. (A)0.04
  2. (B)0.03
  3. (C)0.02
  4. (D)0.01

Correct answer: (C)

Step-by-step solution →
Q184·PhysicsSingle correctJEE Advanced 2017
A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT=0.01\delta T = 0.01δT=0.01 seconds and he measures the depth of the well to be L=20L = 20L=20 meters. Take the acceleration due to gravity g=10 ms−2g = 10\ \text{ms}^{-2}g=10 ms−2 and the velocity of sound is 300 ms−1300\ \text{ms}^{-1}300 ms−1. Then the fractional error in the measurement, δL/L\delta L/LδL/L, is closest to
  1. (A)0.2 %0.2\ \%0.2 %
  2. (B)1 %1\ \%1 %
  3. (C)3 %3\ \%3 %
  4. (D)5 %5\ \%5 %

Correct answer: (B)

Step-by-step solution →
Q185·PhysicsMultiple correctJEE Advanced 2016
In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T=2π7(R−r)5gT = 2\pi\sqrt{\frac{7(R-r)}{5g}}T=2π5g7(R−r)​​. The values of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?
  1. (A)The error in the measurement of r is 10%
  2. (B)The error in the measurement of T is 3.57%
  3. (C)The error in the measurement of T is 2%
  4. (D)The error in the determined value of g is 11%

Correct answer: (A), (B), (D)

Step-by-step solution →
Q186·PhysicsMultiple correctJEE Advanced 2016
A length-scale (ℓ\ellℓ) depends on the permittivity (ε\varepsilonε) of a dielectric material, Boltzmann constant (kBk_BkB​), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for ℓ\ellℓ is(are) dimensionally correct?
  1. (A)ℓ=(nq2εkBT)\ell = \sqrt{\left(\frac{nq^{2}}{\varepsilon k_B T}\right)}ℓ=(εkB​Tnq2​)​
  2. (B)ℓ=(εkBTnq2)\ell = \sqrt{\left(\frac{\varepsilon k_B T}{nq^{2}}\right)}ℓ=(nq2εkB​T​)​
  3. (C)ℓ=(q2εn2/3kBT)\ell = \sqrt{\left(\frac{q^{2}}{\varepsilon n^{2/3} k_B T}\right)}ℓ=(εn2/3kB​Tq2​)​
  4. (D)ℓ=(q2εn1/3kBT)\ell = \sqrt{\left(\frac{q^{2}}{\varepsilon n^{1/3} k_B T}\right)}ℓ=(εn1/3kB​Tq2​)​

Correct answer: (B), (D)

Step-by-step solution →
Q187·PhysicsMultiple correctJEE Advanced 2015
Planck's constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are)
  1. (A)M∝cM \propto \sqrt{c}M∝c​
  2. (B)M∝GM \propto \sqrt{G}M∝G​
  3. (C)L∝hL \propto \sqrt{h}L∝h​
  4. (D)L∝GL \propto \sqrt{G}L∝G​

Correct answer: (A), (C), (D)

Step-by-step solution →
Q188·PhysicsMultiple correctJEE Advanced 2015
In terms of potential difference V, electric current I, permittivity ε0\varepsilon_{0}ε0​, permeability μ0\mu_{0}μ0​ and speed of light c, the dimensionally correct equation(s) is(are)
  1. (A)μ0I2=ε0V2\mu_{0}I^{2} = \varepsilon_{0}V^{2}μ0​I2=ε0​V2
  2. (B)ε0I=μ0V\varepsilon_{0}I = \mu_{0}Vε0​I=μ0​V
  3. (C)I=ε0cVI = \varepsilon_{0}cVI=ε0​cV
  4. (D)μ0cI=ε0V\mu_{0}cI = \varepsilon_{0}Vμ0​cI=ε0​V

Correct answer: (A), (C)

Step-by-step solution →
Q189·PhysicsIntegerJEE Advanced 2015
The energy of a system as a function of time t is given as E(t)=A2exp⁡(−αt)E(t) = A^{2}\exp(-\alpha t)E(t)=A2exp(−αt), where α=0.2\alpha = 0.2α=0.2 s−1^{-1}−1. The measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the percentage error in the value of E(t)E(t)E(t) at t=5t = 5t=5 s is

Correct answer: 4

Step-by-step solution →
Q190·PhysicsIntegerJEE Advanced 2014
To find the distance ddd over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density ρ\rhoρ of the fog, intensity (power/area) SSS of the light from the signal and its frequency fff. The engineer finds that ddd is proportional to S1/nS^{1/n}S1/n. The value of nnn is

Correct answer: 3

Step-by-step solution →
Q191·PhysicsSingle correctJEE Advanced 2013
The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th^{th}th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
  1. (A)5.112 cm
  2. (B)5.124 cm
  3. (C)5.136 cm
  4. (D)5.148 cm

Correct answer: (B)

Step-by-step solution →
Q192·PhysicsSingle correctJEE Advanced 2013
Match List I with List II and select the correct answer using the codes given below the lists:
List IList II
P.Boltzmann Constant1.[ML2T−1][\mathrm{ML^{2}T^{-1}}][ML2T−1]
Q.Coefficient of viscosity2.[ML−1T−1][\mathrm{ML^{-1}T^{-1}}][ML−1T−1]
R.Plank Constant3.[MLT−3K−1][\mathrm{MLT^{-3}K^{-1}}][MLT−3K−1]
S.Thermal conductivity4.[ML2T−2K−1][\mathrm{ML^{2}T^{-2}K^{-1}}][ML2T−2K−1]
  1. (A)P-3, Q-1, R-2, S-4
  2. (B)P-3, Q-2, R-1, S-4
  3. (C)P-4, Q-2, R-1, S-3
  4. (D)P-4, Q-1, R-2, S-3

Correct answer: (C)

Step-by-step solution →
Q193·PhysicsMultiple correctJEE Advanced 2013
Using the expression 2dsin⁡θ=λ2d\sin\theta = \lambda2dsinθ=λ, one calculates the values of ddd by measuring the corresponding angles θ\thetaθ in the range 000 to 90∘90^{\circ}90∘. The wavelength λ\lambdaλ is exactly known and the error in θ\thetaθ is constant for all values of θ\thetaθ. As θ\thetaθ increases from 0∘0^{\circ}0∘,
  1. (A)the absolute error in ddd remains constant.
  2. (B)the absolute error in ddd increases
  3. (C)the fractional error in ddd remains constant.
  4. (D)the fractional error in ddd decreases.

Correct answer: (D)

Step-by-step solution →

Units and Measurements — frequently asked

How many questions from Units and Measurements appear in JEE?

Units and Measurements has appeared in 149 of the last 186 JEE Main and JEE Advanced papers — about 80% of them — contributing 193 questions in total across those papers.

Is Units and Measurements an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 80% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Units and Measurements questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201

All 28 Physics chapters →

Practise Units and Measurements until it stops costing you marks.

Build a timed test from these 193 questions in one click. Jarvis marks it, names the specific misconception behind each wrong answer, and brings the ones you failed back at the right interval.

Practise Units and Measurements freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS