Q3037. If force [F], acceleration [a] and time [T] are chosen as the fundamental physical quantities, find the dimensions of energy.
Year: NEET 2021
✅ Correct Answer: [F][a][T²]
🧠 Solution:
Let [E] = [Fᵃ][aᵇ][Tᶜ]
We know [E] = [ML²T⁻²], [F] = [MLT⁻²], [a] = [LT⁻²]
⇒ [ML²T⁻²] = [MLT⁻²]ᵃ [LT⁻²]ᵇ [T]ᶜ
Comparing exponents: a = 1, b = 1, c = 2
Hence, [E] = [F][a][T²].
Q3025. A screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading = 0 mm, Circular scale reading = 52 divisions.
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
Year: NEET 2021
✅ Correct Answer: 0.052 cm
🧠 Solution:
Least count = Pitch / number of divisions = 1 mm / 100 = 0.01 mm = 0.001 cm.
Diameter D = MSR + (CSR × LC) = 0 + (52 × 0.001) = 0.052 cm.
Q3030. A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of −0.004 cm, the correct diameter of the ball is
Year: NEET 2018
✅ Correct Answer: 0.529 cm
🧠 Solution:
Given, least count (LC) = 0.001 cm
Main scale reading (MSR) = 5 mm = 0.5 cm
Circular scale reading (VSR) = 25
Zero error = −0.004 cm
Final reading = MSR + (VSR × LC) − (zero error)
= 0.5 + (25 × 0.001) − (−0.004)
= 0.5 + 0.025 + 0.004 = 0.529 cm
Thus, diameter of the ball = 0.529 cm.
Q3040. A physical quantity of the dimensions of length that can be formed out of c, G and e² / (4πε₀) is [c is velocity of light, G is universal constant of gravitation and e is charge]
Year: NEET 2017
✅ Correct Answer: (a) (1 / 4πε₀)¹/² (cG / e²)¹/²
🧠 Solution:
As force F = e² / (4πε₀ r²) ⇒ e² / (4πε₀) = F r². Putting dimensions of F and r, ⇒ [e² / (4πε₀)] = [ML³T⁻²] ...(i). Also, F = Gm₁m₂ / r² ⇒ [G] = [M⁻¹L³T⁻²] ...(ii) and [c] = [LT⁻¹] ...(iii). Now checking optionwise: (1 / 4πε₀)¹/² (cG / e²)¹/² = [L].
Q3031. In an experiment, four quantities a, b, c, and d are measured with percentage errors 1%, 2%, 3% and 4% respectively. Quantity P is calculated as P = (a³b²)/(cd). Error in P is
Year: NEET 2013
✅ Correct Answer: 14%
🧠 Solution:
Given P = (a³b²)/(cd)
Percentage error in P = (3Δa/a + 2Δb/b + Δc/c + Δd/d) × 100
= (3×1 + 2×2 + 3 + 4)% = 14%.
Q3039. Dimensions of stress are
Year: NEET (Sep.) 2020
✅ Correct Answer: [ML⁻¹T⁻²]
🧠 Solution:
Stress = Force / Area
⇒ Dimensions = [MLT⁻²] / [L²] = [ML⁻¹T⁻²].
Q3028. Taking into account significant figures, what is the value of 9.99 m − 0.0099 m?
Year: NEET (Sep.) 2020
✅ Correct Answer: 9.98 m
🧠 Solution:
9.99 − 0.0099 = 9.9801 m.
Considering two significant figures after decimal → final answer = 9.98 m.
Q3027. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is
Year: NEET (Sep.) 2020
✅ Correct Answer: 0.5 mm
🧠 Solution:
Pitch = Least count × number of divisions = 0.01 × 50 = 0.5 mm.
Q3029. The main scale of a vernier calliper has n divisions/cm. n divisions of the vernier scale coincide with (n − 1) divisions of main scale. The least count of the vernier callipers is
Year: NEET (Odisha) 2019
✅ Correct Answer: 1 / n² cm
🧠 Solution:
Given n(VSD) = (n − 1)MSD ⇒ 1VSD = (n − 1)/n MSD.
Least Count = 1MSD − 1VSD = MSD / n².
Hence, LC = 1/n² cm.
Q3026. Time intervals measured by a clock give the following readings: 1.25 s, 1.24 s, 1.27 s, 1.21 s, and 1.28 s. What is the percentage relative error of the observations?
The rate of heat flow is given by:
dQ/dt = (K A ΔT) / L
where K = coefficient of thermal conductivity.
∴ Unit of K = watt × metre / (metre² × kelvin) = W m⁻¹ K⁻¹.
Q3042. If dimensions of critical velocity (vₐ) of a liquid flowing through a tube are expressed as [ηˣ ρʸ rᶻ], where η, ρ, and r are the coefficient of viscosity, density of liquid, and radius of the tube respectively, then the values of x, y, and z are
Year: CBSE AIPMT 2015
✅ Correct Answer: (a) 1, −1, −1
🧠 Solution:
η = [ML⁻¹T⁻¹], ρ = [ML⁻³], r = [L], v_c = [LT⁻¹]. So [LT⁻¹] = [ML⁻¹T⁻¹]ˣ [ML⁻³]ʸ [L]ᶻ. Comparing powers of M, L, T: x + y = 0, −x − 3y + z = 1, −x = −1. Hence x = 1, y = −1, z = −1.
Q3041. If energy (E), velocity (v) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be
Year: CBSE AIPMT 2015
✅ Correct Answer: (c) [E v⁻² T⁻²]
🧠 Solution:
Surface tension (S) = Force / Length = [MLT⁻²] / [L] = [MT⁻²]. Energy (E) = Force × displacement ⇒ [E] = [ML²T⁻²]. Velocity (v) = displacement / time ⇒ [v] = [LT⁻¹]. Assume S ∝ Eᵃ vᵇ Tᶜ. Then [MT⁻²] = [ML²T⁻²]ᵃ [LT⁻¹]ᵇ [T]ᶜ. Comparing exponents: a = 1, b = −2, c = −2. ∴ [S] = [E v⁻² T⁻²].
Q3043. If force (F), velocity (v) and time (T) are taken as fundamental units, then the dimensions of mass are
Year: CBSE AIPMT 2014
✅ Correct Answer: (d) [F v⁻¹ T¹]
🧠 Solution:
We know F = ma = m(v / T) ⇒ m = F T / v. Hence [M] = [F v⁻¹ T].
Q3044. The dimensions of (μ₀ε₀)⁻¹/² are
Year: CBSE AIPMT 2012
✅ Correct Answer: (c) [L T⁻¹]
🧠 Solution:
(μ₀ε₀)⁻¹/² represents the velocity of light. As c = 1 / √(μ₀ε₀), the dimension of c is [LT⁻¹].
Q3045. The dimensions of (½ ε₀E²), where ε₀ is the permittivity of free space and E is electric field, are
Year: CBSE AIPMT 2010
✅ Correct Answer: (b) [ML⁻¹T⁻²]
🧠 Solution:
Dimension of ε₀ = [M⁻¹L⁻³T⁴A²], Dimension of E = [MLT⁻³A⁻¹]. Hence (½ ε₀E²) = [M⁻¹L⁻³T⁴A²] × [MLT⁻³A⁻¹]² = [ML⁻¹T⁻²].
Q3046. If the dimensions of a physical quantity are given by [MLᵃTᵇ], then the physical quantity will be
Year: CBSE AIPMT 2009
✅ Correct Answer: (a) Pressure if a = 1, b = −2
🧠 Solution:
Velocity → [ML⁰T⁻¹], Acceleration → [ML⁰T⁻²], Force → [MLT⁻²], Pressure → [ML⁻¹T⁻²]. Hence for pressure, a = 1, b = −2.
Q3047. Which two of the following five physical parameters have the same dimensions? (i) Energy density (ii) Refractive index (iii) Dielectric constant (iv) Young’s modulus (v) Magnetic field
Year: CBSE AIPMT 2008
✅ Correct Answer: (c) (i) and (iv)
🧠 Solution:
Energy density = Energy / Volume = [ML⁻¹T⁻²]. Refractive index and dielectric constant are dimensionless. Young’s modulus = Stress / Strain = [ML⁻¹T⁻²]. Magnetic field = Force / (Charge × Velocity) = [MT⁻²A⁻¹]. Hence, (i) and (iv) have same dimensions.
Q3032. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be
Q3048. Dimensions of resistance (R) in terms of mass (M), length (L), time (T) and current (I) are
Year: CBSE AIPMT 2007
✅ Correct Answer: (d) [ML²T⁻³I⁻²]
🧠 Solution:
V = IR ⇒ R = V / I. Potential difference = Work / Charge. So [R] = [Work] / ([Current] × [Time]) = [ML²T⁻²] / ([I] × [T]) = [ML²T⁻³I⁻²].
Q3049. The velocity (v) of a particle at time t is given by v = a t² + b / t + c, where a, b and c are constants. The dimensions of a, b and c are respectively
Year: CBSE AIPMT 2006
✅ Correct Answer: (a) [LT⁻²], [L], [T]
🧠 Solution:
Given v = a t² + b / t + c. From homogeneity principle, [a t²] = [v] ⇒ [a] = [v] / [t²] = [LT⁻¹] / [T²] = [LT⁻³]. [b / t] = [v] ⇒ [b] = [v][t] = [LT⁻¹][T] = [L]. [c] = [v] = [LT⁻¹].
Q3050. The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of
Year: CBSE AIPMT 2005
✅ Correct Answer: (a) frequency
🧠 Solution:
Energy carried by photon is given by E = hν ⇒ h = E / ν. Hence, [h] = [ML²T⁻²] / [T⁻¹] = [ML²T⁻¹]. Moment of inertia, I = MR² ⇒ [I] = [ML²]. Therefore, [h/I] = [ML²T⁻¹] / [ML²] = [T⁻¹], which is the dimension of frequency. Alternatively, h/I = E/(Iν) = ET/I = (kg·m²/s²)(s)/(kg·m²) = 1/s = frequency. Thus, dimensions of (h/I) are same as that of frequency.
Q3051. The dimensions of universal gravitational constant are
Year: CBSE AIPMT 2004, 1992
✅ Correct Answer: (a) [M⁻¹L³T⁻²]
🧠 Solution:
From Newton’s law of gravitation, F = G m₁m₂ / r² ⇒ G = F r² / (m₁m₂). Hence, [G] = [MLT⁻²][L²] / [M²] = [M⁻¹L³T⁻²].
Q3022. The unit of permittivity of free space (ε₀) is
Year: CBSE AIPMT 2004
✅ Correct Answer: coulomb² / newton·metre²
🧠 Solution:
From Coulomb’s law: F = (1 / 4πε₀) × (q₁q₂ / r²)
⇒ ε₀ = q₁q₂ / (4πr²F)
Unit of ε₀ = coulomb² / (newton·metre²).
Q3052. Planck’s constant has the dimensions of
Year: CBSE AIPMT 2001
✅ Correct Answer: (b) angular momentum
🧠 Solution:
E = hν ⇒ h = E / ν ⇒ [h] = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]. Linear momentum p = mv = [MLT⁻¹]. Angular momentum L = Iω = [ML²T⁻¹]. Energy [E] = [ML²T⁻²]. Power [P] = [ML²T⁻³]. Hence, Planck’s constant has the same dimension as angular momentum.
Q3053. A pair of physical quantities having same dimensional formula is
Year: CBSE AIPMT 2000
✅ Correct Answer: (b) work and energy
🧠 Solution:
Force F = ma = [MLT⁻²], torque τ = Iα = [ML²T⁻²], work W = Fd = [ML²T⁻²], energy E = ½mv² = [ML²T⁻²], impulse = Ft = [MLT⁻¹], linear momentum = mv = [MLT⁻¹], angular momentum = Iω = [ML²T⁻¹]. Hence, work and energy have same dimensional formula.
Q3054. The dimensional formula for magnetic flux is
Year: CBSE AIPMT 1999
✅ Correct Answer: (a) [ML²T⁻²A⁻¹]
🧠 Solution:
Magnetic flux φ = BA. Also, B = F / (i l). Therefore, φ = (F / (i l)) × A ⇒ [φ] = [MLT⁻²][L²] / [A·L] = [ML²T⁻²A⁻¹].
Q3055. The force F on a sphere of radius r moving in a medium with velocity v is given by F = 6πηrv. The dimensions of η are
Year: CBSE AIPMT 1997
✅ Correct Answer: (d) [ML⁻¹T⁻¹]
🧠 Solution:
F = 6πηrv ⇒ η = F / (r v). Hence, [η] = [MLT⁻²] / ([L][LT⁻¹]) = [ML⁻¹T⁻¹].
Q3057. An equation is given as p + a / V² = bVθ, where p = pressure, V = volume and θ = absolute temperature. If a and b are constants, then dimensions of a will be
Q3056. Which of the following will have the dimensions of time?
Year: CBSE AIPMT 1996
✅ Correct Answer: (c) L / R
🧠 Solution:
L / R is the time constant of an RL circuit, which has the dimensions of time. Alternatively, [L/R] = [ML²T⁻²A⁻²] / [ML²T⁻³A⁻²] = [T].
Q3033. The density of a cube is measured by measuring its mass and length of sides. If the maximum error in the measurement of mass and length are 4% and 3% respectively, the maximum error in density will be
Q3058. Which of the following is a dimensional constant?
Year: CBSE AIPMT 1995
✅ Correct Answer: (d) Gravitational constant
🧠 Solution:
A quantity which has dimensions and also has a constant value is called a dimensional constant. Gravitational constant (G) is a dimensional constant.
Q3034. The percentage errors in the measurement of mass and speed are 2% and 3% respectively. The error in kinetic energy obtained by measuring mass and speed will be
Q3035. In a vernier callipers N divisions of vernier scale coincide with N−1 divisions of main scale (1 division = 1 mm). The least count of the instrument should be
Year: CBSE AIPMT 1994
✅ Correct Answer: 1 / 10N
🧠 Solution:
Given N(VSD) = (N − 1)MSD
⇒ 1 VSD = ((N − 1)/N) MSD
Least count (LC) = 1 MSD − 1 VSD = MSD/N² = 1/(10N) cm.
Q3024. In a particular system, the unit of length, mass and time are chosen to be 10 cm, 10 g and 0.1 s respectively. The unit of force in this system will be equivalent to
Year: CBSE AIPMT 1994
✅ Correct Answer: 0.1 N
🧠 Solution:
Force F = MLT⁻².
Using given units:
F = (10 g)(10 cm)(0.1 s)⁻².
In SI: F = (10⁻² kg)(10⁻¹ m)(10⁻¹ s)⁻² = 10⁻¹ N = 0.1 N.
Q3059. Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is p. The viscosity of oil is given by η = (p r²x) / (l × 4v), where v is the velocity of oil at distance x from the axis of the tube. The dimensions of η are
Year: CBSE AIPMT 1993
✅ Correct Answer: (d) [ML⁻¹T⁻¹]
🧠 Solution:
Pressure p = Force / Area = [ML⁻¹T⁻²]. Velocity v = [LT⁻¹]. Hence, η = [p r²x] / [l v] = [ML⁻¹T⁻²][L³] / ([L][LT⁻¹]) = [ML⁻¹T⁻¹].
Q3061. If p represents radiation pressure, c represents speed of light, and S represents radiation energy striking unit area per second, the non-zero integers x, y, z such that pˣSʸcᶻ is dimensionless are
Year: CBSE AIPMT 1992
✅ Correct Answer: (c) x = 1, y = 1, z = −1
🧠 Solution:
Radiation pressure, p = [ML⁻¹T⁻²]; velocity of light, c = [LT⁻¹]; energy striking unit area per second, S = [MT⁻³]. For pˣSʸcᶻ to be dimensionless: [M⁰L⁰T⁰] = [Mˣ⁺ʸ L⁻ˣ⁺ᶻ T⁻²ˣ⁻³ʸ⁻ᶻ]. Solving gives x = 1, y = 1, z = −1.
Q3060. The time dependence of a physical quantity p is given by p = p₀ exp(−αt²), where α is a constant and t is the time. The constant α
Year: CBSE AIPMT 1992
✅ Correct Answer: (b) has dimensions [T⁻²]
🧠 Solution:
p = p₀ exp(−αt²). The power of an exponential quantity must be dimensionless, so αt² is dimensionless ⇒ αt² = 1 ⇒ α = 1/t² = [T⁻²].
Q3062. The dimensional formula for permeability of free space (μ₀) is
Year: CBSE AIPMT 1991
✅ Correct Answer: (a) [ML²T⁻²A⁻²]
🧠 Solution:
From Biot–Savart law: dB = (μ₀/4π)·(Idl·sinθ)/r². Hence, μ₀ = (4πr²dB)/(Idl·sinθ). ⇒ [μ₀] = [L²][MT⁻²A⁻¹]/([A][L]) = [ML²T⁻²A⁻²].
Q3036. A certain body weighs 22.42 g and has a measured volume of 4.7 cc. The possible errors in mass and volume are 0.01 g and 0.1 cc. Then, maximum error in density will be
Year: CBSE AIPMT 1991
✅ Correct Answer: 2%
🧠 Solution:
ρ = m / V
Relative error: (Δρ/ρ) = (Δm/m + ΔV/V)
= (0.01/22.42 + 0.1/4.7) × 100 = 2%.
Q3065. The dimensional formula of pressure is
Year: CBSE AIPMT 1990
✅ Correct Answer: (c) [ML⁻¹T⁻²]
🧠 Solution:
Pressure = Force / Area ⇒ [MLT⁻²] / [L²] = [ML⁻¹T⁻²].
Q3064. According to Newton, the viscous force acting between liquid layers of area A and velocity gradient dv/dz is given by F = −ηA(dv/dz), where η is a constant called
Year: CBSE AIPMT 1990
✅ Correct Answer: (d) [ML⁻¹T⁻¹]
🧠 Solution:
F = −ηA(dv/dz) ⇒ η = F / (A·dv/dz). Since F = [MLT⁻²], A = [L²], dv/dz = [T⁻¹], we get [η] = [ML⁻¹T⁻¹].
Q3063. The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by f = C·mˣ·kʸ, where C is a dimensionless constant. The values of x and y are
Year: CBSE AIPMT 1990
✅ Correct Answer: (d) x = −½, y = ½
🧠 Solution:
f = C·mˣ·kʸ. Dimensions: [T⁻¹] = [Mˣ][ML⁻¹T⁻²]ʸ ⇒ [Mˣ⁺ʸL⁻ʸT⁻²ʸ]. Equating powers: x + y = 0, −2y = −1 ⇒ y = ½, x = −½.
Q3069. Of the following quantities, which one has dimensions different from the remaining three?