JEE Previous Year Questions With Solutions Of Introduction To Physics [Unit, Measurement & Dimension] Part - 3 | IIT JEE | AIEEE | NEET | JEE Main & Advance | AIIMS | WBJEE | KCET | PMT | OTHER JE EXAMS

It is very important to have the idea of any examination to get success in that. The best way to have that idea is to have a look at previous year Questions of that examination. That's why we have collected some previous year Questions with Solutions of JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams so that you can get clear pattern and level of question that will be asked in forthcoming JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams.

JEE Previous Year's Questions With Solutions

1. Identify the pair whose dimensions are equal. (AIEEE 2002)

(A) Torque and Work

(B) Stress and Energy

(C) Force and Stress

(D) Force and Work.

Sol: (A)

[Torque] = ML²T^{– 2}; [Work] = ML²T^{– 2}; [Stress] = ML^{– 1}T^{– 2}; [Energy] = ML²T^{– 2}; [Force] = MLT^{– 2}.

So, the pair of option (A) has same dimensions.

2. Dimensions of 1/ (µ₀ϵ₀), where symbols have their usual meaning, are: (AIEEE 2003)

(A) L^{– 1}T

(B) L^{– 2}T²

(C) L²T^{– 2}

(D) LT^{– 1}.

Sol: (C)

We know,

Speed of light, c = 1/√ (µ₀ϵ₀)

Or, c² = 1/ (µ₀ϵ₀)

Or, [1/ (µ₀ϵ₀)] = [c]² 

Or, [1/ (µ₀ϵ₀)] = (LT^{– 1})² = L²T^{– 2}.

3. The physical quantities not having same dimensions are: (AIEEE 2003)

(A) Torque and Work

(B) Momentum and Planck’s constant

(C) Stress and Young’s modulus

(D) Speed and 1/√ (µ₀ϵ₀).

Sol: (B)

[Momentum] = MLT^{– 1} and

[Planck’s constant] = ML²T^{– 1} 

So, the pair of option (B) do not have same dimensions.

4. Which one of the following represents the correct dimensions of the coefficient of viscosity? (AIEEE 2004, WBJEE 2007)

(A) ML^{– 1}T^{– 2}

(B) MLT^{– 1}  

(C) ML^{– 1}T^{– 1}

(D) ML^{– 2}T^{– 2}.

Sol: (C)

According Newton,

Viscous force, F = – ηA (ΔV/ΔZ)

Where, η = coefficient of viscosity, A = area, V = velocity, Z = distance of fluid layers.

Now, [F] = [η] [A] ([ΔV]/ [ΔZ])

Or, MLT^{– 2} = [η] (L²) (LT^{– 1}/L)

Or, [η] = ML^{– 1}T^{– 1}.

5. A quantity X is given by ϵ₀L(ΔV/Δt) where ϵ₀ is the permittivity of the free space, L is the length, ΔV is a potential difference and Δt is a time interval. The dimensional formula for X is same as that of: (IIT JEE 2001)

(A) Resistance

(B) Charge

(C) Voltage

(D) Current.

Sol: (D)

[ϵ₀L] = [Capacitance, C]

So, ϵ₀L(ΔV/Δt) = C(ΔV/Δt)

Or, ϵ₀L(ΔV/Δt) = Δq/Δt = I (current).

6. A cube has a side of length 1.2 * 10^{– 2} m. Calculate its volume. (IIT JEE 2003)

(A) 1.7 * 10^{– 6} m³

(B) 1.73 * 10^{– 6} m³

(C) 1.70 * 10^{– 6} m³

(D) 1.732 * 10^{– 6} m³.

Sol: (A)

Side of cube, L = 1.2 * 10^{– 2} m (Two significant figure)

Volume = L³ = (1.2 * 10^{– 2})³ = 1.728 * 10^{– 6} = 1.7 * 10^{– 6} m³.

The side has two significant figure. Hence, volume is expressed in two significant figures.

7. A wire has a mass (0.3 ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is: (IIT JEE 2004)

(A) 1

(B) 2

(C) 3

(D) 4.

Sol: (D)

Here, (Δm/m)*100 = (0.003/0.3)*100 = 1%

(Δr/r)*100 = (0.005/0.5)*100 = 1%

(ΔL/L)*100 = (0.06/6)*100 = 1%

We know, density, ρ = mass/volume

Or, ρ = m/ (πr²L)

Taking log and differentiate and added for errors

We get, Δρ/ρ = (Δm/m) + 2(Δr/r) + (ΔL/L)

Or, (Δρ/ρ) * 100 = 1 + 2 * 1 + 1

Or, (Δρ/ρ) * 100 = 4%.

8. Pressure depends on distance as, P = (α/β) exp (- αz/kθ) where α, β are constants, z is distance, k is Boltzmann’s constant and θ is temperature. The dimension of β are: (IIT JEE 2004)

(A) M⁰L⁰T⁰

(B) M^{– 1}L^{– 1}T^{– 1} 

(C) ML^{– 1}T^{– 1}

(D) M⁰L²T⁰.

Sol: (D)

[Boltzmann’s constant, k] = joule/kelvin

Or, [k] = ML²T^{– 2}θ^{– 1} (MP PMT 2002)

Since, exponent expressions are always dimensionless.

Therefore, [- αz/kθ] = M⁰L⁰T⁰

Or, [α] = MLT^{– 2}.

Now, [P] = [α/β] = ML^{– 1}T^{– 2} 

Or, [β] = (MLT^{– 2})/ (ML^{– 1}T^{– 2}) = M⁰L²T⁰. 

9. The dimensional formula of a/b in the equation P = (a – t²)/bx where P is pressure, x is distance and t is time: (Karnataka CET 2003)

(A) LT^{– 3}

(B) ML³T^{– 1} 

(C) MT^{– 2}

(D) M^{– 1}L²T¹.

Sol: (C)

Given: P = (a – t²)/bx

Or, P = (a/bx) – (t²/bx)

Appling the principle of homogeneity

We get, [P] = [a/bx]

Or, ML^{– 1}T^{– 2} = [a/b] * L^{– 1} 

Or, [a/b] = MT^{– 2}.

10. The physical quantity having the same dimensions as Planck’s constant h is: (Karnataka CET 2004)

(A) Linear momentum

(B) Angular momentum

(C) Boltzmann’s constant

(D) Force.

Sol: (C)

Energy, E = hν

Where, h = Planck’s constant, ν = frequency.

→ [h] = ML²T^{– 1}.

[Linear momentum] = MLT^{– 1}

[Angular momentum] = ML²T^{– 1}.

[Boltzmann’s constant, k] = joule/kelvin

Or, [k] = ML²T^{– 2}θ^{– 1} 

[Force] = MLT^{– 2}.

So, [Planck’s constant] = [Angular momentum].

11. The dimensional formula for inductance is: (Karnataka CET 2004, UPSEE 2008)

(A) ML²T^{– 2}A^{– 2}

(B) ML²T^{– 1}A^{– 2} 

(C) ML²T^{– 2}A^{– 1}

(D) ML²TA^{– 1}.

Sol: (A)

We know, Emf, e = L(di/dt)

Or, L = e(dt/di)

Now, [L] = [e] [dt/di]

Or, [L] = (ML²T^{– 2}/AT) (T/A)

Or, [L] = ML²T^{– 2}A^{– 2}.

12. The unit of permittivity of free space, ϵ₀ is: (Karnataka CET 2004)

(A) Newton-metre²/Coulomb²

(B) Coulomb²/ Newton-metre²

(C) Coulomb²/ (Newton-metre)²

(D) Coulomb/ (Newton-metre).

Sol: (B)  

From coulomb’s Law:

We have, F = (1/4πϵ₀) (q₁q₂/r²)

Or, ϵ₀ = (1/4πF) (q₁q₂/r²)

So, the unit of ϵ₀ is Coulomb²/ Newton-metre².

13. If the energy (E), velocity (v) and force (F) be taken as fundamental quantity, then the dimensional formula of mass will be: (BHU 2002)

(A) Fv^{– 1}

(B) Fv^{– 2} 

(C) Ev²

(D) Ev^{– 2}.

Sol: (D)

Let, mass, m = E^aF^bv^c

Or, [m] = [E]^a [F]^b [v]^c

Or, ML⁰T⁰ = (ML²T^{– 2})^a (MLT^{– 2})^b (LT^{– 1})^c

Or, ML⁰T⁰ = M^a+b L^{2a + b + c} T^{– 2a– 2b– c}

Equating the exponents of similar quantities,

We get, a + b = 1; 2a + b + c = 0 and – 2a – 2b – c = 0.

Or, a = 1, b = 0 and c = – 2.

Therefore, m = Ev^{– 2}.

14. The dimensional formula for Young’s modulus is: (BHU 2002)

(A) ML^{– 1}T^{– 2}

(B) M⁰LT^{– 2}

(C) ML²T^{– 2}

(D) ML²T^{– 2}.

Sol: (A)

We know,

Stress = Young’s modulus * strain

→ F/A = Y * (ΔL/L)

Or, [Y] = [F]/[A] = MLT^–2/L² = ML^{– 1}T^–2  

So, [Young’s modulus] = ML^{– 1}T^–2.

15. 1 Wb-m^–2 is equal to: (UPSEE 2005)

(A) 10 ⁴ gauss

(B) 10^{– 4} gauss

(C) 10² gauss

(D) 4 * 10^{– 3} gauss.

Sol: (A)

We know, 

Magnetic field, B = F/(qv) and SI unit is Wb-m^–2.

Dimensional formula of Magnetic field,

[Magnetic field] = [F]/ [qv] = (MT^–2A^–1)

So, 1 Wb-m^–2 = (1 kg) (1 s) ^–2 (1 A) ^–1

And, 1 gauss (CGS) = (1 g) (1 s) ^–2 (1 Biot) ^–1

Thus, 1 N-s /1 dyne-s = (1 kg/1 g) (1 s/1 s) ^–2 (1 A/1 Biot)

= (1000 g/1 g) (1 s/1 s) ^–2 (10 Biot /1 Biot)

= (10)³ (10)

= 10⁴ 

ஃ 1 Wb-m^–2 = 10⁴ gauss.

16. Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be: (UPSEE 2007, Kerala CET 2002)

(A) ML²T^{– 3}I^{– 1}

(B) ML²T^{– 2}

(C) ML²T^{– 1}I^{– 1}

(D) ML²T^{– 3}I^{– 2}.

Sol: (D)

We have, resistance = voltage/current

Or, resistance = Work/ (charge * current)

Or, [resistance] = [Work]/ ([charge] * [current])

Or, [resistance] = (ML²T^{– 2})/ (IT * I)

Or, [resistance] = ML²T^{– 3}I^{– 2}. 

17. The dimensions of permittivity ϵ₀ are: (AIIMS 2004, BVP 2006)

(A) M^{– 1}L^{– 3}T^{– 4}A ²

(B) M^{– 2}L^{– 3}T⁴A^{– 2}

(C) ML^{– 3}T^{– 3}A ²

(D) M^{– 1}L^{– 3}T⁴A ².

Sol: (D)

We have, F = (1/4πϵ₀) (q₁q₂/r²)

Or, [F] = [q₁] [q₂]/ ([ϵ₀] [r]²)

Or, [ϵ₀] = [q₁] [q₂]/ ([F] [r]²)

Or, [ϵ₀] = {(AT)(AT)}/{(MLT^{– 2})(L²)}

Or, [ϵ₀] = M^{– 1}L^{– 3}T⁴A ². 

18. If 3.8 * 10^{– 6} is added to 4.2 * 10^{– 5} giving due regard to significant figures, then the result will be: (UPSEE 2009)

(A) 4.6 * 10^{– 5}

(B) 4.58 * 10^{– 5}

(C) 4.5 * 10^{– 5}

(D) None of these.

Sol: (A)

3.8 * 10^{– 6} = 0.38 * 10^{– 5} and 4.2 * 10^{– 5}

4.2 * 10^{– 5} + 0.38 * 10^{– 5} = 4.58 * 10^{– 5}.

As least no. of significant figures in given values are 2, so we round off the result to 4.6 * 10^{– 5}.

19. Which of the following is not a unit of Young’s modulus? (Karnataka CET 2005)

(A) N-m^{– 1}

(B) N-m^{– 2}

(C) Dyne-cm^{– 2}

(D) Mega Pascal.

Sol: (A)

We have, σ = Yϵ

Or, Y = σ/ϵ = FΔL/AL

Therefore, unit of Young’s modulus are N-m^{– 2} = Pascal (SI unit); Dyne-cm^{– 2} (CGS unit).

So, N-m^{– 1} is not a unit of Young’s modulus.

20. Unit of Stefan’s constant is: (CBSE 2002, Karnataka CET 2006)

(A) Wm^{– 2}K^{– 1}

(B) WmK^{– 4}

(C) Wm^{– 2}K^{– 4}

(D) Wm^{– 2}K⁴.

Sol: (C)

According to Stefan’s Law:

Energy per unit time, E = σAT⁴

Where σ = Stefan’s constant, A = area, T = temperature.

Or, σ = E/ AT⁴ 

So, unit of Stefan’s constant is Wm^{– 2}K^{– 4}.

21. The velocity of a particle (v) at an instant ‘t’ is given by v = at + bt², the dimension of b is: (WBJEE 2008)

(A) L

(B) LT^{– 1}

(C) LT^{– 2}

(D) LT^{– 3}.

Sol: (D)

The equation contains three terms. All of them should have the same dimensions. Since [v] = LT^{– 1}, each of the remaining two must have the dimension of velocity (v).

So, [bt²] = LT^{– 1} 

Or, [b] = LT^–3.

22. The equation of state for n moles of an ideal gas is PV = nRT, where R is a constant. The SI unit for R is: (WBJEE 2009)

(A) JK^–1 per molecule    

(B) JK^–1 mol^–1     

(C) J Kg^–1 K^–1       

(D) JK^–1 g^–1.

Sol: (B)

Given: PV = nRT

Or, R = PV/nT

SI unit of R = {(N-m^–2) (m³)}/ {(mol) (K)}

SI unit of R = N-m- mol^–1K^–1 = J- mol^–1K^–1.

23. Out of the following pairs which one does not have identical dimensions is: (AIEEE 2005)

(A) Moment of inertia and Moment of force 

(B) Work and Torque

(C) Angular momentum and Planck’s constant

(D) Impulse and Momentum.

Sol: (A)

Moment of inertia, I = mr²

[Moment of inertia] = ML²

Moment of force = Fr

[Moment of force] = ML²T^–2.

Therefore, [Moment of inertia] ≠ [Moment of force].

24. Which of the following units denotes the dimensions ML²/Q², where Q denotes the electric charge? (AIEEE 2006)

(A) Weber (Wb) 

(B) Wb/m²

(C) Henry (H)

(D) H/m².

Sol: (C)

ML²/Q² = ML²A^–2T^–2

[Wb] = ML²A^–1T^–2

[Wb/m²] = MA^–1T^–2

[Henry] = ML²A^–2T^–2

[H/m²] = MA^–2T^–2.

Therefore, option (c) is correct.

25. The dimension of magnetic field in M, L, T and C (Coulomb) is given as: (AIEEE 2008)

(A) MT^–2C^{– 1}

(B) MLT^–1C^{– 1}

(C) MT²C^{– 2}

(D) MT^–1C^{– 1}.

Sol: (D)

We know,

Magnetic force, F = qvB sin θ

Where, q = charge (unit coulomb), v = velocity (unit ms^-1), B = magnetic field.

Or, B = F/ (qv sin θ)

Or, [B] = (MLT^–2)/(C)(LT^–1)

Or, [B] = MT^–1C^{– 1}.

26. Which of the following sets have different dimensions? (IIT JEE 2005)

(A) Pressure, Young’s modulus, stress

(B) Emf, potential difference, electric potential

(C) Heat, work done, energy

(D) Dipole moment, electric flux, electric field.

Sol: (D)

[Pressure] = [Young’s modulus] = [stress] = ML^–1T^–2.

[Emf] = [potential difference] = [electric potential] = ML²T^–3A^–1.

[Heat] = [work done] = [energy] = ML²T^–2.

[Dipole moment] = LTA ≠ [electric flux] = ML³T^–3A^–1 ≠ [electric field] = MLT^–3A^–1.

27. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 * 10^-3 are: (AIEEE 2010)

(A) 5, 5, 2

(B) 5, 1, 2

(C) 4, 4, 2

(D) 5, 1, 5.

Sol: (B)

23.023 has 5 significant figures. (All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all).

0.0003 has 1 significant figure. (All the zero to the right of a decimal point and to the left of non-zero digit are not significant).

2.1 * 10^-3 has 2 significant figures. (The powers of ten are not counted as significant figure).

28. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is: (AIEEE 2012)

(A) 6%

(B) 3%

(C) 1%

(D) Zero.

Sol: (A)

Here, (ΔI/I)*100 = 3%

(ΔV/V)*100 = 3%

We know, voltage, V = current (I)*resistance (R)

Or, V = IR

Or, R = V/I

Taking log and differentiate and added for errors

We get, ΔR/R = (ΔV/V) + (ΔI/I)

Or, (ΔR/R) * 100 = 3 + 3

Or, (ΔR/R) * 100 = 6%.

29. If n denotes a positive integer, h the Planck’s constant, q the charge and B the magnetic field, then the quantity (nh/2πqB) has the dimension of: (WBJEE 2014)

(A) Area

(B) Length

(C) Speed

(D) Acceleration

Sol: (A)

→ [nh/2πqB] = [h/qB]

Or, [nh/2πqB] = [ML²T^–1/MT^–1]

Or, [nh/2πqB] = L² = [Area].

30. If x = at + bt² where x is in metre (m) and t is in hour (hr) then unit of b will be: (WBJEE 2016)

(A) m²/hr

(B) m

(C) m/hr

(D) m/hr².

Sol: (D)

Appling the principle of homogeneity

We get, [x] = [bt²]

Or, L = [b] * T²

Or, [b] = LT^{– 2}.

So, unit of b = m/hr².

Discussion - If you have any Query or Feedback comment below.

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