JEE Previous Year Questions With Solutions Of Introduction To Physics [Unit, Measurement & Dimension] Part - 2 | 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. The pairs of physical quantities that have the same dimensions in (are): (IIT JEE 1995)

(A) Reynolds number and coefficient of friction.

(B) Curie and frequency of a light wave.

(C) Latent heat and gravitational potential.

(D) Planck’s constant and torque.

Sol: (A), (B), (C)

(A) Reynolds number and coefficient of friction both are dimensionless quantities.

(B) Curie = No. of atoms/time

Frequency = No. of vibrations/time

[Curie] = T^{– 1} and [Frequency] = T^{– 1}

So, both have same dimension.

(C) [Latent heat] = [gravitational potential] = ML²T^{– 2}.

2. The sides of a rectangle are 6.01 m and 12 m. Taking the significant figures into account, the area of the rectangle is: (Karnataka CET 1994)

(A) 72.00 m²

(B) 72.1 m²

(C) 72 m²

(D) 72.12 m²

Sol: (C)

Sides of rectangle are a = 6.01 m (three significant figures) and b = 12 m (two significant figures).

Area of rectangle is = ab = 6.01 * 12 = 72.12 m².

In case of multiplication, the no. of significant figures in the result is same as the smallest no. of significant figures in any of the factors.

Therefore, Area of rectangle is = 72 m².

3. The velocity of a body which has fallen freely under gravity varies as g^ah^b where g is acceleration due to gravity at a place and h is the height through which the body has fallen. The value of a and b given as: (EAMCET 1994, UPSEAT 1999)

(A) a = ½ , b = ½

(B) a = – ½ , b = – ½

(C) a = – ½ , b = ½

(D) a = ½ , b = – ½

Sol: (A)

Dimensionally, velocity = g^ah^b

Or, [velocity] = [g]^a [h]^b

Or, LT^{– 1} = (LT^{– 2})^a (L)^b

Or, LT^{– 1} = L^a+b T ^{– 2b}  

Equating the exponents of similar quantities,

We get, a + b = 1 and – 1 = – 2b

Or, a = ½ and b = ½.

4. The equation of state for real gas is given by (P + a/V²)(V – b) = RT or constant. Where P is pressure, T is the absolute temperature and a, b, R, are constant. The dimensional formula of ‘a’ is: (IIT JEE 1997, J&K CET 2002, WBJEE 2013)

(A) ML⁵T^{– 2}

(B) ML^{– 1}T^{– 2}

(C) L³

(D) L⁶.

Sol: (A)

Quantities with similar dimensions only can be added.

So, [P] = [a/V²]

Or, ML^{– 1}T^{– 2} = [a]/ L⁶ 

Or, [a] = ML⁵T^{– 2}.

And [b²] = [V] = L³

Or, [b] = L^3/2.

5. The displacement of a particle moving along x-axis with respect to time is x = at + bt² – ct³. The dimensions of c are: (KEAM 2012)

(A) T ^{– 3}

(B) LT ^{– 2}

(C) LT ^{– 3}

(D) LT³

(E) LT².

Sol: (C)

The equation contains four terms. All of them should have the same dimensions. Since [x] = L, each of the remaining three must have the dimension of length (L).

So, [cT³] = L

Or, [c] = LT^–3.

6. The dimensions of electrical conductivity is ______: (IIT JEE 1997, J&K CET 2012)

Sol:  

We know, conductivity, σ = 1/ρ = L/RA

Where, L = length; R = resistance; A = area of cross section.

Resistance, R = V/I = (W/q)/I = W/I²t

Or, [R] = ML²T^{– 3}A^{– 2}

Therefore, [σ] = [L/RA]

Or, [σ] = M^{– 1}L^{– 3}T³A ².

7. Let [ϵ₀] denote of the permittivity of the vacuum, and [µ₀] denote the permeability of the vacuum. If M = mass, L = length, T = time, and I = electric current, then the dimensional formula of ϵ₀ and µ₀ are: (IIT JEE 1998)

(A) [ϵ₀] = M^{– 1}L^{– 3}T²I

(B) [ϵ₀] = M^{– 1}L^{– 3}T⁴I²

(C) [µ₀] = MLT^{– 2}I^{– 2}

(D) [µ₀] = ML²T^{– 1}I.

Sol: (B), (C)

We know, Force, F = (1/4πϵ₀) (q₁q₂/r²)

Or, [F] = [q₁] [q₂]/[r²][ϵ₀]

Or, [MLT^{– 2}] = [IT] [IT]/[L²][ϵ₀]

Or, [ϵ₀] = M^{– 1}L^{– 3}T⁴I².

We know,

Speed of light, c = 1/√ (µ₀ ϵ₀)

Or, c² = 1/ (µ₀ ϵ₀)

Or, [µ₀] = 1/ [ϵ₀][c²]

Or, [µ₀] = 1/ {(M^{– 1}L^{– 3}T⁴I²)(L²T^{– 2})}

Or, [µ₀] = MLT^{– 2}I^{– 2}.

8. The dimensional formula of the ratio of angular to linear momentum is: (MNR 1994)

(A) M⁰L¹T⁰

(B) M¹L¹T^{– 1}

(C) M¹L²T^{– 1}

(D) M^{– 1}L^{– 1}T^{– 1}.

Sol: (A)

[Angular momentum] = ML²T^{– 1}.

[Linear momentum] = MLT^{– 1}.

Therefore, the dimensional formula of the ratio of angular to linear momentum is M⁰L¹T⁰.

9. The pair having the same dimensions is: (MP PET 1994)

(A) Angular momentum, Work

(B) Work, Torque

(C) Potential energy, linear momentum

(D) Kinetic energy, velocity

Sol: (B)

[Angular momentum] = ML²T^{– 1}.

[Work] = ML²T^{– 2}.

[Torque] = ML²T^{– 2}.

[Potential energy] = ML²T^{– 2}.

[Linear momentum] = MLT^{– 1}.

[Kinetic energy] = ML²T^{– 2}.

[Velocity] = LT^{– 1}.

Pair of option (B) has same dimension.

10. The unit of magnetic induction is: (MP PMT 1994)

(A) Coulomb volt^{– 1}

(B) Newton (ampere-metre) ^{– 1}

(C) Volt second (ampere) ^{– 1}

(D) Coulomb² Joule^{– 1}

Sol: (B)

We know,

Magnetic force, F = qvB sin θ

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

Or, B = F/ (qv sin θ)

Therefore, unit of B = N/(C-ms^-1)

Or, unit of B = N/ (A-m) [since, ampere = C-s^-1].

11. The dimensional formula of magnetic flux is: (MP PMT 1994, CBSE 1999, BVP 2007, EAMCET 2003,J&K CET 2015)

(A) MLT^{– 2}A^{– 2}

(B) ML²T^{– 2}A^{– 2}

(C) ML²T^{– 1}A^{– 2}

(D) ML²T^{– 2}A^{– 1}.

Sol: (D)

We know,

Magnetic flux, φ = BA = FA/ (qv sin θ)

Or, [φ] = [F][A]/ [qv sin θ]

Or, [φ] = ML²T^{– 2}A^{– 1}. 
Unit of magnetic flux = Newton-metre/ampere. (AMU 1998)

12. The unit of electric field is not equivalent to: (MP PMT 1994)

(A) N/C

(B) J/C

(C) V/m

(D) J/Cm.

Sol: (D)

We know,

Electric field, E = Force/charge

Unit of E = N/C = Nm/Cm = J/Cm

Since, J/C = V → Unit of E = V/m.

13. The unit of magnetic moment is: (MP PMT 1995)

(A) Wb/m

(B) Wb-m²

(C) A-m

(D) A-m².

Sol: (D)

We know,

Magnetic moment, µ = IA

Where, I = current and A = area.

So, unit of magnetic moment is A-m². 

14. The dimensional formula of pressure is: (CBSE 1994, Kerala CET 2003)

(A) MLT^{– 2}

(B) ML^{– 1}T²

(C) ML^{– 1}T^{– 2}

(D) MLT².

Sol: (C)

We know, pressure = Force/area

Or, [pressure] = [Force]/ [area]

Or, [pressure] = MLT^{– 2}/L²

Or, [pressure] = ML^{– 1}T^{– 2}. 

15. The density of a cube is measure by measuring its mass and length of its sides. If the maximum error in measurement of mass and lengths are 4% and 3% respectively, the maximum error in the measurement of density would be: (CBSE 1996)

(A) 9%

(B) 13%

(C) 12%

(D) 7%.

Sol: (B)

Density, ρ = Mass/volume = m/L³

Taking log and differentiate and added for errors 

We get, Δρ/ρ = (Δm/m) + 3 (ΔL/L)

Or, Δρ/ρ = 4 + 3 * 3 = 13.

So, the maximum error in the measurement of density is 13%.

16. Which one of the following groups have quantities that do not have the same dimensions? (CBSE 2000)

(A) Velocity, Speed

(B) Pressure, stress

(C) Force, Impulse

(D) Work, Energy

Sol: (C)

[Velocity] = [Speed] = LT^{– 1}.

[Pressure] = [stress] = ML^{– 1}T^{– 2}.

[Work] = [Energy] = ML²T^{– 2}.
[Force] = MLT^{– 2} 
[Impulse] = MLT^{– 1} (Karnataka CET 2007)
Therefore, [Force] ≠ [Impulse].

17. A force is given by F = at + bt², where t is time. What are dimensional formulae of a and b? (BHU 1998)

(A) MLT^{– 3} and ML³T⁴

(B) MLT^{– 3} and MLT^{– 4}

(C) MLT^{– 1} and MLT⁰

(D) MLT^{– 4} and MLT¹.

Sol: (B)

The equation contains three terms. All of them should have the same dimensions. Since [F] = MLT^–2, each of the remaining two must have the dimension of Force (MLT^–2).

So, [a] = MLT^–2T^–1 = MLT^–3, [b] = MLT^–2T^–2 = MLT^–4.

18. If force, length, and time are taken as fundamental units, then dimensional formula of mass is: (BHU 2000)

(A) F¹L^{– 1}T ²

(B) F⁰L⁰T ²

(C) F ¹L²T ^{– 2}

(D) F¹L¹T ^{– 2}.

Sol: (A)

We know,

Force, F = mass (m) * acceleration (a)

Or, m = F/a

Or, [m] = F/ (LT^{– 2}) = F¹L^{– 1}T ². 

19. Numerical value of magnitude of a physical quantity is: (BHU 2000)

(A) Independent of system of units.

(B) Directly proportional to magnitude of the unit

(C) Inversely proportional to magnitude of the unit

(D) Either (B) or (C).

Sol: (C)

Example: 10 km = 10000 m

If we decrease the size of unit (u) from km to m. we see numerical value (n) increase from 10 to 10000. Therefore, n ∝ 1/u. 

20. L, C, and R represent physical quantities inductance, capacitance, and resistance respectively. The combinations which does not have the dimensions of frequency is: (AMU 1996)

(A) 1/(RC)

(B) R/L

(C) 1/√(CL)

(D) C/L

Sol: (D)

We know,

Dimensional formula of

Resistance, R = ML²T^–3A^–2

Inductance, L = ML²T^–2A^–2

Capacitance, C = M^–1L^–2T⁴A²

Now,

[1/RC] = 1/T = [frequency]

[R/L] = 1/T = [frequency] (BVP 2005)

[1/√(CL)] = 1/T = [frequency]

[C/L] = M^–2L^–4T⁶A⁴ ≠ 1/T = [frequency].

21. The dimension formula of coefficient of thermal conductivity (k) is: (AMU 1996, UPSEE 2006)

(A) MLT^{– 3}K^{– 1}

(B) M³L^{– 1}T^{– 1}K

(C) ML^{– 3}T^{– 1}K^{– 1}

(D) ML^{– 1}T^{– 3}K.

Sol: (D)

From the Fourier's law, Q = k A (ΔT/ΔL)

Where, k = coefficient of thermal conductivity, A = area, Q = rate of heat transfer, ΔT = change in temperature.

Now, [Q] = [k] [A] ([ΔT]/[ΔL])

Or, ML²T^{– 3} = [k] (L²) (K/L)

Or, [k] = MLT^{– 3}K^{– 1}.

22. The SI unit of inductance, the henry can written as: (IIT JEE 1998)

(A) weber/ampere

(B) Volt-sec/amp

(C) Joule/ (ampere)²

(D) Ohm-second

Sol: (A), (B), (C), (D)

A) Inductance = Flux/current

→ Henry, H = Wb/A

B) ϵ = L (di/dt)

Or, L = ϵ/ (di/dt) → H = Vs/A

C) U = ½ LI²

Or, L = 2U/I² → H = J/A²

D) L = ϵ/ (di/dt) = (ϵ/di) dt

→ H = ohm-sec.

23. The dimensional formula of ½ ϵ₀E² (ϵ₀ = permittivity of free space and E = electric field) is: (IIT JEE 2000, Odisha JEE 2016)

(A) MLT^{– 1}

(B) ML²T^{– 2}

(C) ML^{– 1}T^{– 2}

(D) ML²T^{– 1}.

Sol: (C)

We know, Energy per unit volume = ½ ϵ₀E²

So, [½ ϵ₀E²] = [Energy]/ [volume]

Or, [½ ϵ₀E²] = ML^{– 1}T^{– 2}.

24. Among the following the dimensional formula of impulse is: (Odisha JEE 2015, J&K CET 2011)

(A) MLT ^{– 1}  

(B) MLT^{– 2}

(C) ML^{– 1}T^{– 2}

(D) ML²T^{– 2}.

Sol: (A)

Impulse = F * t

Therefore, [Impulse] = [F] [t]

Or, [Impulse] = MLT^{– 2} T

Or, [Impulse] = MLT ^{– 1}.  

25. The flux density of mass is defined as the amount of mass crossing unit area per unit time. The dimension of this quantity is: (J&K CET 2012)

(A) ML^{– 2}T ^{– 1}  

(B) ML²T^{– 1}

(C) MLT^{– 1}

(D) M^{– 1}L^{– 2}T.

Sol: (A)

Given: flux density of mass = m/(At)

Where m = mass, A = area, t = time.

So, [flux density of mass] = ML^{– 2}T^{– 1}.

26. The ratio of the dimensions of Planck constant and that of moment of inertia has the dimensions of: (Karnataka CET 2015)

(A) Time  

(B) Frequency

(C) Angular momentum

(D) Velocity.

Sol: (B)

[Planck constant, h] = ML²T ^{– 1}.

[Moment of inertia, I] = ML².

[h]/ [I] = ML²T ^{– 1}/ ML² = T ^{– 1}

Therefore, [h/I] = T ^{– 1} = [Frequency].

27. A physical quantity Q is found to depend on observables x, y and z, obeying relation Q = (x³y²/z). The % error in the measurements of x, y and z are 1%, 2% and 4% respectively. What is percentage error in the quantity Q? (Karnataka CET 2014)

(A) 4%

(B) 3%

(C) 11%

(D) 1%

Sol: (C)

Given: Q = x³y²/z

Taking log and differentiate and added for errors

We get, ΔQ/Q = 3 * (Δx/x) + 2 * (Δy/y) + (Δz/z)

Or, ΔQ/Q = 3 * 1 + 2 * 2 + 4 = 11.

So, the error in Q is 11%.

28. The dimensional formula of physical quantity is M^aL^bT^c. Then that physical quantity is: (Karnataka CET 2012)

(A) Surface tension if a = 1, b = 1, c = – 2

(B) Force if a = 1, b = 1, c = 2

(C) Angular frequency if a = 0, b = 0, c = – 1

(D) Spring constant if a = 1, b = – 1, c = – 2.

Sol: (C)

[Surface tension] = MT^{– 2} → a = 1, b = 0, c = – 2.

[Force] = MLT^{– 2} → a = 1, b = 1, c = – 2.

[Angular frequency] = T^{– 1} → a = 0, b = 0, c = – 1.

[Spring constant] = MT^{– 2} → a = 1, b = 0, c = – 2.

So, option (C) is correct.

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


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