JEE Previous Year Questions With Solutions Of Introduction To Physics [Unit, Measurement & Dimension] Part - 1 | 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, 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. Velocity of a particle depends upon t as V = A + Bt + Ct². If velocity is in m/s, the unit of A will be: (CPMT 1971)

(A) m/s

(B) m/s²

(C) m.s

(D) m²/s  

Sol: (A)

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

So, [A] = LT^–1 and unit of A is m/s.

2. The SI unit of heat is:  (CPMT 1979)

(A) calorie

(B) Horse power

(C) Joule

(D) Watt

Sol: (C)

SI unit of heat is joule.

3. The volume of a liquid of density ρ and viscosity η flowing in time t through a capillary tube of length l and radius R, with a pressure difference p, across its ends is proportional to: (CPMT 1982)

(A) p²R²t/ηl²

(B) pR⁴/ηlt

(C) pR⁴t/ηl

(D) ηR⁴/lt

Sol: (C)

Suppose the formula is 

→ Volume, V = k η^a l^b R^c p^d t^e

Where k is proportionality constant.

Then, L³ = [ML^–1T^–1]^a L^b L^c [M L^–1T^–2]^d T^e

Or, L³ = M^{a + d} L^{–a + b + c – d} T^{–a – 2d + e}

Equating the exponents of similar quantities,

We get, a + d = 0; ----------- (i)

– a + b + c – d = 3; ---------- (ii)

– a – 2d + e = 0. ------------- (iii)

Equation (i) gives, a = – d

Equation (ii) gives, b + c = 3

Equation (i) gives, a + e = 0 → a = – e

Therefore, e = d

Power of p and t are same and it is possible in option (c) only.

4. The velocity v of a particle at time t is given by: v = at + b/(t + c). The dimensional formula of a, b, and c are respectively: (CPMT 1990)

(A) L², T, and LT²

(B) LT², LT, and T

(C) LT^–2, L and T

(D) L, LT, and T^–2

Sol: (C)

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 (LT^–1).

So, [a] = LT^–1T^–1 = LT^–2, [b] = L, and [c] = T.

5. The dimensional formula for angular momentum is: (CBSE 1988, CPMT 1999, WBJEE 2012)

(A) [M⁰L²T^–2]

(B) [ML²T^–1]

(C) [MLT^–1]

(D) [ML²T^–2]   

Sol: (B)

We know, angular momentum = Iω

Where, I = mass moment of inertia

And, ω = angular velocity

→ [Angular momentum] = [I] [ω]

Or, [Angular momentum] = ML²T^–1.

6. The dimensional formula for torque is: (CBSE 1990, AIIMS 2002)

(A) [ML²T^–2]

(B) [MLT^–2]

(C) [ML^–1T^–2]

(D) [ML^–2T^–2]

Sol: (A)

We know, Torque = F * L

Where, F = Force

And, L = length

→ [Torque] = [F] [L]

Or, [Torque] = MLT^–2L = ML²T^–2.

7. The motion without consideration of its cause is studied in: (CBSE 1991)

(A) Kinematics

(B) Mechanics

(C) Statics

(D) Modern Physics

Sol: (A) 

8. The dimensional formula of gravitational constant G is: (CBSE 1992, 2004 MPCEE 1999, Punjab PMT 2002, 2003, Karnataka CET 2008, WBJEE 2017)

(A) MLT^–2

(B) ML³T^–2

(C) M^–1L³T^–2

(D) M^–1LT^–2

Sol: (C)

We know,

Gravitational force, F_g = G * (m₁m₂)/r²

Now, [F_g] = [G] [(m₁m₂)/r²]

Or, MLT^–2 = [G] M²L^–2

Or, [G] = M^–1L³T^–2.

9. Dimensional formula of self-inductance is: (CBSE 1992)

(A) MLT^–2A^–2

(B) ML²T^–1A^–2

(C) ML²T^–2A^–2

(D) ML²T^–2A^–1

Sol: (C)

We know, Energy, E = ½ SI²

Where, S = self-inductance and I = current.

Now, [E] = [S] [I²]

Or, ML²T^–2 = [S] A²

Or, [S] = ML²T^–2A^–2.

10. Dyne second stands for the unit of: (IIT JEE 1975)

(A) Force

(B) Work

(C) Momentum

(D) Angular momentum

Sol: (C)

We know, Force, F = Δp/Δt

Or, Δp = F * Δt

Unit of Momentum (p) in cgs is Dyne second.

11. If L, C, R represent induction, capacitance and resistance respectively, the combinations having dimensions of frequency are: (IIT JEE 1984, Odisha JEE 2016)

(A) 1/√(CL)

(B) L/C

(C) R/L

(D) R/C

Sol: (A), (C)

We know,

Dimensional formula of

Resistance = ML²T^–3A^–2

Inductance = ML²T^–2A^–2

Capacitance = M^–1L^–2T⁴A²

Now, [1/√(CL)] = 1/T = [frequency]

And, [R/L] = 1/T = [frequency]

12. Which of the following combinations have the dimensions of time? L, C, R represent induction, capacitance and resistance respectively: (IIT JEE 1986)

(A) RC

(B) √ (LC)

(C) R/L

(D) C/L

Sol: (A), (B)

We know,

Dimensional formula of

Resistance = ML²T^–3A^–2

Inductance = ML²T^–2A^–2

Capacitance = M^–1L^–2T⁴A²

Now, [RC] = [ML²T^–3A^–2] * [M^–1L^–2T⁴A²]

Or, [RC] = T 

And, [√ (LC)] = T (Karnataka CET 2005).
[L/R] = T and unit is second. (Odisha JEE 2002)

13. The dimensions of the quantities in one (or more) of the following pairs are the same. Identify the pair(s): (IIT JEE 1986)

(A) Torque and Work

(B) Angular momentum and Work

(C) Energy and Young’s modulus

(D) Light year and wavelength

Sol: (A), (D)

Torque and Work both have dimensions [ML²T^–2].

Light year and wavelength both have dimensions [L].

14. The dimensions of length are expressed as G^xc^yh^z. Where G, c, h are the universal gravitational constant, speed of light, and Planck’s constant respectively, then (IIT JEE 1992, EAMCET 2014)

(A) x = ½ , y = ½

(B) x = ½ , z = ½

(C) y = – 3/2 , z = ½

(D) y = ½, z = 3/2

Sol: (B), (C)

Given: L = G^xc^yh^z

Or, [L] = [G]^x [c]^y [h]^z

Or, M⁰LT⁰ = [M^–1L³T^–2]^x [LT^–1]^y [ML²T^–1]^z

Or, M⁰LT⁰ = M^{–x +z} L^{3x +y + 2z} T^{–2x–y–z}

Comparing the power of M, L and T we get,

→ 0 = – x + z; 1 = 3x + y + 2z and 0 = – 2x – y – z.

Solving these equation we get, x = ½, y = –3/2, z = 1/2.

15. Which of the following quantities can be written in SI units in kg-m²-A^–2s^–3. (IIT JEE 1993)

(A) Resistance

(B) Inductance

(C) Capacitance

(D) Magnetic flux

Sol: (A)

We know,

Dimensional formula of

Resistance = ML²T^–3A^–2

Inductance = ML²T^–2A^–2

Capacitance = M^–1L^–2T⁴A²

Magnetic flux = ML⁰T^–2A^–1

So, SI unit of Resistance can be written in kg-m²-A^–2s^–3.

16. If C and R denote capacitance and resistance, the dimensional formula of CR is: (CBSE 1992)

(A) M⁰L⁰T¹

(B) M⁰L⁰T⁰

(C) M¹L²T¹

(D) Not expressible in terms of MLT

Sol: (A)

We know,

Dimensional formula of

Resistance, R = ML²T^–3A^–2

Capacitance, C = M^–1L^–2T⁴A²

Now, Dimensional formula of RC is M⁰L⁰T¹.

17. The dimensional formula for modulus of rigidity is: (MNR 1986, Haryana CEET 2002)

(A) ML^–1T^–1

(B) ML^–2T²

(C) MLT^–1

(D) ML^–1T^–2

Sol: (D)

We know,

Shear stress, τ = Gγ

Where, G = modulus of rigidity and γ = shear strain.

Now, [τ] = [G] [γ]

Or, ML^–1T^–2 = [G] (since, [γ] = 1 γ is dimensionless quantity)

Or, [G] = ML^–1T^–2.

18. Which of the following is equal to: (Odisha 1990)

(Joule * ohm)/ (volt * second)

(A) Ampere

(B) Volt

(C) Watt

(D) Tesla

Sol: (B)

We have, JΩ/Vs = (J * V/A)/Vs = J/As = J/C = V.

19. If x = at + bt², where x is the distance travelled by the body in kilometres while t is the time in seconds, then the units of b is: (CBSE 1993)

(A) km/s

(B) km-s

(C) km/s² 

(D) km-s² 

Sol: (C)

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

So, [bt²] = L

Or, [b] = LT^–2 

Therefore, unit of b is km/s².

20. The dependence of a physical quantity P is given by P = P₀ exp (– αt²) [where α is a constant and t is time]. The constant α: (CBSE 1993)

(A) is dimensionless

(B) Has dimension (T^–2)

(C) Has dimension (T^–1)

(D) Has dimension P

Sol: (B)

The equation contains two terms. All of them should have the same dimensions.

→ [P] = [P₀] [exp (– αt²)]

Since, dimension formula of P and P₀ is same.

Therefore, [αt²] = 1

Or, [α] = T^–2.

21. A force F is applied on a square plate of side L. If the percentage error in the determination of L is 2% and that in F is 4%, what is the permissible error in pressure? (CPMT 1993)

(A) 2%

(B) 4%

(C) 6%

(D) 8%

Sol: (D)

We know, pressure = force/area

Or, p = F/A = F/L²

Taking log and differentiate and added for errors

We get, Δp/p = (ΔF/F) + 2 (ΔL/L)

Or, Δp/p = 4 + 2 * 2 = 8.

So, the permissible error in pressure is 8%.

22. Planck’s constant has dimensions______: (IIT JEE 1985, CPMT 1999, Kerala CET 2002, Haryana CEET 2002)

Sol:

We know,

Energy, E = hν

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

So, [h] = [E]/ [ν]

Or, [h] = ML²T^–2/T^–1 = ML²T^–1.

23. In the formula X = 3YZ², X and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are ______: (IIT JEE 1988, AMU 1998)

Sol:

We know,

[X] = [capacitance] = M^–1L^–2T²Q².

[Z] = [magnetic induction] = MT^–1Q^–1.

Therefore, [Y] = [X]/ [3Z]²

Or, [Y] = M^–1L^–2T²Q²/ (MT^–1Q^–1)²

Or, [Y] = M^–3L^–2T⁴Q⁴.

24. The dimensions of 1/√(µ₀ϵ₀) are the same as that of: (CPMT 1993)

(A) Velocity

(B) Time

(C) Capacitance

(D) Distance

Sol: (A)

We know,

Speed of light, c = 1/√(µ₀ϵ₀)

So, the dimensions of 1/√(µ₀ϵ₀) is same as velocity.

25. According to Newton, the viscous force acting between liquid layers of area A and velocity gradient ΔV/ΔZ is given by F = – ηA(ΔV/ΔZ) where η is a constant called coefficient of viscosity. The dimensional formula of η is: (AIIMS 1993)

(A) ML^–2T^–2

(B) M⁰L⁰T⁰

(C) ML²T^–2

(D) ML^–1T^–1

Sol: (D)

Given: F = – ηA(ΔV/ΔZ)

Or, [F] = [η] [A] ([ΔV]/[ΔZ])

Or, MLT^–2 = [η] (L²) (LT^–1/L)

Or, [η] = ML^–1T^–1.

26. What is the approximate percentage error in the measurement of time period of a simple pendulum if maximum errors in the measurement of length L and gravitational acceleration g are 3% and 7% respectively? (UPSEE 2016)

(A) 2%

(B) 3%

(C) 5%

(D) 10%

Sol: (D)

We know, time period, T = 2π √ (L/g)

Taking log and differentiate and added for errors

We get, ΔT/T = ½ * (ΔL/L) + ½ * (Δg/g)

Or, ΔT/T = ½ * 3 + ½ * 7 = 5.

So, the error in time period is 5%.

27. Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit mass M. Then the correct option(s) is (are): (JEE Advanced 2015)

(A) M ∝ √c

(B) M ∝ √G

(C) L ∝ √h

(D) L ∝ √G

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

Let length, L = G^xc^yh^z

Or, [L] = [G]^x [c]^y [h]^z

Or, M⁰LT⁰ = [M^–1L³T^–2]^x [LT^–1]^y [ML²T^–1]^z

Or, M⁰LT⁰ = M^{–x +z} L^{3x +y + 2z} T^{–2x–y–z}

Comparing the power of M, L and T we get,

→ 0 = – x + z; 1 = 3x + y + 2z and 0 = – 2x – y – z.

Solving these equation we get, x = ½, y = – 3/2, z = 1/2.

So, L ∝ √G and L ∝ √h.

Let mass, m = G^xc^yh^z

Or, [m] = [G]^x [c]^y [h]^z

Or, M¹L⁰T⁰ = [M^–1L³T^–2]^x [LT^–1]^y [ML²T^–1]^z

Or, M¹L⁰T⁰ = M^{–x +z} L^{3x +y + 2z} T^{–2x–y–z}

Comparing the power of M, L and T we get,

→ 1 = – x + z; 0 = 3x + y + 2z and 0 = – 2x – y – z.

Solving these equation we get, x = – ½, y = 1/2, z = 1/2.

So, M ∝ √c and M ∝ √h.

28. The mass and volume of a body are found to be 5.00 ± 0.05 kg and 1.00 ± 0.05 m³ respectively. Then the maximum possible percentage error in its density is: (KEAM 2011)

(A) 6%

(B) 3%

(C) 10%

(D) 5%

(E) 7%

Sol: (C)

Error in mass = 5%.

Error in volume = 5%.

We have density, ρ = m/V

Taking log and differentiate and added for errors

We get, Δρ/ρ = (Δm/m) + (ΔV/V)

Or, Δρ/ρ = 5 + 5 = 10.

So, the error in density is 10%. 

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

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HCVERMA SOLUTION (CHAPTERWISE), JEE Main and advance questions and solutions.